To integrate $(\sin x)^3$, we can use a trigonometric identity and substitution. Here's the step-by-step solution:

Step 1: Use the trigonometric identity

We begin by using the identity: $\sin^3 x = \sin x (1 - \cos^2 x)$ So the integral becomes: $\int \sin^3 x \, dx = \int \sin x (1 - \cos^2 x) \, dx$

Step 2: Substitution

Let $u = \cos x$, which implies $du = -\sin x \, dx$. The integral becomes: $\int \sin x (1 - \cos^2 x) \, dx = -\int (1 - u^2) \, du$

Step 3: Simplify and integrate

Now, integrate the expression: $-\int (1 - u^2) \, du = -\left(u - \frac{u^3}{3}\right) + C$

Step 4: Substitute back $u = \cos x$

Now, substitute back $u = \cos x$: $-(\cos x - \frac{\cos^3 x}{3}) + C$ which simplifies to: $\frac{\cos^3 x}{3} - \cos x + C$

Thus, the integral of $(\sin x)^3$ is: $\int \sin^3 x \, dx = \frac{\cos^3 x}{3} - \cos x + C$

Would you like more details on any step or have any questions?

Here are 5 related questions to extend your understanding:

1. How would you integrate $\sin^5 x$?
2. What if you were to integrate $(\cos x)^3$?
3. How can you use a similar method to integrate $(\sin x)^2$?
4. What is the integral of $(\sin x)^n$ for general $n$?
5. Can you derive the reduction formula for $\int (\sin x)^n dx$?

Tip: Trigonometric identities are very useful for simplifying integrals involving powers of sine and cosine!