# Find #int \ (sin2x+1)/(1-sin2x) \ dx #?

# int \ (sin2x+1)/(1-sin2x) \ dx = - x -2/(tanx-1) + C #

We seek:

We can write this as:

The first integral is standard, and for the second we have to perform some further analysis:

Using a standard Weierstraß trigonometric identity we have:

We can now easily integrate this if we perform a substitution:

Hence, we can now integrate [A] to get:

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To find the integral of (\frac{\sin(2x) + 1}{1 - \sin(2x)}) with respect to (x), you can use the substitution method. Let (u = \sin(2x)), then (du = 2\cos(2x)dx). Rewriting (\sin(2x)) in terms of (u), we get:

[\sin(2x) = u]

Now, rewrite the integral using (u):

[\int \frac{1 + u}{1 - u} \cdot \frac{1}{2} du]

This simplifies to:

[\frac{1}{2} \int \frac{1 + u}{1 - u} du]

Now, split the fraction:

[\frac{1}{2} \int \left(1 + \frac{2u}{1 - u}\right) du]

This integrates to:

[\frac{1}{2} \left(u + 2\ln|1 - u|\right) + C]

Replace (u) with (\sin(2x)) and simplify:

[\frac{1}{2} \left(\sin(2x) + 2\ln|1 - \sin(2x)|\right) + C]

So, the integral of (\frac{\sin(2x) + 1}{1 - \sin(2x)}) with respect to (x) is:

[\frac{1}{2} \left(\sin(2x) + 2\ln|1 - \sin(2x)|\right) + 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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