Integration Techniques: Partial Fractions and Special Forms

Integration by Partial Fractions: Method for Rational Functions

Integration by Partial Fractions is a powerful technique used to integrate rational functions, which are functions expressed as the ratio of two polynomials, say $\frac{P(x)}{Q(x)}$. The core idea is to decompose the rational function into a sum of simpler fractions, called partial fractions, whose integrals are known or easier to find.

The method primarily applies to proper rational functions, where the degree of the numerator polynomial $P(x)$ is strictly less than the degree of the denominator polynomial $Q(x)$ (i.e., degree$(P(x)) <$ degree$(Q(x))$).

If the given rational function is improper (degree$(P(x)) \ge$ degree$(Q(x))$), the first step is to perform polynomial long division. This divides the improper rational function into a polynomial part and a remainder part, where the remainder divided by the original denominator is a proper rational function:

$\frac{P(x)}{Q(x)} = \text{Quotient Polynomial}(x) + \frac{\text{Remainder Polynomial}(x)}{Q(x)}$

The integral of the original function is then the integral of the quotient polynomial (which is straightforward) plus the integral of the proper rational function part. We apply the partial fraction decomposition method to this proper rational function $\frac{\text{Remainder}(x)}{Q(x)}$.

Steps for Partial Fraction Decomposition

To decompose a proper rational function $\frac{P(x)}{Q(x)}$ into partial fractions:

1. Factor the Denominator: Completely factor the denominator polynomial $Q(x)$ into a product of linear factors of the form $(ax+b)^m$ and irreducible quadratic factors of the form $(ax^2+bx+c)^n$, where $b^2-4ac < 0$ for the quadratic factors (meaning they have no real roots).
2. Set Up the Partial Fraction Form: Write down the template for the partial fraction decomposition based on the factors of $Q(x)$. Each type of factor contributes specific terms to the decomposition:
   • For each distinct linear factor $(ax+b)$ in $Q(x)$: Include a term of the form $\frac{A}{ax+b}$, where $A$ is a constant to be determined.
   • For each repeated linear factor $(ax+b)^k$ in $Q(x)$, where $k \ge 2$: Include a sum of $k$ terms of the form $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_k}{(ax+b)^k}$, where $A_1, A_2, \dots, A_k$ are constants to be determined.
   • For each distinct irreducible quadratic factor $(ax^2+bx+c)$ in $Q(x)$: Include a term of the form $\frac{Ax+B}{ax^2+bx+c}$, where $A$ and $B$ are constants to be determined.
   • For each repeated irreducible quadratic factor $(ax^2+bx+c)^k$ in $Q(x)$, where $k \ge 2$: Include a sum of $k$ terms of the form $\frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + \dots + \frac{A_kx+B_k}{(ax^2+bx+c)^k}$, where $A_1, B_1, \dots, A_k, B_k$ are constants to be determined.
3. Solve for the Unknown Coefficients: The goal is to find the values of the constants (A, B, $A_i$, $B_i$, etc.) in the partial fraction form.
   • Start by multiplying both sides of the equation from Step 2 by the original denominator $Q(x)$. This clears the denominators, leaving an equation where the original numerator $P(x)$ (or the remainder polynomial) is equal to a sum of polynomials involving the unknown coefficients.
   • Two common methods to solve for the coefficients are:
     • Equating Coefficients: Expand the right side of the equation and collect terms by powers of $x$. Equate the coefficients of corresponding powers of $x$ on both sides of the equation. This results in a system of linear equations in terms of the unknown constants. Solve this system.
     • Substituting Convenient Values: Substitute specific, convenient values of $x$ into the equation obtained after clearing the denominators. The most convenient values are often the real roots of the denominator factors (if any), as these values will make certain terms zero, simplifying the equation and often directly yielding the value of a coefficient. This is sometimes called the Heaviside Cover-Up Method for distinct linear factors.
   • A combination of these methods is often the most efficient way to solve for all coefficients.

