How does the method of partial fractions enable many rational functions to be antidifferentiated?
How does the method of trigonometric substitutions help us find antiderivatives of functions that may include expressions like \(\sqrt{a^2-x^2}\text{,}\) \(a^2 + x^2 \text{,}\) and \(\sqrt{a^2 + x^2} \) where \(a \) is any real number?
So far, we've learned two integration techniques: \(u\)-substitution and integration by parts. The former is used to reverse the chain rule, while the latter is used to reverse the product rule. But we have seen that each works only in specialized circumstances. For example, while \(\int \frac{2x+4}{x^2+4x+13} \, dx\) may be evaluated by \(u\)-substitution and \(\int x e^x \, dx\) by integration by parts, neither method provides a route to evaluate \(\int \frac{1}{x^2+4x+13} \, dx\text{.}\)
In this section, we'll study two more integration techniques: the method of partial fractions and trigonometric substitution. Before introducing them, we'll walk through an example to help build intuition about the techniques we've already learned, and where they might fall short.
For each of the indefinite integrals below, decide whether the integral can be evaluated using \(u\)-substitution, integration by parts, a combination of the two, or neither. For integrals for which your answer is affirmative, state the substitution(s) you would use. It is not necessary to actually evaluate any of the integrals completely, unless the integral can be evaluated immediately using a familiar basic antiderivative.
\(\int x^2 \sin(x^3) \, dx\text{,}\) \(\int x^2 \sin(x) \, dx\text{,}\) \(\int \sin(x^3) \, dx\text{,}\) \(\int x^5 \sin(x^3) \, dx\)
\(\int \frac{1}{1+x^2} \, dx\text{,}\) \(\int \frac{x}{1+x^2} \, dx\text{,}\) \(\int \frac{2x+3}{1+x^2} \, dx\text{,}\) \(\int \frac{e^x}{1+(e^x)^2} \, dx\text{,}\)
\(\int x \ln(x) \, dx\text{,}\) \(\int \frac{\ln(x)}{x} \, dx\text{,}\) \(\int \ln(1+x^2) \, dx\text{,}\) \(\int x\ln(1+x^2) \, dx\text{,}\)
\(\int x \sqrt{1-x^2} \, dx\text{,}\) \(\int \frac{1}{\sqrt{1-x^2}} \, dx\text{,}\) \(\int \frac{x}{\sqrt{1-x^2}}\, dx\text{,}\) \(\int \frac{1}{x\sqrt{1-x^2}} \, dx\text{,}\)
The first integral may be solved by substitution using \(u = x^3 \ , du = 3x^2 \, dx \text{.}\) The second may be solved using IBP with \(u = x^2 \ , dv = \sin(x) \, dx \) The third can not be solved using substitution or IBP. The last one can be solve using a combination. First do a substitution with \(w = x^3 \ , dw = 3x^2 \, dx \) to convert the integral to \(\frac{1}{3} \int w \sin(w) \, dw \text{.}\) This may be solved using IBP with \(u = w \) and \(dv = \sin(w) \, dw. \)
The first integral can be evaluated by following Table5.16 which has antiderivatives of some basic functions. The second integral can be solved using the \(u \)-substitution \(u=1+x^2 \ , du=2x \, dx \text{.}\) This transforms the integral to \(\frac{1}{2} \int \frac{1}{u} \, du \text{.}\) The third integral requires algebraic manipulation before you can successfully use a substitution. We can evaluate \(\int \frac{2x+3}{1+x^2} \, dx \) by recognizing that \(\frac{2x+3}{1+x^2} =\frac{2x}{1+x^2} + \frac{3}{1+x^2}\) so we can split the original integral into two integrals. Then, evaluate \(\int \frac{2x}{1+x^2} \, dx \) with the substitution \(u=1+x^2 \ , du=2x \, dx \) and evaluate \(\int \frac{3}{1+x^2} \, dx \) by following Table5.16. The fourth integral can be found by using the substitution \(u=e^x \ , du=e^x \, dx \) which transforms the integral to \(\int \frac{1}{1+u^2} \, du \text{.}\) The antiderivative of \(\frac{1}{1+u^2} \, du \) is in Table5.16.
The first integral can be evaluated by using integration by parts with \(u= \ln(x) \) and \(dv=x \, dx \text{.}\) The second integral is evaluated by using the substitution \(u=\ln(x) \ , du= \frac{1}{x} \, dx \text{.}\) The third integral cannot be solved by only using the methods of integration by parts or substitution. In fact, one solution to evaluate it involves first using integration by parts then using a method called partial fractions. The fourth integral can be evaluated by using both a substitution and integration by parts. First, perform the substitution \(w=1+x^2 \ , dw= 2x \, dx \) which transforms the integral to \(\frac{1}{2} \int \ln(w) \, dw \text{.}\) Integration by parts can be used to evaluate this with \(u=\ln(w) \) and \(dv=dx \text{.}\)
The first integral can be found by using the substitution \(u=1-x^2 \ , du=-2x \, dx \text{.}\) The second integral can be evaluated by following Table5.16. The third integral can be evaluated by using the substitution \(u=1-x^2 \ , du=-2x\, dx \text{.}\) The fourth integral cannot be solved with standard substitution, integration by parts, or a combination of both. We will need a new style of substitution called trigonometric substitution which will be discussed in this section.
The method of partial fractions is used to integrate rational functions, which are functions that can we written as a quotient of polynomials.
