Integration Techniques: Substitution and By Parts

Integration by Substitution: Method and Applications

Integration by substitution, often referred to as u-substitution, is a fundamental and powerful technique used to find indefinite integrals (and definite integrals, which we'll see later). It is essentially the counterpart of the Chain Rule from differential calculus. Just as the Chain Rule helps differentiate composite functions, u-substitution helps integrate functions that are (or can be related to) the result of a differentiation via the Chain Rule.

The method is applicable when the integrand (the function being integrated) can be recognized as a composite function multiplied by the derivative of the inner function, possibly off by a constant factor.

The Method of Integration by Substitution

The core idea behind integration by substitution is to simplify the integrand by introducing a new variable, typically $u$. This transformation makes the integral easier to evaluate using standard integration formulas.

Let's say we want to evaluate an integral of the form $\int f(x) dx$. The steps for using substitution are:

1. Choose a Substitution ($u$): Select a part of the integrand to set equal to the new variable $u$. A good choice for $u$ is often:
   • The "inner function" of a composite function (e.g., the argument of a trigonometric, exponential, or logarithmic function, or the base of a power).
   • A quantity whose derivative appears elsewhere in the integrand, possibly up to a constant factor.

   Write down the substitution equation: $u = g(x)$, where $g(x)$ is the chosen part of the integrand.

2. Find the Differential $du$: Differentiate the substitution equation $u = g(x)$ with respect to $x$.

   $\frac{du}{dx} = g'(x)$

   Then, write this relationship in differential form: $du = g'(x) dx$. This equation relates the differentials $du$ and $dx$.

3. Substitute into the Integral: Replace all occurrences of $g(x)$ in the original integral with $u$. Use the relationship $du = g'(x) dx$ (or rearrange it to solve for $dx$, $dx = \frac{du}{g'(x)}$) to replace $dx$ and any remaining parts of the integrand involving $x$. The goal is to transform the entire integral into an expression solely in terms of $u$ and $du$. There should be no $x$'s left in the integrand or the differential. If any $x$'s remain, the substitution was either incorrect or incomplete, or the integral cannot be solved by this simple substitution.
4. Integrate with Respect to $u$: Evaluate the transformed integral $\int h(u) du$ using the standard integration formulas you know (e.g., Power Rule, trigonometric integrals, exponential integrals). Remember to add the constant of integration, $+ C$, at the end of the integration step.
5. Substitute Back to the Original Variable ($x$): The result from Step 4 is a function of $u$. Replace $u$ with its original expression in terms of $x$ (i.e., $g(x)$). The final answer should be a function of $x$ plus the constant of integration.

Key Idea: Reversing the Chain Rule

The method of substitution is directly based on reversing the Chain Rule for differentiation.

Recall the Chain Rule: If $F$ is an antiderivative of $f$ ($F'=f$), then the derivative of the composite function $F(g(x))$ is:

Since integration is the inverse of differentiation, integrating both sides of this equation with respect to $x$ should give us back the original function on the left side (plus a constant):

Now, let's look at the integrand $F'(g(x)) \cdot g'(x)$. If we make the substitution $u = g(x)$, then the differential is $du = g'(x) dx$. Also, let $f(u) = F'(u)$. The integral can be written as $\int f(g(x)) g'(x) dx$.

Substituting $u = g(x)$ and $du = g'(x) dx$ transforms the integral on the left side into:

We know that the integral of $f(u)$ with respect to $u$ is its antiderivative $F(u)$ plus a constant: $\int f(u) du = F(u) + C$.

Substituting $u=g(x)$ back into $F(u)+C$ gives $F(g(x)) + C$, which matches the result from integrating the Chain Rule. The substitution method provides a systematic way to make this transformation $\int f(g(x)) g'(x) dx \longrightarrow \int f(u) du$.

Applications of Integration by Substitution

Substitution is applicable in a wide variety of integrals, particularly those involving composite functions. Some common patterns where substitution is useful include:

• $\int [g(x)]^n g'(x) dx$ (Substitute $u = g(x)$, $du = g'(x) dx \implies \int u^n du$)
• $\int \sin(g(x)) g'(x) dx$ (Substitute $u = g(x)$, $du = g'(x) dx \implies \int \sin u du$)
• $\int e^{g(x)} g'(x) dx$ (Substitute $u = g(x)$, $du = g'(x) dx \implies \int e^u du$)
• $\int \frac{g'(x)}{g(x)} dx$ (Substitute $u = g(x)$, $du = g'(x) dx \implies \int \frac{1}{u} du$)
• Integrals involving arguments of inverse trigonometric functions and their derivatives (e.g., $\int \frac{g'(x)}{\sqrt{1 - [g(x)]^2}} dx$)

The key is identifying the "inner function" $u=g(x)$ such that its derivative $g'(x)$ (or a constant multiple of it) is present elsewhere in the integrand to form the $du$ term.

