Contour Integral of irrational polynomial from -1 to 1 I've been stuck at htis contour integral problem for a few hours now, and seem to be hitting brick walls.
$$
\int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^4}dx\,,
$$
I tried a trig substitution $x=\cos{\theta}$ but noticed all the poles were the unit circle and I didn't know how to proceed. 
$$
\frac{1}{2}\int_{0}^{2\pi} \frac{\sin^2{\theta}}{1+\cos^4{\theta}}d\theta\,,
$$
Next I thought maybe a rectangular contour but the contours that go vertically from $1$ to $1+i\infty$ and $-1+i\infty$ to $-1$ didn't seem to cancel, or I wasn't able to show that it does. I manipulated it for awhile.
$$
\int_{0}^\infty \frac{\sqrt{1-(1+iy)^2}}{1+(1+iy)^4}dy + \int_{0}^{-\infty} \frac{\sqrt{1-(1+iy)^2}}{1+(1+iy)^4}dy
$$
I naively wrote the trig integral in terms of z, but the denominator is an 8th order polynomial.
$$
i\oint \frac{2z^5-4z^3+2z}{z^8+4z^6+22z^4+4z^2+1} dz
$$
I think this might be getting closer to a solution, but how can I find the poles by hand? 
 A: As pointed out by @paul garrett, dog-bone contour (dumbbell contour) works perfectly. Indeed, consider the contour $\mathcal{C}$ given as follows:
$\hspace{12em}$
With the principal branch cut, as the radius/band-width of $\mathcal{C}$ goes to zero, 
\begin{align*}
\oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{i(z^4+1)} \, dz
&\longrightarrow
\int_{-1-0^+i}^{1-0^+i} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz
- \int_{-1+0^+i}^{1+0^+i} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz \\
&\quad = 2\int_{-1}^{1} \frac{\sqrt{1-x^2}}{x^4+1} \, dx.
\end{align*}
On the other hand, Residue Theorem tells that
\begin{align*}
\oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz
&= - 2\pi i \sum_{a \ : \ a^4 + 1 = 0} \underset{z=a}{\mathrm{Res}} \, \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \\
&= \pi \sqrt{2(\sqrt{2}-1)}.
\end{align*}
(In order to use Residue Theorem, consider a large circle, apply Residue Theorem to the region enclosed by this circle and $\mathcal{C}$, and then let the radius of the circle go to $\infty$.) Therefore
$$ \int_{-1}^{1} \frac{\sqrt{1-x^2}}{x^4+1} \, dx = \frac{\pi}{2} \sqrt{2(\sqrt{2}-1)}. $$
A: Here is a way to compute 
$$\int_{-\pi/2}^{\pi/2} \frac{\sin^2{\theta}}{1+\cos^4{\theta}}d\theta\,.$$
Let $t=\tan\theta$. Then 
$$\sin^2\theta=\frac{t^2}{1+t^2},\ \cos^2\theta=\frac{1}{1+t^2},\ d\theta=\frac{dt}{1+t^2}.$$
Plugging them in we get
$$\int_{-\infty}^\infty\frac{t^2}{(1+t^2)^2+1}dt.$$
You can find the roots of the denominator easily. So you can use any method you like, say residues or partial fractions, to compute the integral. 
