How do you find #int lnsec^2x#?

It would take a whole day to enter the answer into Socratic's page, so please use the following link and enter the function to see the answer. This is not exactly an easy integral.

https://tutor.hix.ai

To find ( \int \ln(\sec^2(x)) ), use integration by substitution.

Let ( u = \sec(x) ) and ( du = \sec(x) \tan(x) dx ).

Then the integral becomes ( \int \ln(u^2) \sec(x) \tan(x) dx ).

Now, integrate by parts:

( \int \ln(u^2) \sec(x) \tan(x) dx = \frac{1}{2}u^2 \ln(u^2) - \int \frac{u^2}{u} \sec(x) \tan(x) dx ).

Simplify and substitute back:

( \int \ln(u^2) \sec(x) \tan(x) dx = \frac{1}{2}\sec^2(x) \ln(\sec^2(x)) - \int \sec(x) \tan(x) dx ).

The integral ( \int \sec(x) \tan(x) dx ) is a standard integral, equal to ( \sec(x) ).

So, the final result is ( \frac{1}{2}\sec^2(x) \ln(\sec^2(x)) - \sec(x) + C ), where ( C ) is the constant of integration.