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7.6: Trigonometric Substitutions

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Integrals containing one of the terms

\sqrt{a^{2}+x^{2}}, \quad \sqrt{a^{2}-x^{2}}, \quad \text { or } \quad \sqrt{x^{2}-a^{2}} \nonumber

can often be integrated by a trigonometric substitution. The idea is to take x, a, and the square root as the three sides of a right triangle and use one of its acute angles as a new variable \theta. The three kinds of trigonometric substitutions are shown in Figure 7.6.1. These figures do not have to be memorized. Just remember that the sides must be labeled so that

(\text { opposite })^{2}+(\text { adjacent })^{2}=(\text { hypotenuse })^{2} \nonumber

These substitutions frequently give an integral of powers of trigonometric functions discussed in the preceding section.

image

x=a \tan \theta

image

x=a \sin \theta

image

x=a \sec \theta

Figure 7:6.1

Example 1

Find \int\left(a^{2}+x^{2}\right)^{-3 / 2} d x.

Solution

Let \theta=\arctan (x / a). Then from Figure 7.6.2,

x=a \tan \theta, \quad d x=a \sec ^{2} \theta d \theta, \quad \sqrt{a^{2}+x^{2}}=a \sec \theta \nonumber

So

\begin{aligned} \int\left(a^{2}+x^{2}\right)^{-3 / 2} d x & =\int(a \sec \theta)^{-3} a \sec ^{2} \theta d \theta \\ & =\frac{1}{a^{2}} \int(\sec \theta)^{-1} d \theta=\frac{1}{a^{2}} \int \cos \theta d \theta \\ & =\frac{1}{a^{2}} \sin \theta+C=\frac{1}{a^{2}} \frac{\tan \theta}{\sec \theta}+C=\frac{x}{a^{2} \sqrt{a^{2}+x^{2}}}+C \end{aligned} \nonumber

Example 2

Find \int \sqrt{x^{2}-a^{2}} d x.

Solution

Let \theta=\operatorname{arcsec}(x / a) (Figure 7.6.3), so

x=a \sec \theta, \quad d x=a \tan \theta \sec \theta d \theta, \quad \sqrt{x^{2}-a^{2}}=a \tan \theta \nonumber

So

\begin{aligned} \int \sqrt{x^{2}-a^{2}} d x & =\int a \tan \theta a \tan \theta \sec \theta d \theta=a^{2} \int \tan ^{2} \theta \sec \theta d \theta \\ & =a^{2} \int\left(\sec ^{2} \theta-1\right) \sec \theta d \theta \\ & =a^{2} \int \sec ^{3} \theta d \theta-a^{2} \int \sec \theta d \theta \\ & =\left(\frac{1}{2} a^{2} \sec ^{2} \theta \sin \theta+\frac{1}{2} a^{2} \int \sec \theta d \theta\right)-a^{2} \int \sec \theta d \theta \\ & =\frac{1}{2} a^{2} \sec ^{2} \theta \sin \theta-\frac{1}{2} a^{2} \int \sec \theta d \theta \\ & =\frac{1}{2} x \sqrt{x^{2}-a^{2}}-\frac{1}{2} a^{2} \int \sec \theta d \theta \end{aligned} \nonumber

This is as far as we can go on this problem until we find out how to integrate \int \sec \theta d \theta in the next chapter.

 

image
Figure 7.6.3

 

image

Example 3

\int \frac{1}{x^{2} \sqrt{a^{2}-x^{2}}} d x. Let \theta=\arcsin (x / a) (Figure 7.6.4). Then

Solution

\begin{aligned} & x=a \sin \theta, \quad d x=a \cos \theta d \theta, \quad \sqrt{a^{2}-x^{2}}=a \cos \theta \\ & \int \frac{1}{x^{2} \sqrt{a^{2}-x^{2}}} d x=\int \frac{1}{a^{2} \sin ^{2} \theta a \cos \theta} a \cos \theta d \theta=\int \frac{1}{a^{2} \sin ^{2} \theta} d \theta \\ &=\frac{1}{a^{2}} \int \csc ^{2} \theta d \theta=-\frac{1}{a^{2}} \cot \theta+C \\ &=-\frac{1}{a^{2}} \frac{\sqrt{a^{2}-x^{2}}}{x}+C \end{aligned} \nonumber

Example 4

\int \frac{\sqrt{x^{2}}-a^{2}}{x} d x. Put \theta=\operatorname{arcsec}(x / a) (Figure 7.6.5). Then

Solution

\begin{aligned} & x=a \sec \theta, \quad d x=a \tan \theta \sec \theta d \theta, \quad \sqrt{x^{2}-a^{2}}=a \tan \theta \\ & \begin{aligned} \int \frac{\sqrt{x^{2}-a^{2}}}{x} d x & =\int \frac{a \tan \theta}{a \sec \theta} a \tan \theta \sec \theta d \theta=a \int \tan ^{2} \theta d \theta \\ \cdot & =a \int \sec ^{2} \theta d \theta-a \int d \theta=a \tan \theta-a \theta+C \\ & =\sqrt{x^{2}-a^{2}}-a \operatorname{arcsec}(x / a)+C \end{aligned} \end{aligned} \nonumber

Figure 7.6 .4

image

 

image
Figure 7.6.5

 

To keep track of a trigonometric substitution, it is a good idea to actually draw the triangle and label the sides.

