Chapter 7 Integrals (Additional Questions)

The process of finding an integral is the inverse process of differentiation.

If $F(x)$ is an antiderivative of $f(x)$, then $\int f(x) dx = F(x) + C$. This means that differentiating $F(x)$ gives $f(x)$.

Mathematically, if $\frac{d}{dx}(F(x)) = f(x)$, then $\int f(x) dx = F(x) + C$.

Therefore, the correct option is (B) Differentiation.

To find the integral of $x^n$ with respect to $x$, we use the power rule for integration.

The power rule for integration states that for any real number $n \neq -1$:

Where $C$ is the constant of integration.

Let's verify this by differentiating the result:

This matches the integrand.

Option (A) $nx^{n-1} + C$ is the result of differentiation of $x^n$, not integration.

Option (C) $\frac{x^n}{n} + C$ is incorrect.

Option (D) $\log |x| + C$ is the integral of $x^{-1}$ (or $\frac{1}{x}$), not $x^n$ for $n \neq -1$.

Therefore, the correct answer is $\frac{x^{n+1}}{n+1} + C$.

We need to evaluate the integral $\int (\sin x + \cos x) dx$.

Using the linearity property of integrals, we can split this into two separate integrals:

We know the standard integrals:

Combining these results:

Let $C = C_1 + C_2$, which is also an arbitrary constant.

Comparing this with the given options:

(A) $\cos x - \sin x + C$ is incorrect.

(B) $-\cos x + \sin x + C$ is correct.

(C) $-\sin x + \cos x + C$ is incorrect.

(D) $\sin x - \cos x + C$ is incorrect.

Therefore, the correct answer is (B).

Explanation:

We need to evaluate the integral $\int \frac{1}{\sqrt{1-x^2}} dx$.

This is a standard integral form. We know that the derivative of $\sin^{-1} x$ with respect to $x$ is $\frac{1}{\sqrt{1-x^2}}$.

Therefore, the integral of $\frac{1}{\sqrt{1-x^2}}$ with respect to $x$ is $\sin^{-1} x + C$, where C is the constant of integration.

So, we have:

$\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C$

Let's check the given options:

(A) $\tan^{-1} x + C$: The derivative of $\tan^{-1} x$ is $\frac{1}{1+x^2}$.

(B) $\sin^{-1} x + C$: The derivative of $\sin^{-1} x$ is $\frac{1}{\sqrt{1-x^2}}$.

(C) $\cos^{-1} x + C$: The derivative of $\cos^{-1} x$ is $-\frac{1}{\sqrt{1-x^2}}$.

(D) $\sec^{-1} x + C$: The derivative of $\sec^{-1} x$ is $\frac{1}{|x|\sqrt{x^2-1}}$.

Comparing our result with the options, we find that option (B) matches our result.

Thus, $\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C$.

The correct option is (B) $\sin^{-1} x + C$.

Explanation:

We are given the substitution $u = x^2+1$. We need to find the differential $du$.

To find $du$, we differentiate $u$ with respect to $x$:

$u = x^2+1$

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x^2+1)$

Using the sum rule for differentiation, $\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)$:

$\frac{du}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(1)$

Using the power rule, $\frac{d}{dx}(x^n) = nx^{n-1}$, and the derivative of a constant is 0:

$\frac{d}{dx}(x^2) = 2x^{2-1} = 2x$

$\frac{d}{dx}(1) = 0$

So, we have:

$\frac{du}{dx} = 2x + 0$

$\frac{du}{dx} = 2x$

Now, to find $du$, we multiply both sides by $dx$:

$du = 2x \, dx$

Let's compare this result with the given options:

(A) $x \, dx$

(B) $2x \, dx$

(C) $(2x+1) \, dx$

(D) $x^2 \, dx$

Our result $du = 2x \, dx$ matches option (B).

The correct option is (B) $2x \, dx$.

Explanation:

We need to evaluate the integral $\int e^{2x} dx$.

We can use the standard integration formula for exponential functions, which states that:

$\int e^{ax} dx = \frac{1}{a} e^{ax} + C$

where $a$ is a constant and $C$ is the constant of integration.

In this problem, we have $a=2$. Applying the formula:

$\int e^{2x} dx = \frac{1}{2} e^{2x} + C$

Let's compare this result with the given options:

(A) $2e^{2x} + C$

(B) $\frac{1}{2}e^{2x} + C$

(C) $e^{2x} + C$

(D) $e^x + C$

Our result $\frac{1}{2}e^{2x} + C$ matches option (B).

The correct option is (B) $\frac{1}{2}e^{2x} + C$.

Explanation:

We need to evaluate the integral $\int \frac{dx}{x^2+a^2}$.

This is a standard integral form. The formula for integrating $\frac{1}{x^2+a^2}$ with respect to $x$ is:

$\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$

where $a$ is a constant and $C$ is the constant of integration.

Let's compare this standard formula with the given options:

(A) $\tan^{-1}\left(\frac{x}{a}\right) + C$

(B) $\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$

(C) $\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C$ (This is the integral of $\frac{1}{x^2-a^2}$)

(D) $\sin^{-1}\left(\frac{x}{a}\right) + C$ (This is related to the integral of $\frac{1}{\sqrt{a^2-x^2}}$)

The standard integral formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$ directly matches option (B).

The correct option is (B) $\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$.

Explanation:

We are asked to determine the first step to integrate $\int \frac{P(x)}{Q(x)} dx$, where $P(x)$ and $Q(x)$ are polynomials and the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$.

When integrating a rational function $\frac{P(x)}{Q(x)}$:

If the degree of the numerator $P(x)$ is greater than or equal to the degree of the denominator $Q(x)$ (this is called an improper rational function), the first step is to perform polynomial long division.

By dividing $P(x)$ by $Q(x)$, we can express the rational function as:

$\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}$

where $S(x)$ is the quotient (a polynomial) and $R(x)$ is the remainder (a polynomial whose degree is less than the degree of $Q(x)$).

Then the integral becomes:

$\int \frac{P(x)}{Q(x)} dx = \int S(x) dx + \int \frac{R(x)}{Q(x)} dx$

The integral $\int S(x) dx$ can be easily evaluated since $S(x)$ is a polynomial. The integral $\int \frac{R(x)}{Q(x)} dx$ involves a proper rational function (degree of numerator is less than the degree of denominator), which can then be tackled using methods like partial fraction decomposition.

Let's consider the given options:

(A) Use substitution: Substitution might be used later for parts of the integral, but it's not the primary first step for an improper rational function.

(B) Use integration by parts: Integration by parts is generally used for products of functions, not directly as the first step for improper rational functions.

(C) Divide $P(x)$ by $Q(x)$: This is the correct first step to convert an improper rational function into a polynomial plus a proper rational function.

(D) Use partial fractions: Partial fractions are used when the degree of the numerator is less than the degree of the denominator (a proper rational function). If the function is improper, division must be performed first.

Therefore, the first step is to divide $P(x)$ by $Q(x)$.

The correct option is (C) Divide $P(x)$ by $Q(x)$.

Explanation:

We need to evaluate the integral $\int \frac{dx}{x^2-a^2}$.

This integral can be solved using the method of partial fractions or by recognizing it as a standard integral form.

Method 1: Using Partial Fractions

The denominator can be factored as $x^2-a^2 = (x-a)(x+a)$.

We can write the integrand as a sum of partial fractions:

$\frac{1}{x^2-a^2} = \frac{1}{(x-a)(x+a)} = \frac{A}{x-a} + \frac{B}{x+a}$

To find the constants A and B, we multiply both sides by $(x-a)(x+a)$:

$1 = A(x+a) + B(x-a)$

To find A, let $x=a$:

$1 = A(a+a) + B(a-a)$

$1 = A(2a) + B(0)$

$1 = 2aA \implies A = \frac{1}{2a}$

To find B, let $x=-a$:

$1 = A(-a+a) + B(-a-a)$

$1 = A(0) + B(-2a)$

$1 = -2aB \implies B = -\frac{1}{2a}$

So, the partial fraction decomposition is:

$\frac{1}{x^2-a^2} = \frac{1}{2a(x-a)} - \frac{1}{2a(x+a)} = \frac{1}{2a}\left(\frac{1}{x-a} - \frac{1}{x+a}\right)$

Now, we integrate this expression:

$\int \frac{dx}{x^2-a^2} = \int \frac{1}{2a}\left(\frac{1}{x-a} - \frac{1}{x+a}\right) dx$

$= \frac{1}{2a} \left( \int \frac{1}{x-a} dx - \int \frac{1}{x+a} dx \right)$

We know that $\int \frac{1}{u} du = \log|u| + C$.

So, $\int \frac{1}{x-a} dx = \log|x-a|$ and $\int \frac{1}{x+a} dx = \log|x+a|$.

Therefore, the integral becomes:

$= \frac{1}{2a} \left( \log|x-a| - \log|x+a| \right) + C$

Using the logarithm property $\log M - \log N = \log \frac{M}{N}$, we get:

$\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log\left|\frac{x-a}{x+a}\right| + C$

Method 2: Using Standard Integral Formula

The integral $\int \frac{dx}{x^2-a^2}$ is a standard integral, and its formula is:

$\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log\left|\frac{x-a}{x+a}\right| + C$

where $a$ is a non-zero constant and $C$ is the constant of integration.

Let's compare this result with the given options:

(A) $\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C$

(B) $\frac{1}{2a}\log\left|\frac{x+a}{x-a}\right| + C$ (This is the integral of $\frac{1}{a^2-x^2}$)

(C) $\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$ (This is the integral of $\frac{1}{x^2+a^2}$)

(D) $\frac{1}{\sqrt{a^2-x^2}} + C$ (This is not an integral, but $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$)

Our result $\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C$ matches option (A).

The correct option is (A) $\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C$.

Explanation:

The formula for integration by parts is a technique used to find the integral of a product of two functions. It is derived from the product rule for differentiation.

The product rule for differentiation states:

$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

In differential form, this can be written as:

$d(uv) = u \, dv + v \, du$

Integrating both sides with respect to their respective variables, we get:

$\int d(uv) = \int u \, dv + \int v \, du$

$uv = \int u \, dv + \int v \, du$

To find the formula for $\int u \, dv$, we rearrange the equation:

$\int u \, dv = uv - \int v \, du$

This is the standard formula for integration by parts. An arbitrary constant of integration is usually added after evaluating the final integral $\int v \, du$.

Let's compare this formula with the given options:

(A) $uv - \int v \, du$

(B) $uv + \int v \, du$

(C) $\int v \, du - uv$

(D) $uv \int v \, du$

The formula $\int u \, dv = uv - \int v \, du$ matches option (A).

The correct option is (A) $uv - \int v \, du$.

Explanation:

The LIATE rule is a heuristic (rule of thumb) used in integration by parts to choose which function should be designated as 'u' (the function to be differentiated) and which should be 'dv' (the function to be integrated). The goal is to choose 'u' such that its derivative $du$ is simpler than $u$, and $dv$ can be easily integrated to find $v$.

The acronym LIATE stands for the preferred order of choosing 'u':

L: Logarithmic functions (e.g., $\log x$, $\log(x^2+1)$)

I: Inverse trigonometric functions (e.g., $\sin^{-1} x$, $\tan^{-1} x$)

A: Algebraic functions (e.g., $x^2$, $3x$, $\sqrt{x}$)

T: Trigonometric functions (e.g., $\sin x$, $\cos x$, $\sec^2 x$)

E: Exponential functions (e.g., $e^x$, $2^x$)

When integrating a product of two functions from different categories in this list, you generally choose 'u' to be the function type that appears earlier in LIATE.

Let's examine the given options in the context of LIATE:

(A) Logarithmic functions: This corresponds to 'L' in LIATE. These are chosen as 'u' if present.

(B) Inverse trigonometric functions: This corresponds to 'I' in LIATE. If no logarithmic functions are present, these are chosen as 'u'.

(C) Algebraic functions: This corresponds to 'A' in LIATE. If no logarithmic or inverse trigonometric functions are present, these are chosen as 'u'.

(D) Trigonometric functions: This corresponds to 'T' in LIATE. If no logarithmic, inverse trigonometric, or algebraic functions are present, these are chosen as 'u' (when paired with exponential functions, for example).

All the listed function types (A, B, C, D) are components of the LIATE rule used for selecting 'u' in integration by parts. The question asks to "Select all that apply".

Therefore, all options (A), (B), (C), and (D) are part of the LIATE mnemonic for choosing 'u'.

The correct options are (A) Logarithmic functions, (B) Inverse trigonometric functions, (C) Algebraic functions, and (D) Trigonometric functions.

Explanation:

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): $\int \frac{1}{\sqrt{x^2+a^2}} dx = \log\left|x + \sqrt{x^2+a^2}\right| + C$.

This is indeed a standard formula for the integral of $\frac{1}{\sqrt{x^2+a^2}}$. This integral can also be expressed using the inverse hyperbolic sine function as $\text{sinh}^{-1}\left(\frac{x}{a}\right) + C'$, which is equivalent to $\log\left|\frac{x + \sqrt{x^2+a^2}}{a}\right| + C'$. Since $\log a$ is a constant, it can be absorbed into the constant of integration, making $\log\left|x + \sqrt{x^2+a^2}\right| + C$ a valid form. Thus, Assertion (A) is true.

Reason (R): This is a standard integral formula.

The statement in Assertion (A) is one of the fundamental integration formulas taught in calculus. These formulas are derived and then memorized or looked up for use in solving more complex integrals. Therefore, stating that it is a "standard integral formula" is a correct characterization of its status in calculus. Thus, Reason (R) is true.

Now, we need to determine if R is the correct explanation of A.

Assertion (A) presents a mathematical identity: an integral and its result. Reason (R) states that this identity is a "standard integral formula." The fact that it is a standard integral formula is precisely why we accept the assertion as true without re-deriving it every time. The reason provides the context or justification for the assertion's validity in common practice. If it weren't a standard formula, we would need to derive it to confirm its truth. So, R explains why A is presented as a known fact.

Therefore, both A and R are true, and R is the correct explanation of A because A's acceptance as a direct result comes from its status as a standard, recognized integral formula.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Explanation:

We need to evaluate the definite integral $\int_0^1 x^2 dx$.

First, we find the indefinite integral of $x^2$. Using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$.

For $x^2$, $n=2$. So,

$\int x^2 dx = \frac{x^{2+1}}{2+1} + C = \frac{x^3}{3} + C$

Now, we apply the Fundamental Theorem of Calculus to evaluate the definite integral. If $F(x)$ is an antiderivative of $f(x)$, then $\int_a^b f(x) dx = F(b) - F(a)$.

Here, $f(x) = x^2$, and an antiderivative is $F(x) = \frac{x^3}{3}$. The limits of integration are $a=0$ and $b=1$.

$\int_0^1 x^2 dx = \left[ \frac{x^3}{3} \right]_0^1$

Now, substitute the upper limit and subtract the result of substituting the lower limit:

$= F(1) - F(0)$

$= \frac{(1)^3}{3} - \frac{(0)^3}{3}$

$= \frac{1}{3} - \frac{0}{3}$

$= \frac{1}{3} - 0$

$= \frac{1}{3}$

Let's compare this result with the given options:

(A) $1/3$

(B) $1$

(C) $0$

(D) $1/2$

Our result $\frac{1}{3}$ matches option (A).

The correct option is (A) $1/3$.

Explanation:

The question refers to the Fundamental Theorem of Calculus, specifically the part that relates definite integrals to antiderivatives. This is often called the Second Fundamental Theorem of Calculus.

The theorem states that if $f$ is a continuous function on the interval $[a, b]$ and $F$ is an antiderivative of $f$ on $[a, b]$ (meaning $F'(x) = f(x)$ for all $x$ in $[a, b]$, or equivalently, $F(x) = \int f(x) dx$), then the definite integral of $f(x)$ from $a$ to $b$ is given by:

$\int_a^b f(x) dx = F(b) - F(a)$

Let's examine the given options:

(A) $F(b) + F(a)$

(B) $F(b) - F(a)$

(C) $f(b) - f(a)$

(D) $F(a) - F(b)$

Comparing the standard statement of the Fundamental Theorem of Calculus with the options, we see that option (B) matches the theorem.

The correct option is (B) $F(b) - F(a)$.

Explanation:

We need to evaluate the definite integral $\int_{-1}^1 |x| dx$.

The absolute value function, $|x|$, is defined as:

$|x| = \begin{cases} x & , & x \geq 0 \\ -x & , & x < 0 \end{cases}$

The interval of integration is from -1 to 1. Since the definition of $|x|$ changes at $x=0$, we need to split the integral into two parts:

$\int_{-1}^1 |x| dx = \int_{-1}^0 |x| dx + \int_0^1 |x| dx$

For the first integral, $\int_{-1}^0 |x| dx$:

In the interval $[-1, 0)$, $x < 0$, so $|x| = -x$.

