Question

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

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

Answer 1

Answer: (B)

$0$

Let

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

using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

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

$\left(1\right)+\left(2\right)$

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

$2I=0\Rightarrow I=0$