Example 5

The basic integrals:

Solution

(a) 

\int \frac{1}{\sqrt{1-x^{2}}} d x=\arcsin x+C \nonumber

(b)

\int \frac{d x}{1+x^{2}} d x=\arctan x+C \nonumber

(c)

\int \frac{d x}{x \sqrt{x^{2}-1}}=\operatorname{arcsec} x+C, \quad x>1 \nonumber

can be evaluated very easily by a trigonometric substitution.

(a)

\int \frac{1}{\sqrt{1-x^{2}}} d x \nonumber

 

 

image
Figure 7.6.6

 

 

image
Figure 7.6.7

 

 

image
Figure 7.6.8

 

Let \theta=\arcsin x (Figure 7.6.6). Then x=\sin \theta, d x=\cos \theta d \theta, \sqrt{1-x^{2}}= \cos \theta.

\begin{aligned} & \int \frac{1}{\sqrt{1-x^{2}}} d x=\int \frac{\cos \theta d \theta}{\cos \theta}=\int d \theta=\theta+C \\ & \int \frac{1}{\sqrt{1-x^{2}}} d x=\arcsin x+C \end{aligned} \nonumber

(b)

\int \frac{d x}{1+x^{2}} \nonumber

Let \theta=\arctan x (Figure 7.6.7). Then x=\tan \theta, d x=\sec ^{2} \theta d \theta, \sqrt{1+x^{2}}= \sec \theta.

\begin{aligned} & \int \frac{d x}{1+x^{2}}=\int \frac{\sec ^{2} \theta}{\sec ^{2} \theta} d \theta=\int d \theta=\theta+C \\ & \int \frac{d x}{1+x^{2}}=\arctan x+C \end{aligned} \nonumber

(c)

\int \frac{d x}{x \sqrt{x^{2}-1}}, \quad x>1 \nonumber

Let \theta=\operatorname{arcsec} x (Figure 7.6.8). Then x=\sec \theta, d x=\tan \theta \sec \theta d \theta, \sqrt{x^{2}-1}=\tan \theta.

\begin{aligned} & \int \frac{d x}{x \sqrt{x^{2}-1}}=\int \frac{\tan \theta \sec \theta}{\sec \theta \tan \theta} d \theta=\int d \theta=\theta+C \\ & \int \frac{d x}{x \sqrt{x^{2}+1}}=\operatorname{arcsec} x+C, \quad x>1 \end{aligned} \nonumber

It is therefore more important to remember the method of trigonometric substitution than to remember the integration formulas (a), (b), (c).

PROBLEMS FOR SECTION 7.6

Draw the appropriate triangle and evaluate using trigonometric substitutions

1 \quad \int \frac{d x}{\sqrt{1}-4 x^{2}}
2 \quad \int \sqrt{a^{2}-x^{2}} d x
3 \quad \int \frac{x^{3} d x}{\sqrt{9+\overline{x^{2}}}}
4 \int \frac{\sqrt{x^{2}-1}}{x} d x
5 \int\left(4-x^{2}\right)^{-3 / 2} d x
6 \quad \int\left(1-3 x^{2}\right)^{3 / 2} d x
7 \int \frac{\sin \theta d \theta}{\sqrt{2-\cos ^{2} \theta}}
8 \quad \int \frac{d x}{\sqrt{x} \sqrt{1-x}}
9 \quad \int \frac{d x}{x^{2}\left(1+x^{2}\right)}
10 \int \frac{x^{4} d x}{9+x^{2}}
11 \int \frac{x^{2} d x}{\sqrt{4-x^{2}}}
12 \int \frac{\sqrt{9-4 x^{2}}}{x^{4}} d x
13 \int x^{2} \sqrt{1-x^{2}} d x
14 \int \sqrt{x} \sqrt{1-x} d x
15
\int \sqrt{4 x-x^{2}} d x
16 \int \frac{\sqrt{x}}{\sqrt{1-x}} d x
17 \int \frac{x^{3}}{\sqrt{4 x^{2}-1}} d x
18 \int-\frac{d x}{x^{3} \sqrt{x^{2}-3}}
19 \int \frac{x^{3}}{\sqrt{a^{2}-x^{2}}} d x
20 \int x^{3} \sqrt{1+a^{2} x^{2}} d x
21 \int \frac{\sqrt{x^{4}-1}}{x} d x
22 \int x \sqrt{1-x^{4}} d x
23 \int \frac{x^{3}}{\left(a^{2}+x^{2}\right)^{3 / 2}} d x
24

\int_{0}^{2} \sqrt{4-x^{2}} d x \nonumber

25 \int_{-1}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} d x

\int_{0}^{4} \frac{d x}{\left(9+x^{2}\right)^{3 / 2}} \tag{26}

27 \quad \int_{0}^{\infty} \frac{d x}{\left(9+x^{2}\right)^{3 / 2}}

\int_{2}^{4} \frac{\sqrt{x^{2}-2}}{x} d x \tag{28}

29 \int_{2}^{x} \frac{\sqrt{x^{2}-2}}{x} d x
30 \quad \int_{0}^{x} \frac{x^{3}}{\sqrt{1+x^{2}}} d x
31 \int x \arcsin x d x
32 \int x \arccos x d x
33 \int x^{2} \arcsin x d x
34 \int x^{3} \arctan x d x
35 \int x^{-3} \arcsin x d x
36 \int x^{-3} \arctan x d x

37 Find the surface area generated by rotating the ellipse x^{2}+4 y^{2}=1 about the x-axis.


This page titled 7.6: Trigonometric Substitutions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by H. Jerome Keisler.

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