$\int_{-1}^0 |x| dx = \int_{-1}^0 (-x) dx$

Using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1}$:

$\int (-x) dx = -\frac{x^2}{2}$

So, $\int_{-1}^0 (-x) dx = \left[ -\frac{x^2}{2} \right]_{-1}^0$

$= \left( -\frac{(0)^2}{2} \right) - \left( -\frac{(-1)^2}{2} \right)$

$= (0) - \left( -\frac{1}{2} \right)$

$= \frac{1}{2}$

For the second integral, $\int_0^1 |x| dx$:

In the interval $[0, 1]$, $x \geq 0$, so $|x| = x$.

$\int_0^1 |x| dx = \int_0^1 x dx$

Using the power rule for integration:

$\int x dx = \frac{x^2}{2}$

So, $\int_0^1 x dx = \left[ \frac{x^2}{2} \right]_0^1$

$= \left( \frac{(1)^2}{2} \right) - \left( \frac{(0)^2}{2} \right)$

$= \frac{1}{2} - 0$

$= \frac{1}{2}$

Now, add the results of the two integrals:

$\int_{-1}^1 |x| dx = \frac{1}{2} + \frac{1}{2} = 1$

Let's compare this result with the given options:

(A) $0$

(B) $1/2$

(C) $1$

(D) $2$

Our result $1$ matches option (C).

The correct option is (C) $1$.

Solution:

We need to evaluate the definite integral $\int_0^{\pi/2} \cos x dx$.

The integral of $\cos x$ with respect to $x$ is $\sin x$.

So, the antiderivative of $\cos x$ is $\sin x + C$, where $C$ is the constant of integration. For definite integrals, the constant $C$ cancels out, so we can use $\sin x$ as the antiderivative.

According to the Fundamental Theorem of Calculus, Part 2, if $F(x)$ is an antiderivative of $f(x)$, then $\int_a^b f(x) dx = F(b) - F(a)$.

Here, $f(x) = \cos x$, $F(x) = \sin x$, $a = 0$, and $b = \pi/2$.

We need to evaluate $\sin x$ at the upper limit $x = \pi/2$ and at the lower limit $x = 0$, and then subtract the value at the lower limit from the value at the upper limit.

Value at upper limit: $\sin(\pi/2)$

Value at lower limit: $\sin(0)$

We know that $\sin(\pi/2) = 1$ and $\sin(0) = 0$.

Therefore, the definite integral is:

$\int_0^{\pi/2} \cos x dx = [\sin x]_0^{\pi/2} = \sin(\pi/2) - \sin(0)$

$\int_0^{\pi/2} \cos x dx = 1 - 0$

$\int_0^{\pi/2} \cos x dx = 1$

Comparing this result with the given options, we find that the value of the integral is $1$, which corresponds to option (B).

Thus, the correct answer is (B) 1.

Solution:

We are asked to identify the correct property of definite integrals from the given options.

Let's examine each option based on the known properties of definite integrals:

(A) $\int_a^b f(x) dx = \int_b^a f(x) dx$

This statement is incorrect. The order of the limits of integration is important. Reversing the limits changes the sign of the integral.

(B) $\int_a^b f(x) dx = -\int_b^a f(x) dx$

This statement is a fundamental property of definite integrals. Reversing the limits of integration indeed results in the negation of the integral's value. This is a correct property.

(C) $\int_a^a f(x) dx = 1$

This statement is incorrect. The definite integral of a function from a point to itself is always $0$, provided the function is defined at that point. The correct property is $\int_a^a f(x) dx = 0$.

(D) $\int_a^b (f(x) + g(x)) dx = \int_a^b f(x) dx \cdot \int_a^b g(x) dx$

This statement is incorrect. The integral of a sum of functions is the sum of their integrals, not the product. The correct property is $\int_a^b (f(x) + g(x)) dx = \int_a^b f(x) dx + \int_a^b g(x) dx$, assuming $f(x)$ and $g(x)$ are integrable on $[a, b]$.

Based on the analysis of each option, the only correct property listed is (B).

Thus, the correct answer is (B) $\int_a^b f(x) dx = -\int_b^a f(x) dx$.

Solution:

We need to evaluate the indefinite integral $\int \sin^2 x dx$.

To integrate $\sin^2 x$, it is helpful to use a trigonometric identity to express it in terms of $\cos(2x)$. The double angle identity for cosine is $\cos(2x) = 1 - 2\sin^2 x$.

From this identity, we can solve for $\sin^2 x$:

$2\sin^2 x = 1 - \cos(2x)$

$\sin^2 x = \frac{1 - \cos(2x)}{2}$

Now, substitute this expression for $\sin^2 x$ into the integral:

$\int \sin^2 x dx = \int \frac{1 - \cos(2x)}{2} dx$

We can take the constant $\frac{1}{2}$ out of the integral:

$\int \sin^2 x dx = \frac{1}{2} \int (1 - \cos(2x)) dx$

Using the property that the integral of a sum (or difference) is the sum (or difference) of the integrals:

$\int \sin^2 x dx = \frac{1}{2} \left( \int 1 dx - \int \cos(2x) dx \right)$

Now, we evaluate each integral separately:

$\int 1 dx = x$

For the integral of $\cos(2x)$, we can use a substitution. Let $u = 2x$. Then the differential $du = 2 dx$, which means $dx = \frac{1}{2} du$.

$\int \cos(2x) dx = \int \cos(u) \left(\frac{1}{2} du\right) = \frac{1}{2} \int \cos(u) du$

The integral of $\cos(u)$ is $\sin(u)$. So,

$\frac{1}{2} \int \cos(u) du = \frac{1}{2} \sin(u) + C'$

Substitute back $u = 2x$:

$\int \cos(2x) dx = \frac{1}{2} \sin(2x) + C'$

Now, substitute these results back into the main integral expression:

$\int \sin^2 x dx = \frac{1}{2} \left( x - \left(\frac{1}{2} \sin(2x)\right) \right) + C$

where $C$ is the constant of integration (combining $\frac{1}{2}C'$ and any constant from the $\int 1 dx$).

Distribute the $\frac{1}{2}$:

$\int \sin^2 x dx = \frac{1}{2} x - \frac{1}{4} \sin(2x) + C$

This can be written as:

$\int \sin^2 x dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$

Now, we compare our result with the given options:

(A) $\frac{\sin^3 x}{3} + C$ (Incorrect)

(B) $\frac{x}{2} - \frac{\sin 2x}{4} + C$ (Matches our result)

(C) $\frac{1-\cos 2x}{2} + C$ (This is the integrand $\sin^2 x$, not its integral)

(D) $\cos^2 x + C$ (Incorrect)

Therefore, the correct option is (B).

Thus, the correct answer is (B) $\frac{x}{2} - \frac{\sin 2x}{4} + C$.

Solution:

We need to evaluate the integral $\int \frac{x}{(x-1)(x-2)} dx$.

The integrand is a rational function, which is a ratio of two polynomials.

The denominator $(x-1)(x-2)$ is already factored into distinct linear factors.

Let's consider the given methods:

(A) Substitution: This method is useful when the integrand contains a function and its derivative. In this case, there isn't an obvious substitution that simplifies the entire expression $\frac{x}{(x-1)(x-2)}$ significantly.

(B) Integration by parts: This method is typically used for integrals of products of functions, based on the formula $\int u \, dv = uv - \int v \, du$. The integrand is a single fraction, not a product of functions that would be easily handled by this method.

(C) Partial fractions: This method is specifically designed for integrating rational functions. When the denominator of a rational function can be factored, the function can be decomposed into a sum of simpler fractions whose integrals are easier to evaluate (typically involving logarithms or arctangents). The integrand $\frac{x}{(x-1)(x-2)}$ is a proper rational function (degree of numerator is less than degree of denominator) with a factored denominator, making it an ideal candidate for partial fraction decomposition.

The decomposition would be of the form:

$\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$

where $A$ and $B$ are constants that can be determined. Once $A$ and $B$ are found, the integral becomes:

$\int \left(\frac{A}{x-1} + \frac{B}{x-2}\right) dx = A \int \frac{1}{x-1} dx + B \int \frac{1}{x-2} dx$

These resulting integrals are standard forms involving logarithms.

(D) Trigonometric identities: This method is used for integrals involving trigonometric functions. The given integrand does not involve trigonometric functions, so this method is not applicable.

Based on the nature of the integrand (a rational function with a factored denominator), the most appropriate method for evaluation is Partial Fractions.

Thus, the correct answer is (C) Partial fractions.

Solution:

We need to evaluate the indefinite integral $\int \tan x dx$.

We can rewrite $\tan x$ as $\frac{\sin x}{\cos x}$.

So, the integral becomes $\int \frac{\sin x}{\cos x} dx$.

We can use the method of substitution to evaluate this integral.

Let $u = \cos x$.

Differentiating both sides with respect to $x$, we get $\frac{du}{dx} = -\sin x$.

This implies $du = -\sin x \, dx$, or $\sin x \, dx = -du$.

Now, substitute $u$ and $du$ into the integral:

$\int \frac{\sin x}{\cos x} dx = \int \frac{-du}{u} = -\int \frac{1}{u} du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log |u|$.

So, $-\int \frac{1}{u} du = -\log |u| + C$, where $C$ is the constant of integration.

Substitute back $u = \cos x$:

$-\log |\cos x| + C$

Using the logarithmic property $-\log a = \log(a^{-1}) = \log(\frac{1}{a})$, we have:

$-\log |\cos x| = \log \left|\frac{1}{\cos x}\right|$

Since $\frac{1}{\cos x} = \sec x$, we can write the result as:

$\log |\sec x| + C$

Comparing our result with the given options, we find that the value of the integral is $\log |\sec x| + C$, which corresponds to option (B).

Thus, the correct answer is (B) $\log |\sec x| + C$.

Solution:

We need to identify which of the given integrals is not a standard integral form. Standard integral forms are integrals that have well-known formulas or procedures for evaluation, often found in tables of integrals.

Let's analyze each option:

(A) $\int \frac{dx}{\sqrt{a^2-x^2}}$: This is a standard integral form. Its value is $\sin^{-1}\left(\frac{x}{a}\right) + C$.

(B) $\int \frac{dx}{ax^2+bx+c}$: This is a standard type of integral involving a quadratic in the denominator. It can be evaluated by completing the square in the denominator ($ax^2+bx+c$) and then using standard formulas related to $\int \frac{du}{u^2 \pm k^2}$ or $\int \frac{du}{k^2 - u^2}$, which result in terms involving $\tan^{-1}$ or $\log$. So, this is a standard integral form.

(C) $\int \sqrt{a^2-x^2} dx$: This is a standard integral form. It is typically evaluated using trigonometric substitution ($x = a \sin \theta$) and results in the formula $\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$.

(D) $\int \frac{e^x}{\sin x} dx$: This integral involves a combination of exponential and trigonometric functions. While it is possible to write down series expansions or numerical approximations for this integral, it does not have a simple closed-form expression in terms of elementary functions (polynomials, rational functions, exponentials, logarithms, trigonometric and inverse trigonometric functions). Therefore, this is not considered a standard integral form that can be evaluated easily using basic integration techniques or found in a standard table of integrals.

Comparing the options, the integral in (D) is the one that does not have a readily available elementary function as its antiderivative and is not listed as a standard integral form in typical calculus courses.

Thus, the correct answer is (D) $\int \frac{e^x}{\sin x} dx$.

Solution:

Let's analyze the given Assertion (A) and Reason (R).

Assertion (A): $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ is a useful property of definite integrals.

This statement is a well-known and fundamental property of definite integrals. It is often used to simplify the evaluation of many integrals, especially those involving trigonometric functions over intervals like $[0, \pi]$ or $[0, \pi/2]$. Therefore, Assertion (A) is true.

Reason (R): Substituting $t = a-x$ in the integral $\int_0^a f(x) dx$ leads to this property.

Let's perform the substitution $t = a-x$ in the integral $\int_0^a f(x) dx$.

We have $t = a-x$. Differentiating with respect to $x$, we get $\frac{dt}{dx} = -1$, which means $dx = -dt$.

When $x = 0$, the new lower limit for $t$ is $t = a - 0 = a$.

When $x = a$, the new upper limit for $t$ is $t = a - a = 0$.

Also, from $t = a-x$, we have $x = a-t$.

Substituting these into the integral $\int_0^a f(x) dx$, we get:

$\int_0^a f(x) dx = \int_a^0 f(a-t) (-dt)$

Using the property $\int_b^a g(t) dt = -\int_a^b g(t) dt$, we can write $\int_a^0 f(a-t) (-dt) = - \int_a^0 f(a-t) dt = \int_0^a f(a-t) dt$.

Since $t$ is a dummy variable of integration, we can replace it with $x$:

$\int_0^a f(a-t) dt = \int_0^a f(a-x) dx$

Thus, we have shown that $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ using the substitution $t = a-x$. Therefore, Reason (R) is also true.

Now, we need to determine if Reason (R) is the correct explanation for Assertion (A).

As shown above, the substitution $t = a-x$ is precisely the method used to derive the property stated in Assertion (A). Therefore, Reason (R) correctly explains why the property in Assertion (A) holds true.

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Thus, the correct answer is (A) Both A and R are true and R is the correct explanation of A..

Solution:

We need to determine the type of function $f(x)$ for which the definite integral $\int_{-a}^a f(x) dx$ is equal to $0$. The interval of integration $[-a, a]$ is symmetric about $0$.

A key property of definite integrals over symmetric intervals $[-a, a]$ relates to the parity of the function (whether it is even or odd).

Recall the definitions of even and odd functions:

A function $f(x)$ is even if $f(-x) = f(x)$ for all $x$ in its domain.

A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain.

The property of definite integrals over a symmetric interval is:

$\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$

By using the substitution $y = -x$ in the integral $\int_{-a}^0 f(x) dx$, we get $\int_{-a}^0 f(x) dx = \int_0^a f(-x) dx$.

So, $\int_{-a}^a f(x) dx = \int_0^a f(-x) dx + \int_0^a f(x) dx = \int_0^a [f(-x) + f(x)] dx$.

Let's apply this to even and odd functions:

If $f(x)$ is an even function, then $f(-x) = f(x)$.

So, $f(-x) + f(x) = f(x) + f(x) = 2f(x)$.

In this case, $\int_{-a}^a f(x) dx = \int_0^a 2f(x) dx = 2 \int_0^a f(x) dx$. This is generally not $0$ unless $f(x) = 0$ on $[0, a]$.

If $f(x)$ is an odd function, then $f(-x) = -f(x)$.

So, $f(-x) + f(x) = -f(x) + f(x) = 0$.

In this case, $\int_{-a}^a f(x) dx = \int_0^a 0 dx = 0$.

Therefore, the value of $\int_{-a}^a f(x) dx$ is $0$ if $f(x)$ is an odd function.

Continuity (C) ensures the integral exists, but doesn't guarantee it's zero. Periodicity (D) is a property related to the repetition of function values and does not inherently make the integral over a symmetric interval zero unless the function also happens to be odd.

Thus, the correct answer is (B) Odd.

Solution:

We need to evaluate the indefinite integral $\int x \cos x dx$.

This integral is of the form $\int u \, dv$, which can be solved using integration by parts.

The formula for integration by parts is $\int u \, dv = uv - \int v \, du$.

We need to choose appropriate functions for $u$ and $dv$. A common guideline is the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential). In the integrand $x \cos x$, we have an algebraic function ($x$) and a trigonometric function ($\cos x$). According to LIATE, we choose the algebraic function as $u$.

Let $u = x$.

Then $dv = \cos x \, dx$.

Now we need to find $du$ and $v$.

Differentiate $u$ with respect to $x$ to find $du$: $du = \frac{d}{dx}(x) dx = 1 \, dx = dx$.

Integrate $dv$ to find $v$: $v = \int dv = \int \cos x \, dx = \sin x$. (We omit the constant of integration at this step).

Apply the integration by parts formula $\int u \, dv = uv - \int v \, du$:

$\int x \cos x dx = (x)(\sin x) - \int (\sin x) (dx)$

$\int x \cos x dx = x \sin x - \int \sin x \, dx$

Now, evaluate the remaining integral $\int \sin x \, dx$. The integral of $\sin x$ is $-\cos x$.

$\int x \cos x dx = x \sin x - (-\cos x) + C$

where $C$ is the constant of integration.

Simplify the expression:

$\int x \cos x dx = x \sin x + \cos x + C$

Now, compare our result with the given options:

(A) $x \sin x + \cos x + C$

(B) $x \sin x - \cos x + C$

(C) $-x \sin x + \cos x + C$

(D) $-x \sin x - \cos x + C$

Our result matches option (A).

Thus, the correct answer is (A) $x \sin x + \cos x + C$.

Solution:

We need to evaluate the indefinite integral $\int \frac{dx}{\sqrt{9-x^2}}$.

This integral is in the form of a standard integral $\int \frac{dx}{\sqrt{a^2-x^2}}$.

Comparing the given integral $\int \frac{dx}{\sqrt{9-x^2}}$ with the standard form $\int \frac{dx}{\sqrt{a^2-x^2}}$, we can identify $a^2 = 9$.

