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a
πsinα
b
π2cosα
c
πcosα
d
π2sinα
answer is D.
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Detailed Solution
Putting tanx2=t the given integral reduces to=2∫0tanα2 dt(1−cosα)+t2(1+cosα)=1cos2α2∫0tanα/2 dtt2+tan2α2=1cos2α2tanα2tan−1ttanα20tanα/2=1sinα2cosα2⋅π4=π2sinα