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Question

Evaluate the integral
π20(cosxsinx)exdx

A
0
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B
1
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C
1
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D
2
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Solution

The correct option is C 1

π20(cosxsinx)exdx
=ex(cosxsinx)dx

It is in the form of
ex(f(x)+f1(x))dx=exf(x)+c

Thus ex(cosxsinx)dx=excosx+c
π20ex(cosxsinx)dx=(excosx+c)π/20
=(eπ2cosπ2+c)(e0cos0+c)
=1


Thus π20ex(cosxsinx)dx=1


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