Integrating Square Roots of Binomials

Learning Outcomes

Solve integration problems involving the square root of a sum or difference of two squares

Integrals Involving $\sqrt{{a}^{2}-{x}^{2}}$

Before developing a general strategy for integrals containing $\sqrt{{a}^{2}-{x}^{2}}$ , consider the integral $\displaystyle\int \sqrt{9-{x}^{2}}dx$ . This integral cannot be evaluated using any of the techniques we have discussed so far. However, if we make the substitution $x=3\sin\theta$ , we have $dx=3\cos\theta d\theta$ . After substituting into the integral, we have

\int \sqrt{9 - x^{2}} d x = \int^{} \sqrt{9 - {(3 \sin θ)}^{2}} 3 \cos θ d θ

$\displaystyle\int \sqrt{9-{x}^{2}}dx={\displaystyle\int }^{\text{ }}\sqrt{9-{\left(3\sin\theta \right)}^{2}}3\cos\theta d\theta$ .

After simplifying, we have

\int^{} \sqrt{9 - x^{2}} d x = \int^{} 9 \sqrt{1 - \sin^{2} θ} \cos θ d θ

${\displaystyle\int }^{\text{ }}\sqrt{9-{x}^{2}}dx={\displaystyle\int }^{\text{ }}9\sqrt{1-{\sin}^{2}\theta }\cos\theta d\theta$ .

Letting $1-{\sin}^{2}\theta ={\cos}^{2}\theta$ , we now have

\int^{} \sqrt{9 - x^{2}} d x = \int^{} 9 \sqrt{\cos^{2} θ} \cos θ d θ

${\displaystyle\int }^{\text{ }}\sqrt{9-{x}^{2}}dx={\displaystyle\int }^{\text{ }}9\sqrt{{\cos}^{2}\theta }\cos\theta d\theta$ .

Assuming that $\cos\theta \ge 0$ , we have

\int^{} \sqrt{9 - x^{2}} d x = \int^{} 9 \cos^{2} θ d θ

${\displaystyle\int }^{\text{ }}\sqrt{9-{x}^{2}}dx={\displaystyle\int }^{\text{ }}9{\cos}^{2}\theta d\theta$ .

At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions. Before completing this example, let’s take a look at the general theory behind this idea.

To evaluate integrals involving $\sqrt{{a}^{2}-{x}^{2}}$ , we make the substitution $x=a\sin\theta$ and $dx=a\cos\theta$ . To see that this actually makes sense, consider the following argument: The domain of $\sqrt{{a}^{2}-{x}^{2}}$ is $\left[\text{-}a,a\right]$ . Thus, $\text{-}a\le x\le a$ . Consequently, $-1\le \frac{x}{a}\le 1$ . Since the range of $\sin{x}$ over $\left[\text{-}\frac{\pi}{2},\frac{\pi}{2}\right]$ is $\left[-1,1\right]$ , there is a unique angle $\theta$ satisfying $\text{-}\frac{\pi}{2}\le \theta \le \frac{\pi}{2}$ so that $\sin\theta =x\text{/}a$ , or equivalently, so that $x=a\sin\theta$ . If we substitute $x=a\sin\theta$ into $\sqrt{{a}^{2}-{x}^{2}}$ , we get

\begin{array}{ccccc} \sqrt{a^{2} - x^{2}} & = \sqrt{a^{2} - {(a \sin θ)}^{2}} & Let x = a \sin θ where - \frac{π}{2} \leq θ \leq \frac{π}{2} . Simplify. \\ = \sqrt{a^{2} - a^{2} \sin^{2} θ} & Factor out a^{2} . \\ = \sqrt{a^{2} (1 - \sin^{2} θ)} & Substitute 1 - \sin^{2} x = \cos^{2} x . \\ = \sqrt{a^{2} \cos^{2} θ} & Take the square root. \\ = | a \cos θ | \\ = a \cos θ . \end{array}

$\begin{array}{ccccc}\hfill \sqrt{{a}^{2}-{x}^{2}}& =\sqrt{{a}^{2}-{\left(a\sin\theta \right)}^{2}}\hfill & & & \text{Let }x=a\sin\theta \text{ where }-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}.\text{Simplify.}\hfill \\ & =\sqrt{{a}^{2}-{a}^{2}{\sin}^{2}\theta }\hfill & & & \text{Factor out }{a}^{2}.\hfill \\ & =\sqrt{{a}^{2}\left(1-{\sin}^{2}\theta \right)}\hfill & & & \text{Substitute }1-{\sin}^{2}x={\cos}^{2}x.\hfill \\ & =\sqrt{{a}^{2}{\cos}^{2}\theta }\hfill & & & \text{Take the square root.}\hfill \\ & =|a\cos\theta |\hfill & & & \\ & =a\cos\theta .\hfill & & & \end{array}$

Since $\cos\theta \ge 0$ on $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ and $a>0$ , $|a\cos\theta |=a\cos\theta$ . We can see, from this discussion, that by making the substitution $x=a\sin\theta$ , we are able to convert an integral involving a radical into an integral involving trigonometric functions. After we evaluate the integral, we can convert the solution back to an expression involving $x$ . To see how to do this, let’s begin by assuming that [latex]0

Recall: Right Triangle Trigonometry Relationships

Given a right triangle with an acute angle of $\theta$ ,

$\begin{align}&\sin \left(\theta \right)=\frac{\text{opposite}}{\text{hypotenuse}} \\ &\cos \left(\theta \right)=\frac{\text{adjacent}}{\text{hypotenuse}} \\ &\tan \left(\theta \right)=\frac{\text{opposite}}{\text{adjacent}} \end{align}$

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.”

It can be shown that this triangle actually produces the correct values of the trigonometric functions evaluated at $\theta$ for all $\theta$ satisfying $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ . It is useful to observe that the expression $\sqrt{{a}^{2}-{x}^{2}}$ actually appears as the length of one side of the triangle. Last, should $\theta$ appear by itself, we use $\theta ={\sin}^{-1}\left(\frac{x}{a}\right)$ .

