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a
−logCosx+12+Cos2x+Cosx+C
b
logCosx+12+Cos2x+Cosx+C
c
−logCosx−12+Cos2x+Cosx+C
d
logCosx+12−Cos2x+Cosx+C
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Correct option is A
I=∫secx−1dx=∫1−cosxcosxdx=∫(1−cosx)cosx×(1+cosx)(1+cosx)dx=∫1−cos2xcosx+cos2xdx=∫sinxcos2x+cosxdx Let cosx=t . Then d(cosx)=dt or-sinx dx=dt . Therefore, I=∫−dtt2+t=−∫−dtt+122−122=−logt+12+t+122−122∣+C=−logt+12+t2+t+C=−logcosx+12+cos2x+cosx∣+CSimilar Questions
is equal to
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