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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.

Learn how to integrate sin^3(x) step-by-step by using trigonometric identities and substitution. This method simplifies the integral, leading to a clear solution for students in Grades 11-12 and university.

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