Calculus 2nd Edition Rogawski

SECTION 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions 437
Exercises
In Exercises 1–16, calculate the integral.
∫
∫
1. cosh(3x)dx 2. sinh(x + 1)dx
∫
∫
3. x sinh(x 2 + 1)dx 4. sinh 2 x cosh xdx
∫
∫
5. sech 2 (1 − 2x)dx 6. tanh(3x)sech(3x)dx
∫
∫
7. tanh x sech 2 xdx 8.
cosh x
3 sinh x + 4 dx
∫
∫
9. tanh xdx 10. x csch(x 2 ) coth(x 2 )dx
∫
∫ cosh x
cosh x
11.
sinh x dx 12. sinh 2 x dx
∫
∫
13. sinh 2 (4x − 9)dx 14. sinh 3 x cosh 6 xdx
∫
∫
15. sinh 2 x cosh 2 xdx 16. tanh 3 xdx
In Exercises 17–30, calculate the integral in terms of the inverse hyperbolic
functions.
∫
∫
dx
dx
17. √ 18. √
x 2 − 1
9x 2 − 4
∫
∫
dx
dx
19. √ 20. √
16 + 25x 2
1 + 3x 2
∫ √ ∫
21. x 2 − 1 dx 22.
∫ 1/2
23.
−1/2
∫ 1
25.
0
∫
dx
5
1 − x 2 24.
4
∫ 10
dx
√
1 + x 2
26.
Further Insights and Challenges
41. Show that if u = tanh(x/2), then
2
x 2 dx
√
x 2 + 1
dx
1 − x 2
dx
4x 2 − 1
cosh x = 1 + u2
2u
2du
, sinh x = , dx =
1 − u2 1 − u2 1 − u 2
Hint: For the first relation, use the identities
(
sinh 2 x
)
2
= 1 ( 2 (cosh x − 1), x
)
cosh2 = 1 (cosh x + 1)
2 2
Exercises 42 and 43: evaluate using the substitution of Exercise 41.
∫
∫
dx
42. sech xdx 43.
1 + cosh x
44. Suppose that y = f (x) satisfies y ′′ = y. Prove:
(a) f (x) 2 − (f ′ (x)) 2 is constant.
(b) If f(0) = f ′ (0) = 0, then f (x) is the zero function.
∫ −1
∫
dx
0.8
27.
−3 x √ dx
28.
x 2 + 16
0.2 x √ 1 − x 2
∫ √ x
29.
2 ∫
− 1 dx
9 dx
x 2 30.
1 x √ x 4 + 1
31. Verify the formulas
√
sinh −1 x = ln |x + x 2 + 1|
cosh −1 x = ln |x +
√
x 2 − 1| (for x ≥ 1)
32. Verify that tanh −1 x = 1 ∣ ∣∣∣
2 ln 1 + x
1 − x ∣ for |x| < 1.
∫ √
33. Evaluate x 2 + 16 dx using trigonometric substitution. Then
use Exercise 31 to verify that your answer agrees with the answer in
Example 3.
34. Evaluate
∫ √
x 2 − 9 dx in two ways: using trigonometric substitution
and using hyperbolic substitution. Then use Exercise 31 to verify
that the two answers agree.
35. Prove the reduction formula for n ≥ 2:
∫
cosh n xdx= 1 n coshn−1 x sinh x + n − 1 ∫
n
cosh n−2 xdx 2
∫
36. Use Eq. (2) to evaluate cosh 4 xdx.
In Exercises 37–40, evaluate the integral.
∫ tanh −1 ∫
xdx
37.
x 2 38. sinh −1 xdx
− 1
∫
∫
39. tanh −1 xdx 40. x tanh −1 xdx
(c) f (x) = f(0) cosh x + f ′ (0) sinh x.
Exercises 45–48 refer to the function gd(y) = tan −1 (sinh y), called
the gudermannian. In a map of the earth constructed by Mercator projection,
points located y radial units from the equator correspond to
points on the globe of latitude gd(y).
45. Prove that
d
gd(y) = sech y.
dy
46. Let f (y) = 2 tan −1 (e y ) − π/2. Prove that gd(y) = f (y). Hint:
Show that gd ′ (y) = f ′ (y) and f(0) = g(0).
47. Let t(y) = sinh −1 (tan y) Show that t(y) is the inverse of gd(y)
for 0 ≤ y