what is the integral of sin^5(6x)dx

Answer 1

first, note that if u = 6x, you have

1/6 ∫ sin^5 u du
= 1/6 ∫sin^4(u) sinu du
= -1/6 ∫(1-cos^2 u)^2 (-sinu du)
If v = cos u, then dv = -sinu du, and you have

-1/6 ∫(1-v^2)^2 dv

just expand that and it's a regular old polynomial.