$\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x=$

Question

Aicharya · Accepted Answer

Let $I=\int_0^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cos x} d x$ $\begin{aligned} & 	herefore I=\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-xight)-\cos \left(\frac{\pi}{2}-xight)}{1-\sin \left(\frac{\pi}{2}-xight) \cos \left(\frac{\pi}{2}-xight)} d x \ & =\int_0^{\pi / 2} \frac{\cos x-\sin x}{1-\cos x \sin x} d x \end{aligned}$ Eq. (1) $+(2)$ gives $2 \mathrm{I}=\int_0^{\frac{\pi}{2}} 0 \mathrm{dx} \Rightarrow \mathrm{I}=0$

$\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x=$

Solution