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 $\displaystyle \int_c=\int_{\Gamma}-\int_{c_1}-\int_{c_2}=\pi-0-\frac {\pi}{4}+1=\frac {3\pi}{4}+1$

$\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$$
\partiald{F}{x}=\sin\pars{y}\quad\imp\quad F = x\sin\pars{y} + \phi\pars{y}
$$

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

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