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled a, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (a^2 – x^2). To the left of the triangle is the equation sin(theta) = x/a.

The essential part of this discussion is summarized in the following problem-solving strategy.

Problem-Solving Strategy: Integrating Expressions Involving $\sqrt{{a}^{2}-{x}^{2}}$

It is a good idea to make sure the integral cannot be evaluated easily in another way. For example, although this method can be applied to integrals of the form $\displaystyle\int \frac{1}{\sqrt{{a}^{2}-{x}^{2}}}dx$ , $\displaystyle\int \frac{x}{\sqrt{{a}^{2}-{x}^{2}}}dx$ , and $\displaystyle\int x\sqrt{{a}^{2}-{x}^{2}}dx$ , they can each be integrated directly either by formula or by a simple u-substitution.
Make the substitution $x=a\sin\theta$ and $dx=a\cos\theta d\theta$ . Note: This substitution yields $\sqrt{{a}^{2}-{x}^{2}}=a\cos\theta$ .
Simplify the expression.
Evaluate the integral using techniques from the section on trigonometric integrals.
Use the reference triangle from Figure 1 to rewrite the result in terms of $x$ . You may also need to use some trigonometric identities and the relationship $\theta ={\sin}^{-1}\left(\frac{x}{a}\right)$ .

Example: Integrating an Expression Involving $\sqrt{{a}^{2}-{x}^{2}}$

Evaluate ${\displaystyle\int }^{\text{ }}\sqrt{9-{x}^{2}}dx$ .

Show Solution

Begin by making the substitutions $x=3\sin\theta$ and $dx=3\cos\theta d\theta$ . Since $\sin\theta =\frac{x}{3}$ , we can construct the reference triangle shown in the following figure.

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled 3, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (9 – x^2). To the left of the triangle is the equation sin(theta) = x/3.

Thus,

\begin{array}{ccccc} \int^{} \sqrt{9 - x^{2}} d x & = \int^{} \sqrt{9 - {(3 \sin θ)}^{2}} 3 \cos θ d θ & Substitute x = 3 \sin θ and d x = 3 \cos θ d θ . \\ = \int^{} \sqrt{9 (1 - \sin^{2} θ)} 3 \cos θ d θ & Simplify. \\ = \int^{} \sqrt{9 \cos^{2} θ} 3 \cos θ d θ & Substitute \cos^{2} θ = 1 - \sin^{2} θ . \\ = \int^{} 3 | \cos θ | 3 \cos θ d θ & Take the square root. \\ = \int^{} 9 \cos^{2} θ d θ & \begin{matrix} Simplify. Since - \frac{π}{2} \leq θ \leq \frac{π}{2}, \cos θ \geq 0 and \\ | \cos θ | = \cos θ . \end{matrix} \\ = \int^{} 9 (\frac{1}{2} + \frac{1}{2} \cos (2 θ)) d θ & \begin{matrix} Use the strategy for integrating an even power \\ of \cos θ . \end{matrix} \\ = \frac{9}{2} θ + \frac{9}{4} \sin (2 θ) + C & Evaluate the integral. \\ = \frac{9}{2} θ + \frac{9}{4} (2 \sin θ \cos θ) + C & Substitute \sin (2 θ) = 2 \sin θ \cos θ . \\ = \frac{9}{2} \sin^{- 1} (\frac{x}{3}) + \frac{9}{2} \cdot \frac{x}{3} \cdot \frac{\sqrt{9 - x^{2}}}{3} + C & \begin{matrix} Substitute \sin^{- 1} (\frac{x}{3}) = θ and \sin θ = \frac{x}{3} . Use \\ the reference triangle to see that \\ \cos θ = \frac{\sqrt{9 - x^{2}}}{3} and make this substitution. \end{matrix} \\ = \frac{9}{2} \sin^{- 1} (\frac{x}{3}) + \frac{x \sqrt{9 - x^{2}}}{2} + C . & Simplify. \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int }^{\text{ }}\sqrt{9-{x}^{2}}dx& ={\displaystyle\int }^{\text{ }}\sqrt{9-{\left(3\sin\theta \right)}^{2}}3\cos\theta d\theta \hfill & & & \text{Substitute }x=3\sin\theta \text{ and }dx=3\cos\theta d\theta .\hfill \\ & ={\displaystyle\int }^{\text{ }}\sqrt{9\left(1-{\sin}^{2}\theta \right)}3\cos\theta d\theta \hfill & & & \text{Simplify.}\hfill \\ & ={\displaystyle\int }^{\text{ }}\sqrt{9{\cos}^{2}\theta }3\cos\theta d\theta \hfill & & & \text{Substitute }{\cos}^{2}\theta =1-{\sin}^{2}\theta .\hfill \\ & ={\displaystyle\int }^{\text{ }}3|\cos\theta |3\cos\theta d\theta \hfill & & & \text{Take the square root.}\hfill \\ & ={\displaystyle\int }^{\text{ }}9{\cos}^{2}\theta d\theta \hfill & & & \begin{array}{c}\text{Simplify. Since}-\frac{\pi }{2}\le \theta \le \frac{\pi }{2},\cos\theta \ge 0\text{ and }\hfill \\ |\cos\theta |=\cos\theta .\hfill \end{array}\hfill \\ & ={\displaystyle\int }^{\text{ }}9\left(\frac{1}{2}+\frac{1}{2}\cos\left(2\theta \right)\right)d\theta \hfill & & & \begin{array}{c}\text{Use the strategy for integrating an even power}\hfill \\ \text{of }\cos\theta .\hfill \end{array}\hfill \\ & =\frac{9}{2}\theta +\frac{9}{4}\sin\left(2\theta \right)+C\hfill & & & \text{Evaluate the integral.}\hfill \\ & =\frac{9}{2}\theta +\frac{9}{4}\left(2\sin\theta \cos\theta \right)+C\hfill & & & \text{Substitute }\sin\left(2\theta \right)=2\sin\theta \cos\theta .\hfill \\ & =\frac{9}{2}{\sin}^{-1}\left(\frac{x}{3}\right)+\frac{9}{2}\cdot \frac{x}{3}\cdot \frac{\sqrt{9-{x}^{2}}}{3}+C\hfill & & & \begin{array}{c}\text{Substitute }{\sin}^{-1}\left(\frac{x}{3}\right)=\theta \text{ and }\sin\theta =\frac{x}{3}.\text{Use}\hfill \\ \text{the reference triangle to see that}\hfill \\ \cos\theta =\frac{\sqrt{9-{x}^{2}}}{3}\text{and make this substitution.}\hfill \end{array}\hfill \\ & =\frac{9}{2}{\sin}^{-1}\left(\frac{x}{3}\right)+\frac{x\sqrt{9-{x}^{2}}}{2}+C.\hfill & & & \text{Simplify.}\hfill \end{array}$