For example, the function \(f(x) = \frac{1}{x(x-1)} \) is the quotient of the polynomial funtions \(p(x) = 1 \) and \(q(x)=x(x-1) \text{.}\) While it is not immediately clear how to compute \(\int \frac{1}{x(x-1)} \, dx \) as written, it turns out that a simple algebraic maneuver will make the task much simpler. In particular, before attempting to integrate we rewrite the integrand as
One can check that this equality holds by finding a common denominator for the right-hand side and simplifying. The integral can then be calculated as
The expression \(\frac{1}{x}-\frac{1}{x-1}\) is a partial fraction decomposition of \(\frac{1}{x(x-1)} \text{.}\)
A partial fraction decomposition can be thought of as reversing the process of finding a common denominator. Thus, the method is applicable when the denominator of a rational function can be factored into smaller degree polynomials. In the examples that follow, we'll demonstrate how to find more complicated partial fraction decompositions and use them to integrate rational functions.
Evaluate
If we factor the denominator of the integrand, we can see how \(\frac{5x}{x^2-x-2}=\frac{5x}{(x-2)(x+1)}\) might be the sum of two fractions of the form \(\frac{A}{x-2} + \frac{B}{x+1}\text{,}\) so we suppose that
and look for the constants \(A\) and \(B\text{.}\)
Multiplying both sides of this equation by the denominator of the left hand side, \((x-2)(x+1)\text{,}\) we find that
To find \(A \) and \(B\text{,}\) we will rearrange the right hand side of the equation so that similar terms are grouped.
Notice that if this equation is going to hold, then the coefficients for \(x \) on each side of the equation need to match, and we need the constant terms on each side of the equation to match. This gives us the following system of equations:
We find that \(A=\frac{10}{3} \) and \(B=\frac{5}{3} \text{.}\) Thus, using Equation(5.11), we can rewrite the original integral as
This integral can be evaluated by using the fact that \(\int \frac{1}{u} \, du =\ln|u| + C \) and by using substitutions, and hence we find that
Evaluate
If we factor the denominator of the integrand, we can see that \(\frac{2x^2+5}{x^3-2x^2+x} = \frac{2x^2+5}{x(x-1)^2} \text{.}\) Notice that the factor \((x-1)\) shows up twice in the denominator. We call \((x-1)^2 \) a repeated linear root. The repeated linear root slightly changes our partial fraction decomposition. \(\frac{2x^2+5}{x(x-1)^2} \) can be decomposed into the sum of fractions \(\frac{A}{x}+ \frac{B}{x-1}+\frac{C}{(x-1)^2}\text{.}\) We need both \(\frac{B}{x-1} \) and \(\frac{C}{(x-1)^2} \) since both \(x-1 \) and \((x-1)^2 \) divide \(x(x-1)^2 \text{.}\) Suppose
and look for constants \(A \text{,}\) \(B\text{,}\) and \(C\text{.}\)
Multiplying both sides of this equation by the denominator of the left hand side, \(x(x-1)^2 \text{,}\) we find that
To find \(A \text{,}\) \(B \text{,}\) and \(C \text{,}\) we will rearrange the right hand side of the equation so that similar terms are grouped.
Like the previous example, we need the coefficients for \(x^2 \) to be the same on each side of the equation, the coeffiecients for \(x \) to be the same on each side of the equation, and the constant terms to be the same on each side of the equation. Thus, we obtain the system of equations
We find that \(A=5 \text{,}\) \(B=-3 \text{,}\) and \(C=7 \text{.}\) Thus, using Equation(5.12), we can rewrite the original integral as
This integral can be evaluated by using the fact that \(\int \frac{1}{u} \, du =\ln|u|+C\) and by using appropriate substitutions, and hence we find that
Evaluate
The denominator of the integrand is already factored, and the quadratic polynomial \(x^2 + 2x + 3 \) cannot be factored anymore. We call this factor an irreducible quadratic factor. The irreducible quadratic factor slightly changes our partial fraction decomposition. \(\frac{3x^2-1}{(x^2+2x+3)(x-2)} \) can be decomposed into the sum of fractions \(\frac{Ax+B}{x^2+2x+3}+ \frac{C}{x-2}\text{.}\) We need the term \(\frac{Ax+B}{x^2+2x+3} \) to include all possibilities where the degree of the numerator is less than the degree of the denominator. Suppose
and look for constants \(A \text{,}\) \(B\text{,}\) and \(C\text{.}\)
Multiplying both sides of this equation by the denominator of the left hand side, \((x^2+2x+3)(x-2) \text{,}\) we find that
To find \(A \text{,}\) \(B \text{,}\) and \(C \text{,}\) we will rearrange the right hand side of the equation so that similar terms are grouped.
Like the first example, we need the coefficients for \(x^2 \) to be the same on each side of the equation, the coeffiecients for \(x \) to be the same on each side of the equation, and the constant terms to be the same on each side of the equation. Thus, we obtain the system of equations
We find that \(A=2 \text{,}\) \(B=2 \text{,}\) and \(C=1 \text{.}\) Thus, using Equation(5.13), we can rewrite the original integral as
This integral can be evaluated by using the fact that \(\int \frac{1}{u} \, du =\ln|u|+C\) and by using appropriate substitutions, and hence we find that
Evaluate
Before decomposing the integrand into partial fractions, we need to perform polynomial long division because the degree of the numerator, \(3\text{,}\) is greater than the degree of the denominator, \(2\text{.}\) We need to do this because partial fraction decompsition leads to a sum of rational functions where the degree of the numerator is less than the degree of the denominator.
After completing polynomial long division, we find that
hence, we can change the original integral
We'll proceed by using the method of partial fraction decomposition on the integrand \(\frac{7-3x}{-x^2+3x-2} \text{.}\)
Notice that the denominator factors as \(-x^2+3x-2= (x-1)(2-x) \text{,}\) so we'll suppose the integrand decomposes as
and we'll look for constants \(A \) and \(B \text{.}\)
Multiplying both sides of the equation by the denominator of the left hand side, \((x-1)(2-x)\text{,}\) we find that
To find \(A\) and \(B\text{,}\) we will rearrange the right hand side of the equation so that similar terms are grouped.
