<s>[INST] Line Integral with Change of Variables I'm a bit rusty on my computational math, and genuinely can't solve this question which is frustrating me.
QUESTION:
$$\int_Csin(y)dx+xcos(y)dy$$ where C is the ellipse defined as follows: $x^2+xy+y^2=1$
MY ATTEMPT:
Define variable $u= \sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y}$
Define variable $v= \sqrt{\frac{1}{4}}x-\sqrt{\frac{1}{4}}y$
Hence:
$u^2+v^2=(\sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y})^2+(\sqrt{\frac{1}{4}}x-\sqrt{\frac{1}{4}}y)^2$
$=(\frac{3}{4}x^2+\frac{(2)(3)}{4}xy+\frac{3}{4}y^2)+(\frac{1}{4}x^2-\frac{2}{4}xy+\frac{1}{4}y^2)$
$=(\frac{3}{4}+\frac{3}{4})x^2+(\frac{3}{2}-\frac{1}{2})xy+(\frac{3}{4}+\frac{1}{4})y^2$
$=(x^2+xy+y^2) = C$
Therefore, we have Jacobian:
$$     
       \begin{bmatrix}
       \frac{\partial{u}}{\partial{x}} & \frac{\partial{u}}{\partial{y}} \\
       \frac{\partial{v}}{\partial{x}} & \frac{\partial{v}}{\partial{y}} \\
       \end{bmatrix}
$$
$$     
       \begin{bmatrix}
       \sqrt\frac{3}{4} & \sqrt\frac{3}{4} \\
       \sqrt\frac{1}{4} & -\sqrt\frac{1}{4} \\
       \end{bmatrix}
$$
$$ = \frac{\sqrt{3}}{2} $$
Beyond this point, I must incorporate the original dot product with the differential $(dx,dy)$ but don't know how
 [/INST] Hint:
\begin{align*}
\frac{\partial}{\partial x}\left(x\cos\left(y\right)\right)-\frac{\partial}{\partial y}\left(\sin\left(y\right)\right)&=\cos\left(y\right)-\cos\left(y\right)=0
\end{align*}
</s>