Example: Integrating an Expression Involving $\sqrt{{a}^{2}-{x}^{2}}$

Evaluate $\displaystyle\int \frac{\sqrt{4-{x}^{2}}}{x}dx$ .

Show Solution

First make the substitutions $x=2\sin\theta$ and $dx=2\cos\theta d\theta$ . Since $\sin\theta =\frac{x}{2}$ , we can construct the reference triangle shown in the following figure.

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The vertical leg is labeled x, and the horizontal leg is labeled as the square root of (4 – x^2). To the left of the triangle is the equation sin(theta) = x/2.

Thus,

\begin{array}{ccccc} \int \frac{\sqrt{4 - x^{2}}}{x} d x & = \int \frac{\sqrt{4 - {(2 \sin θ)}^{2}}}{2 \sin θ} 2 \cos θ d θ & Substitute x = 2 \sin θ and = 2 \cos θ d θ . \\ = \int \frac{2 \cos^{2} θ}{\sin θ} d θ & Substitute \cos^{2} θ = 1 - \sin^{2} θ and simplify. \\ = \int \frac{2 (1 - \sin^{2} θ)}{\sin θ} d θ & Substitute \sin^{2} θ = 1 - \cos^{2} θ . \\ = \int (2 \csc θ - 2 \sin θ) d θ & \begin{matrix} Separate the numerator, simplify, and use \\ \csc θ = \frac{1}{\sin θ} . \end{matrix} \\ = 2 ln | \csc θ - \cot θ | + 2 \cos θ + C & Evaluate the integral. \\ = 2 ln | \frac{2}{x} - \frac{\sqrt{4 - x^{2}}}{x} | + \sqrt{4 - x^{2}} + C . & \begin{matrix} Use the reference triangle to rewrite the \\ expression in terms of x and simplify. \end{matrix} \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int \frac{\sqrt{4-{x}^{2}}}{x}dx}& ={\displaystyle\int \frac{\sqrt{4-{\left(2\sin\theta \right)}^{2}}}{2\sin\theta }2\cos\theta d\theta} \hfill & & & \text{Substitute}x=2\sin\theta \text{and}=2\cos\theta d\theta .\hfill \\ & ={\displaystyle\int \frac{2{\cos}^{2}\theta }{\sin\theta }d\theta} \hfill & & & \text{Substitute}{\cos}^{2}\theta =1-{\sin}^{2}\theta \text{and simplify.}\hfill \\ & ={\displaystyle\int \frac{2\left(1-{\sin}^{2}\theta \right)}{\sin\theta }d\theta} \hfill & & & \text{Substitute}{\sin}^{2}\theta =1-{\cos}^{2}\theta .\hfill \\ & ={\displaystyle\int \left(2\csc\theta -2\sin\theta \right)d\theta} \hfill & & & \begin{array}{c}\text{Separate the numerator, simplify, and use}\hfill \\ \csc\theta =\frac{1}{\sin\theta }.\hfill \end{array}\hfill \\ & =2\text{ln}|\csc\theta -\cot\theta |+2\cos\theta +C\hfill & & & \text{Evaluate the integral.}\hfill \\ & =2\text{ln}|\frac{2}{x}-\frac{\sqrt{4-{x}^{2}}}{x}|+\sqrt{4-{x}^{2}}+C.\hfill & & & \begin{array}{c}\text{Use the reference triangle to rewrite the}\hfill \\ \text{expression in terms of}x\text{and simplify.}\hfill \end{array}\end{array}$

In the next example, we see that we sometimes have a choice of methods.

Example: Integrating an Expression Involving $\sqrt{{a}^{2}-{x}^{2}}$ Two Ways

Evaluate ${\displaystyle\int }^{\text{ }}{x}^{3}\sqrt{1-{x}^{2}}dx$ two ways: first by using the substitution $u=1-{x}^{2}$ and then by using a trigonometric substitution.

Show Solution

Method 1

Let $u=1-{x}^{2}$ and hence ${x}^{2}=1-u$ . Thus, $du=-2xdx$ . In this case, the integral becomes

\begin{array}{ccccc} \int^{} x^{3} \sqrt{1 - x^{2}} d x & = - \frac{1}{2} \int^{} x^{2} \sqrt{1 - x^{2}} (- 2 x d x) & Make the substitution. \\ = - \frac{1}{2} \int^{} (1 - u) \sqrt{u} d u & Expand the expression. \\ = - \frac{1}{2} \int (u^{1 / 2} - u^{3 / 2}) d u & Evaluate the integral. \\ = - \frac{1}{2} (\frac{2}{3} u^{3 / 2} - \frac{2}{5} u^{5 / 2}) + C & Rewrite in terms of x . \\ = - \frac{1}{3} {(1 - x^{2})}^{3 / 2} + \frac{1}{5} {(1 - x^{2})}^{5 / 2} + C . \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int ^{\text{ }}{x}^{3}\sqrt{1-{x}^{2}}dx}& =-\frac{1}{2}{\displaystyle\int ^{\text{ }}{x}^{2}\sqrt{1-{x}^{2}}\left(-2xdx\right)}\hfill & & & \text{Make the substitution.}\hfill \\ & =-\frac{1}{2}{\displaystyle\int ^{\text{ }}\left(1-u\right)\sqrt{u}du}\hfill & & & \text{Expand the expression.}\hfill \\ & =-\frac{1}{2}{\displaystyle\int \left({u}^{1\text{/}2}-{u}^{3\text{/}2}\right)du}\hfill & & & \text{Evaluate the integral.}\hfill \\ & =-\frac{1}{2}\left(\frac{2}{3}{u}^{3\text{/}2}-\frac{2}{5}{u}^{5\text{/}2}\right)+C\hfill & & & \text{Rewrite in terms of}x.\hfill \\ & =-\frac{1}{3}{\left(1-{x}^{2}\right)}^{3\text{/}2}+\frac{1}{5}{\left(1-{x}^{2}\right)}^{5\text{/}2}+C.\hfill & & & \end{array}$

