I need help on finding the closed form for $(1)$. Partially I managed to find a part of it.
$$\int_{0}^{\pi}{\sin^{2k-1}(x)\over 1+\cos^2(x)}\mathrm dx=2^{k-2}\pi+F(k)\tag1$$
$k\ge1$
I only managed to work out for
k=1: ${\pi\over 2}$
k=2: $\pi- 2$
k=3: $2\pi-{16\over 3}$
k=4: $4\pi-{176\over 15}$
$$I(k)=\int_{0}^{\pi}\frac{\sin(x)^{2k-1}}{1+\cos^2(x)}\,dx = 2\int_{0}^{\pi/2}\frac{\sin(x)^{2k-1}}{1+\cos^2(x)}\,dx=2\int_{0}^{\pi/2}\frac{\cos(x)^{2k-1}}{1+\sin(x)^2}\,dx$$
leads to:
$$ I(k) = 2 \int_{0}^{1}\frac{(1-t^2)^{k-1}}{1+t^2}\,dt = 2\int_{0}^{1}\frac{(2-(1+t^2))^{k-1}}{1+t^2}\,dt $$
where:
$$ (2-(1+t^2))^{k-1} = \sum_{j=0}^{k-1}\binom{k-1}{j}(1+t^2)^j (-1)^j 2^{k-1-j} $$
gives a way to put $I(k+1)-2\,I(k)$ in a simple closed form. We also have
$$ \sum_{k\geq 1}I(k)\,z^{2k}=\int_{0}^{\pi}\frac{z^2\sin(x)}{1-z^2\sin(x)}\,dx =\frac{2z}{\sqrt{1-z^2}}\,\arctan\left(\frac{z}{\sqrt{1-z^2}}\right)$$
as suggested by Sangchul Lee, so the problem boils down to computing the Taylor series of $\frac{1}{\sqrt{1-z^2}}$ or $\arcsin(z)$ in a neighbourhood of the origin, simple through the extended binomial theorem.
One way or another we get:
$$ \delta_k\stackrel{\text{def}}{=}I(k+1)-2\,I(k) = -\int_{0}^{\pi/2}\sin(x)^{2k-1}\,dx = -\frac{4^k}{k\binom{2k}{k}}$$
and
$$\begin{eqnarray*} \delta_k + 2 \delta_{k-1} + 4\delta_{k-2}+\ldots +2^{k-1} \delta_{1} &=& I(k+1)-2^k I(1)\\&=&-\sum_{j=1}^{k}\frac{4^j 2^{k-j}}{j\binom{2j}{j}}\end{eqnarray*}$$
so:
$$\boxed{ \int_{0}^{\pi}\frac{\sin(x)^{2k-1}}{1+\cos(x)^2}\,dx = \color{red}{\pi 2^{k-2}-\frac{1}{2}\sum_{j=1}^{k-1}\frac{2^{j+k}}{j\binom{2j}{j}}}. }$$
An interesting consequence is that the last identity proves
$$ \sum_{j\geq 1}\frac{2^j}{j\binom{2j}{j}}=\frac{\pi}{2} $$
since $\lim_{k\to +\infty}I(k)=0$, obviously. By Laplace's method,
$$ \boxed{\int_{0}^{\pi}\frac{\sin(x)^{2k-1}}{1+\cos(x)^2}\,dx \approx \color{red}{\frac{2\sqrt{\pi}}{\sqrt{4k+2}}}} $$
for large values of $k$.