3.3 Trigonometric Substitution

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 $x=3\text{sin}\theta ,$ we have $dx=3\text{cos}\theta d\theta .$ After substituting into the integral, we have

After simplifying, we have

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

Assuming that $\text{cos}\theta \ge 0,$ 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 $\sqrt{a^2-x^2},$ we make the substitution $x=a\text{sin}\theta$ and $dx=a\text{cos}\theta .$ To see that this actually makes sense, consider the following argument: The domain of $\sqrt{a^2-x^2}$ is $[\text{−}a,a].$ Thus, $\text{−}a\le x\le a.$ Consequently, $\mathrm{-1}\le \frac{x}{a}\le 1.$ Since the range of $\text{sin}x$ over $\left[\text{−}(\pi \text{/}2),\pi \text{/}2\right]$ is $[\mathrm{-1},1],$ there is a unique angle $\theta$ satisfying $\text{−}(\pi \text{/}2)\le \theta \le \pi \text{/}2$ so that $\text{sin}\theta =x\text{/}a,$ or equivalently, so that $x=a\text{sin}\theta .$ If we substitute $x=a\text{sin}\theta$ into $\sqrt{a^2-x^2},$ we get

Since $\text{cos}\theta \ge 0$ on $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ and $a>0,$ $\left|a\text{cos}\theta \right|=a\text{cos}\theta .$ We can see, from this discussion, that by making the substitution $x=a\text{sin}\theta ,$ 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 $0<x<a.$ In this case, $0<\theta <\frac{\pi }{2}.$ Since $\text{sin}\theta =\frac{x}{a},$ we can draw the reference triangle in Figure 3.4 to assist in expressing the values of $\text{cos}\theta ,$ $\text{tan}\theta ,$ and the remaining trigonometric functions in terms of $x.$ 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 $\theta ={\text{sin}}^{\mathrm{-1}}\left(\frac{x}{a}\right).$

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

The following example demonstrates the application of this problem-solving strategy.

Example 3.21

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

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

Solution

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

Thus,

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

Example 3.22

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

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

Solution

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

Thus,

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

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

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 $x=a\text{tan}\theta$ or $x=a\text{cot}\theta .$ Either of these substitutions would actually work, but the standard substitution is $x=a\text{tan}\theta$ or, equivalently, $\text{tan}\theta =x\text{/}a.$ With this substitution, we make the assumption that $\text{−}(\pi \text{/}2)<\theta <\pi \text{/}2,$ so that we also have $\theta ={\text{tan}}^{\mathrm{-1}}\left(x\text{/}a\right).$ The procedure for using this substitution is outlined in the following problem-solving strategy.

Example 3.24

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

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

Solution

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

Thus,

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

To check the solution, differentiate:

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

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