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.

Answer

\( \frac{1}{2} \ln|\sec(2x)| + C \)
Answer for screen readers

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 )

More Information

The antiderivative of the tangent function is connected to the logarithm function, through the secant function.

Tips

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.

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