Evaluate ∫0π xlog⁡sin⁡xdx

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12π2log⁡(1/2)

2π2log⁡(1/2)

π2log⁡(1/2)

π2log⁡(2)

answer is A.

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Detailed Solution

Let I=∫0π xlog⁡sin⁡xdx--------(1)I=∫0π (π−x)log⁡sin⁡(π−x)dx=∫0π (π−x)log⁡sin⁡xdx---------(2) Adding equations (1) and (2), we get 2I=π∫0π log⁡sin⁡xdx or 2I=2π∫0π/2 log⁡sin⁡xdx I=π∫0π/2 log⁡sin⁡xdx ----(3) I=π∫0π/2 log⁡sin(π2-⁡x)dx =π∫0π/2 log⁡cos⁡xdx ----(4)(3) + (4)2I=π∫0π/2 log⁡sin⁡x+logcosxdx (take logsinxcosx=log2sinxcosx2)=π∫0π/2 log⁡sin⁡2xdx-π∫0π/2 log⁡2dxput 2x=t, dx=dt2 UL=π, LL=02I=π2∫0π log⁡sin⁡tdt-πlog2π22I=π∫0π2 log⁡sin⁡tdt-π22log22I=I-π22log2I=-π22log2=π22log⁡(1/2)

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