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Correct option is A
Let I=∫0π xlogsinxdx--------(1)I=∫0π (π−x)logsin(π−x)dx=∫0π (π−x)logsinxdx---------(2) Adding equations (1) and (2), we get 2I=π∫0π logsinxdx or 2I=2π∫0π/2 logsinxdx I=π∫0π/2 logsinxdx ----(3) I=π∫0π/2 logsin(π2-x)dx =π∫0π/2 logcosxdx ----(4)(3) + (4)2I=π∫0π/2 logsinx+logcosxdx (take logsinxcosx=log2sinxcosx2)=π∫0π/2 logsin2xdx-π∫0π/2 log2dxput 2x=t, dx=dt2 UL=π, LL=02I=π2∫0π logsintdt-πlog2π22I=π∫0π2 logsintdt-π22log22I=I-π22log2I=-π22log2=π22log(1/2)Similar Questions
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