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Question

Integrate 2π01+sinxdx

A
4
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B
6
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C
8
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D
None of these
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Solution

The correct option is A 4
2π01+sinxdx
=2π0sin2x2+cos2x2+2sinx2cosx2dx (sin2x+cos2x=1 and sin2x=2sinxcosx)
=2π0(sinx2+cosx2)2dx
=2π0(sinx2+cosx2)dx
=[2cosx2+2sinx2]2π0
=2[sinx2cosx2]2π0
=2[1+1]
=4

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