Solução_Calculo_Stewart_6e

F.
298 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
41. Let u = 1 2 x2 , dv =2x √ 1+x 2 dx ⇒ du = xdx, v = 2 3 (1 + x2 ) 3/2 .
Then
√
x
3
1+x 2 dx = 1 2 x2 2
(1 +
3 x2 ) 3/2 − 2 3 x(1 + x 2 ) 3/2 dx
= 1 3 x2 (1 + x 2 ) 3/2 − 2 3 · 2
5 · 1
2 (1 + x2 ) 5/2 + C
= 1 3 x2 (1 + x 2 ) 3/2 − 2
15 (1 + x2 ) 5/2 + C
Another method: Use substitution with u =1+x 2 to get 1 5 (1 + x2 ) 5/2 − 1 3 (1 + x2 ) 3/2 + C.
43. (a) Take n =2in Example 6 to get
0
sin 2 xdx= − 1 2 cos x sin x + 1 2
1 dx = x 2
−
sin 2x
4
(b) sin 4 xdx= − 1 cos x
4 sin3 x + 3 4 sin 2 xdx= − 1 cos x 4 sin3 x + 3 x − 3 sin 2x + C.
8 16
45. (a) From Example 6, sin n xdx= − 1 n cos x sinn−1 x + n − 1
sin n−2 xdx. Using (6),
n
π/2
sin n xdx= − cos x π/2
sinn−1 x
+ n − 1 π/2
sin n−2 xdx
n
n
=(0− 0) + n − 1
n
0
π/2
0
sin n−2 xdx= n − 1
n
(b) Using n =3in part (a), we have π/2
sin 3 xdx= 2 π/2
sin xdx= − 2 cos x π/2
0 3 0 3
Using n =5in part (a), we have π/2
0
sin 5 xdx= 4 5
π/2
0
sin 3 xdx= 4 5 · 2
3 = 8 15 .
0
π/2
0
0
= 2 3 .
+ C.
sin n−2 xdx
(c) The formula holds for n =1(that is, 2n +1=3) by (b). Assume it holds for some k ≥ 1. Then
π/2
0
sin 2k+1 xdx=
π/2
0
2 · 4 · 6 ·····(2k)
3 · 5 · 7 ·····(2k +1) .ByExample6,
sin 2k+3 xdx=
=
2k +2
2k +3
π/2
0
sin 2k+1 xdx=
2 · 4 · 6 ·····(2k)[2 (k +1)]
3 · 5 · 7 ·····(2k +1)[2(k +1)+1] ,
so the formula holds for n = k +1. By induction, the formula holds for all n ≥ 1.
47. Let u =(lnx) n , dv = dx ⇒ du = n(ln x) n−1 (dx/x), v = x. By Equation 2,
(ln x) n dx = x(ln x) n − nx(ln x) n−1 (dx/x) =x(ln x) n − n (ln x) n−1 dx.
2k +2
2k +3 · 2 · 4 · 6 ·····(2k)
3 · 5 · 7 ·····(2k +1)
49.
tan n xdx= tan n−2 x tan 2 xdx= tan n−2 x (sec 2 x − 1) dx = tan n−2 x sec 2 xdx− tan n−2 xdx
= I − tan n−2 xdx.
Let u =tan n−2 x, dv =sec 2 xdx ⇒ du =(n − 2) tan n−3 x sec 2 xdx, v =tanx. Then, by Equation 2,
I =tan n−1 x − (n − 2) tan n−2 x sec 2 xdx
1I =tan n−1 x − (n − 2)I
(n − 1)I =tan n−1 x
I = tann−1 x
n − 1
Returning to the original integral, tan n xdx= tann−1 x
n − 1
− tan n−2 xdx.