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

what i tried to do was

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

$\displaystyle =\frac{-1}{sinx}e^{cosx}sin2x-\int \frac{-1}{sinx} e^{cosx}(2)(cos2x)$ $\displaystyle dx$

change $\displaystyle sin2x$ to $\displaystyle 2sinxcosx$ and $\displaystyle cos 2x$ to $\displaystyle -2sin^2x+1$, then simplify.

$\displaystyle =-2e^{cosx}cosx+4\int (-sinx)(e^{cosx})$$\displaystyle dx$ $\displaystyle -2\int \frac{1}{sinx}e^{cosx} $$\displaystyle dx$

$\displaystyle =-2e^{cosx}cosx+4e^{cosx}-2[\frac{-1}{sinx}e^{cosx}(\frac{1}{sinx})-\int(\frac{-1}{sinx}e^{cosx}(-cosec x cotx) $$\displaystyle dx]$

it's like never going to end.....

did i make a mistake somewhere?

or this approach is wrong?