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

Since
$$
\tan3x=\tan(x+2x)=\dfrac{\tan x+\tan2x}{1-\tan x\tan2x},
$$
we have
$$
\tan3x-\tan x\tan2x\tan3x=\tan x+\tan2x,
$$
i.e.
$$
\tan x\tan 2x\tan3x=\tan 3x-\tan x-\tan 2x.
$$
For $a\ne 0$ we have
$$
\int\tan(ax)\,dx=-\dfrac{1}{a}\ln|\cos(ax)|+c, 
$$
and therefore
\begin{eqnarray}
\int \tan x\tan 2x\tan3x\,dx&=&\int(\tan 3x-\tan x-\tan 2x)\,dx\\
&=&\ln|\cos x|+\dfrac{1}{2}\ln|\cos2x|-\dfrac13\ln|\cos3x|+c
\end{eqnarray}
