Section 3.3 Trigonometric Substitutions

Subsection 3.3.1 Overview

This section treats indefinite integrals involving square roots of quadratic polynomials. A trigonometric substitution converts the square root into a trigonometric expression, using a Pythagorean identity; the resulting integral can then be evaluated by the methods of Section 3.2.

After completing the square and factoring, we reduce to one of three standard forms: $\sqrt{a^2-u^2}\text{,}$ $\sqrt{u^2-a^2}\text{,}$ or $\sqrt{u^2+a^2}\text{,}$ where $a > 0\text{.}$ The table below pairs each form with its substitution and a right triangle for recovering the original variable.

$Right triangle with hypotenuse a and opposite side u.$

$u=a\sin(\theta)$

$-\pi/2 \leq \theta \leq \pi/2$

$Right triangle with hypotenuse u and adjacent side a.$

$u=a\sec(\theta)$

$0 \leq \theta \leq \pi,\ \theta \neq \pi/2$

$Right triangle with opposite side u and adjacent side a.$

$u=a\tan(\theta)$

$-\pi/2 \lt \theta \lt \pi/2$

The trigonometric choices are not unique. For example, $u=a\cot(\theta)$ also works for $\sqrt{a^2+u^2}\text{.}$ We use the substitutions above because the accompanying identities are especially direct.

The substitution $u = a\sin(\theta)$ transforms $\sqrt{a^2-u^2}$ into

\begin{equation*} a|\cos(\theta)| = a\cos(\theta)\quad (-\pi/2 \le \theta \le \pi/2). \end{equation*}

Similarly, $u = a\tan(\theta)$ converts $\sqrt{a^2 + u^2}$ to

\begin{equation*} a|\sec(\theta)| = a\sec(\theta)\quad (-\pi/2 \lt \theta \lt \pi/2) \end{equation*}

and $u=a\sec(\theta)$ transforms $\sqrt{u^2-a^2}$ into $a|\tan(\theta)|\text{.}$ Note that:

\begin{equation*} \begin{aligned} |\tan(\theta)| = \begin{cases} \phantom{-}\tan(\theta) & 0 \le \theta \lt \pi/2 \\ -\tan(\theta) & \pi/2 \lt \theta \le \pi. \end{cases} \end{aligned} \end{equation*}

We begin with an integral involving $\sqrt{x^2+a^2}\text{.}$

Example 3.29.

To compute $\ds \int \frac{dx}{(x^2+9)^2}\text{,}$ let $x = 3\tan \theta\text{.}$ Then $dx = 3\sec^2(\theta)d\theta$ and $x^2 + 9 = 9\sec^2(\theta)\text{.}$ Thus,

\begin{equation*} \begin{aligned} \int \frac{dx}{(x^2+9)^2} &= \int \frac{3\sec^2\theta}{(9\sec^2\theta)^2} d\theta = \frac{1}{27}\int \cos^2(\theta)d\theta. \end{aligned} \end{equation*}

Using the half-angle identity from Section 3.2, we obtain

\begin{equation*} \begin{aligned} \int \cos^2(\theta)d\theta &= \int \frac{\cos(2\theta) +1}{2} d\theta \\ &= \frac{1}{2}\left( \frac{1}{2}\sin(2\theta) + \theta \right) +C \\ &= \frac{1}{2}\left( \sin\theta\cos\theta + \theta \right) + C. \end{aligned} \end{equation*}

From the tangent-substitution triangle above (with $a=3$), $\theta = \arctan(x/3)$ and:

\begin{equation*} \sin(\theta)\cos(\theta) = \frac{x}{\sqrt{x^2+9}}\frac{3}{\sqrt{x^2+9}} = \frac{3x}{x^2+9}. \end{equation*}

Hence,

\begin{equation*} \int \frac{dx}{(x^2+9)^2} = \frac{1}{54}\left(\frac{3x}{x^2+9} + \arctan \left( \frac{x}{3} \right)\right) + C. \end{equation*}

Consider an integral involving $\sqrt{x^2-a^2}\text{.}$

Example 3.30.

To compute $\ds \int \frac{dx}{x\sqrt{x^2-1}}\text{,}$ we consider the two intervals of the domain. For $x \gt 1\text{,}$ let $x = \sec(\theta)$ with $0 \lt \theta \lt \pi/2\text{,}$ using the secant-substitution triangle above:

\begin{equation*} \begin{aligned} \int \frac{dx}{x\sqrt{x^2-1}} & = \int \frac{d\sec(\theta)}{\sec(\theta)|\tan(\theta)|} = \int \frac{\sec(\theta)\tan(\theta)d\theta}{\sec(\theta)\tan(\theta)} \\ &= \int d\theta = \theta +C \\ &= \asec(x) +C. \quad (x \gt 1) \end{aligned} \end{equation*}

For $x \lt -1\text{,}$ where $\pi/2 \lt \theta \lt \pi\text{,}$ we have $|\tan(\theta)| = -\tan(\theta)\text{,}$ yielding $-\asec(x)+C\text{.}$ Combining these:

\begin{equation*} \int \frac{dx}{x\sqrt{x^2-1}} = g(x) +C \end{equation*}

where

\begin{equation*} \begin{aligned} g(x) &= \begin{cases} \phantom{-}\asec(x) & x \gt 1 \\ -\asec(x) & x \lt -1. \end{cases} \end{aligned} \end{equation*}

The graph below shows that $g(x)$ and $\arctan(\sqrt{x^2-1})$ differ by a constant on each connected component of the domain: $0$ for $x \ge 1$ and $-\pi$ for $x \le -1\text{.}$

See Exercise 3.5.7 for an alternative substitution.

