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sin3xdxcos4x+3cos2x+1tan1secx+cosx

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a
logTan−1Secx+Cosx+C
b
logCot−1Cosex+Cotx+C
c
logTan−1Secx−Tanx+C
d
logCot−1Cosex−Cotx+C

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detailed solution

Correct option is A

I=∫sin3⁡xdxcos4⁡x+3cos2⁡x+1tan−1⁡(sec⁡x+cos⁡x) Put  tan−1⁡(sec⁡x+cos⁡x)=t or  11+(sec⁡x+cos⁡x)2(sec⁡xtan⁡x−sin⁡x)dx=dt or cos2⁡xcos2⁡x+1+cos2⁡x2sin⁡x1cos2⁡x−1dx=dtor   cos2⁡xsin⁡x1−cos2⁡xcos2⁡xcos2⁡x+1+cos4⁡x+2cos2⁡xdx=dtor   sin3⁡xcos4⁡x+3+cos2⁡x+1dx=dtI=∫dtt=log⁡|t|+c =log⁡tan−1⁡(sec⁡x+cos⁡x)+c


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