Taking the square root of $a^2=9$, we get $a = \sqrt{9} = 3$ (since $a$ is typically taken as positive in this formula context).

The standard formula for the integral $\int \frac{dx}{\sqrt{a^2-x^2}}$ is $\sin^{-1}\left(\frac{x}{a}\right) + C$, where $C$ is the constant of integration.

Substitute the value of $a=3$ into the standard formula:

$\int \frac{dx}{\sqrt{9-x^2}} = \int \frac{dx}{\sqrt{3^2-x^2}} = \sin^{-1}\left(\frac{x}{3}\right) + C$

Now, compare our result with the given options:

(A) $\sin^{-1}\left(\frac{x}{3}\right) + C$

(B) $\cos^{-1}\left(\frac{x}{3}\right) + C$

(C) $\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right) + C$

(D) $\log\left|x + \sqrt{9-x^2}\right| + C$

Our calculated result $\sin^{-1}\left(\frac{x}{3}\right) + C$ matches option (A).

Thus, the correct answer is (A) $\sin^{-1}\left(\frac{x}{3}\right) + C$.

Solution:

We need to find the appropriate method to evaluate the integral $\int \frac{\log x}{x} dx$.

Let's examine the integrand $\frac{\log x}{x}$. This can be written as $(\log x) \cdot \left(\frac{1}{x}\right)$.

Let's consider the given methods:

(A) Integration by parts: This method is useful for integrating products of functions using the formula $\int u \, dv = uv - \int v \, du$. While the integrand is a product of $\log x$ and $\frac{1}{x}$, let's see if another method is more direct.

(B) Partial fractions: This method is used for integrating rational functions (ratios of polynomials). The integrand $\frac{\log x}{x}$ is not a rational function because $\log x$ is not a polynomial.

(C) Substitution ($u = \log x$): Let's try the substitution $u = \log x$. Differentiating both sides with respect to $x$, we get $\frac{du}{dx} = \frac{1}{x}$. This means $du = \frac{1}{x} dx$. The integral $\int \frac{\log x}{x} dx$ can be rewritten as $\int (\log x) \cdot \left(\frac{1}{x} dx\right)$. Substituting $u = \log x$ and $du = \frac{1}{x} dx$, the integral transforms into $\int u \, du$. This is a very simple integral that can be easily evaluated.

$\int u \, du = \frac{u^{1+1}}{1+1} + C = \frac{u^2}{2} + C$.

Substituting back $u = \log x$, we get the result $\frac{(\log x)^2}{2} + C$.

This shows that the substitution $u = \log x$ directly simplifies the integral into a standard power rule form.

(D) Standard formulas: The integral $\int \frac{\log x}{x} dx$ itself is not typically listed as one of the most basic standard integral forms, although the result obtained after substitution is based on a standard formula ($\int u^n du$). The question asks for the method used to obtain the integral, not if the resulting integral form is standard.

Comparing the methods, substitution with $u = \log x$ is the most direct and appropriate technique to evaluate $\int \frac{\log x}{x} dx$ because the derivative of $\log x$, which is $\frac{1}{x}$, is present in the integrand.

Thus, the correct answer is (C) Substitution ($u = \log x$).

Solution:

We need to evaluate the definite integral $\int_1^e \log x dx$.

The integral of $\log x$ is typically found using integration by parts. The formula is $\int u \, dv = uv - \int v \, du$.

Let $u = \log x$ and $dv = dx$.

Then, $du = \frac{1}{x} dx$ and $v = \int dx = x$.

Applying the integration by parts formula:

$\int \log x dx = (\log x)(x) - \int x \left(\frac{1}{x} dx\right)$

$\int \log x dx = x \log x - \int 1 dx$

$\int \log x dx = x \log x - x + C$

So, the indefinite integral of $\log x$ is $x \log x - x$.

Now, we evaluate the definite integral using the Fundamental Theorem of Calculus, which states that $\int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)$, where $F(x)$ is an antiderivative of $f(x)$.

Here, $f(x) = \log x$ and $F(x) = x \log x - x$. The limits of integration are $a=1$ and $b=e$.

Evaluate $F(x)$ at the upper limit $x=e$:

$F(e) = e \log e - e$

Since $\log e = 1$ (natural logarithm),

$F(e) = e(1) - e = e - e = 0$

Evaluate $F(x)$ at the lower limit $x=1$:

$F(1) = 1 \log 1 - 1$

Since $\log 1 = 0$,

$F(1) = 1(0) - 1 = 0 - 1 = -1$

Now, subtract $F(1)$ from $F(e)$:

$\int_1^e \log x dx = F(e) - F(1) = 0 - (-1) = 0 + 1 = 1$

Comparing our result with the given options, we find that the value of the integral is $1$, which corresponds to option (A).

Thus, the correct answer is (A) 1.

Solution:

The marginal cost function is given by $MC(x) = 2x + 5$.

The marginal cost is the rate of change of the total cost with respect to the number of units produced, i.e., $MC(x) = \frac{d}{dx}(C(x))$.

To find the total cost function $C(x)$, we need to integrate the marginal cost function with respect to $x$.

$C(x) = \int MC(x) dx$

$C(x) = \int (2x + 5) dx$

Now, we evaluate the indefinite integral:

$\int (2x + 5) dx = \int 2x dx + \int 5 dx$

Using the power rule for integration ($\int x^n dx = \frac{x^{n+1}}{n+1} + C$, for $n \neq -1$) and the integral of a constant ($\int k dx = kx + C$):

$\int 2x dx = 2 \int x^1 dx = 2 \cdot \frac{x^{1+1}}{1+1} + C_1 = 2 \cdot \frac{x^2}{2} + C_1 = x^2 + C_1$

$\int 5 dx = 5x + C_2$

So, $C(x) = (x^2 + C_1) + (5x + C_2) = x^2 + 5x + (C_1 + C_2)$.

Let the constant of integration be $K'$. Then $C(x) = x^2 + 5x + K'$.

The constant of integration $K'$ represents the total cost when zero units are produced, which is the fixed cost.

We are given that the fixed cost is $\textsf{₹} 100$.

So, when $x=0$, $C(0) = 100$.

Using our integrated function, $C(0) = (0)^2 + 5(0) + K' = 0 + 0 + K' = K'$.

Therefore, $K' = 100$.

Substituting the value of $K'$ back into the total cost function, we get:

$C(x) = x^2 + 5x + 100$

Now, we compare this result with the given options:

(A) $x^2 + 5x$

(B) $2x^2 + 5x + 100$

(C) $x^2 + 5x + 100$

(D) $2 + 100$

Our derived total cost function $C(x) = x^2 + 5x + 100$ matches option (C).

Thus, the correct answer is (C) $x^2 + 5x + 100$.

Solution:

We need to evaluate the definite integral $\int_0^{\pi/2} \sin^3 x dx$.

We can rewrite $\sin^3 x$ using the identity $\sin^2 x = 1 - \cos^2 x$:

$\sin^3 x = \sin^2 x \cdot \sin x = (1 - \cos^2 x) \sin x$

So the integral becomes:

$\int_0^{\pi/2} (1 - \cos^2 x) \sin x dx$

Let's use the method of substitution. Let $u = \cos x$.

Differentiating both sides with respect to $x$, we get $\frac{du}{dx} = -\sin x$.

This implies $du = -\sin x \, dx$, or $\sin x \, dx = -du$.

We also need to change the limits of integration according to the substitution:

When the original lower limit $x = 0$, the new lower limit for $u$ is $u = \cos(0) = 1$.

When the original upper limit $x = \frac{\pi}{2}$, the new upper limit for $u$ is $u = \cos(\frac{\pi}{2}) = 0$.

Substituting $u = \cos x$, $du = -\sin x \, dx$, and the new limits into the integral:

$\int_{x=0}^{x=\pi/2} (1 - \cos^2 x) \sin x dx = \int_{u=1}^{u=0} (1 - u^2) (-du)$

$\int_1^0 -(1 - u^2) du = \int_1^0 (u^2 - 1) du$

Using the property $\int_a^b f(u) du = - \int_b^a f(u) du$, we can reverse the limits and change the sign:

$\int_1^0 (u^2 - 1) du = - \int_0^1 (u^2 - 1) du = \int_0^1 (1 - u^2) du$

Now, we evaluate the definite integral with respect to $u$ from $0$ to $1$:

$\int_0^1 (1 - u^2) du = \left[ u - \frac{u^{2+1}}{2+1} \right]_0^1$

$\int_0^1 (1 - u^2) du = \left[ u - \frac{u^3}{3} \right]_0^1$

Apply the limits of integration:

$\left( 1 - \frac{1^3}{3} \right) - \left( 0 - \frac{0^3}{3} \right)$

$\left( 1 - \frac{1}{3} \right) - (0 - 0)$

$1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{3-1}{3} = \frac{2}{3}$

So, the value of the definite integral is $\frac{2}{3}$.

Comparing our result with the given options, we find that the value matches option (B).

Thus, the correct answer is (B) $2/3$.

Alternate Solution using Reduction Formula:

For definite integrals of the form $\int_0^{\pi/2} \sin^n x dx$ or $\int_0^{\pi/2} \cos^n x dx$, where $n$ is a positive integer, Wallis's Formulas can be used.

For $n=3$ (odd), the formula is:

$\int_0^{\pi/2} \sin^3 x dx = \frac{(3-1)}{3} \cdot 1 = \frac{2}{3}$

This method confirms the previous result.

Solution:

An indefinite integral of a function $f(x)$ is denoted by $\int f(x) dx = F(x) + C$, where $F(x)$ is any antiderivative of $f(x)$ and $C$ is the constant of integration.

The derivative of $F(x)$ is $f(x)$, i.e., $F'(x) = f(x)$.

The derivative of $F(x) + C$ with respect to $x$ is $\frac{d}{dx}(F(x) + C) = \frac{d}{dx}(F(x)) + \frac{d}{dx}(C) = f(x) + 0 = f(x)$.

This shows that for any value of $C$, $F(x) + C$ is an antiderivative of $f(x)$.

The graph of $y = F(x)$ represents one specific antiderivative.

The graph of $y = F(x) + C$ for different values of $C$ represents a family of curves.

If we consider two curves from this family, say $y_1 = F(x) + C_1$ and $y_2 = F(x) + C_2$ (where $C_1 \neq C_2$), the difference in their y-values at any given $x$ is $(F(x) + C_2) - (F(x) + C_1) = C_2 - C_1$. This difference is a constant.

This means that the curve $y_2 = F(x) + C_2$ is obtained by shifting the curve $y_1 = F(x) + C_1$ vertically by a constant amount $(C_2 - C_1)$.

Curves that are vertical shifts of each other have the same shape and the same slope at corresponding x-coordinates.

Since their slopes are the same at every $x$, their tangent lines at the same $x$-coordinate are parallel.

Therefore, the curves in the family $y = F(x) + C$ are parallel to each other in the sense that they are vertical translations of one base curve $y=F(x)$.

Let's evaluate the options:

(A) Perpendicular: Incorrect, vertical shifts do not result in perpendicular curves.

(B) Parallel (vertically shifted): Correct, the constant $C$ represents a vertical shift, resulting in parallel curves.

(C) Tangent: Incorrect, tangent refers to a line touching a curve.

(D) Intersecting: Incorrect, for different values of $C$, the curves $y=F(x)+C$ are vertical translations and do not intersect each other (unless $F(x)$ is not single-valued, which is not the typical case in this context).

Thus, the correct answer is (B) Parallel (vertically shifted).

Solution:

We need to evaluate the indefinite integral $\int \frac{dx}{1+\sin x}$.

To simplify the integrand, we can multiply the numerator and the denominator by the conjugate of the denominator, which is $1-\sin x$.

$\int \frac{dx}{1+\sin x} = \int \frac{1}{1+\sin x} \cdot \frac{1-\sin x}{1-\sin x} dx$

$\int \frac{dx}{1+\sin x} = \int \frac{1-\sin x}{(1+\sin x)(1-\sin x)} dx$

The denominator is a difference of squares: $(1+\sin x)(1-\sin x) = 1^2 - \sin^2 x = 1 - \sin^2 x$.

Using the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$, we have $1 - \sin^2 x = \cos^2 x$.

So, the integral becomes:

$\int \frac{1-\sin x}{\cos^2 x} dx$

Now, we can split the integrand into two separate fractions:

$\int \left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) dx$

We can rewrite these terms using standard trigonometric identities:

$\frac{1}{\cos^2 x} = \sec^2 x$

$\frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \cdot \sec x = \sec x \tan x$

So, the integral is now:

$\int (\sec^2 x - \sec x \tan x) dx$

We can integrate each term separately:

$\int \sec^2 x \, dx = \tan x + C_1$

$\int \sec x \tan x \, dx = \sec x + C_2$

Combining the results, we get:

$\int (\sec^2 x - \sec x \tan x) dx = \int \sec^2 x \, dx - \int \sec x \tan x \, dx$

$= (\tan x + C_1) - (\sec x + C_2)$

$= \tan x - \sec x + (C_1 - C_2)$

Let $C = C_1 - C_2$ be the constant of integration.

The indefinite integral is $\tan x - \sec x + C$.

Comparing our result with the given options, we find that it matches option (A).

Thus, the correct answer is (A) $\tan x - \sec x + C$.

Solution:

The statement $\int f(x) dx = g(x) + C$ means that the indefinite integral of $f(x)$ with respect to $x$ is $g(x) + C$.

By the definition of the indefinite integral, the expression $g(x) + C$ is an antiderivative of the function $f(x)$.

This means that the derivative of $g(x) + C$ with respect to $x$ is equal to $f(x)$.

Mathematically, this can be written as:

$\frac{d}{dx} [g(x) + C] = f(x)$

Using the properties of differentiation, the derivative of a sum is the sum of the derivatives, and the derivative of a constant is $0$.

$\frac{d}{dx} [g(x) + C] = \frac{d}{dx} [g(x)] + \frac{d}{dx} [C]$

$\frac{d}{dx} [g(x) + C] = \frac{d}{dx} [g(x)] + 0$

$\frac{d}{dx} [g(x) + C] = \frac{d}{dx} [g(x)]$

Combining this with the previous result $\frac{d}{dx} [g(x) + C] = f(x)$, we get:

$\frac{d}{dx} [g(x)] = f(x)$

Therefore, if $\int f(x) dx = g(x) + C$, then the derivative of $g(x)$ is equal to $f(x)$.

Comparing our result with the given options:

(A) $f(x)$

(B) $g(x)$

(C) $f'(x)$

(D) $g'(x) + C$

Our result $\frac{d}{dx} [g(x)] = f(x)$ matches option (A).

Thus, the correct answer is (A) $f(x)$.

Solution:

We need to evaluate the indefinite integral $\int \frac{dx}{\sqrt{x^2-4}}$.

This integral is in the form of a standard integral $\int \frac{dx}{\sqrt{x^2-a^2}}$.

Comparing the given integral $\int \frac{dx}{\sqrt{x^2-4}}$ with the standard form $\int \frac{dx}{\sqrt{x^2-a^2}}$, we can identify $a^2 = 4$.

Taking the square root of $a^2=4$, we get $a = \sqrt{4} = 2$ (assuming $a > 0$).

The standard formula for the integral $\int \frac{dx}{\sqrt{x^2-a^2}}$ is $\log\left|x + \sqrt{x^2-a^2}\right| + C$, where $C$ is the constant of integration.

Substitute the value of $a=2$ into the standard formula:

$\int \frac{dx}{\sqrt{x^2-4}} = \int \frac{dx}{\sqrt{x^2-2^2}} = \log\left|x + \sqrt{x^2-4}\right| + C$

Now, compare our result with the given options:

(A) $\sin^{-1}\left(\frac{x}{2}\right) + C$ (Incorrect, this is for $\int \frac{dx}{\sqrt{a^2-x^2}}$)

(B) $\log\left|x + \sqrt{x^2-4}\right| + C$ (Matches our result)

(C) $\frac{1}{2}\log\left|\frac{x-2}{x+2}\right| + C$ (Incorrect, this is for $\int \frac{dx}{x^2-a^2}$)

(D) $\sec^{-1}\left(\frac{x}{2}\right) + C$ (Incorrect, this is related to $\int \frac{dx}{x\sqrt{x^2-a^2}}$)

Our calculated result $\log\left|x + \sqrt{x^2-4}\right| + C$ matches option (B).

Thus, the correct answer is (B) $\log\left|x + \sqrt{x^2-4}\right| + C$.

Solution:

We are asked to apply the property of definite integrals $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ to the given integral $\int_0^1 x(1-x)^n dx$.

In this integral, the upper limit of integration is $a=1$.

The function being integrated is $f(x) = x(1-x)^n$.

According to the property, we need to substitute $x$ with $a-x$, which in this case is $1-x$, within the integrand $f(x)$.

So, we need to find $f(1-x)$.

