Line Integral with Change of Variables

Question

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

Answer 1

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*}

Answer 2

Hint: Use Green's theorem instead to ease out calculations:

$\int_C Pdx+Qdy=\int\int_R (\partial Q/\partial x-\partial P/\partial y) dx dy$

where $R$ is the region bounded by closed curve $C$ on $xy-$ plane oriented counterclockwise.

Answer 3

The integral is path independent because

$$\nabla(x\sin y) = (\sin y, x\cos y)$$

So, the integral along the closed contour $C$ vanishes:

$$\int_C (\sin y\,dx + x\cos y\,dy) =0$$