Question

Evaluate :
$\int \frac{\tan x}{\sec x+\tan x}dx$

Answer 1

Given,

$\int \frac{\tan x}{\sec x+\tan x}dx$

Now,

$\begin{array}{c}=\int \frac{\sin x}{1+\sin x}dx\\ =\int \frac{\sin x+1-1}{1+\sin x}dx=x-\int \frac{dx}{{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)}^2}\\ =x-\int \frac{dx}{2{\left(\frac{1}{\sqrt{2}}\sin \frac{x}{2}+\frac{1}{\sqrt{2}}\cos \frac{x}{2}\right)}^2}\\ =x-\frac{1}{2}\int \frac{dx}{\sin \left(\frac{x}{2}+\frac{\pi }{4}\right)}\\ =x-\frac{1}{2}\int \cos e{c}^2\left(\frac{x}{2}+\frac{\pi }{4}\right)dx\\ =x+\frac{1}{2}\int x^2\cot \left(\frac{x}{2}+\frac{\pi }{4}\right)+C\\ =x+\cot \left(\frac{x}{2}+\frac{\pi }{4}\right)+C\end{array}$