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B
π2
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C
0
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D
π4
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Solution
The correct option is Bπ2 We have, I=∫π0sin2xdx Using trigonometric identity sin2x=1−cos2x2 I=∫π0[1−cos2x]2dx =12[∫π0dx−∫π0cos2xdx] =12[x−sin2x2]π0 =12[(π−0)−(sin2π−sin0)2] ∴I=π2