To make sure the coefficients of \(x\) match on each side of the equation and that the constant terms are the same on each side of the equation, we obtain the following system of equations:
We find that \(A=4 \) and \(B=1 \text{.}\) Thus, using Equation(5.14), we can rewrite the original integral as
These integrals can be evaluated by using the antidifferentiation rules from Table5.16 and using appropriate substitutions, and hence we find that
Now that we've seen some examples, we give a general description of the method of partial fractions. Note: This method works best on rational functions when \(Q(x)\) can be factored into linear and irreducible quadratic terms. If the degree of \(P(x) \) is greater than or equal to the degree \(Q(x) \text{,}\) then perform polynomial long division to calculate \(\frac{P(x)}{Q(x)}\text{,}\) else move on to the next step. The result gives a new way to write \(R(x)\text{.}\) We know that \(R(x)= S(x)+ \frac{P_2(x)}{Q(x)} \) where \(S(x) \) is the quotient of the polynomial long division and \(P_2 (x) \) is the remainder of the polynomial long division. Now the degree of \(P_2(x) \) is less than the degree of \(Q(x) \text{,}\) so follow the remaining steps to decompose \(\frac{P_2(x)}{Q(x)}\) . If the degree of \(P(x) \) is less than the degree of \(Q(x)\text{,}\) then factor \(Q(x) \) and add a partial fraction for the decomposition based on the following rules: For each distinct linear factor \((x-c)\) of \(Q(x) \text{,}\) add For each repeated linear factor \((x-c)^n\) of \(Q(x)\) where \(n \) is a positive whole number, add For each irreducible quadratic factor \(t(x) \) of \(Q(x) \text{,}\) add For each repeated irreducible quadratic factor \(t(x)^n \) of \(Q(x) \) where \(n \) is a positive whole number, add Now algebraically solve for the value of the constants \(A_1, A_2,..., B_1, B_2,...\) and integrate each piece separately.Method of Partial Fraction Decomposition for a Rational Function \(R(x)=\frac{P(x)}{Q(x)} \)
\begin{equation*}
\frac{A}{x-c}.
\end{equation*}
\begin{equation*}
\frac{A_1}{x-c} + \frac{A_2}{(x-c)^2} + \cdots + \frac{A_n}{(x-c)^n}.
\end{equation*}
\begin{equation*}
\frac{Ax+B}{t(x)}.
\end{equation*}
\begin{equation*}
\frac{A_1x+B_1}{t(x)} + \frac{A_2x+B_2}{t(x)^2} + \cdots + \frac{A_nx+B_n}{t(x)^n}.
\end{equation*}
The smaller blocks that \(R(x) \) is broken into often lead to more manageable antidifferentiation. For example if \(A \) and \(k \) are real numbers, then \(\int \frac{A}{x-k} \, dx = A \ln |x-k| + C\) by using the substitution \(u=x-k \text{.}\) We also know the antiderivative of other common partial fractions like \(\int \frac{A}{x^2+1} \, dx= A \arctan(x) + C \text{.}\)
For each of the following problems, evaluate the integral by finding the partial fraction decomposition of each integrand.
\(\int \frac{1}{x^2 - 2x - 3} \, dx\)
HintThe partial fraction decomposition of the integrand has the form
To use partial fraction decomposition on the integrand, we'll assume that
To solve for \(A \) and \(B\text{,}\) we'll multiply both sides of the equation by the denominator of the left side of the equation and rearrange like the earlier exmaples. We'll have the following sequence of steps:
We know that
so \(A=1/4 \) and \(B=-1/4 \text{.}\) This implies that
\(\int \frac{x^2+1}{x^3 - x^2} \, dx\)
HintThe partial fraction decomposition of the integrand has the form
To use partial fraction decomposition on the integrand, we'll assume that
To solve for \(A \text{,}\) \(B\text{,}\) and \(C \text{,}\) we'll multiply both sides of the equation by the denominator of the left side of the equation and rearrange like the earlier exmaples. We'll have the following sequence of steps:
We know that
so \(A=-1 \text{,}\) \(B=-1\text{,}\) and \(C=2\text{.}\) This implies that
\(\int \frac{x-2}{x^4 + x^2}\, dx\)
HintThe partial fraction decomposition of the integrand has the form
To use partial fraction decomposition on the integrand, we'll assume that
To solve for \(A \text{,}\) \(B\text{,}\) \(C\text{,}\) and \(D\text{,}\) we'll multiply both sides of the equation by the denominator of the left side of the equation and rearrange like the earlier exmaples. We'll have the following sequence of steps:
We know that
so \(A=1 \text{,}\) \(B=-2 \text{,}\) \(C=-1\text{,}\) and \(D=2\text{.}\) Observing that \(\frac{-x+2}{1+x^2} = \frac{-x}{1+x^2}+\frac{2}{1+x^2}\text{,}\) it now follows that
Noting that \(\int \frac{x}{1+x^2} \, dx\) can be evaluated by the \(u\)-substitution \(u=1+x^2\text{,}\) we see that \(\int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln(1+x^2) + C\text{.}\) Thus
\(\int \frac{4x^2+x}{2x^2+x-1} \, dx\)
HintFirst, use polynomial long division then the partial fraction decomposition of the remainder will have the form \(\frac{A}{2x-1}+\frac{B}{x+1}\text{.}\)
\(\int \frac{4x^2+x}{2x^2+x-1} \, dx=2x + \frac{1}{2} \ln|2x-1|-\ln|x+1|+C\)
First, use polynomial long division since the degree of the numerator is the same as the degree of the denominator. We'll find that
To use partial fraction decomposition on the remainder, we'll assume that
To solve for \(A \) and \(B\text{,}\) we'll multiply both sides of the equation by the denominator of the left side of the equation and rearrange like the earlier exmaples. We'll have the following sequence of steps:
We know that
so \(A=1 \) and \(B=-1\text{.}\) This implies that
In Section5.5, we saw that certain integrals could be simplified considerably by making a \(u \)-substitution. Here, we'll see an extension of this idea that will allow us to compute some integrals which don't immediately appear amenable to traditional substitutions. Making a substitution of the form \(x=\sin(\theta) \) or \(x=\tan(\theta) \) and applying of one of the following trigonometric identities will be our modus operandi.
