A question asked by Biplap Ranjan

Evaluate $\int \frac{\sqrt{\cos 2x}}{\sin x} \; dx $

Answer:

$\int \frac{\sqrt{\cos 2x}}{\sin x} \; dx $

$=\int \frac{\cos 2x}{\sin x \sqrt{\cos 2x}} \; dx $

$=\int \frac{1 - 2\sin^2 x}{\sin x \sqrt{\cos 2x}} \; dx $

$=\int \frac{1}{\sin x \sqrt{\cos 2x}} \; dx - 2\int \frac{\sin x}{\sqrt{\cos 2x}} \; dx $

$=\int \frac{1}{\sin x \sqrt{\cos^2 x- \sin^2 x}} \; dx - 2\int \frac{\sin x}{\sqrt{2\cos^2 x - 1}} \; dx $

$=\int \frac{1}{\sin^2 x \sqrt{\cot^2 x- 1}} \; dx - 2\int \frac{\sin x}{\sqrt{2\cos^2 x - 1}} \; dx $

$=\int \frac{\csc^2 x}{\sqrt{\cot^2 x- 1}} \; dx - 2\int \frac{\sin x}{\sqrt{2\cos^2 x - 1}} \; dx $

Substitute $u= \cot x$ and $v = \cos x $, and we we have

$=-\int \frac{du}{\sqrt{u^2- 1}} + 2 \int \frac{dv}{\sqrt{2v^2 - 1}} $

$=-\int \frac{du}{\sqrt{u^2- 1}} + \frac{1}{\sqrt{2}} \int \frac{dv}{\sqrt{v^2 - \frac{1}{2}}} $

To integrate $\int \frac{du}{\sqrt{u^2- a^2}} $, substitute $ u = a\sec t$ and $du = a\sec t \tan t \; dt $

$\int \frac{du}{\sqrt{u^2- a^2}} = \int \sec t \; dt = \ln(\sec t + \tan t)$

$= \ln\left(\frac{u}{a} + \sqrt{\frac{u^2}{a^2} - 1}\right) + C$

$= \ln\left(u + \sqrt{u^2 - a^2}\right) + C$

So, $-\int \frac{du}{\sqrt{u^2- 1}} + \frac{1}{\sqrt{2}} \int \frac{dv}{\sqrt{v^2 - \frac{1}{2}}} $

$= - \ln\left(u + \sqrt{u^2 - 1}\right) + \frac{1}{\sqrt{2}} \ln \left(v + \sqrt{v^2 - \frac{1}{2}}\right) + C $

$= - \ln\left(\cot x + \sqrt{\cot^2 x - 1}\right) + \frac{1}{\sqrt{2}} \ln \left(\cos x + \sqrt{\cos^2 x - \frac{1}{2}}\right) + C $