∫exsecx (1+tanx) dx equals:
(1)
Ans. exsecx+c
Explanation :
Let I=∫exsecx (1+tanx) dx
=∫exsecx dx+∫ex.secx.tanx dx
It is in the form
∫ex[f(x)+f'(x)] dx=exf(x)+c
f(x)=secx
f'(x)=secx-tanx
∫ex(secx+secx.tanx) dx=exsecx+c