can any body explain full solution step wise of0 to \"/2 integration of log sin x dx

Dear Bharat,

I=∫ log(sin x).dx         .............eq 1
I=∫logsin(π/2-x)dx
I=∫logcosx dx            ..................eq2
adding equation 1&2
2I=∫logsinx+logcosx dx
=∫logsinx*cosx dx
=∫log(sin2x)/2 dx
=∫[logsin2x-log2] dx
=∫logsin2xdx - ∫log2dx .................eq 3

therefore 2I=A-B

where,A=∫logsin2xdx
and B=∫log2dx
with limits from 0 to π/2
Solving B,we get B=log2[x],with limits from 0 to π/2
or B=π/2log2

in the first integral A put 2x=t
d.w.t.x
dx=dt/2
where x=0,t=0 where x=π/2 ,t=π
∫logsint dt with limit t= o & t=π
now
2I=1/2∫logsint dt-B
Now applying property

Here F(2a-x)=F(x)

Therefore 2I=1/2.2∫logsint dt-B with limits from 0 to π/2
2I=I-B

I=-B=-π/2log2

or I=π/2log(1/2).

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