First

$$I=2\int\frac{e^{-\sin{x}}\sin{x}\cos{x}}{(1-\sin{x})^2}dx=2\int\frac{e^{-\sin{x}}\sin{x}}{(1-\sin{x})^2}d\sin{x}$$

Then, letting $u=\sin{x}$ leads to

$$I=2\int\frac{e^{-u}u}{(1-u)^2}du=2\int\frac{(u-1+1)e^{-u}}{(1-u)^2}du=2\left(\int\frac{e^{-u}}{u-1}du+\int\frac{e^{-u}}{(u-1)^2}du\right)$$

Now because

$$\int\frac{e^{-u}}{(u-1)^2}du=-\int e^{-u}d\left(\frac{1}{u-1}\right)=-\frac{e^{-u}}{u-1}-\int \frac{e^{-u}}{u-1}du$$

therefore

$$I=2\left(-\frac{e^{-u}}{u-1}\right)+C=\boxed{\frac{e^{-\sin{x}}}{1-\sin{x}}+C}$$