Question

The value of ${\int }_0^{\pi /2}\frac{f\left(x\right)dx}{f\left(x\right)+f(\pi /2-x)}$ is equal to

• A. $\pi /4$
• B. $\pi /2$
• C. $\pi$
• D. None.

Answer 1

The correct option is A $\pi /4$

$I={\int }_0^{\pi /2}\frac{f\left(x\right)dx}{f\left(x\right)+f(\frac{\pi }{2}-x)}$ ...(1)

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

$I={\int }_0^{\pi /2}\frac{f(\frac{\pi }{2}-x)dx}{f(\frac{\pi }{2}-x)+f\left(x\right)}$ ...(2)

Adding (1) and (2), we get
$2I={\int }_0^{\pi /2}dx={\left[x\right]}_0^{\pi /2}=\frac{\pi }{2}$

$\therefore I=\frac{\pi }{4}$