Question

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

Answer 1

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

Multiplying numerator and denominarator $\tan se{c}^{2}x$

$\int \frac{\tan x}{(1+\tan x)se{c}^{2}x}dx=\int \frac{\tan xse{c}^{2}x}{(1+\tan x)(1+{\tan }^{2}x)}dx$

$\tan x=tse{c}^{2}xdx=dt$

$\int \frac{t}{(1+t)(1+t^2)}dt$

Breaking into partial fractions

$=\frac{-1}{2}\int \frac{dt}{t+1}+\frac{1}{2}+\int \frac{tdt}{t^2+1}+\frac{1}{2}\int \frac{d{t}^2}{t^2+1}$

$=\frac{-1}{2}ln(|t+1|)+\frac{1}{4}ln(|t^2+1|)+\frac{1}{2}{\tan }^{-1}t+c$

Put $t=tanx$

$=\frac{-1}{2}ln(\tan x+1)+\frac{1}{4}ln(|1+{\tan }^{2}x|)+\frac{1}{2}{\tan }^{-1}\tan x+c$

$=\frac{1}{2}ln\left(\frac{|secx|}{|1+\tan x|}\right)+\frac{1}{2}x+c$

$=\frac{-1}{2}ln(|\sin x+\cos x|)+\frac{1}{2}x+c$