∫exsecx(1+tanx)dx is equal to
excosx+C
exsecx+C
exsinx+C
extanx+C
Let I=∫exsecx(1+tanx)dx
⇒I=∫exsecxdx+∫exsecxtanxdx -----i
Now ∫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