Question

By using the properties of definite integrals, evaluate the integral ${\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$ ......... (1)
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\sin (\frac{\pi }{2}-x)-\cos (\frac{\pi }{2}-x)}{1+\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)}dx,(\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\cos x-\sin x}{1+\sin x\cos x}dx$ ...... (2)
Adding (1) and (2), we obtain
$2I={\int }_0^{\frac{\pi }{2}}\frac{0}{1+\sin x\cos x}dx=0$
$\Rightarrow I=0$