Question

# The integral π2∫π4ex(ln(cosx))(cosxsinx−1) 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+ln2−2]π2∫π4ex(ln(cosx))(cosxsinx−1) cosec2x dx Let I=∫(ln(cosx))⋅ex(cotx− cosec2x) dx We know that, ∫ex(f(x)+f′(x))=exf(x)+c⇒∫ex(cotx− cosec2x) dx=excotx+c And using by parts, we get ⇒I=ln(cosx)excotx+∫tanx⋅ex⋅cotx dx⇒I=ln(cosx)excotx+∫ex dx+c⇒I=ex[ln(cosx)⋅cotx+1]+c So, π2∫π4ex(ln(cosx))(cosxsinx−1) cosec2x dx=(ex[ln(cosx)⋅cotx+1])π/2π/4=eπ/2[limx→π/2ln(cosx)⋅cotx]+eπ/2−eπ/4[1−ln√2] Now, solving the limit we get limx→π/2ln(cosx)⋅cotx=limx→π/2ln(cosx)tanx Using L'Hospital's Rule, we get =limx→π/2−sinxcosx⋅sec2x=limx→π/2−sinxcosx=0 Therefore, eπ/2[limx→π/2ln(cosx)⋅cotx]+eπ/2−eπ/4[1−ln√2]=eπ/2−eπ/4[1−ln√2]=eπ/42[2eπ/4+ln2−2]

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