Indefinite Integral $\int\frac{ \sin^3 2x-x \cos^3 x}{\cos^2 x}~dx $ How can I calculate $\displaystyle \int\frac{ \sin^3 2x-x \cos^3 x}{\cos^2 x} dx$? Help me I'm quite blur and  how should I start. A little show on working are gladly appreciated.

Hint: Use that $\sin 2x=2 \sin x\cos x$. Note that $\frac{\sin^3 2x}{\cos^2x}=2^3\sin^3x\cos x$ and so changing the variables. Moreover $x\frac{\cos^3 x}{\cos^2x}=x\cos x$ and integrate by parts.
