Integral question

A8.
(a)
∫(3+4sinx)^3 cosxdx
= ∫(3+4sinx)^3 d(3+4sinx)
= (3+4sinx)^4 (1/4) + C

(b)
∫(9+u)u^(-1/2)du
= ∫ 9u^(-1/2) du + ∫u^(1/2)du
= 9(2)u^(1/2) + (2/3) u^(3/2) + C
= 18u^(1/2) + 2u^(3/2) /3 + C

A9.
Sub x = tan u, dx/du = sec^2 u => dx = sec^2 u du
tan u = x/1
cos u = 1/(1+x^2)^(1/2)
sec u = (1+x^2)^(1/2)

∫x(1+x^2)^(-1/2)dx
= ∫ tan u (1+tan^2 u)^(-1/2) sec^2 u du
= ∫ tan u sec^(1/2) u du
= ∫ sec^(-1/2) u dsecu
= 2 sec^(1/2) u + C
= 2 (1+x^2)^(1/4) + C

A10.
y' = 6x^2 - 8x - t
y = 2x^3 - 4x^2 - tx + C
By (0,5) => C = 5
By (3,-1) => -1 = 54 - 36 - 3t + 5
=> -24 = -3t => t = 8
Hence, y = 2x^3 - 4x^2 - 8x + 5


2011-05-31 19:42:55 補充：
Oops... thz for your reminder.

2011-05-31 19:43:36 補充：
Correction:
A8.
(a)
∫(3+4sinx)^3 cosxdx
= (1/4) ∫(3+4sinx)^3 d(3+4sinx)
= (1/4) (3+4sinx)^4 (1/4) + C
= (3+4sinx)^4 (1/16) + C

2011-06-01 07:16:35 補充：
A9.[Alternative Method]
Sub x = tan u, dx/du = sec^2 u => dx = sec^2 u du
tan u = x/1
cos u = 1/(1+x^2)^(1/2)
sec u = (1+x^2)^(1/2)

∫x(1+x^2)^(-1/2)dx
= ∫ tan u (1+tan^2 u)^(-1/2) sec^2 u du
= ∫ tan u sec^(-1/2) u du
= ∫ sin u cos^(-3/2) u du
= -∫ cos^(-3/2) u dcosu
= 2 cos^(-1/2) u + C

2011-06-01 07:19:05 補充：
Oops" 看來我記錯了~_~1些方法

沒錯 A9, 直接 replace dx = 2xd1+x^2 即可

A9
u=1+x^2
xdx = (1/2) du
∫x(1+x^2)^(-1/2)dx
=∫(1/2) u^(-1/2) du
=(1/2) 2u^(1/2) + C
=(1+x^2)^(1/2) + C

d(3+4sinx)=4cosxdx?