What is the antiderivative of cosecant?
Understand the Problem
The question is asking for the antiderivative (or integral) of the cosecant function (csc). The approach to solve this involves using integration techniques and identities related to trigonometric functions.
Answer
$\ln\text{csc}(x) + \text{cot}(x) + C$
Answer for screen readers
The final answer is $\ln\text{csc}(x) + \text{cot}(x) + C$
Steps to Solve

Rewrite $ ext{csc}(x)$ in terms of known integrals
To make the integral easier to solve, rewrite $ ext{csc}(x)$ as $rac{1}{ ext{sin}(x)}$.
$$ ext{csc}(x) = rac{1}{ ext{sin}(x)}$$

Multiply and divide by $ ext{csc}(x) + ext{cot}(x)$
Multiply and divide the integrand by $ ext{csc}(x) + ext{cot}(x)$ to create a more solvable form.
$$\int ext{csc}(x) , dx = \int \text{csc}(x) \cdot \frac{\text{csc}(x) + \text{cot}(x)}{\text{csc}(x) + \text{cot}(x)} , dx$$

Simplify the integral
Simplify the integral to isolate a more straightforward derivative form
$$\int \frac{\text{csc}^2(x) + \text{csc}(x) \text{cot}(x)}{\text{csc}(x) + \text{cot}(x)} , dx$$

Use substitution to solve the integral
Let $u = \text{csc}(x) + \text{cot}(x)$, then calculate $du$.
$$du = (\text{csc}(x) \text{cot}(x) + \text{csc}^2(x)) , dx$$
Hence,
$$\int \frac{1}{u} , du$$

Integrate and solve
Integrate the function with respect to $u$.
$$\lnu + C$$ where $C$ is the constant of integration.

Substitute $u$ back in
Replace $u$ with $\text{csc}(x) + \text{cot}(x)$.
$$\ln\text{csc}(x) + \text{cot}(x) + C$$
The final answer is $\ln\text{csc}(x) + \text{cot}(x) + C$
More Information
This integral is interesting because it is not immediately obvious and requires a clever substitution and manipulation to solve.
Tips
A common mistake is forgetting to use the absolute value inside the logarithm function or not correctly simplifying the expression for $du$ in terms of $dx$.