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Q.

secx1dx

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a

logCosx+12+Cos2x+Cosx+C

b

logCosx+12+Cos2x+Cosx+C

c

logCosx12+Cos2x+Cosx+C

d

logCosx+12Cos2x+Cosx+C

answer is A.

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Detailed Solution

I=secx1dx=1cosxcosxdx=(1cosx)cosx×(1+cosx)(1+cosx)dx=1cos2xcosx+cos2xdx=sinxcos2x+cosxdx Let cosx=t . Then d(cosx)=dt or-sinx dx=dt . Therefore, I=dtt2+t=dtt+122122=logt+12+t+122122+C=logt+12+t2+t+C=logcosx+12+cos2x+cosx+C

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