# How do you evaluate the integral #int tanxdx# from 0 to #pi# if it converges?

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The integral does not converge.

and add the results.

The limit does not exist.

Bonus material

There is a similar notion, called the Cauchy principal value, that has us, in a sense, evaluate the two improper integrals at once.

Second example:

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To evaluate the integral (\int_0^{\pi} \tan(x) , dx), we first note that (\tan(x)) is not defined at (x = \frac{\pi}{2}) where it has an infinite discontinuity. However, within the given bounds ((0, \pi)), the integral converges.

Using the substitution (u = \cos(x)), (du = -\sin(x) , dx), we have:

[\int \tan(x) , dx = \int \frac{\sin(x)}{\cos(x)} , dx]

[= \int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos(x)| + C]

Evaluating the integral from (0) to (\pi):

[\int_0^{\pi} \tan(x) , dx = -\ln|\cos(\pi)| + \ln|\cos(0)|]

[\ln|\cos(0)| = \ln|1| = 0]

[\ln|\cos(\pi)| = \ln|-1| = \ln(1) = 0]

So,

[\int_0^{\pi} \tan(x) , dx = -\ln(1) + \ln(1) = 0]

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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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