The problem presented involves solving the integral

$\int \sin^3(x) e^{\cos(x)} \, dx$

through substitution and the method of integration by parts. Let me break down the key steps and verify the methodology.

Steps Explained:

1. Substitution:

   • Let $u = \cos(x)$, so $du = -\sin(x) dx$.
   • This transforms the integral into: $\int \sin^3(x) e^{\cos(x)} \, dx = -\int (1 - u^2) e^u \, du$ (using $\sin^2(x) = 1 - \cos^2(x)$).

2. Integration by Parts - Step 1:

   • Split $-\int (1 - u^2) e^u \, du$ using integration by parts.
   • Define $f(u) = -(1 - u^2)$ and $g'(u) = e^u$.
   • Then $g(u) = e^u$, and by integration by parts: $\int F(u) \, du = G(u) - \int H(u) \, du$ where: $G(u) = f(u)g(u) = -(1 - u^2)e^u,$ and: $H(u) = f'(u)g(u) = 2ue^u.$

3. Integration by Parts - Step 2:

   • Compute $\int H(u) \, du = \int 2u e^u \, du$ using integration by parts again.
   • Set $f(u) = 2u$ and $g'(u) = e^u$.
   • Then $g(u) = e^u$, and: $\int 2u e^u \, du = (2u)e^u - \int 2e^u \, du$ $= 2u e^u - 2e^u + C.$

4. Combine Results:

   • Substitute back for $\int H(u) \, du$ to get: $\int F(u) \, du = -(1 - u^2)e^u - (2u e^u - 2e^u + C).$

5. Reverting to $x$:

   • Replace $u = \cos(x)$ to return to the original variable: $\int \sin^3(x) e^{\cos(x)} \, dx = \left( \cos(x)^2 - 2\cos(x) + 1 \right)e^{\cos(x)} + C.$

Let me know if you’d like further clarification or detailed computation steps!

Questions for Deeper Understanding:

1. What is the purpose of using substitution in this integral?
2. How does integration by parts simplify this problem?
3. Why is it necessary to split the integral into two parts with $H(u)$?
4. Could this problem be solved with a different substitution?
5. How does the exponential function influence the complexity of the integration?

Tip: When working with trigonometric-exponential integrals, always simplify the trigonometric terms first using identities, as done here with $\sin^2(x) = 1 - \cos^2(x)$. This often reduces complexity significantly.