Question

Evaluate $\int \frac{1}{1+\cot x}dx$

Answer 1

$\int \frac{1}{1+\cot x}dx$

$=\int \frac{\sin x}{\sin x+\cos x}dx$

$=\frac{1}{2}\int \frac{2\sin x}{\sin x+\cos x}dx$

$=\frac{1}{2}\left(\int \frac{\sin x+\cos x}{\sin x+\cos x}dx\right)$$-\frac{1}{2}\left(\int \frac{\cos x-\sin x}{\sin x+\cos x}dx\right)$

$=\frac{1}{2}\left(\int dx\right)$$-\frac{1}{2}\left(\int \frac{d(\cos x+\sin x)}{\sin x+\cos x}\right)$

$=\frac{x}{2}-\frac{1}{2}\log |(\sin x+\cos x)|+c$ [Where $c$ being integrating constant.]