Pythagorean Identity:
Modified Pythagorean Identity:
(Note that this identity follows from the regular Pythagorean identity and dividing both sides by \(\cos^2(\theta) \text{.}\))
First, let's look at the substitution \(x=\sin(\theta) \) where \(-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}\) 8Example5.60 will show why we need these bounds on \(\theta \) to see how it can be used to find the indefinite integral \(\int \frac{1}{\sqrt{1-x^2}} \, dx\text{.}\) Use the substitution \(x=\sin(\theta) \) where \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \) to find the general antiderivative When using the substitution \(x=\sin(\theta) \) where \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \text{,}\) we are regarding \(x \) as a function of \(\theta \text{.}\) The goal is to change the integrand to contain trigonometric functions so that the Pythagorean Identity can be used. To do this, the integral will be changed to be in terms of the variable \(\theta\text{.}\) With this in mind, both \(x \) and \(dx \) will need to be changed. Starting from \(x=\sin(\theta) \text{,}\) take the derivative with respect to \(\theta \) to find that In other words, \(dx=\cos(\theta) \, d\theta \text{.}\) Now, change the original integral to be in terms of the variable \(\theta \text{:}\) Use the Pythagorean Identity, \(1-\sin^2(\theta)=\cos^2(\theta)\) to simplify the denominator of the integrand: Now we need to use the assumption that \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\text{.}\) In general, \(\sqrt{\cos^2(\theta)}=|\cos(\theta)|\text{,}\) but \(\cos(\theta)\geq 0\) on the interval \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\text{.}\) Thus, \(|\cos(\theta)|=\cos(\theta)\) if \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \text{.}\) The denominator can now be further simplified, and the remaining steps lead to a familar integral: While we have computed an antiderivative successfully, we aren't quite done yet because the answer is not written in terms of the original variable \(x \text{.}\) Rearrange the substitution equation \(x=\sin(\theta) \) to find that \(\theta=\arcsin(x) \) and use this to turn \(\theta\) back into a function of \(x \text{.}\) Therefore,
Example5.60
\begin{equation*}
\int \frac{1}{\sqrt{1-x^2}}\, dx.
\end{equation*}
\begin{equation*}
\frac{dx}{d\theta}=\cos(\theta) .
\end{equation*}
\begin{equation*}
\int\frac{1}{\sqrt{1-x^2}}\, dx=\int \frac{\cos(\theta)}{\sqrt{1-\sin^2(\theta)}} \, d\theta
\end{equation*}
\begin{equation*}
\int \frac{\cos(\theta)}{\sqrt{1-\sin^2(\theta)}} \, d\theta = \int\frac{\cos(\theta)}{\sqrt{\cos^2(\theta)}} \, d \theta.
\end{equation*}
\begin{align*}
\int\frac{\cos(\theta)}{\sqrt{\cos^2(\theta)}} \, d\theta \amp = \int\frac{\cos(\theta)}{|\cos(\theta)|} \, d\theta \\
\amp =\int\frac{\cos(\theta)}{\cos(\theta)} \, d\theta \\
\amp =\int 1 \, d\theta \\
\amp =\theta + C
\end{align*}
\begin{equation*}
\int\frac{1}{\sqrt{1-x^2}} \, dx= \arcsin(x) + C
\end{equation*}
From now on, we'll usually omit the explanation that \(\sqrt{\cos^2(\theta)}=|\cos(\theta)|=\cos(\theta)\) since we additionally assume that \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \) in order to force \(\cos(\theta) \geq 0 \text{.}\) It is still incredibly important so that the calculation of the antiderivative is seamless, but since we always assume that \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \) the same reasoning always applies.
The substitution demonstrated in the previous example can be generalized to situations in which the integrand contains \(a^2-x^2 \text{,}\) where \(a \) is a real number and \(x \) is a variable. In this case, setting \(x = a\sin(\theta) \) results in the simplification \(a^{2}-x^{2} = a^{2}(1-\sin^{2}(\theta))=a^{2}\cos^{2}(\theta) \text{.}\) In this example, we'll evaluate the definite integral \(\int_{0}^{4} \sqrt{16-x^{2}} \,dx \text{.}\) Use a trigonometric substitution of the form \(x = a\sin(\theta) \text{,}\) with \(a \) a real number, to find an antiderivative of Leave the antiderivative in terms of \(\theta \text{.}\) Use the substitution \(x=4\sin(\theta) \) with \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \text{.}\) The double angle formula \(\cos^{2}(\theta) = \frac{1+\cos(2\theta)}{2} \) will be useful in evaluating the integral that results from this substitution. With \(x = 4\sin(\theta) \text{,}\) a general antiderivative in terms of \(\theta \) is given by Use the substitution \(x=4\sin(\theta) \text{.}\) Then \(x^2=16\sin^2(\theta) \) and \(dx=4\cos(\theta) d\theta \text{.}\) So From the hint, we may write \(\cos^{2}(\theta) = \frac{1+\cos(2\theta)}{2} \text{.}\) Thus, Using the solution to part (a), evaluate the definite integral \(\int_{0}^{4} \sqrt{16-x^{2}} \,dx \) by writing the bounds of integration in terms of \(\theta \text{.}\) To rewrite the bounds in terms of \(\theta \text{,}\) solve the equations \(0 = 4 \sin(\theta) \) and \(4 = 4 \sin(\theta) \) for \(\theta \in [-\pi/2, \pi/2] \text{.}\) Using \(x = 4\sin(\theta) \text{,}\) when \(x = 0 \) we must have \(0 = 4\sin(\theta) \text{,}\) which implies \(\theta = 0 \) (recall \(-\pi /2 \leq \theta \leq \pi/2 \)). Similarly, when \(x = 4 \) we have \(4=4\sin(\theta) \text{,}\) which has solution \(\theta = \pi/2.\) Thus, using part (a), we may write It is worth noting that the curve \(y = \sqrt{16-x^{2}} \) for \(0 \leq x \leq 4 \) is the part of the circle of radius \(4 \) centered at the origin which lies in the first quadrant of the plane. The definite integral \(\int_{0}^{4} \sqrt{16-x^{2}} \,dx \) thus gives the area of the part of this circle which lies in the first quadrant. Since the circle in question has total area \(16 \pi \) and the portion in the first quadrant compirese precisely \(1/4 \) of the total area, we are assured our solution of \(4\pi \) is correct.
