Trigonometric Integrals: Learn It 4

Giselle Miller

Trigonometric Integrals: Learn It 4

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

You’ve learned many integration techniques, but some expressions still seem impossible to tackle. Take [latex]\int \sqrt{9-{x}^{2}}dx[/latex], for example. None of your usual methods work here—but don’t worry! There’s a clever approach using trigonometric substitution.

The trick is recognizing that expressions like [latex]\sqrt{9-{x}^{2}}[/latex] can be simplified using trigonometric identities. Let’s see how this works with our example.

Let’s see how the substitution [latex]x=3\sin\theta[/latex] transforms our integral [latex]\displaystyle\int \sqrt{9-x^{2}}dx[/latex].

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

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

After simplifying, we have

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

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

[latex]{\displaystyle\int }^{\text{ }}\sqrt{9-{x}^{2}}dx={\displaystyle\int }^{\text{ }}9\sqrt{{\cos}^{2}\theta }\cos\theta d\theta[/latex].

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

[latex]{\displaystyle\int }^{\text{ }}\sqrt{9-{x}^{2}}dx={\displaystyle\int }^{\text{ }}9{\cos}^{2}\theta d\theta[/latex].

At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions.

Here’s the systematic approach for any expression involving [latex]\sqrt{{a}^{2}-{x}^{2}}[/latex]:

the general method for [latex]\sqrt{{a}^{2}-{x}^{2}}[/latex]

The Substitution: When you substitute [latex]x = a\sin\theta[/latex]:

[latex]\sqrt{{a}^{2}-{x}^{2}} = \sqrt{{a}^{2}-{a}^{2}{\sin}^{2}\theta} = \sqrt{{a}^{2}(1-{\sin}^{2}\theta)} = \sqrt{{a}^{2}{\cos}^{2}\theta} = a\cos\theta[/latex]

Why This Works:

The domain of [latex]\sqrt{{a}^{2}-{x}^{2}}[/latex] is [latex][-a,a][/latex]
This means [latex]-1 \leq \frac{x}{a} \leq 1[/latex]
Since [latex]\sin\theta[/latex] has range [latex][-1,1][/latex] over [latex][-\frac{\pi}{2}, \frac{\pi}{2}][/latex], we can always find a [latex]\theta[/latex] where [latex]\sin\theta = \frac{x}{a}[/latex]

After integrating, you need to express your answer in terms of [latex]x[/latex]. This is where you’ll need to recall some basic trigonometry to build a reference triangle. You’ll use these relationships to convert your trigonometric answer back to an expression with [latex]x[/latex].

Right Triangle Trigonometry For a right triangle with acute angle [latex]\theta[/latex]:

[latex]\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}[/latex]
[latex]\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}[/latex]
[latex]\tan\theta = \frac{\text{opposite}}{\text{adjacent}}[/latex]

Mnemonic: SohCahToa (Sine-opposite-hypotenuse, Cosine-adjacent-hypotenuse, Tangent-opposite-adjacent)

Knowing [latex]x = a\sin\theta[/latex], you can build a right triangle where [latex]\sin\theta = \frac{x}{a}[/latex].

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

Since [latex]\sin\theta = \frac{x}{a}[/latex], we know:

Hypotenuse: [latex]a[/latex]
Opposite side: [latex]x[/latex]
Adjacent side: [latex]\sqrt{a^2-x^2}[/latex] (by Pythagorean theorem)
Therefore: [latex]\theta = \sin^{-1}\left(\frac{x}{a}\right)[/latex]

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

Check first: Can this integral be solved more easily another way? For example, although this method can be applied to integrals of the form [latex]\displaystyle\int \frac{1}{\sqrt{{a}^{2}-{x}^{2}}}dx[/latex], [latex]\displaystyle\int \frac{x}{\sqrt{{a}^{2}-{x}^{2}}}dx[/latex], and [latex]\displaystyle\int x\sqrt{{a}^{2}-{x}^{2}}dx[/latex], they can each be integrated directly either by formula or by a simple u-substitution.
Substitute: Make the substitution [latex]x=a\sin\theta[/latex] and [latex]dx=a\cos\theta d\theta[/latex].
Simplify: Use [latex]\sqrt{{a}^{2}-{x}^{2}} = a\cos\theta[/latex]
Integrate: Use trigonometric integration techniques
Convert back: Use the reference triangle to express the result in terms of [latex]x[/latex] You may also need to use some trigonometric identities and the relationship [latex]\theta ={\sin}^{-1}\left(\frac{x}{a}\right)[/latex].

This technique is most useful when you have [latex]\sqrt{{a}^{2}-{x}^{2}}[/latex] that can’t be handled by simpler methods. However, some integrals like [latex]\int \frac{1}{\sqrt{{a}^{2}-{x}^{2}}}dx[/latex] or [latex]\int \frac{x}{\sqrt{{a}^{2}-{x}^{2}}}dx[/latex] are better solved with basic formulas or u-substitution.

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

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.

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,

[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]

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

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