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dxcos3x2sin2x is equal to 

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a
tan⁡x+tan5/2⁡x5+C
b
tan⁡x+25tan5/2⁡x+C
c
2tan⁡x+25tan5/2⁡x+C
d
None of the above

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detailed solution

Correct option is A

Let I=∫dxcos3⁡x2sin⁡2x=∫dxcos3⁡x4sin⁡xcos⁡x=12∫dxcos7/2⁡xsin1/2⁡x=12∫sec4⁡xtan⁡xdx=12∫1+tan2⁡xsec2⁡xtan⁡xdx Put tan⁡x=t⇒sec2⁡xdx=dt∴ I=12∫1+t2tdt =12∫t−1/2dt+12∫t3/2dt =t1/2+t5/25+C =tan⁡x+15tan5/2⁡x+C


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