Example5.61
\begin{equation*}
\int \sqrt{16-x^{2}} \, dx.
\end{equation*}
Answer Solution
\begin{equation*}
8\theta +4\sin(2\theta) + C
\end{equation*}
\begin{align*}
\int \sqrt{16-x^{2}} \, dx \amp = \int \sqrt{16-16\sin^2(\theta)}\cdot 4\cos(\theta) \, d\theta \\
\amp = \int \sqrt{16(1-\sin^2(\theta))}\cdot 4\cos(\theta) \, d\theta \\
\amp = \int \sqrt{16\cos^2(\theta)}\cdot 4\cos(\theta) \, d\theta\\
\amp = \int 4\cos(\theta) \cdot 4\cos(\theta) \, d\theta\\
\amp = 16\int \cos^{2}(\theta) \, d\theta
\end{align*}
\begin{align*}
16\int \cos^{2}(\theta) \, d\theta \amp = 16\int\frac{1+\cos(2\theta)}{2} \, d\theta \\
\amp = 8 \int 1+\cos(2\theta) \,d \theta \\
\amp = 8(\theta+\frac{1}{2}\sin(2\theta))+C \\
\amp = 8\theta+4\sin(2\theta)+C
\end{align*}
Answer Solution
\begin{equation*}
\int_{0}^{4} \sqrt{16-x^{2}} \,dx = 4 \pi
\end{equation*}
\begin{align*}
\int_{0}^{4} \sqrt{16-x^{2}} \,dx \amp = \left. 8\theta+4\sin(2\theta) \right|_{0}^{\pi/2} \\
\amp = 4\pi+4\sin(\pi)-(0+4\sin(0)) \\
\amp = 4\pi
\end{align*}
Use a trigonometric substitution to find the indefinite integral
Notice that \(\sqrt{9-4x^2}=\sqrt{9-(2x)^2}\text{,}\) so we'll want to replace \(2x \) by an appropriate trigonometric function.
Compared to the first example, where the substitution \(x=\sin(\theta) \) worked, we'll need to make two adjustments so that we can still use Pythagoren's Identity. Since \(\sqrt{9-4x^2}=\sqrt{9-(2x)^2}\text{,}\) we'll want to substitute \(2x=3\sin(\theta) \) so that we can factor out the 9 and obtain the expression \(1-\sin^2(\theta) \text{.}\) Assuming \(2x=3\sin(\theta) \) then \((2x)^2=9 \sin^2(\theta)\) and \(2 dx=3 \cos(\theta)d\theta\text{.}\)
We can solve for \(\theta \) in our substitution equation to find that the antiderivative is:
Note that we could have also computed the integral above using a \(u \)-substitution and the fact that \(\int \frac{1}{\sqrt{1-x^{2}}} \, dx = \arcsin(x)+C \text{.}\) What is the right choice for \(u \) in this substitution?
Find the antiderivative
Use the substitution \(x=3\sin(\theta) \) and treat \((9-x^2)^{3/2}\) as \(((9-x^2)^{1/2})^3 \text{.}\)
\(\int \frac{9}{(9-x^2)^{3/2}} \, dx=\dfrac{x}{9-x^2} + C \)
Use the substitution \(x=3\sin(\theta) \text{.}\) Then \(x^2=9\sin^2(\theta)\) and \(dx=3\cos(\theta) d\theta\text{.}\) As stated in the hint, we'll use that \((9-x^2)^{3/2}=((9-x^2)^{1/2})^3\) and simplify the inner most set of parentheses before cubing.
When we solve for \(\theta \) and replace it with a function of \(x \text{,}\) we obtain
In other words, the general antiderivative is the tangent of the angle whose sine is \(\frac{x}{3} \)
. To simplify this expression, we can look at a right triangle that has \(\theta \) as one of its angles and use some trigonometry. Since \(\sin(\theta) = \frac{x}{3} \text{,}\) we'll assume that the hypotenuse has length 3 and the side opposite the angle \(\theta \) has length \(x \text{.}\) The missing side length is found by using Pythagorean's Theorem. Then using this right triangle, we find that \(\tan(\theta)=\frac{x}{\sqrt{9-x^2}} \text{.}\) Therefore,
We summarize the substitution \(x = a \sin(\theta) \) below.
