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.
x=a \tan \theta
x=a \sin \theta
x=a \sec \theta
Figure 7:6.1
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
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.
\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
\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
To keep track of a trigonometric substitution, it is a good idea to actually draw the triangle and label the sides.
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
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.