Question

Integrate ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\sin \left(\mathrm{x}\right)\right)dx$.

Answer 1

Integration of the given Expression:

${\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{a}-\mathrm{x}\right)\mathrm{dx}$.

${\int }_0^{2\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+{\int }_0^{\mathrm{a}}\mathrm{f}\left(2\mathrm{a}-\mathrm{x}\right)\mathrm{dx}$

Let $\mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\sin \left(\mathrm{x}\right)\right)\mathrm{dx}...\left(\mathrm{i}\right)$

By property $1$

$$\begin{gathered} \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)\right)\mathrm{dx}\left[\because \sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)=\cos \left(\mathrm{x}\right)\right] \\ \Rightarrow \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\cos \left(\mathrm{x}\right)\right)\mathrm{dx}...\left(\mathrm{ii}\right) \\ \left(\mathrm{i}\right)+\left(\mathrm{ii}\right) \\ \Rightarrow 2\mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\cos \left(\mathrm{x}\right)\right)+\log \left(\sin \left(\mathrm{x}\right)\right)\mathrm{dx}\left[\because \log \left(\mathrm{ab}\right)=\log \left(\mathrm{a}\right)+\log \left(\mathrm{b}\right)\right] \\ \Rightarrow 2\mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\cos \left(\mathrm{x}\right)\sin \left(\mathrm{x}\right)\right)\mathrm{dx}\left[\because \sin \left(2\mathrm{x}\right)=2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)\right] \\ \Rightarrow 2\mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\frac{\sin \left(2\mathrm{x}\right)}{2}\right)\mathrm{dx} \end{gathered}$$

Let $2\mathrm{x}=\mathrm{t}$

Differentiate both sides

$\Rightarrow 2\mathrm{dx}=\mathrm{dt}$ Replacing the value of $dx$

$\Rightarrow 2\mathrm{I}=\frac{1}{2}{\int }_0^{\mathrm{\pi }}\log \left(\frac{\sin \left(\mathrm{t}\right)}{2}\right)\mathrm{dt}$

By property $2$

$$\begin{gathered} {\int }_0^{\mathrm{\pi }}\log \left(\frac{\sin \left(\mathrm{t}\right)}{2}\right)\mathrm{dt}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\frac{\sin \left(\mathrm{t}\right)}{2}\right)\mathrm{dt}+{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\frac{\sin \left(\mathrm{\pi }-\mathrm{t}\right)}{2}\right)\mathrm{dt}\left[\sin \left(\mathrm{\pi }-\mathrm{t}\right)=\sin \left(\mathrm{t}\right)\right] \\ \Rightarrow 2\mathrm{I}=\frac{1}{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}2\log \left(\frac{\sin \left(\mathrm{t}\right)}{2}\right)\mathrm{dt} \\ \Rightarrow 2\mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\sin \left(\mathrm{t}\right)\right)-{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(2\right)\left[\because \log \left(\frac{\mathrm{a}}{\mathrm{b}}\right)=\log \left(\mathrm{a}\right)-\log \left(\mathrm{b}\right)\right] \\ \Rightarrow 2\mathrm{I}=\mathrm{I}-\frac{\mathrm{\pi }}{2}\log \left(2\right) \\ \Rightarrow \mathrm{I}=-\frac{\mathrm{\pi }}{2}\log \left(2\right) \end{gathered}$$

Hence ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\sin \left(\mathrm{x}\right)\right)$ is $-\frac{\mathrm{\pi }}{2}\log \left(2\right)$.