Question:
$\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
Solution:
Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$ ...(1)
$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x$ $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$
$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x} d x$ ...(2)
Adding (1) and (2), we obtain
$2 I=\int_{0}^{\frac{\pi}{2}} \frac{0}{1+\sin x \cos x} d x$
$\Rightarrow I=0$