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bharat bhushan sharma Grade: Upto college level
```        can any body explain full solution step wise of

0 to "/2 integration of log sin x dx

```
7 years ago

19 Points
```										Dear Bharat,
I=∫ log(sin x).dx         .............eq 1I=∫logsin(π/2-x)dxI=∫logcosx dx            ..................eq2adding equation 1&22I=∫logsinx+logcosx dx=∫logsinx*cosx dx=∫log(sin2x)/2 dx=∫[logsin2x-log2] dx=∫logsin2xdx - ∫log2dx .................eq 3
therefore 2I=A-B
where,A=∫logsin2xdxand B=∫log2dxwith limits from 0 to π/2Solving B,we get B=log2[x],with limits from 0 to π/2or B=π/2log2
in the first integral A put 2x=td.w.t.xdx=dt/2where x=0,t=0 where x=π/2 ,t=π∫logsint dt with limit t= o & t=πnow2I=1/2∫logsint dt-BNow applying property

Here F(2a-x)=F(x)
Therefore 2I=1/2.2∫logsint dt-B with limits from 0 to π/22I=I-B
I=-B=-π/2log2
or I=π/2log(1/2).
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Bharat !!!

Regards,
MOHIT

```
7 years ago
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