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Question

secx1 dx is equal to
(where C is integration constant)

A
ln(cosx+12)+2sin2x+cosx+C
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B
ln(cosx+12)+cos2x+cosx+C
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C
ln(sinx+12)+cos2x+2cosx+C
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D
ln(sinx+12)+2sin2x+sinx+C
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Solution

The correct option is B ln(cosx+12)+cos2x+cosx+C
I=secx1 dx =1cosxcosxdx =(1cosx)cosx×(1+cosx)(1+cosx)dx =1cos2xcosx+cos2xdx =sinxcos2x+cosxdx
Putting cosx=t
sinx dx=dtI=dtt2+t =dt(t+12)2(12)2 =ln∣ ∣(t+12)+(t+12)2(12)2∣ ∣+C =ln(t+12)+t2+t+CI=ln(cosx+12)+cos2x+cosx+C

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