Using the same substitution u = sin(x) enables us to do16 sin4(x) cos(x) dx = 16u4 du.In terms of u, we get + C, which, in terms of sin(x), becomes + C.

To solve the integral $\int 16 \sin^4(x) \cos(x) \, dx$ using the substitution $u = \sin(x)$, we can follow these steps:

1. Break Down the Problem

We need to convert the integral into a function of $u$ and then integrate. The substitution $u = \sin(x)$ leads to $du = \cos(x) \, dx$, which allows us to replace $\cos(x) \, dx$ with $du$.

2. Relevant Concepts

Using the substitution:

• Given $u = \sin(x)$ implies $\sin^4(x) = u^4$.
• The differential becomes $\cos(x) \, dx = du$.

3. Analysis and Detail

Substituting into the integral: $\int 16 \sin^4(x) \cos(x) \, dx = \int 16 u^4 \, du$ Now, we integrate $16 u^4$: $\int 16 u^4 \, du = 16 \cdot \frac{u^5}{5} = \frac{16}{5} u^5 + C$

4. Verify and Summarize

After integrating, we substitute back $u = \sin(x)$: $\frac{16}{5} u^5 + C = \frac{16}{5} \sin^5(x) + C$

Final Answer

Thus, the final result of the integral in terms of $\sin(x)$ is: $\int 16 \sin^4(x) \cos(x) \, dx = \frac{16}{5} \sin^5(x) + C$