I know how to calculate $\displaystyle\int \tan (x) \tan (2x)\tan (3x)\mathrm{d}x$ with the tangent sum formula
$\tan 3x\tan 2x\tan x =\tan3x -\tan 2x -\tan x$. But I don't know how I can solve
$$\displaystyle\int \tan (2x)\tan (3x)\mathrm{d}x$$
I tried u sub, trig sub, integration by parts. But I couldn't make it.
Thanks.
Best Answer
Proceed as follows
\begin{align} \int \tan 2x\tan 3x {d}x &=\int \frac{\sin 2x\sin 3x }{\cos 2x\cos 3x }dx\\ &=\int \frac{(2\sin x\cos x)[\sin x(4\cos^2x-1)] }{(2\cos^2x-1)[\cos x(4\cos^2x-3) ]}dx\\ &=2\int \frac{(1-\cos^2x)(4\cos^2x-1)}{(2\cos^2x-1)(4\cos^2x-3)}dx\\ &=\int \left( -1 -\frac1{2\cos^2x-1} + \frac2{4\cos^2x-3}\right)dx\\ &=-x -\int \frac{d(\tan x)}{1-\tan^2x} +2\int \frac{d(\tan x)}{1-3\tan^2x}\\ &=-x -\tanh^{-1}(\tan x) +\frac2{\sqrt3}\tanh^{-1}(\sqrt3\tan x) \end{align}