Evaluate the integral.

Question

cot3(4x) dx

Answer 1

first off, get rid of the 4x. Let
u = 4x
du = 4dx

then we have

1/4 ∫ cot^3 u du
= 1/4 ∫ cot u cot^2 u
= 1/4 ∫ cot u (csc^2 u - 1) du
= 1/4 ∫ cot u csc^2 u du - 1/4 ∫ cot u du
= -1/8 cot^2 u - 1/4 log(sin u) + C
= -1/8 cot^2(4x) - 1/4 log(sin(4x)) + C