Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 7.1 INTEGRATION BY PARTS ¤ 297
29. Let u =ln(sinx), dv =cosxdx ⇒ du = cos x dx, v =sinx. Then
sin x
I = cos x ln(sin x) dx =sinx ln(sin x) − cos xdx=sinx ln(sin x) − sin x + C.
Another method: Substitute t =sinx,sodt =cosxdx.ThenI = ln tdt = t ln t − t + C (see Example 2) and so
I =sinx (ln sin x − 1) + C.
31. Let u =(lnx) 2 , dv = x 4 dx ⇒ du =2 ln x x5
dx, v =
x 5 .By(6),
2
x
x 4 (ln x) 2 5
2 2
dx =
1
5 (ln x 4
2
x)2 32
− 2 ln xdx=
1 1 5 (ln x 4
5 2)2 − 0 − 2 ln xdx.
1 5
dx, V =
x5
25 .
Let U =lnx, dV = x4
5 dx ⇒ dU = 1 x
2
x 4
x
5 2 2
Then
1 5 ln xdx= 25 ln x x 4
32
x
5
− dx =
1 1 25 ln 2 − 0 − 25
So 2
1 x4 (ln x) 2 dx = 32 (ln 5 2)2 − 2 32
31
25
ln 2 −
125
125
2
1
=
32
(ln 5 2)2 − 64
62
25
ln 2 + . 125
= 32
25 ln 2 − 32
125 − 1
125
.
33. Let y = √ x,sothatdy = 1 2 x−1/2 dx = 1
2 √ x dx = 1
2y dx. Thus, cos √ xdx= cos y (2ydy)=2 y cos ydy.Now
use parts with u = y, dv =cosydy, du = dy, v =siny to get y cos ydy= y sin y − sin ydy= y sin y +cosy + C 1 ,
so cos √ xdx=2y sin y +2cosy + C =2 √ x sin √ x +2cos √ x + C.
√
35. Let x = θ 2 π
,sothatdx =2θdθ.Thus, θ 3 cos θ 2 √ π
dθ = θ 2 cos θ 2 π
· 1 (2θdθ)= 1 x cos xdx.Nowuse
√ √ 2 2
π/2 π/2 π/2
parts with u = x, dv =cosxdx, du = dx, v =sinx to get
π
1
x π
x cos xdx= 1 π
2
2 sin x − sin xdx
π/2
π/2
π/2
= 1 2
= 1 2 (π sin π +cosπ) − 1 2
π
2 sin π 2 +cos π 2
x sin x +cosx
π
π/2
=
1
2 (π · 0 − 1) − 1 2
π
2 · 1+0 = − 1 2 − π 4
37. Let y =1+x, so that dy = dx. Thus, x ln(1 + x) dx = (y − 1) ln ydy.Nowusepartswithu =lny, dv =(y − 1) dy,
du = 1 dy, v = 1 y 2 y2 − y to get
(y − 1) ln ydy= 1
2 y2 − y ln y − 1
y − 1 dy = 1 y(y − 2) ln y − 1 2 2 4 y2 + y + C
= 1 2 (1 + x)(x − 1) ln(1 + x) − 1 4 (1 + x)2 +1+x + C,
which can be written as 1 2 (x2 − 1) ln(1 + x) − 1 4 x2 + 1 2 x + 3 4 + C.
In Exercises 39 – 42, let f(x) denote the integrand and F (x) its antiderivative (with C =0).
39. Let u =2x +3, dv = e x dx ⇒ du =2dx, v = e x .Then
(2x +3)e x dx =(2x +3)e x − 2 e x dx =(2x +3)e x − 2e x + C
=(2x +1)e x + C
We see from the graph that this is reasonable, since F has a minimum where
f changes from negative to positive.