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Question

The integral π2π4ex(ln(cosx))(cosxsinx1) cosec2x dx is equal to

A
eπ/42[2eπ/4+ln22]
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B
eπ/42[eπ/4+ln22]
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C
eπ/42[eπ/4ln2+2]
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D
eπ/42[2eπ/4ln2+2]
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Solution

The correct option is A eπ/42[2eπ/4+ln22]
π2π4ex(ln(cosx))(cosxsinx1) cosec2x dx
Let
I=(ln(cosx))ex(cotx cosec2x) dx
We know that,
ex(f(x)+f(x))=exf(x)+cex(cotx cosec2x) dx=excotx+c
And using by parts, we get
I=ln(cosx)excotx+tanxexcotx dxI=ln(cosx)excotx+ex dx+cI=ex[ln(cosx)cotx+1]+c
So,
π2π4ex(ln(cosx))(cosxsinx1) cosec2x dx=(ex[ln(cosx)cotx+1])π/2π/4=eπ/2[limxπ/2ln(cosx)cotx]+eπ/2eπ/4[1ln2]
Now, solving the limit we get
limxπ/2ln(cosx)cotx=limxπ/2ln(cosx)tanx
Using L'Hospital's Rule, we get
=limxπ/2sinxcosxsec2x=limxπ/2sinxcosx=0

Therefore,
eπ/2[limxπ/2ln(cosx)cotx]+eπ/2eπ/4[1ln2]=eπ/2eπ/4[1ln2]=eπ/42[2eπ/4+ln22]

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