James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

490 Chapter 7 Techniques of Integration
NOTE As Example 5 illustrates, hyperbolic substitutions can be used in place of trigonometric
substitutions and sometimes they lead to simpler answers. But we usually use
trigonometric substitutions because trigonometric identities are more familiar than hyperbolic
identities.
As Example 6 shows, trigonometric
substitution is sometimes a good idea
when sx 2 1 a 2 d ny2 occurs in an integral,
where n is any integer. The same is
true when sa 2 2 x 2 d ny2 or sx 2 2 a 2 d ny2
occur.
Example 6 Find y 3s3y2
0
x 3
s4x 2 3y2
dx.
1 9d
SOLUTION First we note that s4x 2 1 9d 3y2 − ss4x 2 1 9 d 3 so trigonometric substitution
is appropriate. Although s4x 2 1 9 is not quite one of the expressions in the table
of trigonometric substitutions, it becomes one of them if we make the preliminary
substitution u − 2x. When we combine this with the tangent substitution, we have
x − 3 2 tan , which gives dx − 3 2 sec2 d and
s4x 2 1 9 − s9 tan 2 1 9 − 3 sec
When x − 0, tan − 0, so − 0; when x − 3s3 y2, tan − s3 , so − y3.
y 3s3y2
0
x 3
s4x 2 1 9d dx − y y3
3y2 0
27
8 tan3
27 sec 3
3
2 sec2 d
− 3
16 y y3 tan 3
0 sec d − 3
16 y y3
0
− 3
16 y y3 1 2 cos 2
sin d
0 cos 2
sin 3
cos 2 d
Now we substitute u − cos so that du − 2sin d. When − 0, u − 1; when
− y3, u − 1 2 . Therefore
y 3s3y2
0
x 3
3
s4x 2 3y2
dx − 216 y 1y2 1 2 u 2
1 9d 1 u du 2
− 3
16 y 1y2
s1 2 u 22 d du − 3
16 Fu 1 1 1y2
1
uG1
− 3
16 f( 1 2 1 2) 2 s1 1 1dg − 3 32
n
Example 7 Evaluate y
x
s3 2 2x 2 x 2 dx.
SOLUTION We can transform the integrand into a function for which trigonometric
substitution is appropriate by first completing the square under the root sign:
3 2 2x 2 x 2 − 3 2 sx 2 1 2xd − 3 1 1 2 sx 2 1 2x 1 1d
− 4 2 sx 1 1d 2
This suggests that we make the substitution u − x 1 1. Then du − dx and x − u 2 1, so
y
x
s3 2 2x 2 x 2 dx − y
u 2 1
s4 2 u 2 du
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