$\int \tan x \tan 2 x \tan 3 x ~d x$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$\tan 3 x=\tan (2 x+x)$
$\Rightarrow \tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan x \tan 2 x}$
$\Rightarrow \tan x \tan 2 x \tan 3 x=\tan 3 x-\tan 2 x-\tan x$
∴ $\int \tan x \tan 2 x \tan 3 x d x$
$=\int(\tan 3 x-\tan 2 x-\tan x) d x$
$=\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C$