The integral $\int \sqrt{1+2\cot x(\mathrm{cosec}x+\cot x)}dx$ $\left(0<x<\frac{\pi }{2}\right)$ is equal to (where $C$ is a constant of integration) 

Let $I=\int \sqrt{1+2\cot x(\mathrm{cosec}x+\cot x)}dx$
$\begin{array}{l}=\int \sqrt{1+2\cot x\mathrm{cosec}x+2{\cot }^{2}x}dx\\ =\int \sqrt{1+2\frac{\cos x}{{\sin }^{2}x}+2\frac{{\cos }^{2}x}{{\sin }^{2}x}}dx\\ =\int \sqrt{\frac{{\sin }^{2}x+2\cos x+2{\cos }^{2}x}{{\sin }^{2}x}}dx\\ =\int \sqrt{\frac{{\sin }^{2}x+{\cos }^{2}x+2\cos x+{\cos }^{2}x}{{\sin }^{2}x}}dx\end{array}$

$\begin{array}{l}=\int \sqrt{\frac{1+2\cos x+{\cos }^{2}x}{{\sin }^{2}x}}dx=\int \sqrt{\frac{(1+\cos x{)}^2}{{\sin }^{2}x}}dx\\ =\int \frac{(1+\cos x)}{\sin x}dx=\int \frac{2{\cos }^2(x/2)}{2\sin (x/2)\cos (x/2)}dx\\ =\int \cot (x/2)dx=2\log \left|\sin \frac{x}{2}\right|+C\end{array}$