$\int \frac{1}{\tan x+\cot x+\sec x+cosec~x} d x$ is equal to

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

Let $I=\int \frac{1}{\tan x+\cot x+\sec x+cosec~x} d x$. Then,

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

$\Rightarrow I =\frac{1}{2} \int \frac{(1+2 \sin x \cos x-1)}{1+\sin x+\cos x} d x=\frac{1}{2} \int \frac{(\sin x+\cos x)^2-1}{1+\sin x+\cos x}$

$\Rightarrow I =\frac{1}{2} \int \frac{(\sin x+\cos x+1)(\sin x+\cos x-1)}{1+\sin x+\cos x} d x$

$\Rightarrow I =\frac{1}{2} \int(\sin x+\cos x-1) d x$

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