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I do not have much experience evaluating improper integrals and I hope someone will please demonstrate how to evaluate this:

$$\int_{0}^{\frac{\pi}{2}} \log \sin x dx$$

Thanks in advance! P.S. :I had accidentally put the word indefinite instead of improper.Sorry for the error!

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    This is a *definite* integral.2012-10-25
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    WolframAlpha suggests a numerical approximation: http://bit.ly/ThKVai2012-10-25
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    Very similar to: http://www.goiit.com/posts/list/integral-calculus-integrate-log-sin-x-from-limit-0-to-pi-4-1009649.htm2012-10-25
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    @Emmad ...Where already the first line is wrong since over there the integral is from 0 to pi/4, not to pi/2. :-)2012-10-25
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    I am sorry.I meant an improper integral.2012-10-25
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    @did can I ask u a little question?2012-10-25

2 Answers 2

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Hint: (...Since this has been answered several times on the site.) Let $I$ denote the integral to be evaluated, then $$ 2\cdot I=\int_0^{\pi/2}\log\sin x\,\mathrm dx+\int_0^{\pi/2}\log\cos x\,\mathrm dx=\int_0^{\pi/2}\log\left(\tfrac12\sin(2x)\right)\,\mathrm dx=\ldots$$ To check your solution: In the end, you should reach the value $I=-\frac12\pi\log2$.

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    +1, good hint, specially that the limits are accurate :)2012-10-25
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    @did, thanks. My text had used some intimidating language about this problem calling it an improper integral. So, i thought it better to post it here.(It was referring to "improper integrals treated as definite integrals".Can you please tell me what it was trying to communicate?)2012-10-27
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    Try the explanation [here](http://en.wikipedia.org/wiki/Improper_integral#Examples).2012-10-27
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    @did, thanks a lot.I now get it.2012-10-27
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    Great. You are welcome.2012-10-27
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I think first you have to use integration by parts... This is a hint.

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    Not a very useful hint --- integration by parts gets you to $x\cot x$, and how do you propose to integrate that?2012-10-25