$\int \sin³(x)/(\cos²(x) - 4)\;\mathrm{d}x $

 

Let $u=\cos x$. Then the integral can be written as 
\[\int \frac{\sin^3 x}{\cos ^4 x -1}dx = \int \frac{\sin x (1-\cos^2 x)}{\cos ^4 x -4}dx\]
\[=\int \frac{u^2 -1}{u^4-4}du=\int \frac{u^2-1}{(u^2-2)(u^2+2)}du\]
\[=\int \frac{\frac{1}{4}}{u^2-2}+\frac{\frac{3}{4}}{u^2+2}du\]
\[=\frac{1}{4}\int \frac{1}{u^2-2}du+\frac{3}{4}\int \frac{1}{u^2+2}du\]
both integrals above are kind of elemntary and can be explicitly computed. 

Try the substitution $u = \cos x$ and remember that $\sin^2 x = 1- \cos ^2 x$