Question

Evaluate
$\int \frac{dx}{\cos x\cos 2x}$

Answer 1

We have,

$\int \frac{dx}{\cos x\cos 2x}$

On multiply and divide by $\cos x$

Now,

$\int \frac{\cos xdx}{{\cos }^{2}x\cos 2x}$

$\Rightarrow \int \frac{\cos xdx}{(1-{\sin }^{2}x)(1-2{\sin }^{2}x)}$

Let

$\sin x=t$

$\Rightarrow \cos xdx=dt$

$\Rightarrow \int \frac{dt}{(1-t^2)(1-2t^2)}$

Using partial dfraction and we get,

$\Rightarrow \int \frac{dt}{(1-t^2)(1-2t^2)}=\int [\frac{-1}{1-t^2}+\frac{2}{1-2t^2}]dt$

On integrating and we get,

$\frac{1}{2}\ln (t-1)(t+1)+\frac{1}{\sqrt{2}}\ln \left(\frac{1+\sqrt{2}t}{1-\sqrt{2}t}\right)+C$

Put $t=\sin x$

So,

$=\frac{1}{2}\ln (\sin x-1)(\sin x+1)+\frac{1}{\sqrt{2}}\ln \left(\frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}\right)$

Hence, this is the answer.