To solve the integral $\int \cos^3(x) \sin(x) \, dx$, we can use a substitution method. Let's proceed step by step.

1. Let $u = \cos(x)$. Then, the derivative of $u$ with respect to $x$ is: $\frac{du}{dx} = -\sin(x)$ This implies: $du = -\sin(x) \, dx$ or equivalently: $-du = \sin(x) \, dx$

2. Substitute $u = \cos(x)$ and $\sin(x) \, dx = -du$ into the integral: $\int \cos^3(x) \sin(x) \, dx = \int u^3 (-du)$ Simplify the integral: $\int \cos^3(x) \sin(x) \, dx = -\int u^3 \, du$

3. Now, integrate $u^3$: $\int u^3 \, du = \frac{u^4}{4}$

4. Don't forget to include the negative sign from the substitution: $-\int u^3 \, du = -\frac{u^4}{4}$

5. Substitute back $u = \cos(x)$: $-\frac{u^4}{4} = -\frac{(\cos(x))^4}{4} = -\frac{\cos^4(x)}{4}$

6. Add the constant of integration $C$: $\int \cos^3(x) \sin(x) \, dx = -\frac{\cos^4(x)}{4} + C$

So, the solution to the integral is: $\boxed{-\frac{\cos^4(x)}{4} + C}$

Would you like further details or have any questions? Here are some related questions for further practice:

1. $\int \cos^2(x) \sin(x) \, dx$
2. $\int \cos^3(x) \, dx$
3. $\int \sin^3(x) \cos(x) \, dx$
4. $\int \cos^5(x) \sin(x) \, dx$
5. $\int \sin^4(x) \cos(x) \, dx$
6. $\int \cos^2(x) \, dx$
7. $\int \cos(x) \, dx$
8. $\int \sin(x) \, dx$

Tip: When integrating products of trigonometric functions, substitution often simplifies the problem significantly. Look for expressions where substitution can turn the integral into a simpler form.