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

Please see below.

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.

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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