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Question

π20 ex sin x dx= [Roorkee 1978]


A
12(ex21)
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B
12(ex2+1)
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C
12(1ex2)
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D
2(ex2+1)
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Solution

The correct option is B 12(ex2+1)
Let π20 ex sin x dx=
=[ex cos x]π20+π20ex cos x dx
=[ex cos x]π20+[ex sin x]x20π20 ex sin x dx
2I=[ex(sin xcos x)]π20=(eπ2+1)
Hence π20 ex sin x dx=12(eπ2+1)

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