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$$\int\limits_0^{\frac \pi 2}\log \sin x dx$$

After applying ILATE integration did not solve. Please give hint

user270371
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  • The answer is $\frac{-\pi*ln2}{2}$ That's the hint... – imranfat Sep 20 '15 at 19:07
  • Have a look at this: http://math.stackexchange.com/questions/690644/evaluate-int-0-pi-2-log-cosxdx/690693#690693 – Olivier Oloa Sep 20 '15 at 19:08
  • Of course Olivier and Chappers. This is an well known integral. It has been on stack exchange and other websites many times. If even someone simple like ME knows the answer, that says a lot...The user just doesn't want to do the research, let alone the work... – imranfat Sep 20 '15 at 19:09

1 Answers1

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$$I=\int\limits_0^{\frac \pi 2} \log \sin x dx\\\implies I=\int\limits_0^{\frac \pi 2} \log \cos x dx$$

Now, adding these two, we have, $$2I=\int\limits_0^{\frac \pi 2}\log \sin 2x dx -\int\limits_0^{\frac \pi 2}\log 2 dx$$.

Now, solve $$I'=\int\limits_0^{\frac \pi 2}\log \sin 2x dx$$.

HINT: put $2x=z$, and change the corresponding limits,It will look like $$I'=\frac 12\int\limits_0^{\pi}\log \sin z dz \implies \frac 12\int\limits_0^{2\frac {\pi}2}\log \sin z dz \implies \int\limits_0^{\frac \pi 2}\log \sin z dz \\\implies \int\limits_0^{\frac \pi 2}\log \sin x dx=I$$.

Now,can you do this?