integrate tan(2x)

Understand the Problem

The question is asking to perform integration of the function tan(2x). This requires applying integration techniques to find the antiderivative of the given trigonometric function.

$$\frac{1}{2} \ln|\sec(2x)| + C$$

The final answer is ( \frac{1}{2} \ln|\sec(2x)| + C )

Steps to Solve

1. Substitute the inner function

To simplify the integral, let $u = 2x$. Then $du = 2dx$ or $dx = \frac{du}{2}$. Our integral becomes:

$$\int \tan(2x) , dx = \int \tan(u) \cdot \frac{du}{2}$$

1. Simplify the integral

Factor out the constant $\frac{1}{2}$:

$$\int \tan(u) \cdot \frac{du}{2} = \frac{1}{2} \int \tan(u) , du$$

1. Recall the integral of tangent

The antiderivative of $\tan(u)$ is $\ln|\sec(u)|$:

$$\frac{1}{2} \int \tan(u) , du = \frac{1}{2} \ln|\sec(u)| + C$$

1. Substitute back the original variable

Replace $u$ with $2x$ to return to the original variable:

$$\frac{1}{2} \ln|\sec(u)| + C = \frac{1}{2} \ln|\sec(2x)| + C$$

The final answer is ( \frac{1}{2} \ln|\sec(2x)| + C )

A common mistake is forgetting to perform the substitution for $dx$ correctly when $u$ is introduced. Always make sure to change both the variable and the differential.