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?
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 ?
Substitute $t= \frac{\cos x}{\sqrt{\cos 2x}}$ to have $$dt= \frac{\sin x}{ {\cos^{3/2}2x}}dx,\>\>\>\>\> (t^2-1)(2t^2-1)=\frac{\sin^2t}{\cos^2 2x}$$ and \begin{align} &\int\frac{\sqrt{\cos 2x}}{\sin x}\ dx\\ =&\int \frac{\cos^2 2x}{\sin^2x }\cdot\frac{\sin x}{{\cos^{3/2}2x}} dx = \int\frac1{(t^2-1)(2t^2-1)}dt\\ =&\int \left(\frac1{t^2-1}-\frac{2}{2t^2-1} \right) dt =-\coth^{-1}t+\sqrt2 \coth^{-1}\sqrt2t\\ =& - \coth^{-1} \frac{\cos x}{\sqrt{\cos 2x}} +\sqrt2 \coth^{-1}\frac{\sqrt2 \cos x}{\sqrt{\cos 2x}} \end{align} Similarly \begin{align} &\int\frac{\sqrt{\cos 2x}}{\cos x}\ dx =\cot^{-1} \frac{\sin x}{\sqrt{\cos 2x}} -\sqrt2 \cot^{-1}\frac{\sqrt2 \sin x}{\sqrt{\cos 2x}} \end{align}
