# 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.

$-\ln|\text{csc}(x) + \text{cot}(x)| + C$

The final answer is $-\ln|\text{csc}(x) + \text{cot}(x)| + C$

#### Steps to Solve

1. Rewrite $ext{csc}(x)$ in terms of known integrals

To make the integral easier to solve, rewrite $ext{csc}(x)$ as $rac{1}{ ext{sin}(x)}$.

$$ext{csc}(x) = rac{1}{ ext{sin}(x)}$$

2. 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.

$$\int ext{csc}(x) , dx = \int \text{csc}(x) \cdot \frac{\text{csc}(x) + \text{cot}(x)}{\text{csc}(x) + \text{cot}(x)} , dx$$

3. Simplify the integral

Simplify the integral to isolate a more straightforward derivative form

$$\int \frac{\text{csc}^2(x) + \text{csc}(x) \text{cot}(x)}{\text{csc}(x) + \text{cot}(x)} , dx$$

4. 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,

$$\int \frac{-1}{u} , du$$

5. Integrate and solve

Integrate the function with respect to $u$.

$$-\ln|u| + C$$ where $C$ is the constant of integration.

6. 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 $-\ln|\text{csc}(x) + \text{cot}(x)| + C$

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 $dx$.