TITLE: Integral $\int^{ \pi /2}_{0} \ln (\sin x)\ dx$
QUESTION [5 upvotes]: $$\int^{ \pi /2}_{0} \ln (\sin x)\  dx$$
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.

REPLY [3 votes]: 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 $$\displaystyle \int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx = -\frac{\pi}{2}\cdot \ln(2)$$