# How do I evaluate #int(1-sinx)/cosx dx#?

Well, this one is a little bit tricky... I started with the idea that the result must be a logarithm... So I did a little manipulation to get to a friendlier version of your function by multiplying and dividing by:

Which derived gives (1) or your original function!!!!!

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To evaluate ( \int \frac{1 - \sin(x)}{\cos(x)} , dx ), follow these steps:

- Rewrite the integral using the identity ( \frac{1 - \sin(x)}{\cos(x)} = \frac{\cos(x) - \sin(x)\cos(x)}{\cos(x)} ).
- Simplify the integrand to ( \cos(x) - \sin(x) ).
- Integrate ( \cos(x) - \sin(x) ) term by term.
- Evaluate the integral over the given interval, if specified.

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To evaluate ( \int \frac{1 - \sin(x)}{\cos(x)} , dx ), we can use the substitution method.

Let ( u = \cos(x) ). Then, ( du = -\sin(x) , dx ).

Substituting ( u = \cos(x) ), we get:

( \int \frac{1 - \sin(x)}{\cos(x)} , dx = \int \frac{1 - u}{u} , du )

This simplifies to:

( \int \left( \frac{1}{u} - 1 \right) , du )

Now, integrate each term separately:

( = \ln|u| - u + C )

Finally, substitute back ( u = \cos(x) ):

( = \ln|\cos(x)| - \cos(x) + C )

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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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