Question

How do you integrate `{[(secx)^3] tanx} dx`?

Answer 1

`1/3sec^3x+C`.

Explanation:

Let `I=intsec^3xtanxdx`, rewriting it as,

`I=intsec^2x(secxtanx)dx`.

Let us use subst. `secx=t`, so that, `secxtanxdx=dt`. Hence,

`I=intt^2dt=t^3/3=1/3sec^3x+C`.