Integrals Involving [latex]\sqrt{{a}^{2}-{x}^{2}}[/latex]

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

After simplifying, we have

Letting [latex]1-{\sin}^{2}\theta ={\cos}^{2}\theta [/latex], we now have

Assuming that [latex]\cos\theta \ge 0[/latex], we have

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

Since [latex]\cos\theta \ge 0[/latex] on [latex]-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}[/latex] and [latex]a>0[/latex], [latex]|a\cos\theta |=a\cos\theta [/latex]. We can see, from this discussion, that by making the substitution [latex]x=a\sin\theta [/latex], 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 [latex]x[/latex]. To see how to do this, let’s begin by assuming that [latex]0<x<a[/latex]. In this case, [latex]0<\theta <\frac{\pi }{2}[/latex]. Since [latex]\sin\theta =\frac{x}{a}[/latex], we can draw the reference triangle in Figure 1 to assist in expressing the values of [latex]\cos\theta [/latex], [latex]\tan\theta [/latex], and the remaining trigonometric functions in terms of [latex]x[/latex].  Before drawing the reference triangle, we recall the definition of trigonometric functions in terms of ratios of side lengths below.

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

Figure 1. A reference triangle can help express the trigonometric functions evaluated at [latex]\theta [/latex] in terms of [latex]x[/latex].

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

Example: Integrating an Expression Involving [latex]\sqrt{{a}^{2}-{x}^{2}}[/latex]

Evaluate [latex]{\displaystyle\int }^{\text{ }}\sqrt{9-{x}^{2}}dx[/latex].

Show Solution

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

Figure 2. A reference triangle can be constructed for this example.

Thus,

[latex]\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}[/latex]

 

Example: Integrating an Expression Involving [latex]\sqrt{{a}^{2}-{x}^{2}}[/latex]

Evaluate [latex]\displaystyle\int \frac{\sqrt{4-{x}^{2}}}{x}dx[/latex].

Show Solution

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

Figure 3. A reference triangle can be constructed for this example.

Thus,

[latex]\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}[/latex]

 

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

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Integrating Expressions Involving [latex]\sqrt{{a}^{2}+{x}^{2}}[/latex]

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

Figure 4. A reference triangle can be constructed to express the trigonometric functions evaluated at [latex]\theta [/latex] in terms of [latex]x[/latex].

Example: Integrating an Expression Involving [latex]\sqrt{{a}^{2}+{x}^{2}}[/latex]

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

Show Solution

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

Figure 5. The reference triangle for this example.

Thus,

[latex]\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}[/latex]

 

To check the solution, differentiate:

[latex]\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}[/latex]

 

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

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Integrating Expressions Involving [latex]\sqrt{{x}^{2}-{a}^{2}}[/latex]

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

Figure 6. Use the appropriate reference triangle to express the trigonometric functions evaluated at [latex]\theta [/latex] in terms of [latex]x[/latex].

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: [latex]\sin\left(\cos^{−1}\left(\frac{4}{5}\right)\right)[/latex].

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

[latex]\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}[/latex]

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

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

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 [latex]\sin\left(\cos^{−1}\left(\frac{4}{5}\right)\right)=\sin\theta=\frac{3}{5}[/latex].

Example: Finding the Area of a Region

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

Show Solution

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

Figure 7. Calculating the area of the shaded region requires evaluating an integral with a trigonometric substitution.

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

[latex]\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}[/latex]

 

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