Question

The value of ${\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$ is

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

Answer 1

The correct option is B $0$

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

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

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

$=-I$

$\Rightarrow I=-I$

$\Rightarrow 2I=0$

$\Rightarrow I=0$

$\therefore {\int }_0^{\pi /2}\frac{\sin x-\cos x}{1+\sin x\cos x}dx=0$