Question

lf ${\int }_0^{\pi }$ x.f(Sinx)dx $=k{\int }_0^{\pi /2}$ f(Sinx)dx then the value of $k$ is

• A. $\pi /2$
• B. $\pi$
• C. $\pi /3$
• D. 0

Answer 1

The correct option is B $\pi$

${\int }_0^{\pi }xf\left(sinx\right)dx=k{\int }_0^{\frac{\pi }{2}}f\left(sinx\right)dx$
${\int }_0^{\pi }(\pi -x)f\left(sinx\right)dx;k{\int }_0^{\frac{\pi }{2}}f\left(sinx\right)dx$
$\pi {\int }_0^{\pi }f\left(sinxdx\right)-{\int }_0^{\pi }xf\left(sinx\right)dx$
$=2{\int }_0^{\pi }xf\left(sinx\right)dx=\pi {\int }_0^{\pi }f\left(sinx\right)dx$
${\int }_0^{\pi }xf\left(sinx\right)dx=\frac{\pi }{2}f\left(sinx\right)dx$
$=\frac{\pi }{2}\times 2{\int }_0^{\frac{\pi }{2}}f\left(sinx\right)dx$
$=\pi {\int }_0^{\frac{\pi }{2}}f\left(sinx\right)dx$
So, $k=\pi$