I want to calculate the following path integral:
$$\int_\mathcal{C} ye^{-(x^2+y^2)} (1-2x^2) \:\mathrm{d}x + xe^{-(x^2+y^2)} (1-2y^2) \:\mathrm{d}y ,$$ where $\mathcal{C}: x^2+y^2=1$, with $x,y\geq0$, is the upper semi-circle from $A=(1,0)$ to $B=(-1,0)$.
I know that the integrand is an exact differential and consequently the integral is independent of the path, so it would be sufficient to find the parent function of the differential, which in this case is $F(x,y)=xye^{-(x^2+y^2)+c}$, and calculate the difference $F(B)-F(A)$, which for our choice of $A$ and $B$ would be $0$, if I am not mistaken.
To practice, however, I also wanted to calculate the integral explicitly, especially because the exercise contained the following hint:
If necessary, consider the substitution $x=cos(t)$ when solving $\int\frac{1-2x^2}{\sqrt{1-x^2}}$.
This is how far I got: $$\begin{align} &\int_\mathcal{C} ye^{-(x^2+y^2)} (1-2x^2) \:\mathrm{d}x + xe^{-(x^2+y^2)} (1-2y^2) \:\mathrm{d}y \\ &= \int_1^{-1} ye^{-(x^2+y^2)} (1-2x^2) \:\mathrm{d}x + \int_0^0 xe^{-(x^2+y^2)} (1-2y^2) \:\mathrm{d}y \\ &= \int_1^{-1} ye^{-(x^2+y^2)} (1-2x^2) \:\mathrm{d}x \\ &= e^{-1} \int_1^{-1} \sqrt{1-x^2} \cdot (1-2x^2) \:\mathrm{d}x \\ \end{align}$$
Question 1: How to go on from here? I am not quite sure how to make use of the hint and the fact that $\int\frac{1-2x^2}{\sqrt{1-x^2}} = x\sqrt{1-x^2} +c$.
Question 2: If I type this integral into an online integral calculator, I get $\int_1^{-1} \sqrt{1-x^2} \cdot (1-2x^2) \:\mathrm{d}x = -\frac{\pi}{4}$. But, by my reasoning above, the line integral should be equal to 0. Where is my mistake?
I'm not really sure if this solution is correct, but maybe you should try $x=\sin(t)$
*I suppose your steps before are correct...
$$\int\sqrt{1-x^2}\cdot(1-2x^2)dx$$ $$=\int\sqrt{1-\cos^2(t)}\cdot(1-2\cos^2(t))\cdot\cos(t)dt$$ $$=\int\sin(t)\cos(2t)\cos(t)dt$$ $$=\int\sin(t)\cdot(\cos^2(t)-\sin^2(t))\cdot\cos(t)dt$$ $$=\int\sin(t)\cos^3(t)dt-\int\sin^3(t)\cos(t)dt$$ Basically substitute with $\cos(t)$ and $\sin(t)$ $$=\int u^3du-\int v^3dv$$ $$={1\over4}\cos^4(t)-{1\over4}\sin^4(t)$$ $$={1\over4}(\sqrt{1-x^2}-x^4)$$