Kindly consider the following

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{\pi }{\int }}xlogsinxdx}\dots \left(i\right)\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi -x\right)logsin\left(\pi -x\right)dx}\left[{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dx={\displaystyle \underset{0}{\overset{a}{\int }}f\left(a-x\right)dx}}\right]\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi -x\right)logsinxdx}\dots \left(ii\right)\\ Adding\left(i\right)and\left(ii\right)\\ 2I={\displaystyle \underset{0}{\overset{\pi }{\int }}\left[\left(\pi -x\right)logsinx+xlogsinx\right]dx}\\ 2I={\displaystyle \underset{0}{\overset{\pi }{\int }}\pi logsinxdx}\\ 2I=2\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinxdx\left[?{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dx=2{\displaystyle \underset{0}{\overset{a/2}{\int }}f\left(x\right)dx}}\right]}\\ \therefore I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinxdx}\dots \left(iii\right)\\ I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsin\left(\frac{\pi }{2}-x\right)dx}\\ I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logcosxdx}\dots \left(iv\right)\\ Onadding\left(iii\right)and\left(iv\right),weget\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(logsinx+logcosx\right)dx}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinxcosxdx}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{log2sinxcosx}{2}dx}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsin2xdx}-\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}log2dx}\\ Put2x=t\Rightarrow 2dx=dt\Rightarrow dx=\frac{dt}{2}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\pi }{\int }}logsintdt}-\pi .log2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}1dx}\left[Changingthe\text{\hspace{0.17em}limit}\right]\\ 2I=I-\pi .log2{\left[x\right]}_0^{\frac{\pi }{2}}\text{\&}\end{array}$