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2016-08-30T15:28:48+05:30

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The indefinite Integral of Log (Sin x) dx has no simple function form.  That can only be expressed in terms of complex functions.

It can be found by Taylor series expansion or Mclaurin series.

But definite integral from 0 to pi/2 can be found  to be  - pi/2 * Ln 2..

Repetitively\ Apply\ \int{u}\ dv=uv-\int {v}\ du\\\\\int Log\ sinx\ dx=x\ LogSinx-\int {x\ cotx}\ dx\\\\=x\ logSinx-x^2/2\ Cotx+1/2 \int\ x^2\ cosec^2x\ dx\\\\=x\ logSinx-x^2/2\ Cotx+x^3/6\ cosec^2x+1/3\ \int {x^3 cosec^2x\ cotx}\ dx\\\\x\ logSinx-x^2/2\ Cotx+x^3/6\ cosec^2x+x^4/12\ cosec^2x\ cotx+.....

If we want to evaluate the definite integral then:

I=\int \limits_0^{\pi/2} {LogSinx}\ dx\\\\=\int \limits_{\pi/2}^0 {LogSin(\frac{\pi}{2}-x})\ d(\frac{\pi}{2}-x)\\\\So\ I=\int \limits_0^{\pi/2} {LogCosx}\ dx\\\\I+I=\int \limits_0^{\pi/2} {Log\frac{Sin2x}{2}}\ dx\\\\=\frac{1}{2} \int \limits_{2x=0}^{\pi}LogSin2x\ d(2x)-\int \limits_0^{\pi/2}Log2\ dx\\\\=\frac{1}{2} [\int \limits_0^{\pi/2} {LogSinx}\ dx+\int \limits_{\pi/2}^{\pi} {LogSinx}\ dx]-\pi/2*Log2

2I=\frac{1}{2}[I+I]-\pi/2*Log2\\\\I=-\frac{\pi}{2}Log2
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