Evaluate $\oint_{C} e^{-x} \sin y \;dx+e^{-x} \cos y\;dy$ I need to evaluate the following integral using Green's theorem

$$\oint_{C} e^{-x} \sin y \;dx+e^{-x} \cos y\;dy$$
  $C$: from point $E \to F\to G\to H$
  $$E=(0,0)\,,F=(\pi,0)\;,G=(\pi,\frac{\pi}{2}),\;H=(0,\frac{\pi}{2})$$

My attempt:
$$\oint_{C} \underbrace{e^{-x} \sin y}_{P} \;dx+\underbrace{e^{-x} \cos y}_{Q}\;dy$$
$$=\iint\bigg(\frac{\partial(e^{-x} \cos y)}{\partial x} -\frac{\partial(e^{-x} \sin y)}{\partial y}\bigg)dxdy$$
$$=\iint\bigg(-e^{-x}\cos y-e^{-x}\cos y\bigg)dxdy$$
$$=\iint\bigg(-2e^{-x}\cos y\bigg)dxdy$$
$$=\int_{y=0}^{y=\pi/2}\bigg(\int_{0}^{\pi}[-2e^{-x}\cos y] dx\bigg)dy$$
$$=\int_{y=0}^{y=\pi/2}\bigg([2e^{-x}\cos y] \bigg|_0^{\pi}\bigg)dy$$
$$=\int_{y=0}^{y=\pi/2}\bigg([2e^{-\pi}-2] \bigg)dy$$
$$=2\cdot \frac{\pi}{2}e^{-\pi}-2\frac{\pi}{2}$$
$$=\boxed{\pi e^{-\pi}-\pi}$$

Is it correct?

EDIT:
After using @mickep answer
$$=\int_{y=0}^{y=\pi/2}\bigg([2e^{-x}\cos y] \bigg|_0^{\pi}\bigg)dy$$
$$=\int_{y=0}^{y=\pi/2}\bigg([2e^{-\pi}\cos y-2\cos y] \bigg)dy$$
$$2e^{-\pi}\sin y -2\sin y\bigg|_0^{\pi / 2}$$
$$=\boxed{2e^{-\pi}-2}$$
 A: No. It is correct up to the third line from bottom in your calculation, where you insert the bounds for both $x$ and $y$. You should do that only for $x$. Then integrate the $\cos y$. Also, in your last step, the $\pi$ disappears (but that $\pi$ shouldn't be there if you fix the previous mistake).
I get $2e^{-\pi}-2$ as a result.
A: The integration path have to be closed in order to use Green's theorem so I assume you mean that $C$ is the closed rectangle $E\to F\to G\to H \to E$. 
mickep's answer has explained well where the mistake in the calculation. I just want to point out that when in doubt you can check your calculation by performing the line integral. For rectangular paths this is usually fairly straight forward and in this case most of the integrals vanish making it very easy to do so.
Along $E\to F$ we have $y=dy=0$ so both integral are zero. Along $F\to G$ $dx=0$ so the first integral is zero. Along $G\to H$ we have $dy = 0$ so the last integral is $0$ and along $H\to E$ $dx=0$ so the first integral is zero.
This leaves us with only two independent integrals to evaluate
$$\oint_{C} e^{-x}\sin(y)\,{\rm d}x + e^{-x}\cos(y)\,{\rm d}y \\= \int_{0}^{\frac{\pi}{2}} e^{-\pi}\cos(y)\,{\rm d}y + \int_{\pi}^{0} e^{-x}\sin(\pi/2)\,{\rm d}x + \int_{\frac{\pi}{2}}^0e^{-0}\cos(y)\,{\rm d}x  \\= (e^{-\pi}-1)\int_0^{\frac{\pi}{2}}\cos(y)\,{\rm d}y - \int_0^{\pi} e^{-x}\,{\rm d}x=2e^{-\pi}-2$$
