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