Question

What is the integral of `int secxln(secx + tanx) dx`?

Answer 1

The integral equals `1/2ln^2(secx + tanx) + C`

Explanation:

Let `u = ln(secx + tanx)`. Then by the chain rule we have

`du = (secxtanx + sec^2x)/(secx + tanx)dx` and `dx = (secx + tanx)/(secxtanx + sec^2x)du`

We now have, calling the integral `I`:

`I = int secx * u * (secx + tanx)/(secxtanx + sec^2x)du`

If we do a little factoring we get:

`I = int secx * u * (secx + tanx)/(secx(tanx + secx)) du`

`I = int u du`

`I = 1/2u^2 + C`

`I = 1/2ln^2(secx + tanx) + C`

Hopefully this helps!