To solve the integral of $\int \sin^3(x) \, dx$ using integration by parts, we'll start by simplifying it, but in this case, we will also use a reduction formula approach.

We can rewrite $\sin^3(x)$ as: $\sin^3(x) = \sin(x) \cdot \sin^2(x)$ Using the Pythagorean identity $\sin^2(x) = 1 - \cos^2(x)$, we get: $\sin^3(x) = \sin(x)(1 - \cos^2(x)) = \sin(x) - \sin(x) \cos^2(x)$ Thus, the integral becomes: $\int \sin^3(x) \, dx = \int (\sin(x) - \sin(x)\cos^2(x)) \, dx$ We can now split this into two simpler integrals: $\int \sin^3(x) \, dx = \int \sin(x) \, dx - \int \sin(x)\cos^2(x) \, dx$

Step 1: First integral

$\int \sin(x) \, dx = -\cos(x)$

Step 2: Second integral by substitution

To handle $\int \sin(x)\cos^2(x) \, dx$, we use the substitution $u = \cos(x)$, which gives $du = -\sin(x) \, dx$.

This transforms the integral into: $\int \sin(x)\cos^2(x) \, dx = - \int u^2 \, du$ This evaluates to: $-\frac{u^3}{3} = -\frac{\cos^3(x)}{3}$

Final result:

Combining both parts, the final solution to the integral is: $\int \sin^3(x) \, dx = -\cos(x) + \frac{\cos^3(x)}{3} + C$ where $C$ is the constant of integration.

Would you like more details or clarifications?

Here are 5 related questions to deepen understanding:

1. How would you apply this technique to $\int \sin^5(x) \, dx$?
2. What other methods could be used for this type of trigonometric integral?
3. Can you explain how the substitution method works in more general cases?
4. What are the advantages of reduction formulas in solving integrals of powers of sine and cosine?
5. How would you integrate $\cos^3(x) \, dx$?

Tip: When solving integrals involving powers of trigonometric functions, consider using trigonometric identities to simplify the expression before integrating.