Page 316 - Calculus Demystified

Chapter 7
1 2x + 1 303
(c) ·
2
2
1 +[ln(x + x)] 2 x + x
1
2
(d) · sec x
2
| tan x| [tan x] − 1
10. (a) Tan −1 2
x + C
(b) Sin −1 3
x + C

π/2 π
2
(c) Sin −1 (sin x) = Sin −1 1 − Sin −1 0 = .
0 2
√
1 dx 1 2x
(d) √ = √ · Tan −1 √ + C
5 1 +[ 2/5x] 2 10 5
Chapter 7
1. We do (a), (b), (c), (d).
2
(a) Let u = log x and dv = 1 dx. Then
1
2 2
log xdx = log x · x − x · 2 log x · dx
x

2
= x log x − 2 log xdx.
Now let u = log x and dv = 1 dx. Then
1
2 2
log xdx = x log x − 2 log x · x − x · dx
x
2
= x log x − 2x log x + 2x + C.
(b) Let u = x and dv = e 3x dx. Then
3x 3x
3 e e
x · e xdx = x · − · 1 dx
3 3
e 3x e 3x
= x · − + C.
3 9
2
(c) Let u = x and dv = cos xdx. Then

2 2
x cos xdx = x · sin x − sin x · 2xdx.
Now let u = 2x and dv = sin xdx. Then

2 2
x cos xdx = x · sin x − 2x · (− cos x) − (− cos x) · 2 dx
2
= x sin x + 2x cos x − 2 sin x + C.