Question

Solve
${\int }_0^{\pi /2}\log \sin xdx$

Answer 1

$\begin{array}{c}I={\int }_0^{\frac{\pi }{2}}\log \sin dx\to \left(1\right)\\ {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\\ \Rightarrow I={\int }_0^{\frac{\pi }{2}}\log \cos xdx\to \left(2\right)\\ addingequation\left(1\right)and\left(2\right)\\ \Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log \left(\sin x.\cos x\right)dx={\int }_0^{\frac{\pi }{2}}\log \left(\frac{\sin 2x}{2}\right)dx\\ \Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log \left(\sin 2x\right)dx-{\int }_0^{\frac{\pi }{2}}\log 2dx\\ \Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log \left(\sin 2x\right)dx-\frac{\pi }{2}\log 2\\ let{I}_1={\int }_0^{\frac{\pi }{2}}\log \left(\sin 2x\right)dxLett=2x\\ I_1=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\log \sin tdt={\int }_0^{\frac{\pi }{2}}\log \sin tdt=I\\ \Rightarrow 2I=I-\frac{\pi }{2}\log 2\\ \Rightarrow I=\frac{-\pi }{2}\log 2\end{array}$

Hence, this is the answer.