The graph of g x together with the graph of arctan of square of x square minus 1. — Figure 3.31. $g(x)$ and $\arctan(\sqrt{x^2-1})$

Next, an example with $\sqrt{a^2-x^2}\text{.}$

Example 3.32.

To compute $\ds \int \frac{dx}{x\sqrt{1-x^2}}\text{,}$ let $x = \sin(\theta)$ with $-\pi/2 \le \theta \le \pi/2\text{,}$ using the sine-substitution triangle above. Then:

\begin{equation*} \begin{aligned} \int \frac{dx}{x\sqrt{1-x^2}} &= \int \frac{d\sin(\theta)}{\sin(\theta)|\cos(\theta)|} \\ & = \int \frac{\cos(\theta) d \theta}{\sin(\theta)\cos(\theta)} = \int \csc(\theta) d\theta \\ & = \ln|\csc(\theta) - \cot(\theta)| + C = \ln\left|\frac{1 - \sqrt{1-x^2}}{x}\right| + C. \end{aligned} \end{equation*}

Checkpoint 3.33.

Compute $\ds \int \frac{dx}{\sqrt{4x^2+8x-5}}\text{.}$

Solution.

Completing the square gives $4x^2 + 8x -5= 4(x+1)^2 - 9\text{.}$ Let $u=2(x+1)\text{.}$ The integral becomes:

\begin{equation*} \frac{1}{2} \int \frac{1}{\sqrt{u^2-9}} du. \end{equation*}

The integrand $f(u) = 1/\sqrt{u^2-9}$ is even. If $F(u)$ is an antiderivative for $u \gt 3\text{,}$ then $-F(-u)$ is one for $u \lt -3\text{.}$ For $u \gt 3\text{,}$ use $u=3\sec(\theta)\text{:}$

\begin{equation*} \begin{aligned} \frac{1}{2} \int \frac{3\sec(\theta)\tan(\theta)}{3\tan(\theta)} d\theta & = \frac{1}{2}\int\sec(\theta) d\theta \\ & = \frac{1}{2}\ln|\sec(\theta) + \tan(\theta)|+C \\ & = \frac{1}{2}\ln \left| \frac{u}{3} + \frac{\sqrt{u^2-9}}{3}\right| + C \\ & = \frac{1}{2}\ln \left| u + \sqrt{u^2-9} \right| + C. \end{aligned} \end{equation*}

On the other component of the domain, the corresponding antiderivative has the opposite sign. Thus one antiderivative on each component is

\begin{equation*} \begin{aligned} \frac{1}{2}\operatorname{sgn}(u)\ln\left(|u|+ \sqrt{u^2-9}\right)+C \\ &= \,\frac{1}{2}\operatorname{sgn}(x+1) \\ &\quad\ln\left(2|x+1|+\sqrt{4x^2+8x-5}\right)+C. \end{aligned} \end{equation*}

Example 3.34.

To compute

\begin{equation*} \ds \int \frac{x^{10}}{(16-x^2)^{13/2}} dx, \end{equation*}

let $x =4\sin \theta\text{.}$ Then $dx = 4\cos \theta d\theta\text{.}$ From the reference triangle:

\begin{equation*} \frac{x}{\sqrt{16-x^2}} = \tan\theta,\ \text{and}\ \frac{4}{\sqrt{16-x^2}} = \sec\theta. \end{equation*}

Substituting:

\begin{equation*} \begin{aligned} \int \frac{x^{10}}{(16-x^2)^{13/2}} dx & = \int \left(\frac{x}{\sqrt{16-x^2}}\right)^{10} \left(\frac{1}{\sqrt{16-x^2}}\right)^3 dx \\ &= \int (\tan^{10}\theta) \left(\frac{1}{4^3} \sec^3 \theta\right) 4\cos \theta d\theta \\ &= \frac{1}{4^2} \int \tan^{10}\theta \sec^2 \theta d\theta. \end{aligned} \end{equation*}

Using $w = \tan \theta\text{:}$

\begin{equation*} \begin{aligned} \frac{1}{4^2} \int w^{10} dw & = \frac{1}{4^2} \left(\frac{w^{11}}{11}\right) + C = \frac{1}{176} \left( \tan^{11}\theta \right) +C \\ & = \frac{x^{11}}{176(16-x^2)^{11/2}} +C. \end{aligned} \end{equation*}

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