Substitute $(1-x)$ for every $x$ in the expression for $f(x) = x(1-x)^n$:

$f(1-x) = (1-x) \cdot (1 - (1-x))^n$

$f(1-x) = (1-x) \cdot (1 - 1 + x)^n$

$f(1-x) = (1-x) \cdot (x)^n$

$f(1-x) = (1-x)x^n$

Now, apply the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ with $a=1$ and $f(x) = x(1-x)^n$:

$\int_0^1 x(1-x)^n dx = \int_0^1 f(1-x) dx$

Substitute the expression we found for $f(1-x)$:

$\int_0^1 x(1-x)^n dx = \int_0^1 (1-x)x^n dx$

This resulting integral $\int_0^1 (1-x)x^n dx$ is the form obtained after applying the property. This transformed integral is often easier to evaluate by expanding the integrand: $(1-x)x^n = x^n - x^{n+1}$.

Let's compare our result with the given options:

(A) $\int_0^1 x^n (1-x) dx$ (This is the same as $\int_0^1 (1-x)x^n dx$)

(B) $\int_0^1 (1-x) x^n dx$ (This is the same as $\int_0^1 (1-x)x^n dx$)

(C) $\int_0^1 (1-x)^n x dx$ (This is the original integral)

(D) $\int_0^1 x (1-x)^n dx = \int_0^1 (1-x)(1-(1-x))^n dx = \int_0^1 (1-x)x^n dx$. This is the property application.

Options (A) and (B) both represent the transformed integral. Option (D) explicitly shows the application steps leading to the transformed integral $\int_0^1 (1-x)x^n dx$ and states that this is the result of the property application.

Given the structure of option (D), it appears to be the intended answer as it demonstrates the application of the property and the resulting equality.

Thus, the correct answer is (D) $\int_0^1 x (1-x)^n dx = \int_0^1 (1-x)(1-(1-x))^n dx = \int_0^1 (1-x)x^n dx$. This is the property application..

Solution:

Let's analyze each integral in Column I and identify the most appropriate method or form for integration from Column II.

(i) $\int \frac{x}{x^2+1} dx$:

Observe that the derivative of the denominator $x^2+1$ is $2x$, which is proportional to the numerator $x$. This structure suggests using substitution.

Let $u = x^2+1$. Then $du = 2x \, dx$. So, $x \, dx = \frac{1}{2} du$.

The integral becomes $\int \frac{1}{u} \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du$. This is a standard integral of the form $\frac{1}{2} \log|u| + C = \frac{1}{2} \log|x^2+1| + C$.

The method used is Substitution.

(ii) $\int x e^x dx$:

This integral is a product of an algebraic function ($x$) and an exponential function ($e^x$). Integrals of products of different types of functions are often evaluated using Integration by Parts ($\int u \, dv = uv - \int v \, du$).

Using the LIATE rule, we choose $u=x$ and $dv=e^x dx$. Then $du=dx$ and $v=e^x$.

$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C$.

The method used is Integration by Parts.

(iii) $\int \frac{x^2-1}{x^2+1} dx$:

The integrand is a rational function where the degree of the numerator ($2$) is equal to the degree of the denominator ($2$). When the degree of the numerator is greater than or equal to the degree of the denominator, we should perform Polynomial long division first to write the integrand as a polynomial plus a proper rational function.

We can write $\frac{x^2-1}{x^2+1} = \frac{(x^2+1) - 2}{x^2+1} = 1 - \frac{2}{x^2+1}$.

The integral becomes $\int \left(1 - \frac{2}{x^2+1}\right) dx = \int 1 dx - 2 \int \frac{1}{x^2+1} dx = x - 2 \tan^{-1} x + C$.

The initial step is effectively equivalent to polynomial long division, transforming the integrand.

The method used is related to Polynomial long division before integration.

(iv) $\int \frac{dx}{x^2+2x+2}$:

The denominator is a quadratic expression $x^2+2x+2$. The discriminant is $b^2-4ac = 2^2 - 4(1)(2) = 4 - 8 = -4$, which is negative, indicating no real roots. To integrate this form $\int \frac{dx}{ax^2+bx+c}$ where $ax^2+bx+c$ has no real roots, we use the method of Completing the square in the denominator.

$x^2+2x+2 = (x^2+2x+1) + 1 = (x+1)^2 + 1^2$.

The integral becomes $\int \frac{dx}{(x+1)^2+1^2}$. Let $u = x+1$, then $du = dx$. The integral is $\int \frac{du}{u^2+1^2}$, which is a standard integral $\tan^{-1}\left(\frac{u}{1}\right) + C = \tan^{-1}(x+1) + C$.

The method used is Complete the square in the denominator.

Matching the integrals with the methods:

(i) $\int \frac{x}{x^2+1} dx$ corresponds to (b) Substitution.

(ii) $\int x e^x dx$ corresponds to (a) Integration by Parts.

(iii) $\int \frac{x^2-1}{x^2+1} dx$ corresponds to (d) Polynomial long division before integration.

(iv) $\int \frac{dx}{x^2+2x+2}$ corresponds to (c) Complete the square in the denominator.

The correct matching is:

(i) - (b)

(ii) - (a)

(iii) - (d)

(iv) - (c)

Solution:

We need to evaluate the indefinite integral $\int \sqrt{x^2+9} dx$.

This integral is in the form of a standard integral $\int \sqrt{x^2+a^2} dx$.

Comparing the given integral $\int \sqrt{x^2+9} dx$ with the standard form $\int \sqrt{x^2+a^2} dx$, we can identify $a^2 = 9$.

Taking the positive square root of $a^2=9$, we get $a = \sqrt{9} = 3$.

The standard formula for the integral $\int \sqrt{x^2+a^2} dx$ is:

$\int \sqrt{x^2+a^2} dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log\left|x + \sqrt{x^2+a^2}\right| + C$

where $C$ is the constant of integration.

Substitute the value of $a=3$ into the standard formula:

$\int \sqrt{x^2+9} dx = \int \sqrt{x^2+3^2} dx$

$= \frac{x}{2}\sqrt{x^2+3^2} + \frac{3^2}{2}\log\left|x + \sqrt{x^2+3^2}\right| + C$

$= \frac{x}{2}\sqrt{x^2+9} + \frac{9}{2}\log\left|x + \sqrt{x^2+9}\right| + C$

Now, compare our result with the given options:

(A) $\frac{x}{2}\sqrt{x^2+9} + \frac{9}{2}\log\left|x + \sqrt{x^2+9}\right| + C$ (Matches our result)

(B) $\frac{x}{2}\sqrt{x^2+9} - \frac{9}{2}\log\left|x + \sqrt{x^2+9}\right| + C$ (Incorrect sign before the logarithm term)

(C) $\frac{x}{2}\sqrt{x^2+9} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) + C$ (Incorrect formula, $\sin^{-1}$ is for $\sqrt{a^2-x^2}$)

(D) $\frac{x}{2}\sqrt{x^2+9} - \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) + C$ (Incorrect formula and sign)

Our calculated result matches option (A).

Thus, the correct answer is (A) $\frac{x}{2}\sqrt{x^2+9} + \frac{9}{2}\log\left|x + \sqrt{x^2+9}\right| + C$.

Solution:

We need to evaluate the definite integral $\int_0^{2\pi} |\sin x| dx$.

The function $|\sin x|$ is always non-negative. The behavior of $\sin x$ changes sign over the interval $[0, 2\pi]$.

• For $x \in [0, \pi]$, $\sin x \ge 0$, so $|\sin x| = \sin x$.
• For $x \in [\pi, 2\pi]$, $\sin x \le 0$, so $|\sin x| = -\sin x$.

We can split the integral into two parts based on these intervals:

$\int_0^{2\pi} |\sin x| dx = \int_0^{\pi} |\sin x| dx + \int_{\pi}^{2\pi} |\sin x| dx$

$\int_0^{2\pi} |\sin x| dx = \int_0^{\pi} \sin x dx + \int_{\pi}^{2\pi} (-\sin x) dx$

$\int_0^{2\pi} |\sin x| dx = \int_0^{\pi} \sin x dx - \int_{\pi}^{2\pi} \sin x dx$

Now, we evaluate each definite integral.

The antiderivative of $\sin x$ is $-\cos x$.

Evaluate the first integral:

$\int_0^{\pi} \sin x dx = [-\cos x]_0^{\pi}$

$= (-\cos \pi) - (-\cos 0)$

$= (-(-1)) - (-1)$

$= 1 + 1 = 2$

Evaluate the second integral:

$\int_{\pi}^{2\pi} \sin x dx = [-\cos x]_{\pi}^{2\pi}$

$= (-\cos 2\pi) - (-\cos \pi)$

$= (-1) - (-(-1))$

$= -1 - 1 = -2$

Combine the results:

$\int_0^{2\pi} |\sin x| dx = (2) - (-2) = 2 + 2 = 4$

Alternate Method using Periodicity:

The function $|\sin x|$ is periodic with period $\pi$. The integral over $[0, 2\pi]$ is equivalent to the integral over two periods.

$\int_0^{2\pi} |\sin x| dx = \int_0^{\pi} |\sin x| dx + \int_{\pi}^{2\pi} |\sin x| dx$

By the property $\int_a^{a+T} f(x) dx = \int_0^T f(x) dx$ for a periodic function $f(x)$ with period $T$, we have $\int_{\pi}^{2\pi} |\sin x| dx = \int_0^{\pi} |\sin x| dx$.

So, $\int_0^{2\pi} |\sin x| dx = 2 \int_0^{\pi} |\sin x| dx$

Since $|\sin x| = \sin x$ for $x \in [0, \pi]$,

$\int_0^{2\pi} |\sin x| dx = 2 \int_0^{\pi} \sin x dx$

As calculated before, $\int_0^{\pi} \sin x dx = 2$.

So, $\int_0^{2\pi} |\sin x| dx = 2 \cdot 2 = 4$.

Comparing our result with the given options, we find that the value of the integral is $4$, which corresponds to option (C).

Thus, the correct answer is (C) 4.

Solution:

The definite integral $\int_a^b f(x) dx$ is formally defined as the limit of a Riemann sum.

To form a Riemann sum, we partition the interval $[a, b]$ into $n$ subintervals of width $\Delta x = \frac{b-a}{n}$. Let the partition points be $a = x_0 < x_1 < x_2 < \dots < x_n = b$, where $x_i = a + i\Delta x$ for $i=0, 1, \dots, n$.

In each subinterval $[x_{i-1}, x_i]$, we choose a sample point $c_i$. The Riemann sum is then given by $\sum\limits_{i=1}^n f(c_i) \Delta x$.

The definite integral is the limit of this Riemann sum as the number of subintervals $n$ approaches infinity (which implies $\Delta x$ approaches $0$):

$\int_a^b f(x) dx = \lim\limits_{n \to \infty} \sum\limits_{i=1}^n f(c_i) \Delta x$

This definition holds if $f(x)$ is integrable on $[a, b]$. A common condition for integrability is continuity of $f(x)$ on $[a, b]$, or $f(x)$ being bounded with a finite number of discontinuities.

Let's examine the given options, which represent different choices for the sample point $c_i$ within the $i$-th subinterval $[x_{i-1}, x_i]$:

(A) $\lim\limits_{n \to \infty} \sum\limits_{i=1}^n f(a + i\Delta x) \Delta x$, where $\Delta x = \frac{b-a}{n}$. Here, the sample point $c_i$ is chosen as the right endpoint of the $i$-th subinterval, $c_i = x_i = a + i\Delta x$. This is a valid form of the Riemann sum definition (specifically, a right Riemann sum).

(B) $\lim\limits_{n \to \infty} \frac{b-a}{n} \sum\limits_{i=1}^n f(a + i\frac{b-a}{n})$. This is the same as option (A), just with $\Delta x$ explicitly written as $\frac{b-a}{n}$. The sample point is $c_i = a + i\frac{b-a}{n} = a + i\Delta x$, the right endpoint. This is also a valid form.

(C) $\lim\limits_{n \to \infty} \frac{b-a}{n} [f(a) + f(a+\Delta x) + \dots + f(a+(n-1)\Delta x)]$. This can be written as $\lim\limits_{n \to \infty} \sum\limits_{i=1}^n f(a + (i-1)\Delta x) \Delta x$. Here, the sample point $c_i$ is chosen as the left endpoint of the $i$-th subinterval, $c_i = x_{i-1} = a + (i-1)\Delta x$. This is also a valid form of the Riemann sum definition (specifically, a left Riemann sum).

(D) All of the above, under certain conditions on $f(x)$. If the function $f(x)$ is integrable on $[a, b]$ (e.g., continuous), then the limit of the Riemann sum exists and is equal to the definite integral, regardless of the choice of sample points $c_i$ within each subinterval (as long as the maximum width of the subintervals approaches zero). Options (A), (B), and (C) represent specific valid choices for $c_i$ in the limit definition. Therefore, under the conditions for integrability of $f(x)$, all three expressions (A), (B), and (C) are equivalent to the definite integral $\int_a^b f(x) dx$.

Since options (A), (B), and (C) are all valid formulations of the definite integral as a limit of a sum (representing right, right, and left Riemann sums, respectively), option (D) which states "All of the above, under certain conditions on $f(x)$" is the most comprehensive and correct answer, as these forms are equivalent when the integral exists (i.e., when $f(x)$ is integrable).

Thus, the correct answer is (D) All of the above, under certain conditions on $f(x)$..

Solution:

We are given the function $F(x)$ defined as a definite integral:

$F(x) = \int_0^x t^2 dt$

We need to find the derivative of $F(x)$ with respect to $x$, which is $F'(x)$.

This problem can be solved using the Fundamental Theorem of Calculus, Part 1.

The theorem states that if a function $f(t)$ is continuous on an interval $[a, b]$, then the function $F(x)$ defined by $F(x) = \int_a^x f(t) dt$ for $x$ in $[a, b]$ is differentiable, and its derivative is given by $F'(x) = f(x)$.

In our case, the integral is $F(x) = \int_0^x t^2 dt$.

Here, the integrand is $f(t) = t^2$. This function is continuous for all real numbers $t$.

The lower limit of integration is a constant $a=0$, and the upper limit is $x$.

Applying the Fundamental Theorem of Calculus, Part 1, we have:

$F'(x) = \frac{d}{dx} \left( \int_0^x t^2 dt \right) = f(x)$

Substituting $t^2$ for $f(t)$, we get:

$F'(x) = x^2$

Alternatively, we can first evaluate the integral $F(x)$ and then differentiate it.

$F(x) = \int_0^x t^2 dt$

The indefinite integral of $t^2$ is $\frac{t^3}{3}$.

Evaluating the definite integral from $0$ to $x$:

$F(x) = \left[ \frac{t^3}{3} \right]_0^x = \frac{x^3}{3} - \frac{0^3}{3} = \frac{x^3}{3}$

Now, differentiate $F(x)$ with respect to $x$:

$F'(x) = \frac{d}{dx} \left( \frac{x^3}{3} \right) = \frac{1}{3} \frac{d}{dx} (x^3) = \frac{1}{3} (3x^2) = x^2$

Both methods lead to the same result: $F'(x) = x^2$.

Comparing our result with the given options:

(A) $x^2$

(B) $t^2$

(C) $\frac{x^3}{3}$

(D) $2x$

Our result $x^2$ matches option (A).

Thus, the correct answer is (A) $x^2$.

Solution:

We need to evaluate the indefinite integral $\int \frac{\sin x}{\cos^2 x} dx$.

We can rewrite the integrand by separating the terms in the denominator:

$\frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x \cdot \cos x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}$

Using standard trigonometric identities, $\frac{\sin x}{\cos x} = \tan x$ and $\frac{1}{\cos x} = \sec x$.

So, the integrand can be written as $\tan x \sec x$ or $\sec x \tan x$.

The integral becomes $\int \sec x \tan x \, dx$.

This is a standard integral form. The derivative of $\sec x$ with respect to $x$ is $\sec x \tan x$. Therefore, the integral of $\sec x \tan x$ is $\sec x$ plus the constant of integration.

$\int \sec x \tan x \, dx = \sec x + C$

Alternatively, we can use substitution.

Let $u = \cos x$.

Differentiating both sides with respect to $x$, we get $\frac{du}{dx} = -\sin x$.

This means $du = -\sin x \, dx$, or $\sin x \, dx = -du$.

The integral can be written as $\int \frac{1}{\cos^2 x} \cdot \sin x \, dx$.

Substituting $u = \cos x$ and $\sin x \, dx = -du$, the integral becomes:

$\int \frac{1}{u^2} (-du) = -\int u^{-2} du$

Using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C$, for $n \neq -1$):

$-\int u^{-2} du = -\left( \frac{u^{-2+1}}{-2+1} \right) + C = -\left( \frac{u^{-1}}{-1} \right) + C = -(-u^{-1}) + C = u^{-1} + C$

$u^{-1} + C = \frac{1}{u} + C$

Substitute back $u = \cos x$:

$\frac{1}{\cos x} + C = \sec x + C$

Both methods yield the same result.

