Question

Evaluate : $I=\int \frac{\sec x}{\left(\sec x+\mathrm{tanx}\right)}dx$

Answer 1

$\begin{array}{c}I=\int \frac{\sec x}{\left(\sec x+tanx\right)}dx\\ =\int \frac{\sec x}{\left(\sec x+\tan x\right)}\times \frac{\left(\sec x-\tan x\right)}{\left(\sec x-\tan x\right)}\\ =\int \frac{\left({\sec }^{2}x-\sec x\tan x\right)}{{\sec }^{2}x-{\tan }^{2}x}dx\\ =\int \left({\sec }^{2}x-\sec x\tan x\right)dx\\ =\int {\sec }^{2}dx-\int \sec x\tan xdx\\ =\tan x-\sec x+C.\end{array}$