Method 2

Let $x=\sin\theta$ . In this case, $dx=\cos\theta d\theta$ . Using this substitution, we have

\begin{array}{ccccc} \int^{} x^{3} \sqrt{1 - x^{2}} d x & = \int^{} \sin^{3} θ \cos^{2} θ d θ \\ = \int^{} (1 - \cos^{2} θ) \cos^{2} θ \sin θ d θ & Let u = \cos θ . Thus, d u = - \sin θ d θ . \\ = \int^{} (u^{4} - u^{2}) d u \\ = \frac{1}{5} u^{5} - \frac{1}{3} u^{3} + C & Substitute \cos θ = u . \\ = \frac{1}{5} \cos^{5} θ - \frac{1}{3} \cos^{3} θ + C & \begin{matrix} Use a reference triangle to see that \\ \cos θ = \sqrt{1 - x^{2}} . \end{matrix} \\ = \frac{1}{5} {(1 - x^{2})}^{5 / 2} - \frac{1}{3} {(1 - x^{2})}^{3 / 2} + C . \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int }^{\text{ }}{x}^{3}\sqrt{1-{x}^{2}}dx& ={\displaystyle\int }^{\text{ }}{\sin}^{3}\theta {\cos}^{2}\theta d\theta \hfill & & & \\ & ={\displaystyle\int }^{\text{ }}\left(1-{\cos}^{2}\theta \right){\cos}^{2}\theta \sin\theta d\theta \hfill & & & \text{Let }u=\cos\theta .\text{Thus, }du=\text{-}\sin\theta d\theta .\hfill \\ & ={\displaystyle\int }^{\text{ }}\left({u}^{4}-{u}^{2}\right)du\hfill & & & \\ & =\frac{1}{5}{u}^{5}-\frac{1}{3}{u}^{3}+C\hfill & & & \text{Substitute }\cos\theta =u.\hfill \\ & =\frac{1}{5}{\cos}^{5}\theta -\frac{1}{3}{\cos}^{3}\theta +C\hfill & & & \begin{array}{c}\text{Use a reference triangle to see that}\hfill \\ \cos\theta =\sqrt{1-{x}^{2}}.\hfill \end{array}\hfill \\ & =\frac{1}{5}{\left(1-{x}^{2}\right)}^{5\text{/}2}-\frac{1}{3}{\left(1-{x}^{2}\right)}^{3\text{/}2}+C.\hfill & & & \end{array}$

try it

Rewrite the integral $\displaystyle\int \frac{{x}^{3}}{\sqrt{25-{x}^{2}}}dx$ using the appropriate trigonometric substitution (do not evaluate the integral).

Hint

Substitute $x=5\sin\theta$ and $dx=5\cos\theta d\theta$ .

Show Solution

${\displaystyle\int }^{\text{ }}125{\sin}^{3}\theta d\theta$

Watch the following video to see the worked solution to the above Try It

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “3.3 Trigonometric Substitution” here (opens in new window).

Integrating Expressions Involving $\sqrt{{a}^{2}+{x}^{2}}$

For integrals containing $\sqrt{{a}^{2}+{x}^{2},}$ let’s first consider the domain of this expression. Since $\sqrt{{a}^{2}+{x}^{2}}$ is defined for all real values of $x$ , we restrict our choice to those trigonometric functions that have a range of all real numbers. Thus, our choice is restricted to selecting either $x=a\tan\theta$ or $x=a\cot\theta$ . Either of these substitutions would actually work, but the standard substitution is $x=a\tan\theta$ or, equivalently, $\tan\theta =\frac{x}{a}$ . With this substitution, we make the assumption that $\text{-}\frac{\pi}{2}<\theta <\frac{\pi}{2}$ , so that we also have $\theta ={\tan}^{-1}\left(x\text{/}a\right)$ . The procedure for using this substitution is outlined in the following problem-solving strategy.

Problem-Solving Strategy: Integrating Expressions Involving $\sqrt{{a}^{2}+{x}^{2}}$

Check to see whether the integral can be evaluated easily by using another method. In some cases, it is more convenient to use an alternative method.
Substitute $x=a\tan\theta$ and $dx=a{\sec}^{2}\theta d\theta$ . This substitution yields
$\sqrt{{a}^{2}+{x}^{2}}=\sqrt{{a}^{2}+{\left(a\tan\theta \right)}^{2}}=\sqrt{{a}^{2}\left(1+{\tan}^{2}\theta \right)}=\sqrt{{a}^{2}{\sec}^{2}\theta }=|a\sec\theta |=a\sec\theta$ . (Since $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$ and $\sec\theta >0$ over this interval, $|a\sec\theta |=a\sec\theta$ )
Simplify the expression.
Evaluate the integral using techniques from the section on trigonometric integrals.
Use the reference triangle from Figure 4 to rewrite the result in terms of $x$ . You may also need to use some trigonometric identities and the relationship $\theta ={\tan}^{-1}\left(\frac{x}{a}\right)$ . (Note: The reference triangle is based on the assumption that $x>0$ ; however, the trigonometric ratios produced from the reference triangle are the same as the ratios for which $x \le{0}$ )

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled the square root of (a^2+x^2), the vertical leg is labeled x, and the horizontal leg is labeled a. To the left of the triangle is the equation tan(theta) = x/a.

