Question

What's the integral of `int {[(secx)^3] tanx} dx`?

Answer 1

`sec^3x/3+C`

Explanation:

`I=int(sec^3x)(tanx)dx`

We should be aiming to find one of two things:

• An integrand composed of just `secx` functions with one `secxtanx`, the derivative of `secx`.
• An integrand composed of just `tanx` functions with one `sec^2x`, the derivative of `tanx`.

Here, we see we can easily peel of one `secx` from `sec^3x` to combine it into `secxtanx`, leaving just other `secx` functions.

Rewriting the function:

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

We can now use substitution. Let `u=secx` so that `du=(secxtanx)dx`.

Thus, the integral becomes:

`I=intu^2du`

Which, through the power rule for integration, becomes:

`I=u^3/3+C`

`I=sec^3x/3+C`