The actual problem by Vijay Chavan

Evaluate $\int \frac{dx}{(\sec x + \csc x + \cot x + \tan x)^2} $

Answer:

$\int \frac{dx}{(\sec x + \csc x + \cot x +\tan x)^2} $

In the previous post, we had seen that

$\frac{1}{\sec x + \csc x + \cot x +\tan x} = \frac{-1}{2} (1- \sin x - \cos x) $

So, we have

$ \frac{1}{(\sec x + \csc x + \cot x +\tan x)^2} = \frac{-1}{4} (1- \sin x - \cos x)^2 $

$= \frac{-1}{4}(2 - 2\cos x - 2\sin x + 2\sin x \cos x) $

So,

$\int \frac{dx}{(\sec x + \csc x + \cot x +\tan x)^2} $

$= \frac{-1}{4} \int (2 - 2\cos x - 2\sin x + 2\sin x \cos x) \; dx $

$= \frac{-1}{4} ( 2x - 2\sin x + 2\cos x + \sin^2 x) + C$