Indefinite Integral 2

The answer is as follows :



圖片參考：http://dc616.4shared.com/img/FrB6IeUdce/s7/144bc8e7a90/2014-03-14C.png?async&rand=0.7771071035796195


http://dc616.4shared.com/img/FrB6IeUdce/s7/144bc8e7a90/2014-03-14C.png?async&rand=0.7771071035796195

諗住改answer
點知成題delete左 -.-

2014-03-14 00:40:55 補充：
∫ 89 / √(24-16x-64x^2) dx
= 89 ∫ dx / √[24-64(x^2+x/4)]
= 89 ∫ dx / √[25-64(x+1/8)^2]

2014-03-14 00:41:21 補充：
Let x+1/8=5sinθ /8
dx=5cosθ /8 dθ
89 ∫ dx / √[25-64(x+1/8)^2]
= 89 ∫ (5cosθ /8) dθ / √[25-64(5sinθ /8)^2]
= 89 ∫ (5cosθ /8) dθ / √[25(cosθ)^2]
= 89 ∫ (5cosθ /8) dθ / (5cosθ)
= 89/8 ∫ dθ
= 89/8 θ + C , where C is a constant

2014-03-14 00:41:51 補充：
x+1/8=5sinθ /8
sinθ=(8x+1)/5
θ=sin^(-1) [(8x+1)/5]

∴ ∫ 89 / √(24-16x-64x^2) dx
= 89/8 sin^(-1) [(8x+1)/5] + C , where C is a constant