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Question

If I=ππsin2x1+axdx for a>0,


then I is equal to?

A
π
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B
π/2
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C
aπ
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D
aπ/2
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Solution

The correct option is B π/2
I=ππsin2x1+axdx.......................(1)
I=ππ(sin(x))21+axdx

I=ππaxsin2x1+axdx...................(2)

Adding (1) and (2), we get
2I=ππsin2xdx
=2π0sin2xdx
=π0(1cos2x)dx
=(xsin2x2)π0=π

I=π/2

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