The integral of \( \csc(x) \, dx \) can be evaluated using a standard method. Here's the solution:

\[
\int \csc(x) \, dx
\]

We use the following identity for the antiderivative of \( \csc(x) \):

\[
\int \csc(x) \, dx = \ln \left| \csc(x) - \cot(x) \right| + C
\]

where \( C \) is the constant of integration.

Thus, the result is:

\[
\int \csc(x) \, dx = \ln \left| \csc(x) - \cot(x) \right| + C
\]

Would you like further details or have any questions?

Here are 5 related questions:
1. What is the derivative of \( \ln |\csc(x) - \cot(x)| \)?
2. How would you find the integral of \( \csc^2(x) \)?
3. How can you express \( \csc(x) \) in terms of sine?
4. What is the integral of \( \sec(x) \)?
5. Can you derive the result for \( \int \csc(x) \, dx \) from a trigonometric identity?

**Tip**: The functions \( \csc(x) \) and \( \cot(x) \) are closely related, so knowing their derivatives can make solving integrals easier.

Learn how to find the integral of csc(x) using a standard trigonometric identity. This detailed guide explains the process step-by-step, showing that \int csc(x)dx = ln|csc(x) - cot(x)| + C. Suitable for students in Calculus I or similar courses.

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