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a
excosx+C
b
exsecx+C
c
exsinx+C
d
extanx+C
answer is B.
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Detailed Solution
Let I=∫exsecx(1+tanx)dx⇒I=∫exsecxdx+∫exsecxtanxdx -----iNow ∫exsecxdx=secx∫exdx−∫ddxsecx∫exdxdx=exsecx−∫secxtanxexdx ----ii On putting the value from Eq. (ii)inEq. (i),we get I=exsecx−∫exsecxtanxdx+∫secxtanxexdx+C⇒I=exsecx+C