Solução_Calculo_Stewart_6e

F.
296 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
13. Let u = t, dv =sec 2 2tdt ⇒ du = dt, v = 1 2
TX.10
tan 2t. Then
t sec 2 2tdt= 1 2 t tan 2t − 1 2
tan 2tdt=
1
t tan 2t − 1 ln |sec 2t| + C.
2 4
15. First let u =(lnx) 2 , dv = dx ⇒ du =2lnx · 1 dx, v = x.ThenbyEquation2,
x
I = (ln x) 2 dx = x(ln x) 2 − 2 x ln x · 1
x dx = x(ln x)2 − 2 ln xdx.NextletU =lnx, dV = dx
⇒
dU =1/x dx, V = x to get ln xdx = x ln x − x · (1/x) dx = x ln x − dx = x ln x − x + C 1.Thus,
I = x(ln x) 2 − 2(x ln x − x + C 1 )=x(ln x) 2 − 2x ln x +2x + C,whereC = −2C 1 .
17. First let u =sin3θ, dv = e 2θ dθ ⇒ du =3cos3θdθ, v = 1 2 e2θ .Then
I =
e 2θ sin 3θdθ= 1 2 e2θ sin 3θ − 3 2 e 2θ cos 3θdθ.NextletU =cos3θ, dV = e 2θ dθ ⇒ dU = −3sin3θdθ,
V = 1 2 e2θ to get
e 2θ cos 3θdθ= 1 2 e2θ cos 3θ + 3 2 e 2θ sin 3θdθ. Substituting in the previous formula gives
I = 1 2 e2θ sin 3θ − 3 4 e2θ cos 3θ − 9 4
e 2θ sin 3θdθ = 1 2 e2θ sin 3θ − 3 4 e2θ cos 3θ − 9 4 I
⇒
13
I = 1
4 2 e2θ sin 3θ − 3 4 e2θ cos 3θ + C 1 . Hence, I= 1 13 e2θ (2 sin 3θ − 3cos3θ)+C,whereC = 4 C 13 1.
19. Let u = t, dv =sin3tdt ⇒ du = dt, v = − 1 cos 3t. Then
3
π
0
t sin 3tdt= − 1 t cos 3t π
+ 1 π cos 3tdt= 1
π − 0
+ 1 π
3 0 3 0 3 9 sin 3t = π . 0 3
21. Let u = t, dv =coshtdt ⇒ du = dt, v =sinht. Then
1 t cosh tdt= t sinh t 1
− 1
sinh tdt=(sinh1− sinh 0) − cosh t 1
=sinh1− (cosh 1 − cosh 0)
0 0 0 0
=sinh1− cosh 1 + 1.
We can use the definitions of sinh and cosh to write the answer in terms of e:
sinh 1 − cosh 1 + 1 = 1 2 (e1 − e −1 ) − 1 2 (e1 + e −1 )+1=−e −1 +1=1− 1/e.
23. Let u =lnx, dv = x −2 dx ⇒ du = 1 x dx, v = −x−1 .By(6),
2
ln x
1 x dx = − ln x 2 2
+ x −2 dx = − 1 2
x
ln 2 + ln 1 + − 1 2
= − 1 2
1 1
x
ln 2 + 0 − 1 +1= 1 − 1 ln 2.
2 2 2 2
1
25. Let u = y, dv = dy
e = 2y e−2y dy ⇒ du = dy, v = − 1 2 e−2y .Then
1
0
y
e dy = − 1 ye−2y 1
2y 2
0
+ 1 2
1
0
e −2y dy = − 1 2 e−2 +0 − 1 4
27. Let u =cos −1 x, dv = dx ⇒ du = −√ dx , v = x. Then
1 − x
2
e −2y 1
0
= − 1 2 e−2 − 1 4 e−2 + 1 4 = 1 4 − 3 4 e−2 .
I =
1/2
0
cos −1 xdx= x cos −1 x 1/2
1/2
+
0
0
dt = −2xdx.Thus,I = π 6 + 1 2
xdx
√
1 − x
2 = 1 2 · π
3/4
+ t −1/2 − 1 dt ,wheret =1− x 2
3 2
1
3/4 t−1/2 dt = π 6 + √ t 1
3/4 = π 6 +1− √ 3
2 = 1 6
π +6− 3
√
3
.
1
⇒