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B
12(ex2+1)
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C
12(1−ex2)
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D
2(ex2+1)
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Solution
The correct option is B12(ex2+1) Let∫π20exsinxdx= =−[excosx]π20+∫π20excosxdx =−[excosx]π20+[exsinx]x20−∫π20exsinxdx ∴2I=[ex(sinx−cosx)]π20=(eπ2+1) Hence∫π20exsinxdx=12(eπ2+1)