Question

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

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

Answer 1

The correct option is A

$0$

Explanation for Correct answer:

Finding the value for the given integration:

Let $I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{Sinx-\cos x}{1+\sin x\cos x}dx\to \left(i\right)$

We know that , ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$

Replace $x=0+\frac{\mathrm{\pi }}{2}-x\left[\text{where}a=0,b=\frac{\mathrm{\pi }}{2}\right]$

$\begin{array}{rcl}I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{Sin\left(0+{\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)-\cos \left(0+\frac{\mathrm{\pi }}{2}-x\right)}{1+\sin \left(0+\frac{\mathrm{\pi }}{2}-x\right)\cos \left(0+\frac{\mathrm{\pi }}{2}-x\right)}dx\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos x-\sin x}{1+\cos x\sin x}dx\to \left(ii\right)\left[\because \sin \left(\frac{\mathrm{\pi }}{2}-x\right)=\cos x,\cos \left(\frac{\mathrm{\pi }}{2}-x\right)=\sin x\right]\\ & & \end{array}$

Add equation $\left(i\right)$ and $\left(ii\right)$

$\begin{array}{rcl}I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{Sinx-\cos x}{1+\sin x\cos x}+\frac{\cos x-\sin x}{1+\sin x\cos x}dx\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}0dx\\ & =& 0\end{array}$

Hence, option (A) is the correct answer.