To solve the integral $\int 2e^{\sin(x)}\cos(x) \, \mathrm{d}x$, we can use the substitution method. Here's the step-by-step process:

1. Substitution: Let $u = \sin(x)$. Then, differentiate $u$ with respect to $x$: $\frac{du}{dx} = \cos(x) \implies du = \cos(x) \, dx$

2. Substitute: Replace $\sin(x)$ with $u$ and $\cos(x) \, dx$ with $du$: $\int 2e^{\sin(x)}\cos(x) \, \mathrm{d}x = \int 2e^u \, du$

3. Integrate: The integral of $2e^u$ with respect to $u$ is: $\int 2e^u \, du = 2e^u + C$ where $C$ is the constant of integration.

4. Back-substitute: Replace $u$ with $\sin(x)$ to get the final answer: $2e^{\sin(x)} + C$

So the final result is: $\int 2e^{\sin(x)}\cos(x) \, \mathrm{d}x = 2e^{\sin(x)} + C$

Would you like any further details or explanations? Here are some related questions:

1. How do you integrate $e^{\sin(x)} \cos(x)$ without the coefficient of 2?
2. What is the integral of $e^{\cos(x)} \sin(x)$?
3. Can you explain the process of substitution more thoroughly?
4. How do you differentiate $e^{\sin(x)}$?
5. What if the exponent in the integral was $\cos(x)$ instead of $\sin(x)$?

Tip: Substitution is particularly useful when the integral contains a function and its derivative. Identifying the right substitution can simplify the integral significantly.