Let us assume
I = \(\int\limits_{0}^{\pi}
\)x log sin x dx...equation 1
By property, we know that
![](https://www.sarthaks.com/?qa=blob&qa_blobid=12507983070485431569)
![](https://www.sarthaks.com/?qa=blob&qa_blobid=15761565707115237338)
Adding equation 1 and equation 2
![](https://www.sarthaks.com/?qa=blob&qa_blobid=10772786954178374549)
![](https://www.sarthaks.com/?qa=blob&qa_blobid=11851911740091740611)
= \(\int\limits_{0}^{a}
\)f(x)dx if f(2a - x) = f(x)
= 0 if f(2a - x)= -f(x)
Thus equation 3 becomes
![](https://www.sarthaks.com/?qa=blob&qa_blobid=872999117276729270)
since log sin(π – x) = log sinx
By property, we know that
![](https://www.sarthaks.com/?qa=blob&qa_blobid=10982364642752783264)
![](https://www.sarthaks.com/?qa=blob&qa_blobid=9102630125646474344)
Adding equation 4 and equation 5
![](https://www.sarthaks.com/?qa=blob&qa_blobid=2550634522269057629)
We know log m + log n = log m n
thus
![](https://www.sarthaks.com/?qa=blob&qa_blobid=5095086201793666442)
since log(m/n) = log m – log n
![](https://www.sarthaks.com/?qa=blob&qa_blobid=2324092489715241957)
Let I1 = π\(\int\limits_{0}^{\pi/2}
\)log sin 2x dx
Let 2x = y
2dx = dy
dx = dy/2
For x = 0
y = 0
for x = \(\cfrac{\pi}2\)
y = π
thus substituting value in I1
![](https://www.sarthaks.com/?qa=blob&qa_blobid=12814469174962279202)
Thus substituting the value of I1 in equation 6
![](https://www.sarthaks.com/?qa=blob&qa_blobid=4080023135851162165)