Question

How do you integrate `sec^3x (tanx) dx`?

Answer 1

`sec^3x/3+C`

Explanation:

When working with integrals of secant and tangent, it's important to remember the following:

• `d/dxtanx=sec^2x`
• `d/dxsecx=secxtanx`

Here, we see that we can write `sec^3x(tanx)` as `sec^2x(secxtanx)`, which is perfect, since it composed of `sec^2x` and the derivative of secant, `secxtanx`. This indicates to us that we want to use a substitution of `u=secx`.

`intsec^3x(tanx)dx=intsec^2x(secxtanx)dx`

With `u=secx` and `du=(secxtanx)dx`:

`=intu^2du=u^3/3+C=sec^3x/3+C`