0
$\begingroup$

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!

  • 0
    @did can I ask u a little question?2012-10-25

2 Answers 2

4

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 $I=-\frac12\pi\log2$.

  • 0
    Great. You are welcome.2012-10-27
0

I think first you have to use integration by parts... This is a hint.

  • 3
    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