Problem with solving a complicated Integral I need to determine the $ \int \frac{\sin^3(x)}{8-\cos^3(x)} dx$. It's an indefinite integral.

Hint: $\sin^3(x) = \sin(x)(\sin^2(x)) = \sin(x)(1-\cos^2(x))$
Take $u = \cos(x) \Rightarrow du = -\sin(x) \ dx$
So we have, $- \int \frac{1-u^2}{8-u^3} du = -\int \frac{1}{8-u^3} du - \int \frac{u^2}{8-u^3} du $
