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a
π4
b
π3
c
π2
d
π
answer is A.
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Detailed Solution
Let t=sin2u in the first integral and t=cos2u in the second integral. ThenI1+I2=∫0xusin2udu-∫π/2xusin2udu ⇒I1+I2=∫0xusin2udu+∫xπ/2usin2udu=∫0π/2usin2udu =-u2cos2u+14sin2u0π2 = π4