Integral $\int^{ \pi /2}_{0} \ln (\sin x)\ dx$ $$\int^{ \pi /2}_{0} \ln (\sin x)\  dx$$
The answer is $- \frac{\pi}{2} \ln 2$.
I have changed it into $$\frac{1}{2} \int^{1}_{0} t d \ln t^{2}$$
But I didn't get the answer with it.
 A: Hint: Try
\begin{align}\int^{ \pi /2}_{0} \ln \sin x \, dx &= \int^{ \pi /2}_{0} \ln \left(2 \sin \frac{x}{2} \cos \frac{x}{2} \right) \, dx \\ 
&=\int^{ \pi /2}_{0} \ln 2 \, dx + \int^{ \pi /2}_{0}\ln \left( \sin \frac{x}{2} \right) \, dx + \int^{ \pi /2}_{0} \ln \left( \cos \frac{x}{2} \right) \, dx \\
&= \frac{\pi}{2} \ln 2 + \underbrace{2 \int^{ \pi /4}_{0} \ln (\sin u) \, du}_{\text{Let }u=x/2} + \underbrace{2 \int^{ \pi /4}_{0} \ln (\cos u) \, du}_{\text{Let }u=x/2}
\end{align} 
A: Let $u=\ln\sin x$, $dv=1$. Then:
\begin{eqnarray}
\underline{u}&\underline{v}\\
\ln\sin x & 1\\
\searrow^+\\
\tan x& \xleftarrow{-} x
\end{eqnarray}
Thus,
$$\int^{\pi/2}_0 \ln\sin xdx=\left.x\ln\sin x\right|^{\pi/2}_0-\int^{\pi/2}_0 x\tan xdx$$
$\int x\tan xdx$ is not expressible by elementary functions, so for your purposes, @Glacier's answer should work.
A: My Solution:: Given $\displaystyle \int_{0}^{\frac{\pi}{2}}\ln (\sin x)dx$
Now Let $\displaystyle \sin x = \left(\frac{e^{ix}-e^{-ix}}{2i}\right) = \frac{1}{2i}\cdot \left(\frac{e^{2ix}-1}{e^{ix}}\right)$
So Integral Convert into $$\displaystyle \int_{0}^{\frac{\pi}{2}}\ln \left(\frac{e^{2ix}-1}{2i\cdot e^{ix}}\right)dx = \int_{0}^{\frac{\pi}{2}}\ln(e^{2ix}-1)dx-\int_{0}^{\frac{\pi}{2}}\ln(e^{ix})dx-\int_{0}^{\frac{\pi}{2}}\ln(2)dx-\int_{0}^{\frac{\pi}{2}}\ln(i)dx$$
$$\displaystyle = \int_{0}^{\frac{\pi}{2}}\ln(e^{2ix}-1)dx-i\cdot \frac{\pi^2}{8}-\frac{\pi}{2}\cdot \ln(2)-i\cdot \frac{\pi^2}{4}$$
Now we will Solve $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln(e^{2ix}-1)dx$$
Using $\displaystyle e^{2ix}-1=\cos(2x)+i\sin(2x)-1 = i\cdot 2\sin x\cdot \cos x-2\sin^2 (x) = 2i\sin x\cdot (\cos x+i\sin x) = 2i\sin x\cdot e^{ix}$
So $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln(2i\cdot \sin x)dx+\int_{0}^{\frac{\pi}{2}}\ln(e^{ix})dx$$
So $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln(2i\cdot \sin x)dx+i\cdot \frac{\pi^2}{8}$$
So $$\displaystyle \int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx = -\frac{\pi}{2}\cdot \ln(2)+\bf{Imaginary\; quantity.}$$
But we have calculate only for real values.
So $$\displaystyle \int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx = -\frac{\pi}{2}\cdot \ln(2)$$