The substitution \(u=a\sin(\theta) \) where \(u \) is some function of \(x \text{,}\) \(a \) is a real number, and \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \) is often helpful when the integrand contains an expression of the form \(a^2-u^2 \text{.}\) When this substitution is used, then \(u^2=a^2\sin^2(\theta)\) and after factoring out \(a^2 \text{,}\) the Pythagorean Identity 9\(1-\sin^2(\theta)=\cos^2(\theta) \) can be used to simplify part of the integrand:
Since we assumed \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \text{,}\) then \(\cos (\theta) \geq 0 \text{.}\) This implies that \(\sqrt{\cos^2(\theta)}=|\cos(\theta)|=\cos(\theta) \text{.}\)
Another important trigonometric substitution is \(x=\tan(\theta) \) where \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\text{.}\) This substitution can be used to show that \(\int \frac{1}{1+x^2} \, dx= \arctan(x) + C \text{.}\) Use the trigonometric substitution \(x=\tan(\theta) \) to find the antiderivative Proceed in a manner analogous to Example5.60. When using the substitution \(x=\tan(\theta) \) where \(-\frac{\pi}{2} < \theta < \frac{\pi}{2} \text{,}\) we are regarding \(x \) as a function of \(\theta \text{.}\) The goal is to change the integrand to contain trigonometric functions so that a modified version of the Pythagorean Identity can be used. The version of the Pythagorean Identity we will use when simplifying after a tangent substitution is \(\tan^2(\theta) + 1 =\sec^2(\theta) \text{.}\) To achieve this goal, the integral will be changed to be in terms of the variable \(\theta\text{.}\) With this in mind, both \(x \) and \(dx \) will need to be changed. Starting from \(x=\tan(\theta) \text{,}\) take the derivative with respect to \(\theta \) to find that In other words, \(dx=\sec^2(\theta) \, d\theta \text{.}\) Now, change the original integral to be in terms of the variable \(\theta \text{:}\) Use a modified version of the Pythagorean Identity, \(1+\tan^2(\theta)=\sec^2(\theta)\) to simplify the denominator of the integrand: The integrand can now be further simplified, and the remaining steps lead to a familar antiderivative: While we have computed an antiderivative successfully, we aren't quite done yet because the answer is not written in terms of the original variable \(x \text{.}\) Rearrange the substitution equation \(x=\tan(\theta) \) to find that \(\theta=\arctan(x) \) and use this to turn \(\theta\) back into a function of \(x \text{.}\) Therefore,
Example5.65
\begin{equation*}
\int \frac{1}{1+x^2} \, dx \text{.}
\end{equation*}
Hint
Answer
Solution
\begin{equation*}
\int \frac{1}{1+x^2} \, dx = \arctan(x)+C
\end{equation*}
\begin{equation*}
\frac{dx}{d\theta}=\sec^2(\theta) .
\end{equation*}
\begin{equation*}
\int\frac{1}{1+x^2}\, dx=\int \frac{\sec^2(\theta)}{1+\tan^2(\theta)} \, d\theta
\end{equation*}
\begin{equation*}
\int \frac{\sec^2(\theta)}{1+\tan^2(\theta)} \, d\theta = \int\frac{\sec^2(\theta)}{\sec^2(\theta)} \, d \theta.
\end{equation*}
\begin{align*}
\int\frac{\sec^2(\theta)}{\sec^2(\theta)} \, d \theta&= \int 1 \, d \theta\\
&=\theta + C
\end{align*}
\begin{equation*}
\int\frac{1}{1+x^2} \, dx= \arctan(x) + C
\end{equation*}
Just as the substition \(x = a\sin(\theta) \) can be useful when \(a^{2}-x^{2} \) appears in the integrand, \(x = a \tan(\theta) \) may be the right substitution when dealing with terms involving \(a^{2}+x^{2} \text{.}\) The following examples will explore this idea.
Use a trigonometric substitution to find the antiderivative
Try the substitution \(x=2\tan(\theta) \text{.}\) The integral that results from this substitution may require a \(u\)-substitution to compute.
Set \(x=2\tan(\theta) \text{,}\) so that \(dx=2\sec^2(\theta) d\theta\text{.}\) Since \(x^{2}=4\tan^{2}(\theta) \text{,}\) we get
Since \(\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)} \) and \(\sec(\theta) = \frac{1}{\cos(\theta)} \text{,}\) we have
The integral \(\frac{1}{4}\int \frac{\cos(\theta)}{\sin^{2}(\theta)}\, d\theta \) can be evaluated via yet another substitution---this time, a \(u\)-substitution in which we set \(u = \sin(\theta). \) Then \(du = \cos(\theta) d\theta \text{.}\) Finally,
Substituting back for \(u \text{,}\) we reach \(-\frac{1}{4u} = -\frac{1}{4\sin(\theta)}=-\frac{\csc(\theta)}{4} \text{.}\)
Now, one may establish a right triangle relating the angle \(\theta \) and the variable \(x \text{,}\) just as in Example5.63. Doing this yields \(\sin(\theta) = \frac{x}{\sqrt{4+x^{2}}} \text{,}\) and thus \(-\frac{\csc(\theta)}{4} = -\frac{\sqrt{4+x^{2}}}{4x}.\) We conclude \(\int \frac{1}{x^{2}\sqrt{4 + x^{2}}} \, dx = -\frac{\sqrt{4+x^{2}}}{4x}+C.\)
In the following example, we'll see why it is important that we restrict \(\theta \) to be in the interval \(-\frac{\pi}{2} < \theta < \frac{\pi}{2} \) when we use a tangent substitution.
Use a trigonometric substitution to find the integral
The expression \(25 + 9x^2 \) can be rewritten as \(5^2+(3x)^2\) so that it looks like \(u^2+a^2\text{.}\) Try the substitution \(3x=5\tan(\theta) \text{.}\)
The expression \(25 + 9x^2 \) can be rewritten as \(5^2+(3x)^2\) so that it looks like \(u^2+a^2\text{.}\) The substitution \(3x=5\tan(\theta) \) will allow us to use the modified Pythagorean Identity. Then \((3x)^2=25\tan^2(\theta)\) and \(dx=\frac{5}{3} \sec^2(\theta) d\theta \text{.}\) The original integral can be transformed using this substitution:
We must be careful at this step when we simplify \(\sqrt{\sec^2(\theta)}\text{.}\) Because we chose \(-\frac{\pi}{2} < \theta < \frac{\pi}{2} \text{,}\) then \(\sec^2(\theta) \geq 1 \geq 0 \text{.}\) Notice that if we had picked a different interval for \(\theta \text{,}\) \(\sec(\theta) \) could be negative and then we would have to introduce an absolute value in order for the square root to be defined. Since we know \(\sec^2(\theta) \geq 1\geq 0\) on the interval of \(\theta\) we care about then \(\sqrt{\sec^2(\theta)}=\sec(\theta)\text{.}\) Therefore, the integral simplifies further:
This answer can be simplified to a more recognizable form by using a right triangle with angle \(\theta \) and \(\tan(\theta)=\frac{3x}{5} \text{,}\) and finding \(\sec(\theta) \) by using the triangle. 
