Trigonometric Integrals: Learn It 6

Giselle Miller

Trigonometric Integrals: Learn It 6

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

The third type of expression requires yet another approach. Let’s see why [latex]\sqrt{{x}^{2}-{a}^{2}}[/latex] is different from the previous two cases.

The expression [latex]\sqrt{{x}^{2}-{a}^{2}}[/latex] has a restricted domain: [latex](-\infty, -a] \cup [a, +\infty)[/latex]. This means either [latex]x \leq -a[/latex] or [latex]x \geq a[/latex], which gives us [latex]\frac{x}{a} \leq -1[/latex] or [latex]\frac{x}{a} \geq 1[/latex]. The range of [latex]\sec\theta[/latex] on [latex][0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi][/latex] is exactly [latex](-\infty, -1] \cup [1, +\infty)[/latex]—this perfectly matches what we need!

So we use the substitution [latex]x = a\sec\theta[/latex] where [latex]0 \leq \theta < \frac{\pi}{2}[/latex] or [latex]\frac{\pi}{2} < \theta \leq \pi[/latex], with [latex]dx = a\sec\theta \tan\theta , d\theta[/latex].

Why This Substitution Works

When you substitute [latex]x = a\sec\theta[/latex]:

[latex]\sqrt{{x}^{2}-{a}^{2}} = \sqrt{(a\sec\theta)^{2}-{a}^{2}} = \sqrt{{a}^{2}({\sec}^{2}\theta-1)} = \sqrt{{a}^{2}{\tan}^{2}\theta} = |a\tan\theta|[/latex]

Sign Matters Here! Unlike the previous cases, the sign of [latex]|a\tan\theta|[/latex] depends on which part of the domain you’re in:

For [latex]x \geq a[/latex]: [latex]|a\tan\theta| = a\tan\theta[/latex]
For [latex]x \leq -a[/latex]: [latex]|a\tan\theta| = -a\tan\theta[/latex]

This is why you need two reference triangles for this case!

Since [latex]\sec\theta = \frac{x}{a}[/latex], you need different triangles depending on the sign of [latex]x[/latex]:

For [latex]x \geq a[/latex]:

Hypotenuse: [latex]x[/latex]
Adjacent side: [latex]a[/latex]
Opposite side: [latex]\sqrt{x^2-a^2}[/latex]

For [latex]x \leq -a[/latex]:

The triangle setup changes because [latex]\tan\theta[/latex] will be negative
You’ll need to be careful about signs when converting back to [latex]x[/latex]

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. — Use the appropriate reference triangle to express the trigonometric functions evaluated at θ in terms of x.

Problem-Solving Strategy for [latex]\sqrt{{x}^{2}-{a}^{2}}[/latex]

Check first: Can this integral be solved more easily another way? If so, we may wish to consider applying an alternative technique.
Substitute: [latex]x = a\sec\theta[/latex] and [latex]dx = a\sec\theta \tan\theta , d\theta[/latex].
Simplify: Use [latex]\sqrt{{x}^{2}-{a}^{2}} = |a\tan\theta|[/latex] (watch the sign!)
Integrate: Use trigonometric integration techniques
Convert back: Use the appropriate reference triangle and [latex]\theta = \sec^{-1}\left(\frac{x}{a}\right)[/latex]. (Note: We need both reference triangles, since the values of some of the trigonometric ratios are different depending on whether [latex]x\ge a[/latex] or [latex]x\le \text{-}a.[/latex])

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

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.

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 [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].

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

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 [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]