integral of cos(3x)
Understand the Problem
The question is asking for the integral of the function cos(3x). This involves finding the antiderivative of the cosine function multiplied by a constant. To solve it, we'll apply the basic rules of integration, potentially using substitution for the argument of the cosine function.
Answer
The integral of $\cos(3x)$ is $\frac{1}{3} \sin(3x) + C$.
Answer for screen readers
The integral of $\cos(3x)$ is:
$$ \int \cos(3x) , dx = \frac{1}{3} \sin(3x) + C $$
Steps to Solve
- Identify the integral to solve
We need to find the integral of the function $\cos(3x)$. This can be written as:
$$ \int \cos(3x) , dx $$
- Use a substitution method
To solve this integral, we can use substitution. Let ( u = 3x ). Then, the differential ( du ) is:
$$ du = 3 , dx $$
This means that:
$$ dx = \frac{du}{3} $$
- Substitute and rewrite the integral
Now we can substitute ( u ) and ( dx ) into the integral:
$$ \int \cos(3x) , dx = \int \cos(u) \left( \frac{du}{3} \right) $$
This simplifies to:
$$ \frac{1}{3} \int \cos(u) , du $$
- Integrate the cosine function
The integral of $\cos(u)$ is $\sin(u)$:
$$ \int \cos(u) , du = \sin(u) $$
- Combine and back-substitute
Now we substitute back for ( u ):
$$ \frac{1}{3} \sin(u) + C = \frac{1}{3} \sin(3x) + C $$
where ( C ) is the constant of integration.
The integral of $\cos(3x)$ is:
$$ \int \cos(3x) , dx = \frac{1}{3} \sin(3x) + C $$
More Information
The integration of $\cos(3x)$ shows how substitution can simplify the process of finding the antiderivative. The key aspect here is recognizing the need to adjust for the factor of 3 in the argument of the cosine function.
Tips
- Forgetting to include the constant factor from the substitution (e.g., missing the $\frac{1}{3}$).
- Not changing the variable back to the original variable after integration.