A trigonometric substitution was not necessary to find the antiderivative in part (a). In fact, only a standard \(u \)-subsitution was needed. Verify the answer from part (a) using the method of \(u\)-substitution.
HintThe expressions \(3x \) and \(9x^2 + 25\) are nearly a function-derivative pair.
Use the substitution \(u=9x^2 + 25 \) with \(du=18x dx\) to change the original integral.
In our study of the method of partial fractions, we saw how to deal with rational functions whose denominators could be factored into smaller degree pieces. What if the denominator cannot be factored, as is the case with \(\frac{1}{x^2+4x+13}\text{?}\)
If we complete the square in the denominator to recognize it as \((x-a)^{2}+b\) where \(a \) and \(b \) are real numbers, then substitution becomes feasible. (For more on completing the square, see the relevant section in the Precalculus Textbook).
Find the integral
Completing the square in the denominator gives
To complete the square in the denominator, we need to add and substract \(4 \) so that we can factor a portion of the expression into a linear term that is squared:
Now, use the \(u \)-substitution \(u=x+2 \ , du=dx \text{.}\)
Use the tangent subsitution10We could also factor out \(9 \) from the denominator and use a substitution with \(w=\frac{x+2}{3}\) as demonstrated in Example5.66. \(u=3\tan(\theta) \text{,}\) so \(u^2=9\tan^2(\theta) \) and \(du=3\sec^2(\theta) \, d\theta \text{.}\)
The substitution \(u=a\tan(\theta) \) where \(u \) is some function of \(x \text{,}\) \(a \) is a real number, and \(-\frac{\pi}{2} < \theta < \frac{\pi}{2} \) is often helpful when the integrand contains an expression of the form \(a^2+u^2 \text{.}\) When this substitution is used, then \(u^2=a^2\tan^2(\theta)\) and after factoring out \(a^2 \text{,}\) the modified Pythagorean Identity 11\(1+\tan^2(\theta)=\sec^2(\theta) \) can be used to simplify part of the integrand:
Since we assumed \(-\frac{\pi}{2} < \theta < \frac{\pi}{2} \text{,}\) then \(\sec (\theta) \geq 1\geq 0 \text{.}\) This implies that \(\sqrt{\sec^2(\theta)}=|\sec(\theta)|=\sec(\theta) \text{.}\)
Find the following antiderivatives using a trigonometric substitution or some other method. \(\int \frac{3}{(9x^2+1)^{3/2}} \, dx \) \(\int \frac{1}{x^2\sqrt{5-x^2}} \, dx \) \(\int \frac{\sqrt{25-9x^2}}{9x^2} \, dx\) \(\int \frac{2x+7}{x^2+6x+18} \, dx \) There is a sum of squares, \(9x^2 + 1 =(3x)^2+1^2 \text{,}\) so a tangent substitutuion is likely to help. There is a difference of squares since \(5-x^2=(\sqrt{5})^2-x^2\text{,}\) so a sine substitution is likely to help. There is a difference of squares starting with a constant term \(25-9x^2=5^2-(3x)^2 \text{,}\) so a sine substitution is likely to help. First, complete the square in the denominator. Then, split the integrand into two rational expressions and use different methods for each of the resulting rational expressions. Use the substitution \(3x=\tan(\theta) \text{,}\) so \((3x)^2=9x^2=\tan^2(\theta) \) and \(3dx=\sec^2(\theta)d\theta \text{.}\) We can transform the orignal integral using this substitution: To find the antiderivative in terms of \(x\text{,}\) consider a right triangle with angle \(\theta\) and \(\tan(\theta)=3x\text{.}\) Use the substitution \(x=\sqrt{5} \sin(\theta) \text{,}\) then \(x^2=5\sin^2(\theta)\) and \(dx=\sqrt{5} \cos(\theta) d\theta\text{.}\) Use a right triangle with one angle measuring \(\theta \) and with \(\sin(\theta)=\frac{x}{\sqrt{5}} \) to find the antiderivative with respect to \(x \text{.}\) Use the substitution \(3x=5\sin(\theta) \text{,}\) then \(9x^2=25\sin^2(\theta) \) and \(3dx=5\cos(\theta) d\theta \text{.}\) Use a right triangle with one angle measuring \(\theta \) and with \(\sin(\theta) =\frac{3x}{5} \) to find \(\cot(\theta) \text{:}\) To integrate \(\int \frac{2u}{u^2+9} \, du\text{,}\) use the substitution \(w=u^2+9 \) and \(dw= 2u \, du\text{.}\) Thus, To integrate \(\int \frac{4}{u^2+9} \, du \text{,}\) we can use a trigonometric substitution or we can factor out the 9, use the substitution \(v=w/3 \text{ and }dv=dw/3\) and use the antidifferentiation rule \(\int\frac{1}{1+u^2} \, du=\arctan(u) + C\text{.}\) The latter method will be shown: Therefore,