Once the constants are found, the original rational function is expressed as a sum of simpler partial fractions. The integration is then performed term by term.

Integrating the Partial Fractions

After decomposing the rational function, integrate each partial fraction term. The resulting integrals typically fall into forms that can be solved using standard formulas or simple substitutions:

• $\int \frac{A}{ax+b} dx$: Use substitution $u = ax+b$, $du = a dx$. $\int \frac{A}{u} \frac{du}{a} = \frac{A}{a} \int \frac{1}{u} du = \frac{A}{a} \ln|u| + C = \frac{A}{a} \ln|ax+b| + C$.

  $\int \frac{A}{ax+b} dx = \frac{A}{a} \ln|ax+b| + C$

• $\int \frac{A}{(ax+b)^k} dx$ (for $k \ge 2$): Use substitution $u = ax+b$, $du = a dx$. $\int \frac{A}{u^k} \frac{du}{a} = \frac{A}{a} \int u^{-k} du = \frac{A}{a} \frac{u^{-k+1}}{-k+1} + C = \frac{A}{a(1-k)u^{k-1}} + C = \frac{A}{a(1-k)(ax+b)^{k-1}} + C$.
• $\int \frac{Ax+B}{ax^2+bx+c} dx$ (for irreducible quadratic $ax^2+bx+c$): This integral often requires algebraic manipulation. You might split the numerator into two parts: one that is a multiple of the derivative of the denominator ($2ax+b$) and a remaining constant term. Completing the square in the denominator might also be needed to relate it to the $\int \frac{1}{u^2+k^2} du$ form (leading to arctangent).

  $\int \frac{2ax+b}{ax^2+bx+c} dx = \ln|ax^2+bx+c| + C$ (by substitution $u=ax^2+bx+c$).

  $\int \frac{k}{u^2+p^2} du = \frac{k}{p} \arctan\left(\frac{u}{p}\right) + C$.

Finally, combine the integral of the polynomial part (if any) with the integrals of all the partial fractions and add a single constant of integration $C$ to the total result.

Answer:

Integration of Rational Functions of $\sin x$ and $\cos x$

Integrals involving rational functions of trigonometric terms, specifically $\sin x$ and $\cos x$, i.e., integrals of the form $\int R(\sin x, \cos x) dx$, where $R$ is a rational function (a ratio of polynomials) of $\sin x$ and $\cos x$, can often be transformed into integrals of rational functions of a single new variable $t$ using a universal substitution. This technique is called the Weierstrass substitution or the tangent half-angle substitution.

Once transformed into a rational function of $t$, the integral can then (theoretically) be solved using the method of partial fractions.

The Weierstrass Substitution

The standard substitution is based on the tangent of half the angle:

$t = \tan(x/2)$

This substitution allows us to express $\sin x$, $\cos x$, and the differential $dx$ entirely in terms of $t$ and $dt$.

1. Finding $dx$ in terms of $dt$:

   From $t = \tan(x/2)$, we can solve for $x$. If $t = \tan(x/2)$, then $\frac{x}{2}$ is the angle whose tangent is $t$. This is the arctangent of $t$.

   $\frac{x}{2} = \arctan t$

   [Taking arctan of both sides]

   Multiply by 2 to solve for $x$:

   "$x = 2 \arctan t$"

   Now, differentiate $x$ with respect to $t$:

   "$\frac{dx}{dt} = \frac{d}{dt}(2 \arctan t) = 2 \cdot \frac{1}{1+t^2}$"

   [Using standard derivative of arctan]

   Write this in differential form:

   $dx = \frac{2 dt}{1+t^2}$

2. Finding $\sin x$ in terms of $t$:

   We use the double-angle identity for sine: $\sin x = 2 \sin(x/2) \cos(x/2)$. We need to express $\sin(x/2)$ and $\cos(x/2)$ in terms of $t = \tan(x/2)$.

