= Solution
{c}
The <contour integral> along a piecewise smooth path is defined by parametrization:
$$
\int_\gamma f(z)\,dz=\int_a^b f(\gamma(t))\gamma'(t)\,dt,
$$
with a sum over the smooth pieces. The <residue theorem> states that for a positively oriented simple closed contour and a function holomorphic on and inside it apart from finitely many isolated singularities $a_j$ inside,
$$
\oint_\gamma f(z)\,dz=2\pi i\sum_j\operatorname{Res}(f,a_j).
$$
More generally a closed contour uses the <winding number> of each singularity as its weight.
Set $z=e^{it}$, so $dt=dz/(iz)$ and $\cos t=(z+z^{-1})/2$. The required real <integral> becomes
$$
I=\frac1i\oint_{|z|=1}\frac{z}{2z^4+5z^2+2}\,dz.
$$
The denominator factors as $(2z^2+1)(z^2+2)$. Only the simple <poles> $z=\pm i/\sqrt2$ lie inside the <unit circle>. At either one,
$$
\operatorname{Res}\left(\frac z{2z^4+5z^2+2},z_0\right)
=\frac{z_0}{8z_0^3+10z_0}=\frac1{8z_0^2+10}=\frac16.
$$
The <residue theorem> now gives
$$
\boxed{I=\frac1i\,2\pi i\left(\frac16+\frac16\right)=\frac{2\pi}{3}.}
$$
Back to article page
