Integration of $\log(\sin x)$ In trying to integrate $\log(\sin x)$ and I ended up looking for a solution and found the one at this link: http://www.meritnation.com/ask-answer/question/how-to-integrate-f-log-sin-x-dx/math/766517

However, there is one step in the process I don't understand. How do the limits from $0$ to $\pi$ change to be $0$ to $\pi/2$ in the bottom part of the image and the $t$ remains but the $1/2$ disappears? I'm relatively new to this process...


  From equation (3) Now,
  
  $$\int_0^{\pi/2}\left[\log(\sin 2x)dx\right] = \int_0^\pi\left[\frac{1}{2}\log(\sin t) \right]dt=\int_0^{\pi/2}\left[\log(\sin t) \right]dt=\int_0^{\pi/2}\log(\sin t)dt = I$$
Here's a quick proof i could come up with:

$2I=I'=\int^{\pi/2}_{0} \ln(\sin t) dt+\int_{\pi/2}^{\pi} \ln(\sin t) dt $

Consider second integral,

Put $y=-\frac{\pi}{2}+t$

$I_2=\int^{\pi/2}_{0}\ln (\sin \left(y+\frac{\pi}{2}\right)) dy$

$I_2=\int^{\pi/2}_{0}\ln (\sin \left(t+\frac{\pi}{2}\right)) dt$

$I_2=\int^{\pi/2}_{0}\ln \sin (t) dt$

This is applicable to all possible function. Depending on whether odd or even function.
in property $6$ of tbk, after applying upper limit changes from $2a$ to a which implies $2a=\pi$. After applying the property $6, a = \frac\pi2.$