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Integration by Parts Formula and Application

Integration by Parts is a crucial technique for evaluating integrals of products of functions, especially when direct substitution (u-substitution) is not applicable. It is derived directly from the Product Rule for differentiation.

Derivation from the Product Rule

Recall the Product Rule for differentiating the product of two differentiable functions $u(x)$ and $v(x)$:

We can rearrange this equation to isolate one of the terms on the right side:

Now, integrate both sides of this rearranged equation with respect to $x$. Recall that the integral of a derivative of a function gives back the original function (plus a constant of integration, which we will include at the final step):

Evaluate the integral of the derivative term:

This equation is the core of the Integration by Parts formula.

Integration by Parts Formula

The formula derived above is usually written in a more compact form using differentials. Let $u = u(x)$ and $v = v(x)$. Then the differential of $u$ is $du = u'(x) dx$, and the differential of $v$ is $dv = v'(x) dx$. Substituting these into the formula $\int u(x)v'(x) dx = u(x)v(x) - \int v(x)u'(x) dx$ gives the standard Integration by Parts formula:

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

This formula means that if you have an integral of the form $\int u \cdot dv$ (where you identify one part of the integrand as $u$ and the rest, including $dx$, as $dv$), you can evaluate it by computing $uv$ and then subtracting the integral of $v \cdot du$. The goal is for the new integral $\int v du$ to be simpler to evaluate than the original integral $\int u dv$.

How to Use Integration by Parts

To evaluate an integral $\int f(x) dx$ using integration by parts, you need to creatively split the integrand into two parts, one part to be designated as $u$ and the other part (including $dx$) to be designated as $dv$.

Steps for Applying Integration by Parts:

1. Choose $u$ and $dv$: Identify the integrand and decide how to split it into two parts: a function $u$ and a differential $dv$. The success of the method heavily depends on making a good choice. The ideal choice is one where:
   • $u$ is easy to differentiate to find $du$.
   • $dv$ is easy to integrate to find $v$.
   • Most importantly, the new integral $\int v du$ is significantly simpler or easier to evaluate than the original integral $\int u dv$.
2. Calculate $du$ and $v$:
   • Find the differential of $u$ by differentiating $u$ with respect to $x$ and multiplying by $dx$: $du = \frac{du}{dx} dx$.
   • Find $v$ by integrating $dv$: $v = \int dv$. When integrating $dv$ to find $v$, we do not need to add the constant of integration $+C$ at this intermediate step. Adding it would just introduce extra complexity that cancels out later.
3. Apply the Integration by Parts Formula: Substitute the expressions you found for $u, v, du,$ and $dv$ into the formula:

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

4. Evaluate the New Integral: Evaluate the integral $\int v du$ on the right side of the formula. This might be a standard integral, one solvable by substitution, or in some cases, it might require another application of integration by parts.
5. Combine and Add the Constant of Integration: Combine the $uv$ term and the result from evaluating $\int v du$. Add the constant of integration, $+ C$, to the final result.

Guideline for Choosing $u$ and $dv$ (LIATE / ILATE Rule)

When the integrand is a product of two different types of functions, the choice of $u$ and $dv$ can be guided by the LIATE (or ILATE) acronym. This mnemonic suggests the order of preference for choosing the part of the integrand that will be $u$ (the function to differentiate):

L - Logarithmic functions (e.g., $\ln x, \log_{10} x$) - These are often chosen as $u$ because their derivatives are simpler algebraic functions ($1/x$).

I - Inverse Trigonometric functions (e.g., $\arcsin x, \arctan x$) - Their derivatives are algebraic functions, which can be easier to handle in $\int v du$.

A - Algebraic functions (e.g., $x^n$, polynomials) - These become simpler after differentiation (the power decreases). Choose a polynomial as $u$ if the remaining $dv$ (often exponential or trigonometric) is easily integrable.

T - Trigonometric functions (e.g., $\sin x, \cos x$) - These often cycle through derivatives or become slightly simpler, but their antiderivatives are straightforward.

E - Exponential functions (e.g., $e^x, a^x$) - These are often chosen as $dv$ because they are easy to integrate, and their form doesn't change significantly after integration or differentiation.