Example: Integrating an Expression Involving $\sqrt{{a}^{2}+{x}^{2}}$

Evaluate $\displaystyle\int \frac{dx}{\sqrt{1+{x}^{2}}}$ and check the solution by differentiating.

Show Solution

Begin with the substitution $x=\tan\theta$ and $dx={\sec}^{2}\theta d\theta$ . Since $\tan\theta =x$ , draw the reference triangle in the following figure.

Thus,

\begin{array}{ccccc} \int \frac{d x}{\sqrt{1 + x^{2}}} & = \int \frac{\sec^{2} θ}{\sec θ} d θ & \begin{matrix} Substitute x = \tan θ and d x = \sec^{2} θ d θ . This \\ substitution makes \sqrt{1 + x^{2}} = \sec θ . Simplify. \end{matrix} \\ = \int^{} \sec θ d θ & Evaluate the integral. \\ = ln | \sec θ + \tan θ | + C & \begin{matrix} Use the reference triangle to express the result \\ in terms of x . \end{matrix} \\ = ln | \sqrt{1 + x^{2}} + x | + C . \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int \frac{dx}{\sqrt{1+{x}^{2}}}}& ={\displaystyle\int \frac{{\sec}^{2}\theta }{\sec\theta }d\theta }\hfill & & & \begin{array}{c}\text{Substitute}x=\tan\theta \text{and}dx={\sec}^{2}\theta d\theta .\text{This}\hfill \\ \text{substitution makes}\sqrt{1+{x}^{2}}=\sec\theta .\text{Simplify.}\hfill \end{array}\hfill \\ & ={\displaystyle\int ^{\text{ }}\sec\theta d\theta }\hfill & & & \text{Evaluate the integral.}\hfill \\ & =\text{ln}|\sec\theta +\tan\theta |+C\hfill & & & \begin{array}{c}\text{Use the reference triangle to express the result}\hfill \\ \text{in terms of}x.\hfill \end{array}\hfill \\ & =\text{ln}|\sqrt{1+{x}^{2}}+x|+C.\hfill & & & \end{array}$

To check the solution, differentiate:

\begin{array}{cc} \frac{d}{d x} (ln | \sqrt{1 + x^{2}} + x |) & = \frac{1}{\sqrt{1 + x^{2}} + x} \cdot (\frac{x}{\sqrt{1 + x^{2}}} + 1) \\ = \frac{1}{\sqrt{1 + x^{2}} + x} \cdot \frac{x + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}} \\ = \frac{1}{\sqrt{1 + x^{2}}} . \end{array}

$\begin{array}{cc}\hfill \frac{d}{dx}\left(\text{ln}|\sqrt{1+{x}^{2}}+x|\right)& =\frac{1}{\sqrt{1+{x}^{2}}+x}\cdot \left(\frac{x}{\sqrt{1+{x}^{2}}}+1\right)\hfill \\ & =\frac{1}{\sqrt{1+{x}^{2}}+x}\cdot \frac{x+\sqrt{1+{x}^{2}}}{\sqrt{1+{x}^{2}}}\hfill \\ & =\frac{1}{\sqrt{1+{x}^{2}}}.\hfill \end{array}$

Since $\sqrt{1+{x}^{2}}+x>0$ for all values of $x$ , we could rewrite $\text{ln}|\sqrt{1+{x}^{2}}+x|+C=\text{ln}\left(\sqrt{1+{x}^{2}}+x\right)+C$ , if desired.

In the example below, we explore how to use hyperbolic trigonometric functions as an alternative substitution method. First, we briefly review the definition of these functions and their derivatives.

Recall: Definition and Derivatives of Hyperbolic Trigonometric Functions

For any real number $x$ , the hyperbolic sine and hyperbolic cosine are defined as:

$\sinh x = \frac{e^x - e^{-x}}{2} \: \text{and}\:\cosh x = \frac{e^x + e^{-x}}{2}$

Their derivatives are given by:

$\frac{d}{dx} \left( \sinh x \right) = \cosh x \:\text{and}\:\frac{d}{dx} \left( \cosh x \right) = \sinh x$

Example: Evaluating $\displaystyle\int \frac{dx}{\sqrt{1+{x}^{2}}}$ Using a Different Substitution

Use the substitution $x=\text{sinh}\theta$ to evaluate $\displaystyle\int \frac{dx}{\sqrt{1+{x}^{2}}}$ .

Show Solution

Because $\text{sinh}\theta$ has a range of all real numbers, and $1+{\text{sinh}}^{2}\theta ={\text{cosh}}^{2}\theta$ , we may also use the substitution $x=\text{sinh}\theta$ to evaluate this integral. In this case, $dx=\text{cosh}\theta d\theta$ . Consequently,

