I have to calculate the line integral for $\int_{\mathrm{C2}}F\, dr\,$ where $C2$ is given by $$\frac{(cost,sint)}{1+e^t}$$ and $0\le t \lt \infty$ Where $\vec{F} = \big(\frac{x}{\sqrt{1-x^2-y^2}}, \frac{y}{\sqrt{1-x^2-y^2}}\big)$.
I have no idea how to get started, I have never worked with ranges with $\infty$ before. Any tips are appreciated!
Alternatively, as pointed out in the comments, $\vec F$ is conservative, with scalar potential
$$f(x,y)=-\sqrt{1-x^2-y^2}+c$$
since $f_x=\frac x{\sqrt{1-x^2-y^2}}$ and $f_y=\frac y{\sqrt{1-x^2-y^2}}$.
The the integral is
$$\begin{align} \int_{C_2}\vec F\cdot\mathrm d\vec r&=\int_{C_2} \nabla f(x,y)\cdot\mathrm d\vec r\\[1ex] &=\lim_{t\to\infty}f\left(\frac{\cos t}{1+e^t},\frac{\sin t}{1+e^t}\right)-\lim_{t\to0^+}f\left(\frac{\cos t}{1+e^t},\frac{\sin t}{1+e^t}\right) \end{align}$$
We have
$$\begin{align} g(t)=f\left(\frac{\cos t}{1+e^t},\frac{\sin t}{1+e^t}\right)&=-\sqrt{1-\frac{\cos^2t}{(1+e^t)^2}-\frac{\sin^2t}{(1+e^t)^2}}+c\\[1ex] &=-\frac{\sqrt{2e^t+e^{2t}}}{1+e^t}+c\\[1ex] &=-\frac{\sqrt{2e^{-t}+1}}{e^{-t}+1}+c \end{align}$$
As $\lim\limits_{t\to\infty}g(t)=-1+c$ and $\lim\limits_{t\to0^+}g(t)=-\frac{\sqrt3}2+c$, we find that the line integral has a value of $\boxed{\frac{\sqrt3}2-1}$.
Alternatively, you can compute the integral directly, if you feel so bold. Composing $\vec F$ with $x(t)=\frac{\cos t}{1+e^t}$ and $y(t)=\frac{\sin t}{1+e^t}$ gives
$$\begin{align} \vec F(x(t),y(t))&=\left(\frac{\frac{\cos t}{1+e^t}}{\sqrt{1-\left(\frac{\cos t}{1+e^t}\right)^2-\left(\frac{\sin t}{1+e^t}\right)^2}},\frac{\frac{\sin t}{1+e^t}}{\sqrt{1-\left(\frac{\cos t}{1+e^t}\right)^2-\left(\frac{\sin t}{1+e^t}\right)^2}}\right)\\[1ex] &=\frac{\sqrt{e^t(2+e^t)}}{(1+e^t)^2}(\cos t,\sin t) \end{align}$$
and differentiating the parameterization of $C_2$ gives
$$\frac{\mathrm d\vec r}{\mathrm dt}=\frac{\left(-(1+e^t)\sin t-e^t\cos t,(1+e^t)\cos t-e^t\sin t\right)}{(1+e^t)^2}$$
Now, in the integral,
$$\begin{align} \int_{C_2}\vec F\cdot\mathrm d\vec r&=\int_0^\infty \vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r}{\mathrm dt}\,\mathrm dt\\[1ex] &=-\int_0^\infty\frac{e^{\frac t2}}{(1+e^t)^2\sqrt{2+e^t}}\,\mathrm dt\\[1ex] &=-2\int_{\frac\pi4}^{\frac\pi2}\frac{\cos^3v}{\sqrt{1+\cos^2v}}\,\mathrm dv&\text{where }\tan v=e^{\frac t2}\\[1ex] &\vdots \end{align}$$
(The precise steps for finding the antiderivative elude me, but rest assured this integral has the same value as found earlier.)