The integral provided is:

$\int 2 \sin^4(x) \cos(x) \, dx$

To solve this, we can use substitution and trigonometric identities.

Step 1: Substitution

Let: $u = \sin(x)$ Then: $\frac{du}{dx} = \cos(x) \quad \Rightarrow \quad du = \cos(x) \, dx$

Step 2: Rewrite the integral

Using the substitution $u = \sin(x)$, the integral becomes: $\int 2 u^4 \, du$

Step 3: Integrate

Now integrate the expression: $\int 2 u^4 \, du = 2 \cdot \frac{u^5}{5} = \frac{2}{5} u^5$

Step 4: Substitute back $u = \sin(x)$

Now, replace $u$ with $\sin(x)$: $\frac{2}{5} \sin^5(x)$

Thus, the solution to the integral is:

$\frac{2}{5} \sin^5(x) + C$

Let me know if you'd like further details or explanations!

Here are some related questions:

1. How does substitution help in evaluating integrals?
2. Can you provide an example where the substitution method does not work well?
3. How would you approach an integral with higher powers of sine and cosine?
4. How does integration differ from differentiation in terms of complexity?
5. How can you verify the result of an indefinite integral?

Tip: Always double-check your substitution and integration limits (for definite integrals) to avoid errors when reversing substitutions!