Question

How do you find the antiderivative of `int tan^3xsecx dx`?

Answer 1

`inttan^3xsecxdx=1/3sec^3x-secx+C`

Explanation:

`I=inttan^3xsecxdx`

Rewrite this using `tan^2x+1=sec^2x`:

`I=inttan^2xtanxsecxdx`

`I=int(sec^2x-1)(secxtanx)dx`

Let `u=secx`. Then `du=secxtanxdx`:

`I=int(u^2-1)du`

`I=1/3u^3-u+C`

`I=1/3sec^3x-secx+C`