Integration of $2\cos x \ e^{2\sin x}$

Question

I'm currently working my way through last semesters exams and I'm trying to understand an particular equation:

$$\int 2\cos x \ e^{2 \sin x} dx$$

Now, apparently the solution to this one is just $\int_{}{} e^t dt$ with $t = 2\sin x$ and $dt = 2\cos x $. I've spend a fair amount of time on trying to understand why $2 \cos x $ in the original Integral just disappears in the end but wasn't able to find a satisfying answer and I'm just wondering if anybody else could provide me with some clue.

Answer 1

This is because substituting $t=2\sin x$ means that $\dfrac{d}{dx}t=2\cos x$. Multiplying by $dx$ on both sides, $dt=2\cos x\; dx$. Can you continue?

Answer 2

Use the fact that $$\int f(g(x))f'(x)dx=\int f(t)dt , \hspace{5mm}\text{where} \hspace {5mm} t=g(x)$$

We get $$\int2\cos x e^{2\sin x}dx=\int e^tdt, t=2\sin x=e^{2\sin x}+C $$