The integral you've provided is:

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

This integral can be solved using a combination of trigonometric identities and substitution rather than directly using integration by parts. However, let me guide you through it step by step.

Step 1: Rewrite $\sin^3(x)$

We can use the identity: $\sin^3(x) = \sin(x) \cdot \sin^2(x)$ And recall that: $\sin^2(x) = 1 - \cos^2(x)$ Thus: $\sin^3(x) = \sin(x) \cdot (1 - \cos^2(x)) = \sin(x) - \sin(x) \cos^2(x)$

Step 2: Split the integral

Now, split the integral into two parts: $\int \sin^3(x) \, dx = \int \sin(x) \, dx - \int \sin(x) \cos^2(x) \, dx$

Step 3: Solve the first integral

The first integral is straightforward: $\int \sin(x) \, dx = -\cos(x)$

Step 4: Solve the second integral using substitution

For the second integral, use the substitution $u = \cos(x)$, which gives $du = -\sin(x) \, dx$. The integral becomes: $-\int \sin(x) \cos^2(x) \, dx = \int u^2 \, du$ The solution to this is: $\int u^2 \, du = \frac{u^3}{3} = \frac{\cos^3(x)}{3}$

Step 5: Combine results

Putting everything together, the solution is: $\int \sin^3(x) \, dx = -\cos(x) + \frac{\cos^3(x)}{3} + C$

This is the final result.

Do you want more details or have any questions?

Here are five related questions that could expand on this topic:

1. How would you apply integration by parts to trigonometric functions?
2. Can you generalize the method to integrate $\sin^n(x) \, dx$ for other powers of $n$?
3. How would you solve the integral $\int \cos^3(x) \, dx$?
4. What are the main differences between integration by substitution and by parts?
5. How can we integrate other functions like $\sin(x)\cos(x)$ using trigonometric identities?

Tip: Always check for trigonometric identities that can simplify the powers of sine or cosine before attempting direct integration.