\begin{array}{ccccc} \int \frac{d x}{\sqrt{1 + x^{2}}} & = \int \frac{cosh θ}{\sqrt{1 + {sinh}^{2} θ}} d θ & \begin{matrix} Substitute x = sinh θ and d x = cosh θ d θ . \\ Substitute 1 + {sinh}^{2} θ = {cosh}^{2} θ . \end{matrix} \\ = \int \frac{cosh θ}{\sqrt{{cosh}^{2} θ}} d θ & \sqrt{{cosh}^{2} θ} = | cosh θ | \\ = \int \frac{cosh θ}{| cosh θ |} d θ & | cosh θ | = cosh θ since cosh θ > 0 for all θ . \\ = \int \frac{cosh θ}{cosh θ} d θ & Simplify. \\ = \int^{} 1 d θ & Evaluate the integral. \\ = θ + C & Since x = sinh θ, we know θ = {sinh}^{- 1} x . \\ = {sinh}^{- 1} x + C . \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int \frac{dx}{\sqrt{1+{x}^{2}}}}& ={\displaystyle\int \frac{\text{cosh}\theta }{\sqrt{1+{\text{sinh}}^{2}\theta }}d\theta }\hfill & & & \begin{array}{c}\text{Substitute}x=\text{sinh}\theta \text{and}dx=\text{cosh}\theta d\theta .\hfill \\ \text{Substitute}1+{\text{sinh}}^{2}\theta ={\text{cosh}}^{2}\theta .\hfill \end{array}\hfill \\ & ={\displaystyle\int \frac{\text{cosh}\theta }{\sqrt{{\text{cosh}}^{2}\theta }}d\theta }\hfill & & & \sqrt{{\text{cosh}}^{2}\theta }=|\text{cosh}\theta |\hfill \\ & ={\displaystyle\int \frac{\text{cosh}\theta }{|\text{cosh}\theta |}d\theta }\hfill & & & |\text{cosh}\theta |=\text{cosh}\theta \text{since}\text{cosh}\theta >0\text{for all}\theta .\hfill \\ & ={\displaystyle\int \frac{\text{cosh}\theta }{\text{cosh}\theta }d\theta }\hfill & & & \text{Simplify.}\hfill \\ & ={\displaystyle\int ^{\text{ }}1d\theta }\hfill & & & \text{Evaluate the integral.}\hfill \\ & =\theta +C\hfill & & & \text{Since}x=\text{sinh}\theta ,\text{we know}\theta ={\text{sinh}}^{-1}x.\hfill \\ & ={\text{sinh}}^{-1}x+C.\hfill & & & \end{array}$

Analysis

This answer looks quite different from the answer obtained using the substitution $x=\tan\theta$ . To see that the solutions are the same, set $y={\text{sinh}}^{-1}x$ . Thus, $\text{sinh}y=x$ . From this equation we obtain:

\frac{e^{y} - e^{- y}}{2} = x

$\frac{{e}^{y}-{e}^{\text{-}y}}{2}=x$ .

After multiplying both sides by $2{e}^{y}$ and rewriting, this equation becomes:

e^{2 y} - 2 x e^{y} - 1 = 0

${e}^{2y}-2x{e}^{y}-1=0$ .

Use the quadratic equation to solve for ${e}^{y}\text{:}$

e^{y} = \frac{2 x \pm \sqrt{4 x^{2} + 4}}{2}

${e}^{y}=\frac{2x\pm \sqrt{4{x}^{2}+4}}{2}$ .

Simplifying, we have:

e^{y} = x \pm \sqrt{x^{2} + 1}

${e}^{y}=x\pm \sqrt{{x}^{2}+1}$ .

Since $x-\sqrt{{x}^{2}+1}<0$ , it must be the case that ${e}^{y}=x+\sqrt{{x}^{2}+1}$ . Thus,

y = ln (x + \sqrt{x^{2} + 1})

$y=\text{ln}\left(x+\sqrt{{x}^{2}+1}\right)$ .

Last, we obtain

{sinh}^{- 1} x = ln (x + \sqrt{x^{2} + 1})

${\text{sinh}}^{-1}x=\text{ln}\left(x+\sqrt{{x}^{2}+1}\right)$ .

After we make the final observation that, since $x+\sqrt{{x}^{2}+1}>0$ ,

ln (x + \sqrt{x^{2} + 1}) = ln | \sqrt{1 + x^{2}} + x |

$\text{ln}\left(x+\sqrt{{x}^{2}+1}\right)=\text{ln}|\sqrt{1+{x}^{2}}+x|$ ,

we see that the two different methods produced equivalent solutions.

Example: Finding an Arc Length

Find the length of the curve $y={x}^{2}$ over the interval $\left[0,\frac{1}{2}\right]$ .

Show Solution

Because $\frac{dy}{dx}=2x$ , the arc length is given by

\int_{0}^{1 / 2} \sqrt{1 + {(2 x)}^{2}} d x = \int_{0}^{1 / 2} \sqrt{1 + 4 x^{2}} d x

${\displaystyle\int }_{0}^{1\text{/}2}\sqrt{1+{\left(2x\right)}^{2}}dx={\displaystyle\int }_{0}^{1\text{/}2}\sqrt{1+4{x}^{2}}dx$ .

To evaluate this integral, use the substitution $x=\frac{1}{2}\tan\theta$ and $dx=\frac{1}{2}{\sec}^{2}\theta d\theta$ . We also need to change the limits of integration. If $x=0$ , then $\theta =0$ and if $x=\frac{1}{2}$ , then $\theta =\frac{\pi }{4}$ . Thus,

\begin{array}{ccccc} \int_{0}^{1 / 2} \sqrt{1 + 4 x^{2}} d x & = \int_{0}^{π / 4} \sqrt{1 + \tan^{2} θ} \frac{1}{2} \sec^{2} θ d θ & \begin{matrix} After substitution, \\ \sqrt{1 + 4 x^{2}} = \tan θ . Substitute \\ 1 + \tan^{2} θ = \sec^{2} θ and simplify. \end{matrix} \\ = \frac{1}{2} \int_{0}^{π / 4} \sec^{3} θ d θ & \begin{matrix} We derived this integral in the \\ previous section. \end{matrix} \\ = \frac{1}{2} (\frac{1}{2} \sec θ \tan θ + ln | \sec θ + \tan θ |) |_{\begin{matrix} 0 \end{matrix}}^{\begin{matrix} π / 4 \end{matrix}} & Evaluate and simplify. \\ = \frac{1}{4} (\sqrt{2} + ln (\sqrt{2} + 1)) . \end{array}

