How can I find $\int\tan\;x\;\cos\;2x\;\mathrm dx$? My question is ; How can I solve the following integral question?

$$\int\tan\;x\;\cos\;2x\;\mathrm dx$$

Thanks in advance,
HINT

(1) $\cos 2x = \cos^2x-\sin^2x=2\cos^2x-1$.

(2) $2\sin x\cos x = \sin 2x$.

(3) $\frac{d}{dx}\log(f(x))=?$
$\int\tan\;x\;\cos\;2x\;\mathrm dx$

=$\int\frac{sinx}{cosx}\;\cos\;2x\;\mathrm dx$

=$-\int\frac{1}{cosx}\;\cos\;2x\;\mathrm -sinx dx$                    $\frac{d(cosx)}{dx}$ = -sinx.dx

=$-\int\frac{1}{cosx}\;\cos\;2x\;\mathrm d(cosx)$                               

cos2x=2${cos^2x}$-1

=$-\int\frac{1}{cosx}\cdot(2{cos^2x}$-1) d(cosx)$

=$-\int(2{cosx}-\frac{1}{cosx}) d(cosx)$

= $-[\int(2{cosx}d(cosx)$ - $\int\frac{d(cosx)}{cosx}$

= $-[{cos^2x} + C1 - log (cosx) -C2]$

=$log(cosx)-{cos^2x}+C2-C1$ [C2,C1 - Integral Constants]