Comparing our result with the given options:

(A) $\sec x + C$ (Matches our result)

(B) $\tan x + C$ (Incorrect, the derivative of $\tan x$ is $\sec^2 x$)

(C) $\sec x \tan x + C$ (Incorrect, this is the integrand itself)

(D) $\text{cosec } x + C$ (Incorrect, the derivative of $-\text{cosec } x$ is $\text{cosec } x \cot x$)

Our result $\sec x + C$ matches option (A).

Thus, the correct answer is (A) $\sec x + C$.

Solution:

Let's analyze the given Assertion (A) and Reason (R).

Assertion (A): $\int \frac{1}{x^2+x+1} dx$ can be evaluated by completing the square in the denominator.

The integrand is of the form $\frac{1}{ax^2+bx+c}$. When the quadratic denominator $ax^2+bx+c$ cannot be factored into real linear factors (i.e., its discriminant $b^2-4ac < 0$), the standard method for integrating $\frac{1}{ax^2+bx+c}$ is to complete the square in the denominator to transform it into the form $\int \frac{dx}{(x+p)^2 + q^2}$ or $\int \frac{dx}{(x+p)^2 - q^2}$ or $\int \frac{dx}{q^2 - (x+p)^2}$, which then leads to an integral involving $\tan^{-1}$, $\log$, or inverse hyperbolic functions, respectively.

Let's calculate the discriminant of $x^2+x+1$: $a=1, b=1, c=1$. Discriminant $= b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3$. Since the discriminant is negative, the quadratic $x^2+x+1$ has no real roots and cannot be factored into real linear factors.

Therefore, completing the square in the denominator is the appropriate method to evaluate this integral, as it transforms it into a standard inverse tangent form. So, Assertion (A) is true.

Reason (R): The denominator $x^2+x+1$ is a quadratic expression that can be written in the form $(x+p)^2 + q^2$.

Let's complete the square for the quadratic expression $x^2+x+1$.

$x^2+x+1 = x^2 + 2 \left(\frac{1}{2}\right) x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 1$

$= \left( x + \frac{1}{2} \right)^2 - \frac{1}{4} + 1$

$= \left( x + \frac{1}{2} \right)^2 + \left( 1 - \frac{1}{4} \right)$

$= \left( x + \frac{1}{2} \right)^2 + \frac{3}{4}$

This is in the form $(x+p)^2 + q^2$, where $p = \frac{1}{2}$ and $q^2 = \frac{3}{4}$ (so $q = \frac{\sqrt{3}}{2}$).

Since the quadratic $x^2+x+1$ can indeed be written in the form $(x+p)^2 + q^2$, Reason (R) is true.

Now, we need to determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that completing the square is the method to evaluate the integral. Reason (R) states that the denominator can be written in the form $(x+p)^2 + q^2$ by completing the square.

The fact that the denominator can be transformed into the form $(x+p)^2 + q^2$ by completing the square is precisely why completing the square is the correct method to evaluate the integral $\int \frac{dx}{x^2+x+1}$. This form leads to the standard integral $\int \frac{du}{u^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C$ after a substitution.

Therefore, Reason (R) is the correct explanation for Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Thus, the correct answer is (A) Both A and R are true and R is the correct explanation of A..

Solution:

We need to evaluate the definite integral $\int_0^2 [x] dx$, where $[x]$ is the greatest integer function.

The greatest integer function $[x]$ gives the largest integer less than or equal to $x$. The value of $[x]$ changes at each integer value of $x$.

The interval of integration is $[0, 2]$. We need to split the integral into subintervals where the value of $[x]$ is constant.

For $x$ in the interval $[0, 1)$, the greatest integer $[x] = 0$.

For $x$ in the interval $[1, 2)$, the greatest integer $[x] = 1$.

At $x=2$, the value is $[2]=2$, but for the integral over $[1, 2]$, the value up to (but not including) $2$ is relevant.

We can write the integral as the sum of integrals over these subintervals:

$\int_0^2 [x] dx = \int_0^1 [x] dx + \int_1^2 [x] dx$

Now, substitute the value of $[x]$ in each interval:

$\int_0^2 [x] dx = \int_0^1 0 \, dx + \int_1^2 1 \, dx$

Evaluate each integral:

The integral of $0$ from $0$ to $1$ is:

$\int_0^1 0 \, dx = [0x]_0^1 = 0(1) - 0(0) = 0 - 0 = 0$

The integral of $1$ from $1$ to $2$ is:

$\int_1^2 1 \, dx = [x]_1^2 = 2 - 1 = 1$

Add the results of the two integrals:

$\int_0^2 [x] dx = 0 + 1 = 1$

Comparing our result with the given options, we find that the value of the integral is $1$, which corresponds to option (B).

Thus, the correct answer is (B) 1.

Solution:

We are given the integral $\int e^x (\tan x + \sec^2 x) dx$ and told that it is of the form $\int e^x (f(x) + f'(x)) dx$. We need to identify the function $f(x)$.

The standard integral formula is $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.

In the given integral $\int e^x (\tan x + \sec^2 x) dx$, we have $e^x$ multiplied by the expression $(\tan x + \sec^2 x)$.

We need to identify a function $f(x)$ within the parentheses such that the other part is its derivative $f'(x)$.

Let's consider the functions present in the parentheses: $\tan x$ and $\sec^2 x$.

Recall the derivatives of standard trigonometric functions:

• $\frac{d}{dx}(\tan x) = \sec^2 x$
• $\frac{d}{dx}(\sec x) = \sec x \tan x$
• $\frac{d}{dx}(\sec^2 x) = \frac{d}{dx}((\sec x)^2) = 2 \sec x (\sec x \tan x) = 2 \sec^2 x \tan x$

In the expression $(\tan x + \sec^2 x)$, we can see that the derivative of $\tan x$ is $\sec^2 x$.

If we choose $f(x) = \tan x$, then $f'(x) = \sec^2 x$.

The expression in the parentheses is indeed in the form $f(x) + f'(x)$.

So, $f(x) = \tan x$ and $f'(x) = \sec^2 x$.

Therefore, the integral $\int e^x (\tan x + \sec^2 x) dx$ is equal to $\int e^x (\tan x + (\tan x)') dx$.

Using the formula, the value of the integral is $e^x \tan x + C$.

We are asked to identify $f(x)$. Based on our analysis, $f(x) = \tan x$.

Comparing our identification with the given options:

(A) $e^x$ (Incorrect)

(B) $\tan x$ (Matches our identification)

(C) $\sec^2 x$ (Incorrect, this is $f'(x)$ if $f(x) = \tan x$)

(D) $\tan x + \sec^2 x$ (Incorrect, this is the expression $f(x) + f'(x)$)

Thus, the correct answer is (B) $\tan x$.

Solution:

We need to evaluate the indefinite integral $\int \frac{x+1}{\sqrt{x^2+2x+3}} dx$.

Let's examine the relationship between the numerator $(x+1)$ and the expression inside the square root in the denominator $(x^2+2x+3)$.

Consider the derivative of the expression inside the square root:

$\frac{d}{dx}(x^2+2x+3) = 2x + 2 = 2(x+1)$

The numerator $(x+1)$ is proportional to the derivative of the expression inside the square root.

This suggests using the method of substitution.

Let $u = x^2+2x+3$.

Then, the differential $du$ is given by the derivative of $u$ with respect to $x$ multiplied by $dx$:

$du = (2x+2) dx$

$du = 2(x+1) dx$

From this, we can express $(x+1) dx$ in terms of $du$:

$(x+1) dx = \frac{1}{2} du$

Now, substitute $u = x^2+2x+3$ and $(x+1) dx = \frac{1}{2} du$ into the integral:

$\int \frac{x+1}{\sqrt{x^2+2x+3}} dx = \int \frac{\frac{1}{2} du}{\sqrt{u}}$

$\int \frac{x+1}{\sqrt{x^2+2x+3}} dx = \frac{1}{2} \int \frac{1}{\sqrt{u}} du$

We can write $\frac{1}{\sqrt{u}}$ as $u^{-1/2}$.

$\int \frac{x+1}{\sqrt{x^2+2x+3}} dx = \frac{1}{2} \int u^{-1/2} du$

Now, integrate $u^{-1/2}$ using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C$, where $n = -1/2$:

$\int u^{-1/2} du = \frac{u^{-1/2+1}}{-1/2+1} + C' = \frac{u^{1/2}}{1/2} + C' = 2u^{1/2} + C'$

Substitute this back into the integral expression with the factor of $\frac{1}{2}$:

$\int \frac{x+1}{\sqrt{x^2+2x+3}} dx = \frac{1}{2} (2u^{1/2}) + C$

$\int \frac{x+1}{\sqrt{x^2+2x+3}} dx = u^{1/2} + C$

$\int \frac{x+1}{\sqrt{x^2+2x+3}} dx = \sqrt{u} + C$

Finally, substitute back $u = x^2+2x+3$ to express the result in terms of $x$:

$\int \frac{x+1}{\sqrt{x^2+2x+3}} dx = \sqrt{x^2+2x+3} + C$

Comparing our result with the given options:

(A) $\sqrt{x^2+2x+3} + C$ (Matches our result)

(B) $\frac{1}{2}\sqrt{x^2+2x+3} + C$ (Incorrect)

(C) $\log\left|x+1 + \sqrt{x^2+2x+3}\right| + C$ (This is the integral of $\frac{1}{\sqrt{x^2+2x+3}}$)

(D) $\sin^{-1}\left(\frac{x+1}{\sqrt{2}}\right) + C$ (Incorrect form)

Thus, the correct answer is (A) $\sqrt{x^2+2x+3} + C$.

Solution:

We are given the condition $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$.

This is a property related to the integration of functions over a symmetric interval $[-a, a]$.

Recall the properties of definite integrals over symmetric intervals:

1. If $f(x)$ is an even function (i.e., $f(-x) = f(x)$ for all $x$ in the domain), then $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$.

2. If $f(x)$ is an odd function (i.e., $f(-x) = -f(x)$ for all $x$ in the domain), then $\int_{-a}^a f(x) dx = 0$.

The given condition $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$ exactly matches the property for an even function.

Therefore, the function $f(x)$ must be an even function.

Let's verify this:

We know that $\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$.

Using the substitution $u = -x$ in the first integral, we get $du = -dx$. When $x=-a$, $u=a$. When $x=0$, $u=0$.

$\int_{-a}^0 f(x) dx = \int_{a}^0 f(-u) (-du) = -\int_{a}^0 f(-u) du = \int_0^a f(-u) du = \int_0^a f(-x) dx$ (since $u$ is a dummy variable).

So, $\int_{-a}^a f(x) dx = \int_0^a f(-x) dx + \int_0^a f(x) dx = \int_0^a [f(-x) + f(x)] dx$.

The given condition is $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$.

Thus, $\int_0^a [f(-x) + f(x)] dx = 2 \int_0^a f(x) dx = \int_0^a 2f(x) dx$.

This implies that $\int_0^a [f(-x) + f(x) - 2f(x)] dx = 0$, which simplifies to $\int_0^a [f(-x) - f(x)] dx = 0$.

For this equality to hold for any continuous function $f(x)$ on $[0, a]$, the integrand $f(-x) - f(x)$ must be $0$ over the interval $(0, a)$, i.e., $f(-x) - f(x) = 0$, or $f(-x) = f(x)$. This is the definition of an even function.

Comparing our conclusion with the given options:

(A) Odd (Incorrect)

(B) Even (Correct)

(C) Neither odd nor even (Incorrect)

(D) Cannot be determined (Incorrect)

Thus, the correct answer is (B) Even.

Solution:

We need to evaluate the definite integral $I = \int_0^{\pi/2} \log(\tan x) dx$.

We can use the property of definite integrals: $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.

Here, $a = \pi/2$ and $f(x) = \log(\tan x)$.

Applying the property, we substitute $x$ with $\pi/2 - x$ in the integrand:

$I = \int_0^{\pi/2} \log(\tan(\pi/2 - x)) dx$

We know that $\tan(\pi/2 - x) = \cot x$.

So, $I = \int_0^{\pi/2} \log(\cot x) dx$.

We now have two expressions for the integral $I$:

$\int_0^{\pi/2} \log(\tan x) dx$

$\int_0^{\pi/2} \log(\cot x) dx$

Let's add these two expressions:

$2I = \int_0^{\pi/2} \log(\tan x) dx + \int_0^{\pi/2} \log(\cot x) dx$

$2I = \int_0^{\pi/2} [\log(\tan x) + \log(\cot x)] dx$

Using the logarithm property $\log A + \log B = \log(A \cdot B)$, we have:

$\log(\tan x) + \log(\cot x) = \log(\tan x \cdot \cot x)$

Since $\tan x \cdot \cot x = 1$ (for $x \neq n\pi/2$),

$\log(\tan x \cdot \cot x) = \log(1) = 0$.

Substituting this back into the integral:

$2I = \int_0^{\pi/2} 0 \, dx$

$2I = [0x]_0^{\pi/2}$

$2I = 0(\pi/2) - 0(0)$

$2I = 0$

Dividing by 2, we get:

$I = 0$

Comparing our result with the given options, we find that the value of the integral is $0$, which corresponds to option (C).

Thus, the correct answer is (C) 0.

Solution:

We need to identify the most suitable method for evaluating the integral $\int \sin^5 x \cos x dx$.

Let's examine the integrand, which is $\sin^5 x \cos x$. We can observe that the integrand contains a function, $\sin x$, raised to a power, and it also contains the derivative of that function, $\cos x$, as a factor.

This structure, where we have a function raised to a power multiplied by its derivative, is a strong indicator that the method of substitution is appropriate.

Let's try the substitution $u = \sin x$.

Differentiating both sides with respect to $x$, we get $\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$.

From this, we have the differential relationship $du = \cos x \, dx$.

Now, substitute $u = \sin x$ and $du = \cos x \, dx$ into the integral:

$\int \sin^5 x \cos x dx = \int (\sin x)^5 (\cos x \, dx) = \int u^5 du$

This is a simple power rule integral, which can be easily evaluated:

$\int u^5 du = \frac{u^{5+1}}{5+1} + C = \frac{u^6}{6} + C$

Substituting back $u = \sin x$, the result is $\frac{(\sin x)^6}{6} + C = \frac{\sin^6 x}{6} + C$.

Let's consider why the other options are less suitable:

(A) Integration by parts: While this integral can be technically solved using integration by parts (as shown in thought process), it involves setting up $u$ and $dv$ and integrating a new integral, which is usually more complex than the direct substitution approach when the derivative is present.

(C) Partial fractions: This method is used for rational functions (ratios of polynomials). The integrand $\sin^5 x \cos x$ is a product of trigonometric functions, not a rational function, so partial fractions are not applicable.

(D) Trigonometric identity for $\sin^5 x$: Using identities to rewrite $\sin^5 x$ (e.g., using power reduction formulas) would lead to an integral involving powers of cosine or multiple angles, which would likely require integrating individual terms, some of which might still need substitution or other methods. This would be significantly more complicated than the direct substitution $u = \sin x$.

The substitution $u = \sin x$ simplifies the integral directly into a basic power rule form, making it the most suitable method.

Thus, the correct answer is (B) Substitution ($u = \sin x$).

Solution:

We need to evaluate the indefinite integral $\int \sec x dx$.

This is a standard integral. One common method to evaluate it is to multiply the integrand by $\frac{\sec x + \tan x}{\sec x + \tan x}$.

$\int \sec x dx = \int \sec x \cdot \frac{\sec x + \tan x}{\sec x + \tan x} dx = \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} dx$

Now, we can use the method of substitution.

Let $u = \sec x + \tan x$.

Differentiating $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sec x) + \frac{d}{dx}(\tan x) = (\sec x \tan x) + (\sec^2 x)$

So, $du = (\sec x \tan x + \sec^2 x) dx$.

The numerator of the integrand is exactly $du$. The denominator is $u$.

The integral becomes:

$\int \frac{du}{u}$

The integral of $\frac{1}{u}$ is $\log |u|$.

$\int \frac{du}{u} = \log |u| + C$

Substitute back $u = \sec x + \tan x$:

$\int \sec x dx = \log |\sec x + \tan x| + C$

This matches option (A).

Let's consider option (C): $\log |\tan(\frac{x}{2} + \frac{\pi}{4})| + C$.

We can show that $\tan(\frac{x}{2} + \frac{\pi}{4}) = \frac{\tan(x/2) + \tan(\pi/4)}{1 - \tan(x/2)\tan(\pi/4)} = \frac{\tan(x/2) + 1}{1 - \tan(x/2)}$.

Using the half-angle formulas $\tan(x/2) = \frac{1-\cos x}{\sin x}$ and $\tan(x/2) = \frac{\sin x}{1+\cos x}$:

$\frac{1+\tan(x/2)}{1-\tan(x/2)} = \frac{1 + \frac{\sin x}{1+\cos x}}{1 - \frac{\sin x}{1+\cos x}} = \frac{\frac{1+\cos x + \sin x}{1+\cos x}}{\frac{1+\cos x - \sin x}{1+\cos x}} = \frac{1+\cos x + \sin x}{1+\cos x - \sin x}$

This does not immediately look like $\sec x + \tan x$.