$\begin{array}{ccccc}\hfill {\displaystyle\int }_{0}^{1\text{/}2}\sqrt{1+4{x}^{2}}dx& ={\displaystyle\int }_{0}^{\pi \text{/}4}\sqrt{1+{\tan}^{2}\theta }\frac{1}{2}{\sec}^{2}\theta d\theta \hfill & & & \begin{array}{c}\text{After substitution,}\hfill \\ \sqrt{1+4{x}^{2}}=\tan\theta .\text{Substitute}\hfill \\ 1+{\tan}^{2}\theta ={\sec}^{2}\theta \text{ and simplify.}\hfill \end{array}\hfill \\ & =\frac{1}{2}{\displaystyle\int }_{0}^{\pi \text{/}4}{\sec}^{3}\theta d\theta \hfill & & & \begin{array}{c}\text{We derived this integral in the}\hfill \\ \text{previous section.}\hfill \end{array}\hfill \\ & =\frac{1}{2}\left(\frac{1}{2}\sec\theta \tan\theta +\text{ln}|\sec\theta +\tan\theta |\right)|{}_{\begin{array}{c}\\ 0\end{array}}^{\begin{array}{c}\pi \text{/}4\\ \end{array}}\hfill & & & \text{Evaluate and simplify.}\hfill \\ & =\frac{1}{4}\left(\sqrt{2}+\text{ln}\left(\sqrt{2}+1\right)\right).\hfill & & & \end{array}$

try it

Rewrite ${\displaystyle\int }^{\text{ }}{x}^{3}\sqrt{{x}^{2}+4}dx$ by using a substitution involving $\tan\theta$ .

Hint

Use $x=2\tan\theta$ and $dx=2{\sec}^{2}\theta d\theta$ .

Show Solution

${\displaystyle\int }^{\text{ }}32{\tan}^{3}\theta {\sec}^{3}\theta d\theta$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for this segmented clip of “3.3 Trigonometric Substitution” here (opens in new window).

Try It

Integrating Expressions Involving $\sqrt{{x}^{2}-{a}^{2}}$

The domain of the expression $\sqrt{{x}^{2}-{a}^{2}}$ is $\left(\text{-}\infty ,\text{-}a\right]\cup \left[a,\text{+}\infty \right)$ . Thus, either $x\le \text{-}a$ or $x\ge a$ . Hence, $\frac{x}{a}\le -1$ or $\frac{x}{a}\ge 1$ . Since these intervals correspond to the range of $\sec\theta$ on the set $\left[0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right]$ , it makes sense to use the substitution $\sec\theta =\frac{x}{a}$ or, equivalently, $x=a\sec\theta$ , where $0\le \theta <\frac{\pi }{2}$ or $\frac{\pi }{2}<\theta \le \pi$ . The corresponding substitution for $dx$ is $dx=a\sec\theta \tan\theta d\theta$ . The procedure for using this substitution is outlined in the following problem-solving strategy.

Problem-Solving Strategy: Integrals Involving $\sqrt{{x}^{2}-{a}^{2}}$

Check to see whether the integral cannot be evaluated using another method. If so, we may wish to consider applying an alternative technique.
Substitute $x=a\sec\theta$ and $dx=a\sec\theta \tan\theta d\theta$ . This substitution yields

$\sqrt{{x}^{2}-{a}^{2}}=\sqrt{{\left(a\sec\theta \right)}^{2}-{a}^{2}}=\sqrt{{a}^{2}\left({\sec}^{2}\theta -1\right)}=\sqrt{{a}^{2}{\tan}^{2}\theta }=|a\tan\theta |$ .

For $x\ge a$ , $|a\tan\theta |=a\tan\theta$ and for $x\le -a$ , $|a\tan\theta |=\text{-}a\tan\theta$ .
Simplify the expression.
Evaluate the integral using techniques from the section on trigonometric integrals.
Use the reference triangles from Figure 6 to rewrite the result in terms of $x$ . You may also need to use some trigonometric identities and the relationship $\theta ={\sec}^{-1}\left(\frac{x}{a}\right)$ . (Note: We need both reference triangles, since the values of some of the trigonometric ratios are different depending on whether $x\ge a$ or $x\le \text{-}a.$ )

This figure has two right triangles. The first triangle is in the first quadrant of the xy coordinate system and has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled x, the vertical leg is labeled the square root of (x^2-a^2), and the horizontal leg is labeled a. The horizontal leg is on the x-axis. To the left of the triangle is the equation sec(theta) = x/a, x>a. There are also the equations sin(theta)= the square root of (x^2-a^2)/x, cos(theta) = a/x, and tan(theta) = the square root of (x^2-a^2)/a. The second triangle is in the second quadrant, with the hypotenuse labeled –x. The horizontal leg is labeled –a and is on the negative x-axis. The vertical leg is labeled the square root of (x^2-a^2). To the right of the triangle is the equation sec(theta) = x/a, x<-a. There are also the equations sin(theta)= the negative square root of (x^2-a^2)/x, cos(theta) = a/x, and tan(theta) = the negative square root of (x^2-a^2)/a.

For the next example, it is worthwhile to review the technique of how to precisely evaluate a trignometric function whose input is an inverse trigonometric function.

Recall: Evaluating Trigonometric Functions composed with Inverse Trigonometric Functions

Suppose we wanted to evaluate a trigonometric function composed with an inverse trigonometric function, for example: $\sin\left(\cos^{−1}\left(\frac{4}{5}\right)\right)$ .

Beginning with the inside, we can say there is some angle such that $\theta=\cos^{−1}(\frac{4}{5})$ , which means $\cos\theta=\frac{4}{5}$ , and we are looking for $\sin\theta$ . We can use the Pythagorean identity to do this.

$\begin{align} &\sin^{2}\theta+\cos^{2}\theta=1 && \text{Use our known value for cosine.} \\ &\sin^{2}\theta+\left(\frac{4}{5}\right)^{2}=1 && \text{Solve for sine.} \\ &\sin^{2}\theta=1−\frac{16}{25} \\ &\sin\theta=\pm\sqrt{\frac{9}{25}}=\pm\frac{3}{5} \end{align}$

Since $\theta=\cos^{−1}(\frac{4}{5})$ is in quadrant I, $\sin{\theta}$ must be positive, so the solution is $\frac{3}{5}$ . See Figure A below.

An illustration of a right triangle with an angle theta. Oppostie the angle theta is a side with length 3. Adjacent the angle theta is a side with length 4. The hypoteneuse has angle of length 5.

