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Question

For a>0 the value of ππsin2x1+axdx

A
π2
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B
π
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C
aπ2
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D
aπ
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Solution

The correct option is A π2
Iππsin2x1+axdx....(1)
I=ππsin2(x)(1+ax)=ππaxsin2x1+axdx.....(2)
Adding (1) and (2)
2I=ππsin2dx
2I=2π0sin2xdx
=2π0[1cos2x2]dx
=π0(1cos2x)dx
=[xsin2x2]π0
=[π]

Thus I=π2

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