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Question

π2π4 ex (log sin x+cot x)dx=

A
eπ4log 2
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B
eπ4log 2
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C
12eπ4 log 2
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D
12eπ4 log 2
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Solution

The correct option is C 12eπ4 log 2
Let I=π2π4 ex (log sin x+cot x)dx
I=π2π4 ex log sin x dx+π2π4 ex cot x dx
=π2π4 ex log sin xdx+[ex log sin x]π2π4
π2π4 ex log sin x dx
=eπ2 log sin π2eπ4 log sin π4=12eπ4log 2

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