Question

Evaluate: ${\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$

Answer 1

$I={\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx.........\left(i\right)I={\int }_0^{\frac{\pi }{2}}\frac{\sin (\frac{\pi }{2}-x)-\cos (\frac{\pi }{2}-x)}{1+\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)}dxI={\int }_0^{\frac{\pi }{2}}\frac{\cos x-\sin x}{1+\sin x\cos x}dxI=-{\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx.........\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$

$2I=0$

$I=0$