Question

Integrate the following function $\frac{1}{1+\cot x}$

Answer 1

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

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

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

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

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

$=\frac{1}{2}\log |\sin x+\cos x|+\frac{1}{2}x+c$

Formula:

$\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\log f\left(x\right)+c$