Page 318 - Calculus Demystified

Chapter 7
3. We do (a), (b), (c), (d). 305
(a) Let u = sin x, du = cos xdx. Then the integral becomes
2 3
(1 + u )
2 2
(1 + u ) 2udu = + C.
3
Resubstituting x, we obtain the final answer
2 3
2 2 (1 + sin x)
(1 + sin x) 2 sin x cos xdx = + C.
3
√ √
(b) Let u = x, du = 1/[2 x] dx. Then the integral becomes

2 sin udu =−2 cos u + C.

Resubstituting x, we obtain the final answer
√

sin x √
√ dx =−2 cos x + C.
x
(c) Let u = ln x, du =[1/x] dx. Then the integral becomes

1 1
cos u sin udu = sin 2udu =− cos 2u + C.
2 4
Resubstituting x, we obtain the final answer

cos(ln x) sin(ln x) 1
dx =− cos(2ln x) + C.
x 4
2
(d) Let u = tan x, du = sec xdx. Then the integral becomes

u
u
e du = e + C.
Resubstituting x, we obtain the final answer

2
e tan x sec xdx = e tan x + C.
4. We do (a), (b), (c), (d).

(a) Let u = cos x, du =− sin xdx. Then the integral becomes
u 3
2
− u du =− + C.
3
Resubstituting x, we obtain the final answer
3
2 cos x
sin x cos xdx =− + C.
3