Question

If ${\int }_0^{\pi }xf({\sin }^{2}x+{\sec }^{2}x)dx=k{\int }_0^{\pi /2}f({\sin }^{2}x+{\sec }^{2}x)dx$, then the value of $k$ is

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $-\frac{\pi }{2}$
• D. None of the above

Answer 1

The correct option is B $\pi$

We have, ${\int }_0^{\pi }xf({\sin }^{2}x+{\sec }^{2}x)dx=k{\int }_0^{\pi /2}f({\sin }^{2}x+{\sec }^{2}x)dx$
Let $I={\int }_0^{\pi }xf({\sin }^{2}x+{\sec }^{2}x)dx$ .....(i)
$={\int }_0^{\pi }(\pi -x)f({\sin }^2(\pi -x)+{\sec }^2(\pi -x))dx$
$={\int }_0^{\pi }(\pi -x)f({\sin }^{2}x+{\sec }^{2}x)dx$ ....(ii)
On adding equations (i) and (ii), we get
$2I=\pi {\int }_0^{\pi }f({\sin }^{2}x+{\sec }^{2}x)dx$
$\Rightarrow 2I=2\pi {\int }_0^{\pi /2}f({\sin }^{2}x+{\sec }^{2}x)dx$
$\Rightarrow I=\pi {\int }_0^{\pi /2}f({\sin }^{2}x+{\sec }^{2}x)dx$
On comparing with given integral, we get $k=\pi$