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a
2tan−1tanx/2+12+C
b
tan−1tanx/2+12+C
c
2tan−1tanx/22+C
d
None of these
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Correct option is A
Let I=∫dx2+sinx+cosx⇒ I=∫dx2+2tanx21+tan2x2+1−tan2x21+tan2x2=∫sec2x2dx2+2tan2x2+2tanx2+1−tan2x2I=∫sec2x2dxtan2x2+2tanx2+3Put tanx2=t⇒12sec2x2dx=dt∴ I=∫2dtt2+2t+3=2∫dtt2+2t+1+2=2∫dt(t+1)2+(2)2=2⋅12tan−1t+12+C⇒ I=2tan−1tanx/2+12+CSimilar Questions
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