The integral provided is:

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

To solve this, we can use substitution and trigonometric identities.

Step 1: Rewrite the integral

We know that $\sin^3(x)$ can be rewritten as:

$\sin^3(x) = \sin(x) \cdot \sin^2(x)$

Using the identity $\sin^2(x) = 1 - \cos^2(x)$, we get:

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

So, the integral becomes:

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

Step 2: Substitution

Let $u = \sin(x)$, hence $du = \cos(x) \, dx$. The integral simplifies to:

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

Step 3: Simplify and Integrate

Expanding the integrand:

$\int u \cdot (1 - u^2) \, du = \int (u - u^3) \, du$

Now, integrate term by term:

$\int u \, du - \int u^3 \, du = \frac{u^2}{2} - \frac{u^4}{4} + C$

Step 4: Back Substitution

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

$\frac{\sin^2(x)}{2} - \frac{\sin^4(x)}{4} + C$

Thus, the solution to the integral is:

$\boxed{\frac{\sin^2(x)}{2} - \frac{\sin^4(x)}{4} + C}$

Would you like a detailed explanation of any particular step? Here are a few related questions to practice:

1. How would you solve $\int \sin^3(x) \, dx$?
2. What substitution method can be used for integrals involving products of sine and cosine?
3. Can you express $\cos^3(x)$ using trigonometric identities?
4. How does integration by parts apply to integrals like this?
5. How do you handle definite integrals of trigonometric functions?

Tip: When solving integrals involving powers of sine or cosine, using trigonometric identities often simplifies the problem.