The integral $\int \sqrt{1+2 \cot x(\operatorname{cosec} x+\cot x)} d x$ $\left(0

Question

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

Professor PC Thomas Classes · Answer

(a) : Let\ $I=\int \sqrt{1+2 \cot x(\operatorname{cosec} x+\cot x)} d x$\ $=\int \sqrt{1+2 \cot x \operatorname{cosec} x+2 \cot ^{2} x} d x$\ $=\int \sqrt{1+2 \frac{\cos x}{\sin ^{2} x}+2 \frac{\cos ^{2} x}{\sin ^{2} x}} d x$\ $=\int \sqrt{\frac{\sin ^{2} x+2 \cos x+2 \cos ^{2} x}{\sin ^{2} x}} d x$\ $=\int \sqrt{\frac{\sin ^{2} x+\cos ^{2} x+2 \cos x+\cos ^{2} x}{\sin ^{2} x}} d x$\ $=\int \sqrt{\frac{1+2 \cos x+\cos ^{2} x}{\sin ^{2} x}} d x$\ $=\int \sqrt{\frac{(1+\cos x)^{2}}{\sin ^{2} x}} d x=\int \frac{(1+\cos x)}{\sin x} d x$\ $=\int \frac{2 \cos ^{2}(x / 2)}{2 \sin (x / 2) \cos (x / 2)} d x$\ $=\int \cot (x / 2) d x=2 \log \left|\sin \frac{x}{2}ight|+C$\