Question

${\int }_0^{\pi /2}\frac{\sin x-\cos x}{1-\sin x\cdot \cos x}dx$ is equal to

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\pi$

Answer 1

The correct option is A $0$

Let $I={\int }_0^{\pi /2}\frac{\sin x-\cos x}{1-\sin x\cos x}dx$ ....$\left(i\right)$
On putting $x=\left(\frac{\pi }{2}-x\right)$ in eq. $\left(i\right)$, we get
$I={\int }_0^{\pi /2}\frac{\sin \left(\frac{\pi }{2}-x\right)-\cos \left(\frac{\pi }{2}-x\right)}{1-\sin \left(\frac{\pi }{2}-x\right)\cos \left(\frac{\pi }{2}-x\right)}dx$
$={\int }_0^{\pi /2}\frac{\cos x-\sin x}{1-\sin x\cos x}dx$
$=-{\int }^{\pi /2}0\left(\frac{\sin x-\cos x}{1-\sin x\cos x}\right)dx$ $.....\left(ii\right)$
On adding Eqs. $\left(i\right)$ and $\left(ii\right)$, we get
$2I={\int }_0^{\pi /2}0dx=0$
$\Rightarrow I=0$