prove that pie/2 ingrete0 logsinxdx=-pie/2log2

Here, we need to evaluate 0π2log sinx dx.
I=0π2logsinxdx                        ...(1)
First, use the property 0afxdx=0afa-xdx. So,
I=0π2logsinπ2-xdx =0π2logcosxdx                     ...(2)

Adding (1) and (2), we get
Next, add and subtract log 2 from the above integral. So, we get
2I=0π2logsinx+logcosx+log2-log2dx  =0π2log2sinxcosx-log2dx                     ( Using log x+log y=log xy)   =0π2logsin2x-log2dx                               ( Using sin2x=2sinxcosx)  =0π2logsin2x dx-0π2log2 dx

Further, use 2x = t, then 2dx=dt. So, when x=0, t=0 and when x=π2, t=π
2I  =0πlogsint dt2-π2log2    =120πlogsint dt-π2log2

Next, use the property 02afxdx=20afxdx as sinπ-t=sint
2I  =220π2logsint dt-π2log2
Further, changing the variable t to x. We get
                  2I  =0π2logsinx dx-π2log2                       =I-π2log2                                          (Using (1))               2I-I=-π2log2                     I=-π2log20π2logsinx dx=-π2log2
Hence proved.

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