Solução_Calculo_Stewart_6e

F.
316 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
53. From the graph, we see that the integral will be negative, and we guess
that the area is about the same as that of a rectangle with width 2 and
height 0.3,soweestimatetheintegraltobe−(2 · 0.3) = −0.6. Now
1
x 2 − 2x − 3 = 1
(x − 3)(x +1) = A
x − 3 + B ⇔
x +1
TX.10
1=(A + B)x + A − 3B,soA = −B and A − 3B =1 ⇔ A = 1 4
and B = − 1 , so the integral becomes
4
2
dx
0 x 2 − 2x − 3 = 1 2
dx
4 0 x − 3 − 1 2
dx
4 0 x +1 = 1
2
ln |x − 3| − ln |x +1| = 1
ln
4
0 4 x − 3
2 x +1
0
= 1 4 ln
1
− ln 3 = − 1 ln 3 ≈−0.55
3 2
dx
55.
x 2 − 2x = dx
(x − 1) 2 − 1 = du
[put u = x − 1]
u 2 − 1
= 1
2 ln u − 1
u +1 + C [by Equation 6] = 1
2 ln x − 2
x + C
x
57. (a) If t =tan ,then x 2 2 =tan−1 t.Thefigure gives
x
1
x
cos = √
2 and sin t
= √
1+t
2 2
. 1+t
2
(b) cos x =cos 2 · x x
=2cos 2 − 1
2
2
2
1
=2 √ − 1= 2 1 − t2
− 1= 1+t
2 1+t2 1+t 2
(c) x 2
=arctant ⇒ x = 2 arctan t ⇒ dx =
2 1+t dt 2
59. Let t =tan(x/2). Then, using the expressions in Exercise 57, we have
1
3sinx − 4cosx dx = 1
2t 1 − t
2
2 dt
1+t =2 dt
3 − 4
2 3(2t) − 4(1 − t 2 ) = dt
2t 2 +3t − 2
1+t 2 1+t 2
dt
2
=
(2t − 1)(t +2) = 1
5 2t − 1 − 1
1
dt [using partial fractions]
5 t +2
= 1 ln |2t − 1| − ln |t +2| + C = 1
5
5 ln 2t − 1
t +2 + C = 1
5 ln 2tan(x/2) − 1
tan (x/2) + 2 + C
61. Let t =tan(x/2). Then, by Exercise 57,
π/2
sin 2x
π/2
2t
0 2+cosx dx = 2sinx cos x
1 2 ·
0 2+cosx dx = 1+t · 1 − t2
8t(1 − t 2 )
2 1+t 2 2
1
0
2+ 1 − t2 1+t dt = (1 + t 2 ) 2
2 0 2(1 + t 2 )+(1− t 2 ) dt
1+t 2
1
1 − t 2
= 8t ·
(t 2 +3)(t 2 +1) dt = I 2
0