Convergence of $\int_{1}^{\infty} e^{\sin(x)}\cdot\frac{\sin(x)}{x^2} $

$\sin x \leq 1$, so composing by $e^x$, which is a monotone function, gives $e^{\sin x} \leq e^1$. Also, $|\sin x| \leq 1$.

So we have that 

$|\int_1^\infty e^{sin x} \frac{sin x}{x^2}dx| \leq \int_1^\infty |e^{sin x} \frac{sin x}{x^2}dx| \leq \int_1^\infty e^{1} \frac{1}{x^2}dx = e(-x^{-1})|_{1}^{\infty} =\frac{1}{e}<\infty$