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B
π4
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C
0
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D
π
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Solution
The correct option is Bπ4 Let I=π2∫0cos2xdx
we know that, cos2x=2cos2x−1 ⇒cos2x=1+cos2x2 ⇒I=π/2∫0(1+cos2x2)dx=12π2∫0dx+12π2∫0cos2xdx =12[x]π20+12[sin2x2]π20=12(π2−0)+14(0−0)=π4