To integrate $\sin(x) e^{\cos(x)}$, we can use the method of substitution. Here's the step-by-step process:

Step 1: Substitution

Let: $u = \cos(x)$ Then, the derivative of $u$ with respect to $x$ is: $\frac{du}{dx} = -\sin(x)$ This gives us: $du = -\sin(x) \, dx$ or: $-\sin(x) \, dx = du$

Step 2: Rewrite the integral

Using this substitution, we rewrite the integral $\int \sin(x) e^{\cos(x)} \, dx$ as: $\int e^{u} (-du)$ This simplifies to: $-\int e^{u} \, du$

Step 3: Integrate

The integral of $e^u$ with respect to $u$ is simply $e^u$. Therefore: $-\int e^u \, du = -e^u + C$ where $C$ is the constant of integration.

Step 4: Substitute back

Now, substitute $u = \cos(x)$ back into the result: $-e^{\cos(x)} + C$

Final Answer:

$\int \sin(x) e^{\cos(x)} \, dx = -e^{\cos(x)} + C$

Would you like any additional details or clarifications?

Here are 5 related questions to explore this concept further:

1. How do you determine the appropriate substitution in an integral?
2. Can this technique be applied to other trigonometric integrals involving exponentials?
3. How would you solve $\int \cos(x) e^{\sin(x)} \, dx$?
4. How do you handle integrals involving products of different functions?
5. What are the common strategies for integrating functions involving both sine and cosine?

Tip: Always check if substitution is a viable method by looking for a function and its derivative in the integral.