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Question Number 89161 by M±th+et£s last updated on 15/Apr/20
∫0π2log(sin(x))dx
Commented by niroj last updated on 15/Apr/20
I=∫0π2logsinxdx.....(i)=∫0π2logsin(π2−x)dx[∵∫0af(c)dx=∫0af(a−c)dx]I=∫0π2logcosxdx.....(ii)added(i)+(ii)2I=∫0π2(logsinx+logcosx)dx2I=∫0π2logsinx.cosxdx2I=∫0π2log(2sinxcosx2)dx2I=∫0π2logsin2xdx−∫0π2log2dxput2x=t2dx=dtdx=dt2ifx=π2thent=πifx=0thent=0∫0πlogsint.dt2=12∫0πlogsintdt=12×2∫0π2logsintdt∴∫0π2logsintdt=∫0π2logsinxdx=I2I=I−log2∫0π2dx2I−I=−log2[x]0π2I=log2−1[π2−0]I=π2log12//.
Commented by M±th+et£s last updated on 15/Apr/20
thanxforthesolutions
Answered by TANMAY PANACEA. last updated on 15/Apr/20
I=∫0π2log(sin(π2−x))dx2I=∫0π2logsinx+logcosxdx2I=∫0π2log(sin2x2)2I=∫0π2log(sin2x)dx−∫0π2log2dx2I=∫0π2log(sin2x)dx−π2log2nowmainpoint...◼◼t=2x∫0πlogsint×dt212∫0πlogsintdt12×2∫0π2logsintdt=I◼◼using∫02af(x)dx=2∫0af(x)dxwhenf(2a−x)=f(x)herea=π2sosin(2×π2−x)=sinxnow...2I=∫0π2logsin2xdx−π2log22I=I−π2log2I=−π2log2
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