Evaluate $\int \tan^6x\sec^3x \ \mathrm{d}x$ Integrate $$\int \tan^6x\sec^3x \ \mathrm{d}x$$
I tried to split integral to $$\tan^6x\sec^2x\sec x$$ but no luck for me. Help thanks

You have
$$
\tan^6x\sec^3x=\frac{\sin^6x}{\cos^9x}=\frac{\sin^6x}{(1-\sin^2x)^5}\cos x
$$
so by substituting $t=\sin x$
$$
\int\frac{t^6}{(1-t^2)^5}dt
$$
that is long with partial fractions, but simple.
