Compute $\int_C\sin y\,dx+(x\cos y-\sin y)\,dy$ Compute $\int_C\sin y\,dx+(x\cos y-\sin y)\,dy$ where $C$ is $\displaystyle \frac {x^2}{4}+\frac {y^2}{2}=1$ in the first quadrant counter clockwise. 
I set $x=2\cos \theta$ and $y=\sqrt{2}\sin\theta$ where $0\leq\theta\leq \frac {\pi}{2}$.
But when I plug $x$ and $y$ back into the integral I am unable to solve it. Please drop some hints.
Thanks.
EDIT:
$\displaystyle\int_{\Gamma}=\iint_D(\frac {\partial Q}{\partial x}-\frac {\partial P}{\partial y})\,dA=\int_0^2\int_0^{\sqrt{2}}2\cos y\,dy\,dx=\pi$
$\int_{c_1}+\int_{c_2}=\int_0^20+\int_0^\sqrt{2}-\sin t\,dt$, where 
$C_1:$ $x=t,\,dx=dt, y=0,dy=0\,dt, 0\leq t\leq 2$ 
$C_2: y=t, dy=\,dt, x=0, dx=0\,dt, 0\leq t\leq \sqrt{2}$
So $\displaystyle \int_c=\int_{\Gamma}-\int_{c_1}-\int_{c_2}=\pi-0-\frac {\pi}{4}+1=\frac {3\pi}{4}+1$
 A: If you do this by substitution you get
$$I=\int_0^{\pi/2} -2\sin(\sqrt2\sin\theta)\sin\theta+\sqrt2(2\cos\theta\cos(\sqrt2\sin\theta)-\sin(\sqrt2\sin\theta))\cos\theta\,d\theta$$
(or something like that - I didn't check it carefully) which certainly does not look nice.  So you need a different method.
Hints.  Let $C_1$ be the $x$-axis from $0$ to $2$, let $C_2$ be the $y$-axis from $\sqrt2$ to $0$, and let $\Gamma$ be the closed curve consisting of $C$, $C_1$ and $C_2$.  Then


*

*$\displaystyle \int_\Gamma=\int_C+\int_{C_1}+\int_{C_2}$;

*the integrals along $C_1$ and $C_2$ are easy;

*the integral along $\Gamma$ can be done by Green's Theorem.
Good luck!
A: $\newcommand{\+}{^{\dagger}}
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$$
\partiald{F}{x}=\sin\pars{y}\quad\imp\quad F = x\sin\pars{y} + \phi\pars{y}
$$

$$
x\cos\pars{y} - \sin\pars{y} =\partiald{F}{y}=x\cos\pars{y} + \phi'\pars{y}
\quad\imp\quad\phi\pars{y} = \cos\pars{y} + \mbox{constant}
$$

$$
F = x\sin\pars{y} + \cos\pars{y} + \mbox{constant}\quad\imp\quad
\color{#00f}{\large\int_{C}\dd F = 0}
$$
