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B
−eπ4log2
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C
12eπ4log2
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D
−12eπ4log2
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Solution
The correct option is C12eπ4log2 LetI=∫π2π4ex(logsinx+cotx)dx I=∫π2π4exlogsinxdx+∫π2π4excotxdx =∫π2π4exlogsinxdx+[exlogsinx]π2π4 −∫π2π4exlogsinxdx =eπ2logsinπ2−eπ4logsinπ4=12eπ4log2