Question #de65e

1 Answer
Oct 1, 2017

910secxtanx+910ln|secx+tanx|+C

Explanation:

I=9secxsec2xdx
Using integration by parts
=9secxsec2xdx9(ddxsecxsec2xdx)dx
=9secxtanx9(secxtanxtanx)dx
=9secxtanx9(secxtan2x)dx
=9secxtanx9(secx(sec2x1)dx
I=9secxtanx9sec3xdx+9secxdx
I=9secxtanx9I+9secxdx
10I=9secxtanx+9ln|secx+tanx|+C
I=910secxtanx+910ln|secx+tanx|+C