prove that pie/2 ingrete0 logsinxdx=-pie/2log2

Here, we need to evaluate ${\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin x dx$.
$I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin xdx ...\left(1\right)$
First, use the property ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx$. So,
$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin \left(\frac{\mathrm{\pi }}{2}-x\right)dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos xdx ...\left(2\right) \end{gathered}$$

Adding (1) and (2), we get
$2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin x+\log \cos xdx$
Next, add and subtract log 2 from the above integral. So, we get
$$\begin{gathered} 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\log \sin x+\log \cos x+\log 2\right)-\log 2dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\log 2\sin x\cos x\right)-\log 2dx ( \mathrm{Using} \log x+\log y=\log xy\mathit{)} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin 2x-\log 2dx ( \mathrm{Using} \sin 2x=2\sin x\cos x) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin 2x dx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2 dx \end{gathered}$$

Further, use 2x = t, then $2dx=dt$. So, when $x=0, t=0 \mathrm{and} \mathrm{when} x=\frac{\mathrm{\pi }}{2}, t=\mathrm{\pi }$
Thus,
$$\begin{gathered} 2I ={\int }_0^{\mathrm{\pi }}\log \sin t \frac{dt}{2}-\frac{\mathrm{\pi }}{2}\log 2 \\ =\frac{1}{2}{\int }_0^{\mathrm{\pi }}\log \sin t dt-\frac{\mathrm{\pi }}{2}\log 2 \end{gathered}$$

Next, use the property ${\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx \mathrm{as} \sin \left(\mathrm{\pi }-t\right)=\sin t$
$2I =\frac{2}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin t dt-\frac{\mathrm{\pi }}{2}\log 2$
Further, changing the variable t to x. We get
$$\begin{gathered} 2I ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin x dx-\frac{\mathrm{\pi }}{2}\log 2 \\ =I-\frac{\mathrm{\pi }}{2}\log 2 (U\sin g (1\left)\right) \\ 2I-I=-\frac{\mathrm{\pi }}{2}\log 2 \\ I=-\frac{\mathrm{\pi }}{2}\log 2 \\ {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin x dx=-\frac{\mathrm{\pi }}{2}\log 2 \end{gathered}$$
Hence proved.