Figure A. Right triangle illustrating that if $\cos\theta=\frac{4}{5}$ , then $\sin\theta=\frac{3}{5}$

We know that the inverse cosine always gives an angle on the interval [0, π], so we know that the sine of that angle must be positive; therefore $\sin\left(\cos^{−1}\left(\frac{4}{5}\right)\right)=\sin\theta=\frac{3}{5}$ .

Example: Finding the Area of a Region

Find the area of the region between the graph of $f\left(x\right)=\sqrt{{x}^{2}-9}$ and the x-axis over the interval $\left[3,5\right]$ .

Show Solution

First, sketch a rough graph of the region described in the problem, as shown in the following figure.

This figure is the graph of the function f(x) = the square root of (x^2-9). It is an increasing curve that starts on the x-axis at 3 and is in the first quadrant. Under the curve above the x-axis is a shaded region bounded to the right at x = 5.

We can see that the area is $A={\displaystyle\int }_{3}^{5}\sqrt{{x}^{2}-9}dx$ . To evaluate this definite integral, substitute $x=3\sec\theta$ and $dx=3\sec\theta \tan\theta d\theta$ . We must also change the limits of integration. If $x=3$ , then $3=3\sec\theta$ and hence $\theta =0$ . If $x=5$ , then $\theta ={\sec}^{-1}\left(\frac{5}{3}\right)$ . After making these substitutions and simplifying, we have

\begin{array}{ccccc} Area & = \int_{3}^{5} \sqrt{x^{2} - 9} d x \\ = \int_{0}^{\sec^{- 1} (5 / 3)} 9 \tan^{2} θ \sec θ d θ & Use \tan^{2} θ = 1 - \sec^{2} θ . \\ = \int_{0}^{\sec^{- 1} (5 / 3)} 9 (\sec^{2} θ - 1) \sec θ d θ & Expand. \\ = \int_{0}^{\sec^{- 1} (5 / 3)} 9 (\sec^{3} θ - \sec θ) d θ & Evaluate the integral. \\ = (\frac{9}{2} ln | \sec θ + \tan θ | + \frac{9}{2} \sec θ \tan θ) - 9 ln | \sec θ + \tan θ | |_{\begin{matrix} 0 \end{matrix}}^{\begin{matrix} \sec^{- 1} (5 / 3) \end{matrix}} & Simplify. \\ = \frac{9}{2} \sec θ \tan θ - \frac{9}{2} ln | \sec θ + \tan θ | |_{\begin{matrix} 0 \end{matrix}}^{\begin{matrix} \sec^{- 1} (5 / 3) \end{matrix}} & \begin{matrix} Evaluate. Use \sec (\sec^{- 1} \frac{5}{3}) = \frac{5}{3} \\ and \tan (\sec^{- 1} \frac{5}{3}) = \frac{4}{3} . \end{matrix} \\ = \frac{9}{2} \cdot \frac{5}{3} \cdot \frac{4}{3} - \frac{9}{2} ln | \frac{5}{3} + \frac{4}{3} | - (\frac{9}{2} \cdot 1 \cdot 0 - \frac{9}{2} ln | 1 + 0 |) \\ = 10 - \frac{9}{2} ln 3. \end{array}

$\begin{array}{ccccc}\hfill \text{Area}& ={\displaystyle\int }_{3}^{5}\sqrt{{x}^{2}-9}dx\hfill & & & \\ & ={\displaystyle\int }_{0}^{{\sec}^{-1}\left(5\text{/}3\right)}9{\tan}^{2}\theta \sec\theta d\theta \hfill & & & \text{Use }{\tan}^{2}\theta =1-{\sec}^{2}\theta .\hfill \\ & ={\displaystyle\int }_{0}^{{\sec}^{-1}\left(5\text{/}3\right)}9\left({\sec}^{2}\theta -1\right)\sec\theta d\theta \hfill & & & \text{Expand.}\hfill \\ & ={\displaystyle\int }_{0}^{{\sec}^{-1}\left(5\text{/}3\right)}9\left({\sec}^{3}\theta -\sec\theta \right)d\theta \hfill & & & \text{Evaluate the integral.}\hfill \\ & =\left(\frac{9}{2}\text{ln}|\sec\theta +\tan\theta |+\frac{9}{2}\sec\theta \tan\theta \right)-9\text{ln}|\sec\theta +\tan\theta ||{}_{\begin{array}{c}\\ 0\end{array}}^{\begin{array}{c}{\sec}^{-1}\left(5\text{/}3\right)\\ \end{array}}\hfill & & & \text{Simplify.}\hfill \\ & =\frac{9}{2}\sec\theta \tan\theta -\frac{9}{2}\text{ln}|\sec\theta +\tan\theta ||{}_{\begin{array}{c}\\ 0\end{array}}^{\begin{array}{c}{\sec}^{-1}\left(5\text{/}3\right)\\ \end{array}}\hfill & & & \begin{array}{c}\text{Evaluate. Use }\sec\left({\sec}^{-1}\frac{5}{3}\right)=\frac{5}{3}\hfill \\ \text{and }\tan\left({\sec}^{-1}\frac{5}{3}\right)=\frac{4}{3}.\hfill \end{array}\hfill \\ & =\frac{9}{2}\cdot \frac{5}{3}\cdot \frac{4}{3}-\frac{9}{2}\text{ln}|\frac{5}{3}+\frac{4}{3}|-\left(\frac{9}{2}\cdot 1\cdot 0-\frac{9}{2}\text{ln}|1+0|\right)\hfill & & & \\ & =10-\frac{9}{2}\text{ln}3.\hfill & & & \end{array}$

try it

Evaluate $\displaystyle\int \frac{dx}{\sqrt{{x}^{2}-4}}$ . Assume that $x>2$ .

Hint

Substitute $x=2\sec\theta$ and $dx=2\sec\theta \tan\theta d\theta$ .

Show Solution

$\text{ln}|\frac{x}{2}+\frac{\sqrt{{x}^{2}-4}}{2}|+C$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for this segmented clip of “3.3 Trigonometric Substitution” here (opens in new window).

Module 3: Techniques of Integration