Alternatively, let's start from $\sec x + \tan x$ and try to reach $\tan(\frac{x}{2} + \frac{\pi}{4})$ using half-angle formulas:

$\sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1+\sin x}{\cos x}$

Using identities $\sin x = 2 \sin(x/2) \cos(x/2)$ and $\cos x = \cos^2(x/2) - \sin^2(x/2)$:

$\frac{1+\sin x}{\cos x} = \frac{\cos^2(x/2) + \sin^2(x/2) + 2 \sin(x/2) \cos(x/2)}{\cos^2(x/2) - \sin^2(x/2)}$

The numerator is $(\cos(x/2) + \sin(x/2))^2$. The denominator is $(\cos(x/2) - \sin(x/2))(\cos(x/2) + \sin(x/2))$.

So, $\frac{(\cos(x/2) + \sin(x/2))^2}{(\cos(x/2) - \sin(x/2))(\cos(x/2) + \sin(x/2))} = \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)}$ (assuming $\cos(x/2) + \sin(x/2) \neq 0$).

Divide numerator and denominator by $\cos(x/2)$:

$\frac{1 + \tan(x/2)}{1 - \tan(x/2)}$

We know the tangent addition formula: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.

Comparing $\frac{1 + \tan(x/2)}{1 - \tan(x/2)}$ with this formula, if we let $A = x/2$ and $\tan B = 1$. Since $\tan(\pi/4) = 1$, we can let $B = \pi/4$.

So, $\frac{1 + \tan(x/2)}{1 - \tan(x/2)} = \frac{\tan(\pi/4) + \tan(x/2)}{1 - \tan(\pi/4) \tan(x/2)} = \tan(\frac{\pi}{4} + \frac{x}{2}) = \tan(\frac{x}{2} + \frac{\pi}{4})$.

Therefore, $\sec x + \tan x = \tan(\frac{x}{2} + \frac{\pi}{4})$.

Since the arguments of the logarithm are equal, their logarithms are also equal (up to a potential sign difference depending on the range, but the modulus handles that). Thus, $\log |\sec x + \tan x| = \log |\tan(\frac{x}{2} + \frac{\pi}{4})|$.

Both options (A) and (C) represent valid forms of the integral of $\sec x$.

Comparing with the options, option (D) states "Both (A) and (C)". Since we have shown that the results in (A) and (C) are equivalent, option (D) is the correct choice.

Thus, the correct answer is (D) Both (A) and (C).

Solution:

We need to identify the correct formula for the indefinite integral $\int \sqrt{a^2-x^2} dx$.

Integrals of the form $\int \sqrt{a^2-x^2} dx$ often arise when dealing with areas or volumes related to circles, as the expression $\sqrt{a^2-x^2}$ is related to the equation of a circle $x^2 + y^2 = a^2$, or $y = \sqrt{a^2-x^2}$ (for the upper semi-circle).

This integral is a standard result obtained typically through trigonometric substitution ($x = a \sin \theta$).

The standard formula for the integral $\int \sqrt{a^2-x^2} dx$ is:

$\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$

where $C$ is the constant of integration.

Let's compare this standard formula with the given options:

(A) $\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\log\left|x + \sqrt{x^2+a^2}\right| + C$: This is the formula for $\int \sqrt{x^2+a^2} dx$, not $\int \sqrt{a^2-x^2} dx$. (Note the term $\sqrt{x^2+a^2}$ in the logarithm).

(B) $\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$: This matches the standard formula exactly.

(C) $\frac{x}{2}\sqrt{a^2-x^2} - \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$: This formula has a subtraction sign instead of an addition sign before the $\sin^{-1}$ term. It is incorrect.

(D) $\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$: This formula is missing the term $\frac{x}{2}\sqrt{a^2-x^2}$. It is incorrect.

Based on the comparison with the standard integral formula, option (B) is the correct one.

Thus, the correct answer is (B) $\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$.

Solution:

We are given the equality $\int_0^{\pi/2} \sin^n x \cos^m x dx = \int_0^{\pi/2} \sin^m x \cos^n x dx$ and asked to identify the property of definite integrals that leads to this result.

Let the integral be $I = \int_0^{\pi/2} \sin^n x \cos^m x dx$.

Let $f(x) = \sin^n x \cos^m x$. The interval of integration is $[0, \pi/2]$.

Consider the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. In this case, $a = \pi/2$.

Applying this property, we substitute $x$ with $a-x = \pi/2 - x$ in the integrand:

$f(\pi/2 - x) = \sin^n (\pi/2 - x) \cos^m (\pi/2 - x)$

Using the trigonometric identities $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$, we get:

$f(\pi/2 - x) = (\cos x)^n (\sin x)^m = \cos^n x \sin^m x = \sin^m x \cos^n x$

According to the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we have:

$\int_0^{\pi/2} f(x) dx = \int_0^{\pi/2} f(\pi/2 - x) dx$

Substituting the expressions for $f(x)$ and $f(\pi/2 - x)$:

$\int_0^{\pi/2} \sin^n x \cos^m x dx = \int_0^{\pi/2} \sin^m x \cos^n x dx$

This is exactly the equality given in the question. The property used is $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ with $a = \pi/2$.

Now, let's look at the given options:

(A) $\int_a^b f(x) dx = -\int_b^a f(x) dx$: This property is about reversing the limits of integration and changing the sign.

(B) $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$: This is the additive property of integrals over adjacent intervals.

(C) $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: This is the property we applied to derive the result.

(D) $\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$ if $f$ is even, $0$ if $f$ is odd: This property applies to integrals over symmetric intervals $[-a, a]$ based on the parity of the function.

The property that directly leads to the given equality is option (C).

Thus, the correct answer is (C) $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.

To evaluate the integral $\int (x^2 + 3x + 5) dx$, we will use the power rule for integration, which states that $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$. We will also use the property that the integral of a sum is the sum of the integrals.

We can break down the integral into three separate integrals:

Now, we evaluate each integral separately:

1. For the first term, $\int x^2 dx$:

2. For the second term, $\int 3x dx$:

We can pull the constant 3 out of the integral:

Applying the power rule where $n=1$:

3. For the third term, $\int 5 dx$:

We can pull the constant 5 out of the integral:

The integral of a constant is the constant times $x$:

Now, we combine the results of the three integrals:

Combining the constants of integration ($C_1 + C_2 + C_3$) into a single constant $C$:

Thus, the evaluation of the integral is $\frac{x^3}{3} + \frac{3x^2}{2} + 5x + C$.

To evaluate the integral $\int \sin(ax+b) dx$, we can use a substitution method. Let $u = ax+b$. Then, the differential $du$ can be found by differentiating $u$ with respect to $x$:

From this, we can express $dx$ in terms of $du$ and $a$:

Now, substitute $u$ for $ax+b$ and $\frac{du}{a}$ for $dx$ into the integral:

We can pull the constant $\frac{1}{a}$ out of the integral:

The integral of $\sin(u)$ with respect to $u$ is $-\cos(u)$:

Where $C$ is the constant of integration. Now, substitute back $u = ax+b$:

This simplifies to:

Therefore, the evaluation of the integral is $-\frac{1}{a} \cos(ax+b) + C$.

To evaluate the integral $\int \frac{1}{x \log x} dx$, we can use a substitution method. Let $u = \log x$. Then, the differential $du$ can be found by differentiating $u$ with respect to $x$:

From this, we can express $dx$ in terms of $du$ and $x$:

Now, substitute $u$ for $\log x$ and $x \, du$ for $dx$ into the integral:

The $x$ terms cancel out:

The integral of $\frac{1}{u}$ with respect to $u$ is $\log |u|$:

Where $C$ is the constant of integration. Now, substitute back $u = \log x$:

Therefore, the evaluation of the integral is $\log |\log x| + C$.

To evaluate the integral $\int x \cos x dx$, we can use integration by parts. The integration by parts formula is:

We need to choose which part of the integrand will be $u$ and which will be $dv$. A good mnemonic for choosing $u$ is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential). In this case, we have an algebraic part ($x$) and a trigonometric part ($\cos x$). So, we choose:

Let $u = x$ and $dv = \cos x \, dx$.

Now, we need to find $du$ and $v$. Differentiating $u$ with respect to $x$ gives:

Integrating $dv$ gives:

Now, apply the integration by parts formula:

This simplifies to:

Now, evaluate the remaining integral $\int \sin x \, dx$:

Substitute this result back into the equation:

This simplifies to:

Therefore, the evaluation of the integral is $x \sin x + \cos x + C$.

To evaluate the integral $\int \frac{dx}{\sqrt{1-4x^2}}$, we can use a trigonometric substitution or recognize it as a standard integral form. Let's rewrite the expression to match a known integral form.

The integral resembles the form $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C$.

We can rewrite $1-4x^2$ as $1 - (2x)^2$.

Let $u = 2x$. Then, $\frac{du}{dx} = 2$, which means $dx = \frac{du}{2}$.

Substitute $u$ and $dx$ into the integral:

Pull out the constant $\frac{1}{2}$:

Now, this integral is in the standard form $\int \frac{du}{\sqrt{a^2 - u^2}}$ where $a=1$. The integral of $\frac{1}{\sqrt{1-u^2}}$ is $\sin^{-1}(u)$.

Finally, substitute back $u = 2x$:

Therefore, the evaluation of the integral is $\frac{1}{2} \sin^{-1}(2x) + C$.

To evaluate the integral $\int \frac{dx}{x^2 + 2x + 2}$, we need to complete the square in the denominator to transform it into a form that resembles the standard integral $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$.

First, complete the square for the denominator $x^2 + 2x + 2$:

$x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x+1)^2 + 1$

Now, the integral becomes:

Let $u = x+1$. Then, $\frac{du}{dx} = 1$, which means $dx = du$.

Substituting $u$ and $dx$ into the integral:

This integral is in the standard form $\int \frac{du}{u^2 + a^2}$, where $a=1$. The integral of $\frac{1}{u^2 + 1}$ is $\tan^{-1}(u)$.

Finally, substitute back $u = x+1$:

Therefore, the evaluation of the integral is $\tan^{-1}(x+1) + C$.

To evaluate the definite integral $\int_0^1 (x^2 + 1) dx$, we first find the indefinite integral of the function $(x^2 + 1)$ and then apply the Fundamental Theorem of Calculus.

The indefinite integral of $x^2 + 1$ is:

Using the power rule for integration ($\int x^n dx = \frac{x^{n+1}}{n+1}$):

And the integral of a constant:

So, the indefinite integral is:

Now, we apply the Fundamental Theorem of Calculus, which states that $\int_a^b f(x) dx = F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$. In this case, $a=0$, $b=1$, and $F(x) = \frac{x^3}{3} + x$.

Evaluate $F(b)$:

Evaluate $F(a)$:

Now, subtract $F(a)$ from $F(b)$:

Therefore, the value of the definite integral is $\frac{4}{3}$.

To evaluate the definite integral $\int_0^{\pi/2} \sin x dx$, we first find the indefinite integral of $\sin x$. The antiderivative of $\sin x$ is $-\cos x$.

Using the Fundamental Theorem of Calculus, which states that $\int_a^b f(x) dx = F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$. In this case, $a=0$, $b=\frac{\pi}{2}$, and $F(x) = -\cos x$.

Evaluate $F(b)$:

We know that $\cos\left(\frac{\pi}{2}\right) = 0$. So,

Evaluate $F(a)$:

We know that $\cos(0) = 1$. So,

Now, subtract $F(a)$ from $F(b)$:

Therefore, the value of the definite integral is 1.

To evaluate the integral $\int \frac{e^x}{1 + e^x} dx$, we can use a substitution method. This integral is in the form of $\int \frac{f'(x)}{f(x)} dx$, which integrates to $\log|f(x)| + C$.

Let $u = 1 + e^x$. Then, we find the differential $du$ by differentiating $u$ with respect to $x$:

From this, we can express $dx$ in terms of $du$ and $e^x$:

Substitute $u$ for $1 + e^x$ and $\frac{du}{e^x}$ for $dx$ into the integral:

The $e^x$ terms cancel out:

The integral of $\frac{1}{u}$ with respect to $u$ is $\log|u|$:

Where $C$ is the constant of integration. Now, substitute back $u = 1 + e^x$:

Since $e^x$ is always positive, $1 + e^x$ is also always positive, so we can remove the absolute value signs.

Therefore, the evaluation of the integral is $\log(1 + e^x) + C$.

To evaluate the integral $\int \sin^2 x dx$, we need to use a trigonometric identity to reduce the power of $\sin x$. The relevant identity is:

Substitute this identity into the integral:

We can split this into two integrals and pull out the constant $\frac{1}{2}$:

Now, we evaluate each integral:

1. $\int 1 \, dx = x$

2. For $\int \cos(2x) \, dx$, we can use a substitution. Let $u = 2x$, so $du = 2 \, dx$, which means $dx = \frac{du}{2}$.

The integral of $\cos(u)$ is $\sin(u)$. Substituting back $u=2x$:

Combining these results:

Distributing the $\frac{1}{2}$:

Therefore, the evaluation of the integral is $\frac{x}{2} - \frac{1}{4} \sin(2x) + C$.

To evaluate the integral $\int \frac{dx}{9x^2 + 6x + 5}$, we need to complete the square in the denominator and manipulate it to match the standard form for the arctangent integral: $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$.

First, complete the square for the denominator $9x^2 + 6x + 5$. We can factor out 9 from the terms involving $x$:

$9x^2 + 6x + 5 = 9\left(x^2 + \frac{6}{9}x\right) + 5 = 9\left(x^2 + \frac{2}{3}x\right) + 5$

To complete the square inside the parenthesis, we take half of the coefficient of $x$ (which is $\frac{2}{3}$), square it, and add and subtract it. Half of $\frac{2}{3}$ is $\frac{1}{3}$, and its square is $\left(\frac{1}{3}\right)^2 = \frac{1}{9}$.

$9\left(x^2 + \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}\right) + 5$

$9\left(\left(x + \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 5$

$9\left(x + \frac{1}{3}\right)^2 - 9 \cdot \frac{1}{9} + 5$

$9\left(x + \frac{1}{3}\right)^2 - 1 + 5$

$9\left(x + \frac{1}{3}\right)^2 + 4$

Now, rewrite the integral with the completed square:

We can factor out 4 from the denominator:

Pull out the constant $\frac{1}{4}$:

Now, let $u = \frac{3}{2}\left(x + \frac{1}{3}\right)$. Then, find $du$:

$u = \frac{3}{2}x + \frac{3}{2} \cdot \frac{1}{3} = \frac{3}{2}x + \frac{1}{2}$

Differentiating with respect to $x$:

So, $dx = \frac{2}{3} du$.

Substitute $u$ and $dx$ into the integral:

Pull out the constant $\frac{2}{3}$:

This simplifies to:

The integral of $\frac{1}{u^2 + 1}$ is $\tan^{-1}(u)$:

Finally, substitute back $u = \frac{3}{2}\left(x + \frac{1}{3}\right)$:

Or, simplifying the argument of the arctangent:

Therefore, the evaluation of the integral is $\frac{1}{6} \tan^{-1}\left(\frac{3x+1}{2}\right) + C$.

To evaluate the integral $\int \frac{x+2}{\sqrt{x^2 + 4x + 5}} dx$, we can use a substitution method. Notice that the derivative of the expression inside the square root, $x^2 + 4x + 5$, is $2x + 4$. The numerator of the integrand is $x+2$, which is exactly half of $2x+4$. This suggests a substitution.

Let $u = x^2 + 4x + 5$. Then, find the differential $du$ by differentiating $u$ with respect to $x$:

We can factor out 2 from the derivative:

Rearranging this, we get $du = 2(x + 2) dx$, or $(x + 2) dx = \frac{1}{2} du$.

Now, substitute $u$ for $x^2 + 4x + 5$ and $\frac{1}{2} du$ for $(x+2) dx$ into the integral:

Pull out the constant $\frac{1}{2}$:

Rewrite $\frac{1}{\sqrt{u}}$ as $u^{-1/2}$:

Now, use the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1}$):

This simplifies to:

Which is:

Finally, substitute back $u = x^2 + 4x + 5$:

Therefore, the evaluation of the integral is $\sqrt{x^2 + 4x + 5} + C$.

To evaluate the integral $\int \log x dx$, we will use integration by parts. The formula for integration by parts is $\int u \, dv = uv - \int v \, du$.

We need to choose $u$ and $dv$. A common choice for $\log x$ is:

Let $u = \log x$ and $dv = dx$.

Now we find $du$ by differentiating $u$ and $v$ by integrating $dv$:

Apply the integration by parts formula:

Simplify the equation:

Now, evaluate the remaining integral $\int 1 \, dx$:

Substitute this back into the equation:

This simplifies to:

Therefore, the evaluation of the integral is $x \log x - x + C$.

