Page 320 - Calculus Demystified

Chapter 7
We may now solve for the desired integral: 307
1 1
x
e sin xdx = [e · sin 1 − e · cos 1] .
0 2
2
(b) Integrate by parts with u = ln x, dv = x dx. Thus
e 3 e e 3 1

2 x x
x ln xdx = ln x · − · dx
1 3 1 1 3 x
3 e
e 3 1 3 x
= 1 · − 0 · −
3 3 6 1
e 3 e 3 1 3
= − + .
3 9 9
(c) We write
2x + 1 1 1 −1
= + + .
2
x (x + 1) x x 2 x + 1
Thus

4 (2x + 1)dx 4 1 4 1 4 −1
= dx + dx + dx
3
2 x + x 2 2 x 2 x 2 2 x + 1

−1 −1
=[ln 4 − ln 2]+ − +[ln 3 − ln 5]
4 2
6 1
= ln + .
5 4
(d) We write
π 1 π

2 2 2
sin x cos xdx = sin 2xdx
0 4 0
π
1 1 − cos 4x
= dx
4 0 2
π
1 sin 4x
= x −
8 4 0
1
= [(π − 0) − (0 − 0)]
8
π
= .
8