# Integrating $\int\frac{\sqrt{\cos 2x}}{\sin x}\,\text{d}x$.

The integral is $$\int\frac{\sqrt{\cos 2x}}{\sin x}\,\text{d}x.$$ I have tried solving this by taking the sine inside the radical as follows: $$\int\sqrt\frac{\cos 2x}{\sin^2 x}\,\text{d}x$$$$\int\sqrt\frac{\cos^2x-\sin^2x}{\sin^2 x}\,\text{d}x$$$$\int\sqrt{\cot^2x-1}\,\text{d}x.$$ I don't know how to proceed from here, or whether this is even right. Any suggestions?

• Have you tried the Weierstrass substitution ? – Lucian Dec 14 '13 at 6:22
• This leads to the same answer as that by substituting $\cos x = t$, right? Or that's what I got. – Artemisia Dec 15 '13 at 4:42
• – Martin Sleziak Jan 22 at 10:51

You can continue like this: \begin{align*} \int\sqrt{\cot^2(x)-1}\,dx&=\int\frac{\cot^2(x)-1}{\sqrt{\cot^2(x)-1}}\,dx\\ &=\int\frac{\csc^2(x)-2}{\sqrt{\cot^2(x)-1}}\,dx\\ &=\int\frac{\csc^2(x)}{\sqrt{\cot^2(x)-1}}\,dx-2\int\frac{1}{\sqrt{\cot^2(x)-1}}\,dx\\ \end{align*} The first integral can be solved with substitution $u=\cot(x)$ and the second one with substitution $\cot(x)=\cosh(t)$ will be converted to the following one: \begin{align*} \int\frac{1}{\sqrt{\cot^2(x)-1}}\,dx&=-\int\frac{dt}{1+\cosh^2(t)}\\ &=-\int\frac{2}{2+e^t+e^{-t}}\,dt\\ &=-\int\frac{2e^t}{e^{2t}+2e^t+1}\,dt\\ &=-\int\frac{2e^t}{(e^t+1)^2}\,dt\\ &=\frac{2}{e^t+1} \end{align*} Can you continue?( I hope you can)
Suppose you express everything in terms of $\cos(x)$ and later make a change of variables such that $y = \cos(x)$. Remember that $\frac{dx}{\sin(x)}$ is also $\frac{-d(\cos(x)) }{1-\cos^2(x)}$. Does this help and can you continue from here ?