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
- Substitute the inner function
To simplify the integral, let $u = 2x$. Then $du = 2dx$ or $dx = \frac{du}{2}$. Our integral becomes:
$$ \int \tan(2x) , dx = \int \tan(u) \cdot \frac{du}{2} $$
- Simplify the integral
Factor out the constant $\frac{1}{2}$:
$$ \int \tan(u) \cdot \frac{du}{2} = \frac{1}{2} \int \tan(u) , du $$
- Recall the integral of tangent
The antiderivative of $\tan(u)$ is $\ln|\sec(u)|$:
$$ \frac{1}{2} \int \tan(u) , du = \frac{1}{2} \ln|\sec(u)| + C $$
- 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.
