Question

Evaluate the integral
${\int }_0^{\pi /2}\frac{\cos x-\sin x}{1+\cos x.\sin x}dx$

• A. $0$
• B. $-\pi$
• C. $\frac{1}{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $0$

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

${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

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

$2I={\int }_0^{\frac{\pi }{2}}0dx$

Thus $I=0$