Question

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

Answer 1

Let $I={\int }_0^{\pi /2}\frac{\cos x-\sin x}{1+\sin x\cos x}dx$
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$\therefore I={\int }_0^{\pi /2}\frac{\cos (\frac{\pi }{2}-x)-\sin (\frac{\pi }{2}-x)}{1+\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)}dx$
$={\int }_0^{\pi /2}\frac{\sin x-\cos x}{1+\sin x\cos x}dx=-1\therefore 2I=0\Rightarrow I=0$