Question

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

• A. 0
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (A)

0

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

Also

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

$\Rightarrow 2I=0\Rightarrow I=0$