The function type that appears higher in the LIATE list is usually chosen as $u$. The remaining part of the integrand, along with $dx$, is then $dv$. This choice often leads to the integral $\int v du$ being simpler than the original $\int u dv$.

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Standard Integrals solvable by Parts (e.g., $\int e^x (f(x) + f'(x)) dx$)

Beyond applying the integration by parts formula $\int u dv = uv - \int v du$ step-by-step, there are certain forms of integrals that are frequently encountered and can be solved using integration by parts, leading to standard results. Recognizing these forms can significantly speed up the integration process.

Integral of the Form $\int e^x [f(x) + f'(x)] dx$

This is a very important standard integral form that directly arises from applying the Product Rule to a specific combination of functions. The result is remarkably simple.

Theorem: For a differentiable function $f(x)$, the integral of $e^x$ multiplied by the sum of $f(x)$ and its derivative $f'(x)$ is given by:

$\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$

Proof of the Theorem:

We can prove this theorem by differentiating the proposed result, $e^x f(x) + C$, and showing that its derivative is the integrand $e^x [f(x) + f'(x)]$.

Consider the function $y = e^x f(x)$. We differentiate this product using the Product Rule for differentiation, $\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$. Let $u(x) = e^x$ and $v(x) = f(x)$.

We know $\frac{d}{dx}(e^x) = e^x$ and $\frac{d}{dx}(f(x)) = f'(x)$. Substitute these into the equation:

Factor out the common term $e^x$ on the right side:

By the definition of an indefinite integral, if the derivative of $e^x f(x)$ is $e^x [f(x) + f'(x)]$, then the indefinite integral of $e^x [f(x) + f'(x)]$ must be $e^x f(x)$ plus the constant of integration $C$.

This completes the proof. Q.E.D.

Application of the Theorem: When evaluating an integral, look closely at the integrand. If it is of the form $e^x$ multiplied by an expression that is the sum of a function and its derivative, you can immediately apply this theorem to write down the result $e^x f(x) + C$, where $f(x)$ is the function part of the sum whose derivative is also present.

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Other Standard Integrals Solvable by Parts

Integration by parts is the standard technique for evaluating integrals of several other common forms. While you can always derive these results by applying the integration by parts formula, it's useful to be familiar with them.

• $\int \ln x dx = x \ln x - x + C$ (We derived this in Example 2 of the previous section, I2)
• $\int x e^x dx$. Let $u=x, dv=e^x dx$. Then $du=dx, v=e^x$. $\int x e^x dx = xe^x - \int e^x dx = xe^x - e^x + C$.

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

• $\int x \sin x dx$. Let $u=x, dv=\sin x dx$. Then $du=dx, v=-\cos x$. $\int x \sin x dx = x(-\cos x) - \int (-\cos x) dx = -x\cos x + \int \cos x dx = -x\cos x + \sin x + C$.

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

• $\int x \cos x dx$. Let $u=x, dv=\cos x dx$. Then $du=dx, v=\sin x$. $\int x \cos x dx = x\sin x - \int \sin x dx = x\sin x - (-\cos x) + C = x\sin x + \cos x + C$. (We derived this in Example 1 of the previous section, I2)
• $\int \arctan x dx$. Let $u=\arctan x, dv=dx$. Then $du=\frac{1}{1+x^2} dx, v=x$. $\int \arctan x dx = x \arctan x - \int x \frac{1}{1+x^2} dx$. The new integral can be solved by substitution (let $w=1+x^2$, $dw=2x dx$). $\int \frac{x}{1+x^2} dx = \frac{1}{2} \int \frac{1}{w} dw = \frac{1}{2} \ln|w| + C_1 = \frac{1}{2}\ln(1+x^2) + C_1$ (since $1+x^2 > 0$). So, $\int \arctan x dx = x \arctan x - \frac{1}{2} \ln(1+x^2) + C$.

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

• Integrals of the form $\int e^{ax} \cos(bx) dx$ and $\int e^{ax} \sin(bx) dx$ require applying integration by parts twice and then solving for the original integral algebraically.
• Integrals involving square roots of quadratics like $\int \sqrt{a^2-x^2} dx$, $\int \sqrt{x^2+a^2} dx$, $\int \sqrt{x^2-a^2} dx$ can be solved using integration by parts (by writing $\sqrt{\cdot} = \sqrt{\cdot} \cdot 1$ and choosing $u=\sqrt{\cdot}, dv=dx$), or more commonly, using trigonometric substitution (covered in a later section).

Familiarity with these standard forms allows for quicker identification of when integration by parts is needed and sometimes recalling the result directly.