Question

${\int }_0^{\pi }x\log \sin x=\frac{{\pi }^k}{m}\log \frac{1}{n}.$.Find $k+m+n$ ?

Answer 1

Let $I={\int }_0^{\pi }x\log sinx$ ...(1)
using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$\therefore I={\int }_0^{\pi }(\pi -x)\log \left(sin(\pi -x)\right)dx$
$={\int }_0^{\pi }(\pi -x)\log sinxdx$ ...(2)
From (1) and (2)
$2I$$=\pi {\int }_0^{\pi }\log sinxdx==\pi ({\int }_0^{\frac{\pi }{2}}\log sinxdx+{\int }_0^{\frac{\pi }{2}}\log \left(sin(\pi -x)\right)dx)$
using ${\int }_0^{2a}f\left(x\right)dx={\int }_0^{a}f\left(x\right)+f(2aq-x)dx$
$\therefore 2I=2\pi {\int }_0^{\frac{\pi }{2}}\log sinxdx$
$\Rightarrow I=\pi {\int }_0^{\frac{\pi }{2}}\log sinxdx$ ...(3)
Now using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$
$\therefore I=\pi {\int }_0^{\frac{\pi }{2}}\log sin(\frac{\pi }{2}-x)dx$
$=\pi {\int }_0^{\frac{\pi }{2}}\log cosxdx$ ...(4)
From (3) and (4)
$\Rightarrow 2I=\pi {\int }_0^{\frac{\pi }{2}}\log \left(sinxcosx\right)dx=\pi {\int }_0^{\frac{\pi }{2}}\log sin2xdx-\pi {\int }_0^{\frac{\pi }{2}}\log 2dx$
Substitue $2x=t\Rightarrow 2dx=dt$
$\therefore 2I=\frac{\pi }{2}{\int }_0^{\pi }\log sintdt-{\left[\pi x\log 2\right]}_0^{\frac{\pi }{2}}$
$2I=\frac{2\pi }{2}{\int }_0^{\frac{\pi }{2}}\log sintdt-\frac{{\pi }^2}{2}\log 2$
$\Rightarrow 2I=\pi {\int }_0^{\frac{\pi }{2}}\log sinxdx-\frac{{\pi }^2}{2}\log 2$
$\therefore I=\frac{{\pi }^2}{2}\log \frac{1}{2}$
Hence, $k+m+n=2+2+2=6$