Really need help with math

integral (sin10x)((cos10x+1)^(1/8))

Hello,

rewrite the integral as:

∫ [cos(10x) + 1]^(1/8) sin(10x) dx =

let:

[cos(10x) + 1] = u

differentiate both sides:

d[cos(10x) + 1] = du

10 [- sin(10x)] dx = du

- 10sin(10x) dx = du

sin(10x) dx = (-1/10) du

then, substituting:

∫ [cos(10x) + 1]^(1/8) sin(10x) dx = ∫ u^(1/8) (-1/10) du =

(pulling the constant out)

- (1/10) ∫ u^(1/8) du =

- (1/10) {1/[(1/8)+1]} u^[(1/8)+1] + C =

- (1/10) [1/(9/8)] u^(9/8) + C =

- (1/10)(8/9)u^(9/8) + C =

- (4/45)u^(9/8) + C

substitute back [cos(10x) + 1] for u, ending with:


∫ sin(10x) [cos(10x) + 1]^(1/8) dx = - (4/45)[cos(10x) + 1]^(9/8) + C


I hope it helps