To evaluate the integral $\int e^x (\tan x + \sec^2 x) dx$, we can recognize a specific form of integration by parts. Consider the derivative of the product $e^x f(x)$:

Using the product rule for differentiation, $\frac{d}{dx} [e^x f(x)] = e^x f(x) + e^x f'(x) = e^x (f(x) + f'(x))$.

This means that if we have an integral of the form $\int e^x (f(x) + f'(x)) dx$, the result is simply $e^x f(x) + C$.

In our given integral, $\int e^x (\tan x + \sec^2 x) dx$, let's identify $f(x)$ and $f'(x)$.

Let $f(x) = \tan x$.

The derivative of $f(x) = \tan x$ is $f'(x) = \sec^2 x$.

Our integrand matches the form $e^x (f(x) + f'(x))$ where $f(x) = \tan x$ and $f'(x) = \sec^2 x$.

Therefore, by the rule derived above, the integral is:

Substituting $f(x) = \tan x$ back:

Therefore, the evaluation of the integral is $e^x \tan x + C$.

Let the given integral be $I$:

We use the property of definite integrals: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$.

Here, $a=0$ and $b=\frac{\pi}{2}$. So, $a+b-x = 0 + \frac{\pi}{2} - x = \frac{\pi}{2} - x$.

We replace $x$ with $\frac{\pi}{2} - x$ in the integrand:

$\sin\left(\frac{\pi}{2} - x\right) = \cos x$

$\cos\left(\frac{\pi}{2} - x\right) = \sin x$

So the integrand becomes:

$\frac{\sin\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)} = \frac{\cos x}{\cos x + \sin x}$

Therefore, the integral $I$ can be written as:

Now, add equation (i) and equation (ii):

$I + I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx + \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} dx$

$2I = \int_0^{\pi/2} \left(\frac{\sin x}{\sin x + \cos x} + \frac{\cos x}{\sin x + \cos x}\right) dx$

$2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} dx$

$2I = \int_0^{\pi/2} 1 \, dx$

Evaluate the integral of 1:

Apply the limits of integration:

$2I = \frac{\pi}{2}$

Now, solve for $I$:

Therefore, the value of the integral is $\frac{\pi}{4}$.

To evaluate the definite integral $\int_0^1 \frac{dx}{\sqrt{1-x^2}}$, we first recognize the standard integral form for $\frac{1}{\sqrt{a^2 - x^2}}$.

The indefinite integral of $\frac{1}{\sqrt{1-x^2}}$ is $\sin^{-1}(x)$ (or $\arcsin(x)$).

Using the Fundamental Theorem of Calculus, $\int_a^b f(x) dx = F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$.

Here, $a=0$, $b=1$, and $F(x) = \sin^{-1}(x)$.

Evaluate $F(b)$:

The value of $x$ for which $\sin x = 1$ in the interval $[0, \pi]$ is $\frac{\pi}{2}$. So, $\sin^{-1}(1) = \frac{\pi}{2}$.

Evaluate $F(a)$:

The value of $x$ for which $\sin x = 0$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is $0$. So, $\sin^{-1}(0) = 0$.

Now, subtract $F(a)$ from $F(b)$:

So, the value of the integral is $\frac{\pi}{2}$.

To evaluate the integral $\int \frac{\sec^2 x}{\tan x} dx$, we can use a substitution method. This integral is in the form of $\int \frac{f'(x)}{f(x)} dx$, which integrates to $\log|f(x)| + C$.

Let $u = \tan x$. Then, we find the differential $du$ by differentiating $u$ with respect to $x$:

From this, we can express $dx$ in terms of $du$ and $\sec^2 x$:

Substitute $u$ for $\tan x$ and $\frac{du}{\sec^2 x}$ for $dx$ into the integral:

The $\sec^2 x$ terms cancel out:

The integral of $\frac{1}{u}$ with respect to $u$ is $\log|u|$:

Where $C$ is the constant of integration. Now, substitute back $u = \tan x$:

Therefore, the evaluation of the integral is $\log|\tan x| + C$.

To evaluate the integral $\int \sqrt{1 + \sin 2x} dx$, we can use trigonometric identities to simplify the expression under the square root.

Recall the identity $\sin(2x) = 2 \sin x \cos x$. Also, we know that $\sin^2 x + \cos^2 x = 1$.

Substitute these into the expression under the square root:

$1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x$

This expression is a perfect square, specifically the square of $(\sin x + \cos x)$:

$\sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$

Now, substitute this back into the integral:

The square root and the square cancel each other out, but we must consider the absolute value:

The sign of $\sin x + \cos x$ depends on the interval of integration. Without specified limits, we consider the common case where $\sin x + \cos x \ge 0$. We can rewrite $\sin x + \cos x$ as $\sqrt{2} \sin(x + \frac{\pi}{4})$ or $\sqrt{2} \cos(x - \frac{\pi}{4})$.

For simplicity, and assuming the interval where $\sin x + \cos x \ge 0$, we can remove the absolute value:

Now, integrate term by term:

$\int \sin x \, dx = -\cos x$

$\int \cos x \, dx = \sin x$

Combining these results:

Therefore, the evaluation of the integral is $\sin x - \cos x + C$.

Note: If the integration is over an interval where $\sin x + \cos x < 0$, the integral would be $\int -(\sin x + \cos x) dx = -(-\cos x + \sin x) + C = \cos x - \sin x + C$.

To evaluate the integral $\int \frac{x}{x^2 + 1} dx$, we can use a substitution method. This integral is in the form of $\int \frac{f'(x)}{f(x)} dx$, which integrates to $\log|f(x)| + C$.

Let $u = x^2 + 1$. Then, we find the differential $du$ by differentiating $u$ with respect to $x$:

From this, we can express $dx$ in terms of $du$ and $x$:

Substitute $u$ for $x^2 + 1$ and $\frac{1}{2} du$ for $x \, dx$ into the integral:

Pull out the constant $\frac{1}{2}$:

The integral of $\frac{1}{u}$ with respect to $u$ is $\log|u|$:

Where $C$ is the constant of integration. Now, substitute back $u = x^2 + 1$:

Since $x^2 + 1$ is always positive for real values of $x$, we can remove the absolute value signs.

Therefore, the evaluation of the integral is $\frac{1}{2} \log(x^2 + 1) + C$.

To evaluate the definite integral $\int_0^2 x \sqrt{2-x} dx$, we can use a substitution. Let $u = 2-x$.

Then, $du = -dx$, which means $dx = -du$.

Also, we need to express $x$ in terms of $u$. From $u = 2-x$, we get $x = 2-u$.

Next, we need to change the limits of integration:

• When $x=0$, $u = 2 - 0 = 2$.
• When $x=2$, $u = 2 - 2 = 0$.

Now, substitute $u$, $x$, $dx$, and the new limits into the integral:

We can use the property $\int_a^b f(x) dx = -\int_b^a f(x) dx$ to swap the limits and remove the negative sign:

Now, distribute $u^{1/2}$ into the parenthesis:

Now, integrate term by term using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$:

$\int 2u^{1/2} du = 2 \frac{u^{1/2 + 1}}{1/2 + 1} = 2 \frac{u^{3/2}}{3/2} = 2 \cdot \frac{2}{3} u^{3/2} = \frac{4}{3} u^{3/2}$

$\int u^{3/2} du = \frac{u^{3/2 + 1}}{3/2 + 1} = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}$

So, the integral becomes:

Now, apply the limits of integration:

Evaluate at the upper limit ($u=2$):

Let's simplify this: $2^{3/2} = 2\sqrt{2}$ and $2^{5/2} = 4\sqrt{2}$.

$\frac{4}{3} (2\sqrt{2}) - \frac{2}{5} (4\sqrt{2}) = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5}$

Find a common denominator (15):

$\frac{8\sqrt{2} \cdot 5}{15} - \frac{8\sqrt{2} \cdot 3}{15} = \frac{40\sqrt{2} - 24\sqrt{2}}{15} = \frac{16\sqrt{2}}{15}$

Evaluate at the lower limit ($u=0$):

Subtract the lower limit value from the upper limit value:

Therefore, the value of the integral is $\frac{16\sqrt{2}}{15}$.

To evaluate the integral $\int \frac{dx}{x(x+1)}$, we will use the method of partial fraction decomposition.

We want to express $\frac{1}{x(x+1)}$ as a sum of simpler fractions:

To find the constants $A$ and $B$, we multiply both sides by $x(x+1)$:

Now, we can find $A$ and $B$ by substituting convenient values for $x$.

Set $x=0$:

Set $x=-1$:

So, the partial fraction decomposition is:

Now, we can integrate this expression:

We can split this into two integrals:

The integral of $\frac{1}{x}$ is $\log|x|$, and the integral of $\frac{1}{x+1}$ (using a substitution $u=x+1$) is $\log|x+1|$.

Using the logarithm property $\log a - \log b = \log \frac{a}{b}$:

Therefore, the evaluation of the integral is $\log\left|\frac{x}{x+1}\right| + C$.

To evaluate the integral $\int \sin x \sin 2x \sin 3x dx$, we will use product-to-sum trigonometric identities to simplify the integrand.

We use the identity: $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$.

Let's first multiply $\sin x$ and $\sin 3x$:

$2 \sin x \sin 3x = \cos(3x-x) - \cos(3x+x) = \cos(2x) - \cos(4x)$

So, $\sin x \sin 3x = \frac{1}{2} (\cos(2x) - \cos(4x))$.

Now, substitute this back into the integral:

$\frac{1}{2} \int (\cos(2x) \sin 2x - \cos(4x) \sin 2x) dx$

We can use the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$.

For the first term, $\cos(2x) \sin 2x$:

Using $2 \sin A \cos A = \sin(2A)$, we have $2 \sin 2x \cos 2x = \sin(4x)$.

So, $\cos(2x) \sin 2x = \frac{1}{2} \sin(4x)$.

For the second term, $\cos(4x) \sin 2x$:

Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$, with $A=2x$ and $B=4x$:

$2 \sin 2x \cos 4x = \sin(2x+4x) + \sin(2x-4x) = \sin(6x) + \sin(-2x) = \sin(6x) - \sin(2x)$

So, $\cos(4x) \sin 2x = \frac{1}{2} (\sin(6x) - \sin(2x))$.

Substitute these back into the integral expression:

$\frac{1}{4} \int (\sin(4x) - \sin(6x) + \sin(2x)) dx$

Now, we integrate each term:

• $\int \sin(4x) dx = -\frac{1}{4} \cos(4x)$
• $\int \sin(6x) dx = -\frac{1}{6} \cos(6x)$
• $\int \sin(2x) dx = -\frac{1}{2} \cos(2x)$

Substitute these results back:

$\frac{1}{4} \left(-\frac{1}{4} \cos(4x) + \frac{1}{6} \cos(6x) - \frac{1}{2} \cos(2x)\right) + C$

Distribute the $\frac{1}{4}$:

Therefore, the evaluation of the integral is $-\frac{1}{16} \cos(4x) + \frac{1}{24} \cos(6x) - \frac{1}{8} \cos(2x) + C$.

To evaluate the integral $\int \frac{x+2}{\sqrt{4x - x^2}} dx$, we first need to complete the square in the denominator $4x - x^2$.

$4x - x^2 = -(x^2 - 4x) = -(x^2 - 4x + 4 - 4) = -( (x-2)^2 - 4 ) = 4 - (x-2)^2$.

So the integral becomes $\int \frac{x+2}{\sqrt{4 - (x-2)^2}} dx$.

We can split the numerator $x+2$ to match the derivative of the term inside the square root, or part of it.

The derivative of $4 - (x-2)^2$ with respect to $x$ is $-2(x-2) = -2x + 4$. This isn't directly in the numerator, but $x-2$ is related to $x+2$.

Let's rewrite the numerator $x+2$ in terms of $(x-2)$: $x+2 = (x-2) + 4$.

So the integral becomes:

We can split this into two integrals:

For the first integral: $\int \frac{x-2}{\sqrt{4 - (x-2)^2}} dx$

Let $u = 4 - (x-2)^2$. Then $du = -2(x-2) dx$, which means $(x-2) dx = -\frac{1}{2} du$.

The integral becomes: $\int \frac{-\frac{1}{2} du}{\sqrt{u}} = -\frac{1}{2} \int u^{-1/2} du$

Using the power rule: $-\frac{1}{2} \left(\frac{u^{1/2}}{1/2}\right) = -u^{1/2} = -\sqrt{4 - (x-2)^2}$.

For the second integral: $\int \frac{4}{\sqrt{4 - (x-2)^2}} dx$

We can pull out the constant 4: $4 \int \frac{1}{\sqrt{4 - (x-2)^2}} dx$.

This integral is in the form $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\frac{u}{a}\right)$.

Here, let $v = x-2$, so $dv = dx$. And $a^2 = 4$, so $a=2$.

The integral becomes: $4 \int \frac{1}{\sqrt{4 - v^2}} dv = 4 \int \frac{1}{\sqrt{2^2 - v^2}} dv$

This evaluates to $4 \sin^{-1}\left(\frac{v}{2}\right) = 4 \sin^{-1}\left(\frac{x-2}{2}\right)$.

Combining the results of both integrals:

We can also write $\sqrt{4 - (x-2)^2}$ as $\sqrt{4 - (x^2 - 4x + 4)} = \sqrt{4 - x^2 + 4x - 4} = \sqrt{4x - x^2}$.

Therefore, the evaluation of the integral is $-\sqrt{4x - x^2} + 4 \sin^{-1}\left(\frac{x-2}{2}\right) + C$.

To evaluate the integral $\int \frac{x^2}{(x+1)(x-1)^2} dx$, we will use the method of partial fraction decomposition.

Since the denominator has a linear term $(x+1)$ and a repeated linear term $(x-1)^2$, the partial fraction decomposition will be of the form:

To find the constants $A$, $B$, and $C$, multiply both sides by $(x+1)(x-1)^2$:

Now, we find the constants by substituting specific values for $x$.

Set $x=1$:

Set $x=-1$:

To find $B$, we can compare the coefficients of $x^2$ on both sides of the equation $x^2 = A(x-1)^2 + B(x+1)(x-1) + C(x+1)$.

Expand the terms: $x^2 = A(x^2 - 2x + 1) + B(x^2 - 1) + C(x+1)$ $x^2 = Ax^2 - 2Ax + A + Bx^2 - B + Cx + C$ $x^2 = (A+B)x^2 + (-2A + C)x + (A - B + C)$

Comparing the coefficient of $x^2$: $1 = A+B$. Since $A = \frac{1}{4}$, we have $1 = \frac{1}{4} + B$. $B = 1 - \frac{1}{4} = \frac{3}{4}$.

We can also check with the coefficient of $x$: $0 = -2A + C$. $0 = -2(\frac{1}{4}) + \frac{1}{2} = -\frac{1}{2} + \frac{1}{2} = 0$. This confirms our values.

So, the partial fraction decomposition is:

Now, integrate each term:

Using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1}$, where $u=x-1$ and $n=-2$:

Combining all the integrated terms:

Therefore, the evaluation of the integral is $\frac{1}{4} \log|x+1| + \frac{3}{4} \log|x-1| - \frac{1}{2(x-1)} + C$.

To evaluate the integral $\int e^{2x} \sin 3x dx$, we use integration by parts twice. The formula for integration by parts is $\int u \, dv = uv - \int v \, du$.

Let $I = \int e^{2x} \sin 3x dx$.

First Integration by Parts:

Let $u = \sin 3x$ and $dv = e^{2x} dx$.

Then $du = 3 \cos 3x \, dx$ and $v = \int e^{2x} dx = \frac{1}{2} e^{2x}$.

Applying the formula:

$I = \frac{1}{2} e^{2x} \sin 3x - \frac{3}{2} \int e^{2x} \cos 3x dx$

Second Integration by Parts for $\int e^{2x} \cos 3x dx$:

Let $u = \cos 3x$ and $dv = e^{2x} dx$.

Then $du = -3 \sin 3x \, dx$ and $v = \frac{1}{2} e^{2x}$.

Applying the formula to $\int e^{2x} \cos 3x dx$:

$\int e^{2x} \cos 3x dx = \frac{1}{2} e^{2x} \cos 3x + \frac{3}{2} \int e^{2x} \sin 3x dx$

Now substitute this result back into the expression for $I$:

$I = \frac{1}{2} e^{2x} \sin 3x - \frac{3}{2} \left( \frac{1}{2} e^{2x} \cos 3x + \frac{3}{2} \int e^{2x} \sin 3x dx \right)$

$I = \frac{1}{2} e^{2x} \sin 3x - \frac{3}{4} e^{2x} \cos 3x - \frac{9}{4} \int e^{2x} \sin 3x dx$

Notice that the integral on the right side is our original integral $I$. So, we have:

Now, we solve for $I$ by gathering all $I$ terms on one side:

$I + \frac{9}{4} I = \frac{1}{2} e^{2x} \sin 3x - \frac{3}{4} e^{2x} \cos 3x$

$\frac{4}{4} I + \frac{9}{4} I = \frac{1}{2} e^{2x} \sin 3x - \frac{3}{4} e^{2x} \cos 3x$

$\frac{13}{4} I = \frac{1}{2} e^{2x} \sin 3x - \frac{3}{4} e^{2x} \cos 3x$

To isolate $I$, multiply both sides by $\frac{4}{13}$:

$I = \frac{4}{13} \left( \frac{1}{2} e^{2x} \sin 3x - \frac{3}{4} e^{2x} \cos 3x \right) + C$

$I = \frac{2}{13} e^{2x} \sin 3x - \frac{3}{13} e^{2x} \cos 3x + C$

We can factor out $e^{2x}$:

Therefore, the evaluation of the integral is $\frac{e^{2x}}{13} (2 \sin 3x - 3 \cos 3x) + C$.

