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Question Number 89161 by M±th+et£s last updated on 15/Apr/20

∫_0 ^(π/2) log(sin(x))dx

0π2log(sin(x))dx

Commented by niroj last updated on 15/Apr/20

  I=∫_0 ^(π/2) log sin x dx.....(i)     = ∫_0 ^(π/2) log sin((π/2)−x) dx   [∵ ∫_0 ^( a) f(c)dx=∫_0 ^a f(a−c)dx]   I  = ∫_0 ^(π/2) log cos x dx.....(ii)   added (i)+(ii)    2I= ∫_0 ^(π/2) (log sin x+log cos x)dx    2I= ∫_0 ^(π/2)  log sin x.cos xdx     2I= ∫_0 ^(π/2)  log( ((2sin xcos x)/2))dx     2I= ∫^(π/2) _0  log sin2x dx−∫_0 ^(π/2) log 2 dx       put 2x= t               2dx=dt                dx=(dt/2)     if x=(π/2) then t=π    if x=0 then t=0         ∫_0 ^π  log sin t.(dt/2)     =(1/2)∫_0 ^( π) log sint dt        = (1/2)×2∫_0 ^(π/2)  log sint dt    ∴  ∫_0 ^(π/2) log sint dt = ∫_0 ^(π/2) log sin x dx=I    2I= I−log 2∫_0 ^(π/2) dx  2I−I = −log 2[ x]_0 ^(π/2)      I =  log 2^(−1) [ (π/2)−0]    I = (π/2) log (1/2) //.

I=0π2logsinxdx.....(i)=0π2logsin(π2x)dx[0af(c)dx=0af(ac)dx]I=0π2logcosxdx.....(ii)added(i)+(ii)2I=0π2(logsinx+logcosx)dx2I=0π2logsinx.cosxdx2I=0π2log(2sinxcosx2)dx2I=0π2logsin2xdx0π2log2dxput2x=t2dx=dtdx=dt2ifx=π2thent=πifx=0thent=00πlogsint.dt2=120πlogsintdt=12×20π2logsintdt0π2logsintdt=0π2logsinxdx=I2I=Ilog20π2dx2II=log2[x]0π2I=log21[π20]I=π2log12//.

Commented by M±th+et£s last updated on 15/Apr/20

thanx for the solutions

thanxforthesolutions

Answered by TANMAY PANACEA. last updated on 15/Apr/20

I=∫_0 ^(π/2) log(sin((π/2)−x))dx  2I=∫_0 ^(π/2) logsinx+logcosx  dx  2I=∫_0 ^(π/2) log(((sin2x)/2))  2I=∫_0 ^(π/2) log(sin2x)dx−∫_0 ^(π/2) log2 dx  2I=∫_0 ^(π/2) log(sin2x)dx−(π/2)log2  now main point...■■  t=2x  ∫_0 ^π logsint×(dt/2)  (1/2)∫_0 ^π logsint dt  (1/2)×2∫_0 ^(π/2) logsint dt=I      ■■  using ∫_0 ^(2a) f(x)dx=2∫_0 ^a f(x)dx  when f(2a−x)=f(x)  here a=(π/2)  so sin(2×(π/2)−x)=sinx  now...  2I=∫_0 ^(π/2) logsin2x dx−(π/2)log2  2I=I−(π/2)log2  I=−(π/2)log2

I=0π2log(sin(π2x))dx2I=0π2logsinx+logcosxdx2I=0π2log(sin2x2)2I=0π2log(sin2x)dx0π2log2dx2I=0π2log(sin2x)dxπ2log2nowmainpoint...t=2x0πlogsint×dt2120πlogsintdt12×20π2logsintdt=Iusing02af(x)dx=20af(x)dxwhenf(2ax)=f(x)herea=π2sosin(2×π2x)=sinxnow...2I=0π2logsin2xdxπ2log22I=Iπ2log2I=π2log2

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