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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,

$$\int_0^{\pi/2}\left[\log(\sin 2x)dx\right] = \int_0^\pi\left[\frac{1}{2}\log(\sin t) \right]dt=\int_0^{\pi/2}\left[\log(\sin t) \right]dt=\int_0^{\pi/2}\log(\sin t)dt = I$$

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  • $\begingroup$ $f(x)=f(2a-x)$ then the integral is simply equivalent to "?" where $2a$ is upper limit. Check this property out, This definite integral is indeed the most over-exploited integral in indian syllabus. $\endgroup$ – Mann May 7 '15 at 12:01
  • $\begingroup$ On the first row, when $x=\pi/2$, $t=\pi$; then on the second row "fold" with the symmetry $\sin(t)=\sin(\pi-t)$. $\endgroup$ – Yves Daoust May 7 '15 at 12:40
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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.

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He set $2x=t$.

When $x=0,t=0$

When $x=\frac{\pi}{2}, t=\pi$

As for the second change, it is due to the fact that

$\sin(\pi-x)=\sin(x)$

In other words, $\sin(t)$ where $t$ ranges over $(0;\pi)$ is the same as $t$ ranging over $2\sin(t)$ where $t$ ranges over $(0;\frac{\pi}{2})$

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  • $\begingroup$ Sorry, I meant in the second line of the image when the limits go back to 0 and pi/2 but the t remains and the 1/2 disappears. I'm relatively new to this process... $\endgroup$ – Bob May 7 '15 at 12:02
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in property $6$ of tbk, after applying upper limit changes from $2a$ to a which implies $2a=\pi$. After applying the property $6, a = \frac\pi2.$

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