Solução_Calculo_Stewart_6e

F.
15. Let u = e x ,sothatdu = e x dx and e 2x = u 2 .Then
du
e 2x arctan(e x ) dx = u 2 arctan u
u
SECTION 7.6 TX.10 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ¤ 325
92
= u2 +1
2
=
u arctan udu
arctan u − u 2 + C = 1 2 (e2x +1)arctan(e x ) − 1 2 ex + C
17. Let z =6+4y − 4y 2 =6− (4y 2 − 4y +1)+1=7− (2y − 1) 2 , u =2y − 1,anda = √ 7.Thenz = a 2 − u 2 , du =2dy,
and
y 6+4y − 4y2 dy = y √ zdy= 1
(u +1)√ a
2 2 − u 2 1
du = √ √ 1
2 4 u a2 − u 2 du + 1 a2 − u
4
2 du
√ √
= 1 a2 − u
4
2 du − 1 8 (−2u) a2 − u 2 du
This can be rewritten as
6+4y − 4y
2
30
= u √
a2 − u
8
2 + a2 u
8 sin−1 − 1 a 8
√wdw
w = a 2 − u 2 ,
dw = −2udu
= 2y − 1
6+4y − 4y2 + 7 2y − 1
8
8 sin−1 √ − 1
7 8 · 2
3 w3/2 + C
= 2y − 1
6+4y − 4y2 + 7 2y − 1
8
8 sin−1 √ − 1
7 12 (6 + 4y − 4y2 ) 3/2 + C
1
8 (2y − 1) − 1 )
12 (6 + 4y − 4y2 + 7 2y − 1
8 sin−1 √ + C
7
1
=
3 y2 − 1 12 y − 5 6+4y
− 4y2 + 7 2y − 1
8
8 sin−1 √ + C
7
= 1 24 (8y2 − 2y − 15) 6+4y − 4y 2 + 7 8 sin−1 2y − 1
√
7
+ C
19. Let u =sinx. Thendu =cosxdx,so
sin 2 x cos x ln(sin x) dx =
u 2 ln udu 101
= u2+1
(2 + 1) 2 [(2 + 1) ln u − 1] + C = 1 9 u3 (3 ln u − 1) + C
= 1 9 sin3 x [3 ln(sin x) − 1] + C
21. Let u = e x and a = √ 3.Thendu = e x dx and
e x
3 − e dx = du 19
= 1 2x a 2 − u 2 2a ln u + a
u − a + C = 1
2 √ 3 ln e x + √ 3
e x − √ 3
+ C.
23. sec 5 xdx = 77 1 tan x
4 sec3 x + 3 4 sec 3 xdx= 77 1 tan x
4 sec3 x + 3 1 tan x sec x +
1
4 2 2 sec xdx
14
= 1 tan x 4 sec3 x + 3 tan x sec x + 3 ln|sec x +tanx| + C
8 8
25. Let u =lnx and a =2.Thendu = dx/x and
4+(lnx)
2
x
a2
dx = + u 2 du = 21 u
a2 + u
2 2 + a2
2 ln u +
a 2 + u 2 + C
= 1 (ln x) 4+(lnx)
2 2 +2ln
ln x +
4+(lnx) 2 + C