$\int\frac{dx}{\tan x+\cot x+\sec x+cosec x}$ is equal to:

$\frac{1}{2}(\sin x+\cos x+x)+C$ $\frac{1}{2}(\sin x-\cos x-x)+C$ $\frac{1}{2}(\cos x-x+\sin x)+C$ None of these

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

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