The value of ${\int }_0^{\mathrm{\pi }/2} \log \left(\frac{4+3\sin \mathrm{x}}{4+3\cos \mathrm{x}}\right)\mathrm{dx}$ is

Let,

$\begin{array}{l}\mathrm{I}={\int }_0^{\mathrm{\pi }/2} \log \left(\frac{4+3\sin \mathrm{x}}{4+3\cos \mathrm{x}}\right)\mathrm{dx}-----\left(\mathrm{i}\right)\\ \mathrm{I}={\int }_0^{\mathrm{\pi }/2} \log \left[\frac{4+3\sin (\mathrm{\pi }/2-\mathrm{x})}{4+3\cos (\mathrm{\pi }/2-\mathrm{x})}\right]\mathrm{dx}\\ \left[\because {\int }_0^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}\right]\\ \mathrm{I}={\int }_0^{\mathrm{\pi }/2} \log \left(\frac{4+3\cos \mathrm{x}}{4+3\sin \mathrm{x}}\right)\mathrm{dx}-----\left(\mathrm{ii}\right)\\ \left[\because \sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)=\cos \mathrm{x}\text{ and }\cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)=\sin \mathrm{x}\right]\end{array}$

On adding Eqs. (i) and (ii), we get

$\begin{array}{l}2\mathrm{I}={\int }_0^{\mathrm{\pi }/2} \left[\log \left(\frac{4+3\sin \mathrm{x}}{4+3\cos \mathrm{x}}\right)+\log \left(\frac{4+3\cos \mathrm{x}}{4+3\sin \mathrm{x}}\right)\right]\mathrm{dx}\\ \Rightarrow 2\mathrm{I}={\int }_0^{\mathrm{\pi }/2} \log \left(\frac{4+3\sin \mathrm{x}}{4+3\cos \mathrm{x}}\times \frac{4+3\cos \mathrm{x}}{4+3\sin \mathrm{x}}\right)\mathrm{dx}\\ \Rightarrow 2\mathrm{I}={\int }_0^{\mathrm{\pi }/2} \log 1\mathrm{dx}\Rightarrow 2\mathrm{I}={\int }_0^{\mathrm{\pi }/2} 0\mathrm{dx}[\because \log 1=0]\\ \Rightarrow \mathrm{I}=0\end{array}$