prove that $\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$ 
prove that $$\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$$

My attempt:
C is semicircle in upper half complex plane
Simple poles = $e^{i\frac{\pi}{8}}, e^{i\frac{3\pi}{8}},e^{i\frac{5\pi}{8}},e^{i\frac{7\pi}{8}}$ lie in upper semi-circle C and real axis
Given integral value $= 2\pi i \cdot (\text{sum of residues}) = 2 \pi i \left(\frac{-1}{8}\right) \left[e^{i\frac{5\pi}{8}}+e^{i\frac{15\pi}{8}}+e^{i\frac{25\pi}{8}}+e^{i\frac{35\pi}{8}}\right] = 0.27059 \pi$
This is numerically equal to $\frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$. But without using calculator, how to get this expression.

Just extending what you've got so far. Let's note $z=e^{i\frac{5\pi}{8}}$ and recall that $\cos{z}=\frac{e^{iz}+e^{-iz}}{2}$, then:
$$2 \pi i \left(\frac{-1}{8}\right) \left[e^{i\frac{5\pi}{8}}+e^{i\frac{15\pi}{8}}+e^{i\frac{25\pi}{8}}+e^{i\frac{35\pi}{8}}\right]=
2 \pi i \left(\frac{-1}{8}\right) \left[z+z^3+z^5+z^7\right]=\\
2 \pi i \left(\frac{-1}{8}\right) z \left[1+z^2+z^4+z^6\right] = 
2 \pi i \left(\frac{-1}{8}\right) z \left[1+z^2+z^4(1+z^2)\right]=\\
2 \pi i \left(\frac{-1}{8}\right) z (1+z^2)(1+z^4)=
2 \pi i \left(\frac{-1}{8}\right) z^4 (z^{-1}+z)(z^{-2}+z^2)=\\
2 \pi i \left(\frac{-1}{8}\right) e^{i\frac{5\pi}{2}} 2\cos\left(\frac{5\pi}{8}\right)2 \cos\left(\frac{5\pi}{4}\right)=\pi i(-1)i \cos\left(\frac{\pi}{2}+\frac{\pi}{8}\right)\left(-\frac{1}{\sqrt{2}}\right)=\\
\frac{\pi}{\sqrt{2}}\sin\left(\frac{\pi}{8}\right)$$
