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a
tanx+tan5/2x5+C
b
tanx+25tan5/2x+C
c
2tanx+25tan5/2x+C
d
None of the above
answer is A.
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Detailed Solution
Let I=∫dxcos3x2sin2x=∫dxcos3x4sinxcosx=12∫dxcos7/2xsin1/2x=12∫sec4xtanxdx=12∫1+tan2xsec2xtanxdx Put tanx=t⇒sec2xdx=dt∴ I=12∫1+t2tdt =12∫t−1/2dt+12∫t3/2dt =t1/2+t5/25+C =tanx+15tan5/2x+C