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0-pi-2-log-sin-d-




Question Number 313 by Vishal Bhardwaj last updated on 25/Jan/15
∫_0 ^(π/2) log sinθ dθ
0π2logsinθdθ
Answered by prakash jain last updated on 20/Dec/14
I=∫_0 ^(π/2) ln sin θdθ        .....(i)  put θ=(π/2)−α, dθ=−dα  I=∫_(π/2) ^0 ln sin ((π/2)−α)(−dα)  =−∫_(π/2) ^0 ln cos αdα=∫_0 ^(π/2) ln cos αdα  =∫_0 ^(π/2) ln cos θdθ       ....(ii)  add (i) and (ii)  2I=∫_0 ^(π/2) (ln sin θ+ln cos θ)dθ  2I=∫_0 ^(π/2) ln ((sin 2θ)/2) dθ  2I=∫_0 ^(π/2) (ln sin 2θ −ln 2) dθ        2I=∫_0 ^(π/2) ln sin 2θ dθ−∫_0 ^(π/2) ln 2 dθ        In first integral put 2θ=β ,2dθ=dβ so  the equation one now with updated limits  2I=∫_0 ^π ln sin β(dβ/2)−∫_0 ^(π/2) ln 2 dθ    2I=(1/2)∫_0 ^π ln sin β dβ−∫_0 ^(π/2) ln 2 dθ  ....(iii)  Now we will evaluate first integral  ∫_0 ^π ln sin β dβ=∫_0 ^(π/2) ln sin β dβ+∫_(π/2) ^π ln sin β dβ  ∫_0 ^π ln sin β dβ==I+∫_(π/2) ^π ln sin βdβ  put β=π−t dβ=−dt  ∫_0 ^π ln sin β dβ==I+∫_(π/2) ^0 ln sin (π−t)(−dt)  ∫_0 ^π ln sin β dβ==I−∫_(π/2) ^0 ln sin tdt  ∫_0 ^π ln sin β dβ==I+∫_0 ^(π/2) ln sin t dt  ∫_0 ^π ln sin β dβ==I+I=2I     ....(iv)  putting this value in (iii)  2I=(1/2)(2I)−∫_0 ^(π/2) ln 2dθ  2I=I−(π/2)ln 2  I=−(π/2)ln 2
I=0π/2lnsinθdθ..(i)putθ=π2α,dθ=dαI=π/20lnsin(π2α)(dα)=π/20lncosαdα=0π/2lncosαdα=0π/2lncosθdθ.(ii)add(i)and(ii)2I=0π/2(lnsinθ+lncosθ)dθ2I=0π/2lnsin2θ2dθ2I=0π/2(lnsin2θln2)dθ2I=0π/2lnsin2θdθ0π/2ln2dθInfirstintegralput2θ=β,2dθ=dβsotheequationonenowwithupdatedlimits2I=0πlnsinβdβ20π/2ln2dθ2I=120πlnsinβdβ0π/2ln2dθ.(iii)Nowwewillevaluatefirstintegral0πlnsinβdβ=0π/2lnsinβdβ+π/2πlnsinβdβ0πlnsinβdβ==I+π/2πlnsinβdβputβ=πtdβ=dt0πlnsinβdβ==I+π/20lnsin(πt)(dt)0πlnsinβdβ==Iπ/20lnsintdt0πlnsinβdβ==I+0π/2lnsintdt0πlnsinβdβ==I+I=2I.(iv)puttingthisvaluein(iii)2I=12(2I)0π/2ln2dθ2I=Iπ2ln2I=π2ln2