Question

Evaluate :

$\int \tan (x-\theta )\tan (x+\theta )\tan 2xdx$

Answer 1

We have to find $\int \tan (x-\theta )\tan (x+\theta )\tan 2xdx$

We know,

$2x=(x-\theta )+(x+\theta )$

$\tan 2x=\tan \left[\right(x-\theta )+(x+\theta \left)\right]$

$\tan 2x=\frac{\tan (x-\theta )+\tan (x+\theta )}{1-\tan (x-\theta )\tan (x+\theta )}$

$\tan 2x-\tan (x-\theta )\tan (x+\theta )\tan 2x=\tan (x-\theta )+\tan (x+\theta )$

$\tan (x-\theta )\tan (x+\theta )\tan 2x=\tan 2x-\tan (x-\theta )-\tan (x+\theta )$

Therefore,

$I=\int \tan (x-\theta )\tan (x+\theta )\tan 2xdx$

$=\int [\tan 2x-\tan (x-\theta )-\tan (x+\theta \left)\right]dx$

We know that , $\int \left[\tan x\right]=\log \cos x+C$

$\therefore I=-\frac{1}{2}\log |\cos 2x|+\log |\cos (x-\theta )|+\log |\cos (x+\theta )|+C$