Example5.70
Hint
Answer
Solution
\begin{equation*}
\int \frac{3}{(9x^2+1)^{3/2}} \, dx = \frac{3x}{\sqrt{9x^2+1}} + C
\end{equation*}
\begin{equation*}
\int \frac{1}{x^2\sqrt{5-x^2}} \, dx =\frac{-\sqrt{5-x^2}}{5x} + C
\end{equation*}
\begin{equation*}
\int \frac{\sqrt{25-9x^2}}{9x^2} \, dx=- \frac{\sqrt{25x^2-9}}{9x} - \frac{1}{3} \arcsin\left(\frac{3x}{5} \right) + C\
\end{equation*}
\begin{equation*}
\int \frac{2x+7}{x^2+6x+18} \, dx= \ln|x^2+6x+18| + \frac{4}{3}\arctan\left(\frac{x+3}{3} \right) + C
\end{equation*}
\begin{align*}
\int \frac{3}{(9x^2+1)^{3/2}} \, dx&=\int \frac{\sec^2(\theta)}{(\tan^2(\theta)+ 1)^{3/2}} \, d \theta\\
&=\int \frac{\sec^2(\theta)}{(\sec^2(\theta))^{3/2}} \, d \theta\\
&=\int \frac{\sec^2(\theta)}{\sec^3(\theta)} \, d \theta\\
&=\int \cos(\theta) \, d \theta\\
&=\sin(\theta) + C
\end{align*}
\begin{equation*}
\int \frac{3}{(9x^2+1)^{3/2}} \, dx=\frac{3x}{\sqrt{9x^2+1}} + C
\end{equation*}
\begin{align*}
\int \frac{1}{x^2\sqrt{5-x^2}} \, dx&=\int \frac{\sqrt{5}\cos(\theta)}{5\sin^2(\theta) \sqrt{5-5\sin^2(\theta)}} \, d\theta\\
&=\int \frac{\sqrt{5}\cos(\theta)}{5\sin^2(\theta) \sqrt{5\cos^2(\theta)}} \, d\theta\\
&=\frac{\sqrt{5}}{5\sqrt{5}}\int \frac{\cos(\theta)}{\sin^2(\theta) \cos(\theta) } \, d\theta\\
&=\frac{1}{5}\int \frac{1}{\sin^2(\theta)} \, d\theta\\
&=-\frac{1}{5}\cot(\theta) + C
\end{align*}
\begin{equation*}
\int \frac{1}{x^2\sqrt{5-x^2}} \, dx =\frac{-\sqrt{5-x^2}}{5x} + C
\end{equation*}
\begin{align*}
\int\frac{\sqrt{25-9x^2}}{9x^2} \, dx &= \int \frac{\sqrt{25-25\sin^2(\theta)}}{25\sin^2(\theta)} \frac{5}{3} \cos(\theta) \, d\theta \\
&= \int \frac{25\cos^2(\theta)}{75\sin^2(\theta)} \, d\theta \\
&= \frac{1}{3}\int \frac{1-\sin^2(\theta)}{\sin^2(\theta)} \, d\theta \\
&= \frac{1}{3}\int \frac{1}{\sin^2(\theta)} \, d\theta - \frac{1}{3} \int 1 \, d\theta \\
&= -\frac{1}{3}\cot(\theta) -\frac{1}{3} \arcsin(\theta) + C
\end{align*}
\begin{equation*}
\int\frac{\sqrt{25-9x^2}}{9x^2} \, dx= -\frac{\sqrt{25-9x^2}}{9x} -\frac{1}{3}\arcsin\left(\frac{3x}{5} \right) + C
\end{equation*}
\begin{align*}
\int \frac{2x+7}{x^2+6x+18} \, dx&=\int \frac{2x+ 7}{(x+3)^2+9} \, dx \\
\text{use the substitution } u=x+3 &\qquad du=dx \\
&=\int \frac{2(u-3)+ 7}{u^2+9} \, dx \\
&=\int \frac{2u}{u^2+9} \, du + \int \frac{4}{u^2+9} \, du
\end{align*}
\begin{align*}
\int \frac{2u}{u^2+9} \, du&=\int \frac{1}{w} \, dw\\
&=\ln|w| + C_1\\
&=\ln|(x+3)^2 + 9| + C_1
\end{align*}
\begin{align*}
\int \frac{4}{u^2+9} \, du &=\frac{4}{9} \int \frac{1}{(u/3)^2 + 1} \, du\\
&=\frac{4}{3} \int \frac{1}{v^2 + 1} \, dv\\
&=\frac{4}{3} \arctan(v) + C_2 \\
&=\frac{4}{3} \arctan\left(\frac{x+3}{3}\right) + C_2
\end{align*}
\begin{equation*}
\int \frac{2x+7}{x^2+6x+18} \, dx= \ln|(x+3)^2 + 9| + \frac{4}{3} \arctan\left(\frac{x+3}{3}\right) + C
\end{equation*}
We can antidifferentiate any rational function with the method of partial fractions. Any polynomial function can be factored into a product of linear and irreducible quadratic terms, so any rational function may be written as the sum of:
Trigonometric substitutions of the form \(u=\sin(x)\text{,}\) \(u=\tan(x)\text{,}\) or variants thereof let us find integrals like \(\int \frac{1}{\sqrt{1-x^2}} dx\text{,}\) \(\int \frac{1}{1+x^2} dx\text{,}\) and \(\int \frac{1}{\sqrt{1-x^2}} \, dx \text{.}\) A tangent-based substitution is essential for finding integrals of the form \(\frac{Bx+C}{t(x)}\) where \(t(x)\) is an irreducible quadratic.
\(\int \frac{1}{\sqrt{25-x^2}} dx \)
\(\int \frac{1}{1+4x^2} dx \)
\(\int \frac{1}{\sqrt{25-4x^2}} dx \)
\(\int \frac{1}{7+9x^2} dx \)
\(\int \frac{1+x}{1+x^2} dx \)
\(\int \frac{x^2+4x+1}{x^3+x} dx \)
\(\int \frac{1}{x^2+6x+10} dx \)
\(\int \frac{e^x}{e^{2x}+1} dx \)
