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Question

If I=ππsin2x1+axdx,a>0 then I equals

A
π
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B
π2
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C
aπ
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D
aπ2
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Solution

The correct option is A π2
I=ππsin2x1+axdx (i)
I=ππ(sin(x))21+axdx
=ππaxsin2x1+axdx ...(ii)
Adding (i) and (ii), we get
2I=ππsin2xdx
=2π0sin2xdx
=π0(1cos2x)dx
=(xsin2x2)π0=π
I=π2

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