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

Section 7.3 Trigonometric Substitution 491
Figure 5 shows the graphs of the integrand
in Example 7 and its indefinite
integral (with C − 0). Which is which?
3
We now substitute u − 2 sin , giving du − 2 cos d and s4 2 u 2 − 2 cos , so
y
x
dx − y 2 sin 2 1
s3 2 2x 2 x 2 2 cos
2 cos d
_4
2
− y s2 sin 2 1d d
− 22 cos 2 1 C
FIGURE 5
_5
− 2s4 2 u 2 2 sin 21S u 2D 1 C
− 2s3 2 2x 2 x 2 2 sin 21S x 1 1
2
D 1 C n
1–3 Evaluate the integral using the indicated trigonometric
substitution. Sketch and label the associated right triangle.
dx
1. y
x 2 s4 2 x 2
2. y
x 3
sx 2 1 4
dx
3. y sx2 2 4
x
dx
4–30 Evaluate the integral.
4. y
x 2
s9 2 x 2 dx
x − 2 sin
x − 2 tan
x − 2 sec
5. y sx2 2 1
dx 6. y 3
x 4 0
7. y a
0
9. y 3
2
dx
sa 2 1 x 2 d 3y2 , a . 0
dx
sx 2 2 1d 3y2
11. y 1y2
x s1 2 4x 2 dx 12. y 2
0
13. y sx 2 2 9
dx 14. y 1
x 3 0
15. y a
x 2 sa 2 2 x 2 dx
0
x
s36 2 x 2 dx
8. dt
y
t 2 st 2 2 16
10. y 2y3
s4 2 9x 2 dx
0
0
16.
y 2y3
dt
s4 1 t 2
dx
sx 2 1 1d 2
s2y3
dx
x 5 s9x 2 2 1
17. y
x
sx 2 2 7
dx
19. y s1 1 x 2
x
21. y 0.6
0
dx
18. y
20. y
x 2
s9 2 25x 2 dx 22. y 1
dx
23. y
sx 2 1 2x 1 5
24. y 1
25. y x 2 s3 1 2x 2 x 2 dx 26. y
27. y sx 2 1 2x dx 28. y
29. y xs1 2 x 4 dx 30. y y2
dx
fsaxd 2 2 b 2 g 3y2
x
s1 1 x 2 dx
0 sx 2 1 1 dx
0 sx 2 x2 dx
31. (a) Use trigonometric substitution to show that
y
0
x 2
dx
s3 1 4x 2 4x 2 3y2
d
x 2 1 1
sx 2 2 2x 1 2d 2 dx
cos t
s1 1 sin 2 t
dx
sx 2 1 a 2 − ln(x 1 sx 2 1 a 2 ) 1 C
(b) Use the hyperbolic substitution x − a sinh t to show that
y
dx
sx 2 1 a 2 − sinh 21S x aD 1 C
These formulas are connected by Formula 3.11.3.
dt
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