Solução_Calculo_Stewart_6e

F.
3
9.
1 r4 ln rdr
11.
u =lnr,
du = dr
r
dv = r 4 dr,
v = 1 5 r5
TX.10
= 1
5 r5 ln r 3
− 3 1
1 1 5 r4 dr = 243 ln 3 − 0 − 1
r5 3
5 25 1
= 243 ln 3 − 243
− 1
5 25 25 =
243 242
5
ln 3 −
25
x − 1
(x − 2) + 1
u
x 2 − 4x +5 dx = (x − 2) 2 +1 dx = u 2 +1 + 1
du
u 2 +1
SECTION 7.5 STRATEGY FOR INTEGRATION ¤ 319
[u = x − 2, du = dx]
= 1 2 ln(u2 +1)+tan −1 u + C = 1 2 ln(x2 − 4x +5)+tan −1 (x − 2) + C
13. sin 3 θ cos 5 θdθ = cos 5 θ sin 2 θ sin θdθ= − cos 5 θ (1 − cos 2 θ)(− sin θ) dθ
= −
u 5 (1 − u 2 u =cosθ,
) du
du = − sin θdθ
= (u 7 − u 5 ) du = 1 8 u8 − 1 6 u6 + C = 1 8 cos8 θ − 1 6 cos6 θ + C
Another solution:
sin 3 θ cos 5 θdθ = sin 3 θ (cos 2 θ) 2 cos θdθ= sin 3 θ (1 − sin 2 θ) 2 cos θdθ
= u 3 (1 − u 2 ) 2 du
u =sinθ,
du =cosθdθ
= u 3 (1 − 2u 2 + u 4 ) du
= (u 3 − 2u 5 + u 7 ) du = 1 4 u4 − 1 3 u6 + 1 8 u8 + C = 1 4 sin4 θ − 1 3 sin6 θ + 1 8 sin8 θ + C
15. Let x =sinθ,where− π 2 ≤ θ ≤ π 2 .Thendx =cosθdθand (1 − x2 ) 1/2 =cosθ,
so
dx cos θdθ
(1 − x 2 ) = 3/2 (cos θ) = 3
sec 2 θdθ=tanθ + C =
x
√
1 − x
2 + C.
17. x sin 2 xdx
u = x,
du = dx
dv =sin 2 xdx,
v = sin 2 xdx = 1
2 (1 − cos 2x) dx = 1 2 x − 1 2 sin x cos x
= 1 2 x2 − 1 2 x sin x cos x − 1
2 x − 1 2 sin x cos x dx
= 1 2 x2 − 1 2 x sin x cos x − 1 4 x2 + 1 4 sin2 x + C = 1 4 x2 − 1 2 x sin x cos x + 1 4 sin2 x + C
Note: sin x cos xdx= sds= 1 2 s2 + C
[where s =sinx, ds =cosxdx].
A slightly different method is to write x sin 2 xdx= x · 1
2 (1 − cos 2x) dx = 1 2
the second integral by parts, we arrive at the equivalent answer 1 4 x2 − 1 x sin 2x − 1 4 8
cos 2x + C.
19. Let u = e x . Then e x+ex dx = e ex e x dx = e u du = e u + C = e ex + C.
xdx−
1
2 x cos 2xdx.Ifweevaluate
21. Let t = √ x,sothatt 2 = x and 2tdt = dx. Then arctan √ xdx = arctan t (2tdt)=I. Nowusepartswith
u =arctant, dv =2tdt ⇒ du = 1
1+t 2 dt, v = t2 . Thus,
I = t 2 arctan t −
t 2
1+t dt = 2 t2 arctan t − 1 − 1
dt = t 2 arctan t − t +arctant + C
1+t 2
= x arctan √ x − √ x +arctan √ x + C
or (x +1)arctan √ x − √
x + C