Solve indefinite integral $\int\tan(x-a)\tan(x+a)\tan(2x)\ dx$

Question

Tried expanding $\tan$ terms but was not able to reach anywhere with it. How should I proceed ?

Answer 1

Its quite easy.

I would give you hint then you can try a bit further If $$2x = (x+a) + (x-a)$$ $$\tan2x = \tan\left[ (x+a) + (x-a) \right]$$ solving these you will get a result which is as follows : $$\tan(2x)-\tan(x+a)-\tan(x-a)=\tan(2x)\tan(x+a)\tan(x-a)$$

Now integrate $\tan(2x)-\tan(x+a)-\tan(x-a)$ instead of $\tan(2x)\tan(x+a)\tan(x-a)$

Try it from here now you might be able to solve it from here.

Answer 2

HINT:

$$\tan(A+B+C)=\dfrac{\sum\tan A-\prod\tan A}{\cdots}$$

For integer $n,$ if $A+B+C=n\pi,\sum\tan A-\prod\tan A=0$

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

and $$\tan(x+a)\tan(x-a)\tan(-2x)=\tan(x+a)+\tan(x-a)+\tan(-2x)$$

as $x+a+(x-a)+(-2x)=0\cdot\pi$