Question

Integrate ${\int }_{\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)\right)\mathrm{dx}$.

Answer 1

Integrate the given Expression:

Using the property ${\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+{\int }_0^{\mathrm{a}}\mathrm{f}\left(-\mathrm{x}\right)\mathrm{dx}$

Let $\mathrm{I}={\int }_{\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)\right)\mathrm{dx}$

Apply the property written above

$$\begin{gathered} \Rightarrow \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)\right)\mathrm{dx}+{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\cos \left(\mathrm{x}\right)-\sin \left(\mathrm{x}\right)\right)\mathrm{dx}\left[\because \sin \left(-\mathrm{x}\right)=-\sin \left(\mathrm{x}\right)\right] \\ \Rightarrow \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)\left(\cos \left(\mathrm{x}\right)-\sin \left(\mathrm{x}\right)\right)\right)\mathrm{dx}\left[\because \log \left(\mathrm{ab}\right)=\log \left(\mathrm{a}\right)+\log \left(\mathrm{b}\right)\right] \\ \Rightarrow \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left({\cos }^2\left(\mathrm{x}\right)-{\sin }^2\left(\mathrm{x}\right)\right)\mathrm{dx}\left[\because \left({\mathrm{a}}^2-{\mathrm{b}}^2\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right] \\ \Rightarrow \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\cos \left(2\mathrm{x}\right)\right)\mathrm{dx}\left[\because \cos \left(2\mathrm{x}\right)={\cos }^2\left(\mathrm{x}\right)-{\sin }^2\left(\mathrm{x}\right)\right] \end{gathered}$$

Let $2\mathrm{x}=\mathrm{t}$ Differentiating both sides

$$\begin{gathered} \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{2} \\ \Rightarrow \mathrm{I}=\frac{1}{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\cos \left(\mathrm{t}\right)\right)\mathrm{dt} \\ \Rightarrow \mathrm{I}=\frac{1}{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\cos \left(\mathrm{t}\right)\right)\mathrm{dt} \end{gathered}$$

Apply the property

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

Add both $I$

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

Let $2\mathrm{t}=\mathrm{u}$ Differentiating both sides

$$\begin{gathered} \Rightarrow \mathrm{dt}=\frac{\mathrm{du}}{2} \\ \Rightarrow 2\mathrm{I}=\frac{1}{2}{\int }_0^{\mathrm{\pi }}\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}-\frac{\log \left(2\right)}{2} \end{gathered}$$

Apply the property

$$\begin{gathered} {\int }_0^{2\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+{\int }_0^{\mathrm{a}}\mathrm{f}\left(2\mathrm{a}-\mathrm{x}\right)\mathrm{dx} \\ \Rightarrow 2\mathrm{I}=\frac{1}{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\sin \left(\mathrm{u}\right)\right)\mathrm{du}-\frac{1}{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\log \left(2\right)}{2}+\frac{1}{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\log \left(\sin \left(\mathrm{\pi }-\mathrm{u}\right)\right)\mathrm{du}-\frac{1}{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\log \left(2\right)}{2} \end{gathered}$$

Since

$$\begin{gathered} \sin \left(\mathrm{\pi }-\mathrm{u}\right)=\sin \left(\mathrm{u}\right) \\ \Rightarrow 2\mathrm{I}=\mathrm{I}-\frac{\mathrm{\pi }}{4}\log \left(2\right) \\ \Rightarrow \mathrm{I}=-\frac{\mathrm{\pi }}{4}\log \left(2\right) \end{gathered}$$

Hence ${\int }_{\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)\right)\mathrm{dx}$ is $-\frac{\mathrm{\pi }}{4}\log \left(2\right)$