   Consider a right triangle where one of the non-right angles is $x/2$. Since $\tan(x/2) = t = t/1$, we can label the opposite side as $t$ and the adjacent side as 1. By the Pythagorean theorem, the hypotenuse is $\sqrt{t^2 + 1^2} = \sqrt{1+t^2}$.

   From this triangle, we can find $\sin(x/2)$ and $\cos(x/2)$:

   $\sin(x/2) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{t}{\sqrt{1+t^2}}$

   $\cos(x/2) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{1+t^2}}$

   Substitute these into the double-angle identity for $\sin x$:

   "$\sin x = 2 \left(\frac{t}{\sqrt{1+t^2}}\right) \left(\frac{1}{\sqrt{1+t^2}}\right) = \frac{2t}{(\sqrt{1+t^2})^2} = \frac{2t}{1+t^2}$"

   $\sin x = \frac{2t}{1+t^2}$

3. Finding $\cos x$ in terms of $t$:

   We use the double-angle identity for cosine: $\cos x = \cos^2(x/2) - \sin^2(x/2)$.

   Substitute the expressions for $\sin(x/2)$ and $\cos(x/2)$ from the right triangle:

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

   Combine the terms:

   "$= \frac{1-t^2}{1+t^2}$"

   $\cos x = \frac{1-t^2}{1+t^2}$

By using the Weierstrass substitution $t = \tan(x/2)$, an integral $\int R(\sin x, \cos x) dx$ can be transformed into an integral of a rational function of $t$ by substituting $dx = \frac{2 dt}{1+t^2}$, $\sin x = \frac{2t}{1+t^2}$, and $\cos x = \frac{1-t^2}{1+t^2}$. The resulting integral $\int R\left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2}\right) \frac{2 dt}{1+t^2}$ is an integral of a rational function of $t$, which can then be solved using techniques like partial fraction decomposition.

Note: While the Weierstrass substitution is universal for rational functions of $\sin x$ and $\cos x$, it often leads to complex rational functions in $t$. Before resorting to it, always check if a simpler substitution (like $u = \sin x$, $u = \cos x$, or $u = \tan x$) is applicable.

Answer:

Standard Evaluation using Specific Formulas (e.g., $\int \frac{dx}{\sqrt{a^2 \pm x^2}}$)

Beyond the basic power rule, trigonometric, and exponential integrals, there are several standard integral forms that commonly appear, particularly those involving square roots of quadratic expressions like $a^2 \pm x^2$ or $x^2 \pm a^2$. These integrals often have standard results that are typically derived using trigonometric substitution, but they are valuable to recognize and use directly.

In the following formulas, $a$ is a positive constant ($a > 0$), and $C$ is the constant of integration. The domains of the results are typically restricted to where the integrand is defined and the result is real-valued.

Standard Integrals involving $\sqrt{a^2 - x^2}$

These forms are related to inverse trigonometric functions.

• $\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C$ (for $-a < x < a$)

(Derived using substitution $x = a \sin \theta$, where $dx = a \cos \theta d\theta$ and $\sqrt{a^2-x^2} = \sqrt{a^2 - a^2\sin^2\theta} = \sqrt{a^2(1-\sin^2\theta)} = \sqrt{a^2 \cos^2\theta} = a|\cos\theta|$. For $x=a\sin\theta$ with $\theta \in [-\pi/2, \pi/2]$, $\cos\theta \ge 0$, so $\sqrt{a^2-x^2}=a\cos\theta$. The integral becomes $\int \frac{a\cos\theta d\theta}{a\cos\theta} = \int d\theta = \theta + C$. Since $x = a\sin\theta$, $\sin\theta = x/a$, so $\theta = \arcsin(x/a)$.)

• $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right) + C$ (for $-a \le x \le a$)

(This integral can also be derived using the substitution $x=a\sin\theta$ and involves integrating $\cos^2\theta$).