The integral presented is:

$\int 19 \cos^3(x) \, dx$

The first step suggests using a trigonometric identity to simplify $\cos^3(x)$, specifically the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$. To proceed, we express $\cos^3(x)$ as a product of $\cos(x)$ and $\cos^2(x)$, and then replace $\cos^2(x)$ with $1 - \sin^2(x)$:

$\cos^3(x) = \cos(x) \cdot \cos^2(x) = \cos(x)(1 - \sin^2(x))$

Now, the integral becomes:

$\int 19 \cos(x)(1 - \sin^2(x)) \, dx$

At this point, we can use substitution, letting $u = \sin(x)$, so that $du = \cos(x) \, dx$. Substituting these into the integral:

$\int 19 (1 - u^2) \, du$

This integral can be split into two simpler integrals:

$19 \int (1 - u^2) \, du = 19 \left( \int 1 \, du - \int u^2 \, du \right)$

Now, integrate each term:

$19 \left( u - \frac{u^3}{3} \right) + C$

Finally, substitute $u = \sin(x)$ back:

$19 \left( \sin(x) - \frac{\sin^3(x)}{3} \right) + C$

This is the solution to the given integral.

Would you like further clarification, or any specific steps explained in more detail?

Here are 5 related questions you might find useful:

1. How do you apply substitution for trigonometric integrals in general?
2. What are other strategies for integrating odd powers of trigonometric functions?
3. How can we apply integration techniques to $\sin^n(x)$ when $n$ is odd or even?
4. How does substitution work when using other trigonometric identities?
5. Can integration by parts be used for similar integrals?

Tip: When integrating powers of sine and cosine, always check for simplifications using identities like $\sin^2(x) + \cos^2(x) = 1$.