How to integrate $\tan(x) \tan(2x) \tan(3x)$? We have to integrate the following
$$\int \tan x \tan 2x \tan 3x \,\mathrm dx$$
In this I tried as
First split it into sine and cosine terms
Then used $\sin 2x =2\sin x \cos x$
But after that got stuck

In addition to the other answers, we can also do like this:

Use $\tan 3x =\frac {3\cos^2 x \sin x-\sin ^3 x}{\cos^3 x-3\cos x\sin^2 x}$ ,$\tan 2x=\frac {2\cos x\sin x}{\cos^2 x-\sin^2 x}$ and substitute $u=\cos x $ to get $$I =\int \tan x\tan 2x\tan 3x dx =\int -\frac {2(4\cos^4 x -5\cos^2 x+1)}{\cos x (2\cos^2 x-1)(4\cos^2 x-3)} \sin x dx =2 \int \frac {4u^4-5u^2+1}{u (2u^2-1)(4u^2-3)} du = -\frac {8}{3} \int \frac{u}{4u^2-3} du +\int \frac {2u}{2u^2-1} du +\int \frac {2}{3u} du $$ Hope you can take it from here. 
