Question

$\stackrel{\pi /2}{\underset{0}{\int }}\frac{\sin x-\cos x}{1+\sin x.\cos x}dx$

• A. $0$
• B. $\frac{\pi }{2}$
• C. $1$
• D. $-\frac{\pi }{2}$

Answer 1

The correct option is A $0$

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

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

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

$I+I=0$

$\Rightarrow I=0$.