An integral question asked by Vijay Chavan

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

Answer:

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

$= \int \frac{dx}{\frac{1}{\cos x} + \frac{1}{\sin x} + \frac{\sin x}{\cos x} + \frac{\cos x}{ \sin x}} $

$= \int \frac{dx}{ \frac{\sin x}{\cos x \sin x} + \frac{\cos x}{\sin x \cos x} + \frac{\sin^2 x}{\cos x \sin x} + \frac{\cos^2 x}{\sin x \cos x}} $

$= \int \frac{\sin x \cos x \; dx}{\sin x + \cos x + 1} $

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

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

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

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

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