To evaluate the integral $\int \frac{dx}{\sin x + \cos x}$, we can use the $t$-substitution method, where $t = \tan(\frac{x}{2})$.

Recall the following substitutions:

• $\sin x = \frac{2t}{1+t^2}$
• $\cos x = \frac{1-t^2}{1+t^2}$
• $dx = \frac{2 dt}{1+t^2}$

Substitute these into the integral:

Combine the terms in the denominator:

Cancel out the $(1+t^2)$ terms:

Now, we need to complete the square in the denominator $1 + 2t - t^2$.

$1 + 2t - t^2 = -(t^2 - 2t - 1) = -(t^2 - 2t + 1 - 1 - 1) = -( (t-1)^2 - 2 ) = 2 - (t-1)^2$.

The integral becomes:

This is of the form $\int \frac{du}{a^2 - u^2} = \frac{1}{2a} \log\left|\frac{a+u}{a-u}\right| + C$.

Here, $a^2 = 2$, so $a = \sqrt{2}$, and $u = t-1$, so $du = dt$.

Substitute these values:

$2 \left( \frac{1}{2\sqrt{2}} \log\left|\frac{\sqrt{2} + (t-1)}{\sqrt{2} - (t-1)}\right| \right) + C$

$\frac{1}{\sqrt{2}} \log\left|\frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1}\right| + C$

Now, substitute back $t = \tan(\frac{x}{2})$:

Alternatively, we can rewrite $\sin x + \cos x$ as $\sqrt{2} (\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x) = \sqrt{2} (\cos(\frac{\pi}{4}) \sin x + \sin(\frac{\pi}{4}) \cos x) = \sqrt{2} \sin(x + \frac{\pi}{4})$.

So the integral becomes $\int \frac{dx}{\sqrt{2} \sin(x + \frac{\pi}{4})} = \frac{1}{\sqrt{2}} \int \csc(x + \frac{\pi}{4}) dx$.

The integral of $\csc u$ is $-\log|\csc u + \cot u| + C$ or $\log|\csc u - \cot u| + C$.

Let $v = x + \frac{\pi}{4}$, so $dv = dx$. $\frac{1}{\sqrt{2}} \int \csc v \, dv = \frac{1}{\sqrt{2}} (-\log|\csc v + \cot v|) + C$ $= -\frac{1}{\sqrt{2}} \log\left|\csc\left(x+\frac{\pi}{4}\right) + \cot\left(x+\frac{\pi}{4}\right)\right| + C$.

Using the identities $\csc A = \frac{1}{\sin A}$ and $\cot A = \frac{\cos A}{\sin A}$, and the sum formulas for sine and cosine, this simplifies to the same logarithmic expression as above.

Let's express $\sin x + \cos x$ as $\sqrt{2} \cos(x - \frac{\pi}{4})$.

Then the integral is $\int \frac{dx}{\sqrt{2} \cos(x - \frac{\pi}{4})} = \frac{1}{\sqrt{2}} \int \sec(x - \frac{\pi}{4}) dx$.

The integral of $\sec u$ is $\log|\sec u + \tan u| + C$.

Let $w = x - \frac{\pi}{4}$, $dw = dx$. $\frac{1}{\sqrt{2}} \int \sec w \, dw = \frac{1}{\sqrt{2}} \log|\sec w + \tan w| + C$ $= \frac{1}{\sqrt{2}} \log\left|\sec\left(x-\frac{\pi}{4}\right) + \tan\left(x-\frac{\pi}{4}\right)\right| + C$.

This also leads to the same logarithmic form after simplification.

Therefore, the evaluation of the integral is $\frac{1}{\sqrt{2}} \log\left|\frac{\sqrt{2} + \tan(\frac{x}{2}) - 1}{\sqrt{2} - \tan(\frac{x}{2}) + 1}\right| + C$.

Let the given integral be $I$:

We use the property of definite integrals: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$.

Here, $a=0$ and $b=\frac{\pi}{2}$. So, $a+b-x = \frac{\pi}{2} - x$.

Replace $x$ with $\frac{\pi}{2} - x$ in the integrand:

$\sin\left(\frac{\pi}{2} - x\right) = \cos x$.

So, the integral becomes:

Now, add equation (i) and equation (ii):

$I + I = \int_0^{\pi/2} \log(\sin x) dx + \int_0^{\pi/2} \log(\cos x) dx$

$2I = \int_0^{\pi/2} (\log(\sin x) + \log(\cos x)) dx$

Using the logarithm property $\log A + \log B = \log(AB)$:

We know the double angle identity $\sin(2x) = 2 \sin x \cos x$, which means $\sin x \cos x = \frac{1}{2} \sin(2x)$.

Using the logarithm property $\log(\frac{A}{B}) = \log A - \log B$:

$2I = \int_0^{\pi/2} \log(\sin(2x)) dx - \int_0^{\pi/2} \log 2 \, dx$

Let's evaluate the first integral: $\int_0^{\pi/2} \log(\sin(2x)) dx$.

Let $t = 2x$. Then $dt = 2 dx$, so $dx = \frac{1}{2} dt$. When $x=0$, $t=0$. When $x=\frac{\pi}{2}$, $t=\pi$.

We use the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$. Here, $f(t) = \log(\sin t)$. $\sin(\pi - t) = \sin t$, so $\log(\sin(\pi-t)) = \log(\sin t)$. Thus, $\int_0^{\pi} \log(\sin t) dt = 2 \int_0^{\pi/2} \log(\sin t) dt = 2I$.

So, $\frac{1}{2} \int_0^{\pi} \log(\sin t) dt = \frac{1}{2} (2I) = I$.

Now let's evaluate the second integral: $\int_0^{\pi/2} \log 2 \, dx$.

Substituting these back into the equation for $2I$:

Now, solve for $I$:

$2I - I = -\frac{\pi}{2} \log 2$

$I = -\frac{\pi}{2} \log 2$

Using the logarithm property $a \log b = \log b^a$: $I = \log(2^{-\pi/2})$.

Therefore, the value of the integral is $-\frac{\pi}{2} \log 2$.

To evaluate the integral $\int \sqrt{a^2 - x^2} dx$, we use a trigonometric substitution. Let $x = a \sin \theta$.

Then, $dx = a \cos \theta \, d\theta$.

Substitute these into the integral:

Simplify the term under the square root:

$a^2 - a^2 \sin^2 \theta = a^2 (1 - \sin^2 \theta) = a^2 \cos^2 \theta$.

So, $\sqrt{a^2 - x^2} = \sqrt{a^2 \cos^2 \theta} = a |\cos \theta|$. For the standard substitution, we assume $\cos \theta \ge 0$, so $\sqrt{a^2 - x^2} = a \cos \theta$.

The integral becomes:

Pull out the constant $a^2$:

Use the power-reducing identity: $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$.

Integrate term by term:

$\int 1 \, d\theta = \theta$

$\int \cos(2\theta) \, d\theta = \frac{1}{2} \sin(2\theta)$

So the integral is:

Use the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$:

Now, substitute back in terms of $x$. From $x = a \sin \theta$, we have $\sin \theta = \frac{x}{a}$.

Also, $\theta = \sin^{-1}\left(\frac{x}{a}\right)$.

For $\cos \theta$, we use the identity $\cos \theta = \sqrt{1 - \sin^2 \theta}$ (assuming $\cos \theta \ge 0$):

$\cos \theta = \sqrt{1 - \left(\frac{x}{a}\right)^2} = \sqrt{1 - \frac{x^2}{a^2}} = \sqrt{\frac{a^2 - x^2}{a^2}} = \frac{\sqrt{a^2 - x^2}}{a}$.

Substitute these back into the integrated expression:

Simplify:

Distribute $\frac{a^2}{2}$:

Which simplifies to:

Therefore, the evaluation of the integral is $\frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$.

Let the given integral be $I$:

We use the property of definite integrals: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$.

Here, $a=0$ and $b=\pi$. So, $a+b-x = \pi - x$.

Replace $x$ with $\pi - x$ in the integrand:

$\sin(\pi - x) = \sin x$

$\cos(\pi - x) = -\cos x$, so $\cos^2(\pi - x) = (-\cos x)^2 = \cos^2 x$.

The integrand becomes $\frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} = \frac{(\pi - x) \sin x}{1 + \cos^2 x}$.

Therefore, the integral $I$ can be written as:

Now, add equation (i) and equation (ii):

$I + I = \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx + \int_0^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} dx$

$2I = \int_0^{\pi} \left(\frac{x \sin x}{1 + \cos^2 x} + \frac{(\pi - x) \sin x}{1 + \cos^2 x}\right) dx$

$2I = \int_0^{\pi} \frac{x \sin x + \pi \sin x - x \sin x}{1 + \cos^2 x} dx$

$2I = \int_0^{\pi} \frac{\pi \sin x}{1 + \cos^2 x} dx$

Now, let's evaluate this integral. Let $u = \cos x$. Then $du = -\sin x \, dx$, so $\sin x \, dx = -du$.

Change the limits of integration:

• When $x=0$, $u = \cos 0 = 1$.
• When $x=\pi$, $u = \cos \pi = -1$.

The integral becomes:

Use the property $\int_a^b f(x) dx = -\int_b^a f(x) dx$ to swap the limits:

The integral of $\frac{1}{1+u^2}$ is $\tan^{-1}(u)$:

Evaluate at the limits:

So, $2I = -\pi \left(\frac{\pi}{4} - (-\frac{\pi}{4})\right) = -\pi \left(\frac{\pi}{4} + \frac{\pi}{4}\right) = -\pi \left(\frac{2\pi}{4}\right) = -\pi \left(\frac{\pi}{2}\right) = -\frac{\pi^2}{2}$.

Now, solve for $I$:

$I = -\frac{\pi^2}{4}$

Therefore, the value of the integral is $-\frac{\pi^2}{4}$.

To evaluate the integral $\int \frac{dx}{1 + \tan x}$, we can rewrite $\tan x$ in terms of $\sin x$ and $\cos x$.

This integral can be solved using a standard trick. We manipulate the numerator to contain terms related to the derivative of the denominator or a constant.

We want to write $\cos x$ in terms of $(\cos x + \sin x)$ and its derivative. The derivative of $(\cos x + \sin x)$ is $(-\sin x + \cos x)$.

Consider writing the numerator as a linear combination of the denominator and its derivative:

$\cos x = A(\cos x + \sin x) + B(-\sin x + \cos x)$

$\cos x = (A+B)\cos x + (A-B)\sin x$

Equating coefficients:

• For $\cos x$: $A+B = 1$
• For $\sin x$: $A-B = 0 \implies A = B$

Substituting $A=B$ into the first equation: $A+A = 1 \implies 2A = 1 \implies A = \frac{1}{2}$.

Since $A=B$, $B = \frac{1}{2}$.

So, we can rewrite the integrand as:

$= \frac{1}{2} \frac{\cos x + \sin x}{\cos x + \sin x} + \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x}$

$= \frac{1}{2} + \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x}$

Now, integrate the expression:

$= \int \frac{1}{2} dx + \int \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x} dx$

The first part is $\int \frac{1}{2} dx = \frac{1}{2}x$.

For the second part, $\int \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x} dx$, let $u = \cos x + \sin x$. Then $du = (\cos x - \sin x) dx$.

The integral becomes $\frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \log|u| = \frac{1}{2} \log|\cos x + \sin x|$.

Combining the results:

Therefore, the evaluation of the integral is $\frac{x}{2} + \frac{1}{2} \log|\cos x + \sin x| + C$.

To evaluate the integral $\int \sin^3 x \cos^2 x dx$, we can use a substitution by separating one $\sin x$ term.

Rewrite $\sin^3 x$ as $\sin^2 x \cdot \sin x$. Use the identity $\sin^2 x = 1 - \cos^2 x$.

So, $\sin^3 x = (1 - \cos^2 x) \sin x$.

The integral becomes:

Now, let $u = \cos x$. Then $du = -\sin x \, dx$, which means $\sin x \, dx = -du$.

Substitute $u$ and $du$ into the integral:

Pull out the negative sign:

Distribute $u^2$:

Integrate term by term using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$:

Distribute the negative sign:

Finally, substitute back $u = \cos x$:

Therefore, the evaluation of the integral is $\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$.

To evaluate the definite integral $\int_0^1 \frac{\tan^{-1}x}{1+x^2} dx$, we use a substitution. Let $u = \tan^{-1}x$.

Then, we find the differential $du$ by differentiating $u$ with respect to $x$:

This means $du = \frac{1}{1+x^2} dx$.

Now, we need to change the limits of integration:

• When $x=0$, $u = \tan^{-1}(0) = 0$.
• When $x=1$, $u = \tan^{-1}(1) = \frac{\pi}{4}$.

Substitute $u$ and $du$ into the integral, along with the new limits:

Now, integrate with respect to $u$ using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$:

Apply the limits of integration:

This simplifies to:

Therefore, the value of the definite integral is $\frac{\pi^2}{32}$.

To evaluate the integral $\int \frac{dx}{(x^2 + 1)(x^2 + 4)}$, we use partial fraction decomposition. Since the denominator consists of irreducible quadratic factors, the form of the decomposition is:

Multiply both sides by $(x^2 + 1)(x^2 + 4)$:

Expand both sides:

$1 = (Ax^3 + 4Ax + Bx^2 + 4B) + (Cx^3 + Cx + Dx^2 + D)$

$1 = (A+C)x^3 + (B+D)x^2 + (4A+C)x + (4B+D)$

Equate the coefficients of corresponding powers of $x$:

• Coefficient of $x^3$: $A+C = 0 \implies C = -A$
• Coefficient of $x^2$: $B+D = 0 \implies D = -B$
• Coefficient of $x^1$: $4A+C = 0$
• Coefficient of $x^0$ (constant term): $4B+D = 1$

From $A+C=0$ and $4A+C=0$: Subtracting the first from the second gives $(4A+C) - (A+C) = 0 - 0 \implies 3A = 0 \implies A = 0$. Since $C = -A$, we get $C = 0$.

From $B+D=0$ and $4B+D=1$: Subtracting the first from the second gives $(4B+D) - (B+D) = 1 - 0 \implies 3B = 1 \implies B = \frac{1}{3}$. Since $D = -B$, we get $D = -\frac{1}{3}$.

So, the partial fraction decomposition is:

Now, integrate each term:

$\int \frac{1/3}{x^2 + 1} dx = \frac{1}{3} \int \frac{1}{x^2 + 1} dx = \frac{1}{3} \tan^{-1}(x)$

$\int -\frac{1/3}{x^2 + 4} dx = -\frac{1}{3} \int \frac{1}{x^2 + 2^2} dx$

Using the standard integral $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$ with $a=2$:

$-\frac{1}{3} \left(\frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right)\right) = -\frac{1}{6} \tan^{-1}\left(\frac{x}{2}\right)$

Combining the results:

Therefore, the evaluation of the integral is $\frac{1}{3} \tan^{-1}(x) - \frac{1}{6} \tan^{-1}\left(\frac{x}{2}\right) + C$.

Let the given integral be $I$:

We use the property of definite integrals: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$.

Here, $a=0$ and $b=\frac{\pi}{2}$. So, $a+b-x = \frac{\pi}{2} - x$.

Replace $x$ with $\frac{\pi}{2} - x$ in the integrand:

$\cos\left(\frac{\pi}{2} - x\right) = \sin x$

$\sin\left(\frac{\pi}{2} - x\right) = \cos x$

The integrand becomes $\frac{\sin x}{1 + \cos x + \sin x}$.

Therefore, the integral $I$ can be written as:

Now, add equation (i) and equation (ii):

$I + I = \int_0^{\pi/2} \frac{\cos x}{1 + \sin x + \cos x} dx + \int_0^{\pi/2} \frac{\sin x}{1 + \cos x + \sin x} dx$

$2I = \int_0^{\pi/2} \frac{\cos x + \sin x}{1 + \sin x + \cos x} dx$

$2I = \int_0^{\pi/2} 1 \, dx$

Evaluate the integral of 1:

Apply the limits of integration:

$2I = \frac{\pi}{2}$

Now, solve for $I$:

Therefore, the value of the integral is $\frac{\pi}{4}$.