To evaluate the integral of $\cos^3\left(\frac{x}{13}\right)$ with respect to $x$, we can use a trigonometric identity to simplify the integral. Specifically, we use the identity:

$\cos^3(\theta) = \cos(\theta) \cdot \cos^2(\theta) = \cos(\theta) \cdot \left(\frac{1 + \cos(2\theta)}{2}\right)$

Let $\theta = \frac{x}{13}$. Then:

$\cos^3\left(\frac{x}{13}\right) = \cos\left(\frac{x}{13}\right) \cdot \frac{1 + \cos\left(\frac{2x}{13}\right)}{2}$

This expands to:

$\frac{1}{2}\cos\left(\frac{x}{13}\right) + \frac{1}{2}\cos\left(\frac{x}{13}\right)\cos\left(\frac{2x}{13}\right)$

So the integral becomes:

$\int \cos^3\left(\frac{x}{13}\right) \, dx = \frac{1}{2} \int \cos\left(\frac{x}{13}\right) \, dx + \frac{1}{2} \int \cos\left(\frac{x}{13}\right)\cos\left(\frac{2x}{13}\right) \, dx$

Now, we handle each integral separately.

First Integral

The first term is straightforward:

$\frac{1}{2} \int \cos\left(\frac{x}{13}\right) \, dx$

The integral of $\cos\left(\frac{x}{13}\right)$ is:

$13 \sin\left(\frac{x}{13}\right)$

So the first integral becomes:

$\frac{13}{2} \sin\left(\frac{x}{13}\right)$

Second Integral

For the second integral:

$\frac{1}{2} \int \cos\left(\frac{x}{13}\right)\cos\left(\frac{2x}{13}\right) \, dx$

We use the product-to-sum identity:

$\cos(A)\cos(B) = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$

So:

$\cos\left(\frac{x}{13}\right)\cos\left(\frac{2x}{13}\right) = \frac{1}{2}\left[\cos\left(-\frac{x}{13}\right) + \cos\left(\frac{3x}{13}\right)\right]$

Since $\cos(-\theta) = \cos(\theta)$, we simplify to:

$\frac{1}{2}\left[\cos\left(\frac{x}{13}\right) + \cos\left(\frac{3x}{13}\right)\right]$

So the second integral becomes:

$\frac{1}{4} \int \cos\left(\frac{x}{13}\right) \, dx + \frac{1}{4} \int \cos\left(\frac{3x}{13}\right) \, dx$

The integrals are:

$\frac{13}{4} \sin\left(\frac{x}{13}\right) + \frac{13}{12} \sin\left(\frac{3x}{13}\right)$

Final Answer

Combining everything:

$\int \cos^3\left(\frac{x}{13}\right) \, dx = \frac{13}{2} \sin\left(\frac{x}{13}\right) + \frac{13}{8} \sin\left(\frac{x}{13}\right) + \frac{13}{12} \sin\left(\frac{3x}{13}\right) + C$

Simplifying:

$\int \cos^3\left(\frac{x}{13}\right) \, dx = \frac{39}{8} \sin\left(\frac{x}{13}\right) + \frac{13}{12} \sin\left(\frac{3x}{13}\right) + C$

where $C$ is the constant of integration.

Do you have any questions or need further clarification on this?

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Here are 5 related questions to consider:

1. What is the integral of $\cos^2\left(\frac{x}{13}\right)$ with respect to $x$?
2. Can you integrate $\cos^4\left(\frac{x}{13}\right)$ using similar identities?
3. How does the process differ when integrating $\sin^3\left(\frac{x}{13}\right)$?
4. What is the indefinite integral of $\cos^2(x)\sin(x)$ with respect to $x$?
5. How do you apply integration by parts to trigonometric integrals?

Tip: Utilizing trigonometric identities can greatly simplify integrals involving powers of trigonometric functions.