= Solution
Shrink the positive outer <contour> to the <branch cut> $[-1,1]$. On its upper bank the chosen <branch of a multivalued function> has $\sqrt{x^2-1}=i\sqrt{1-x^2}$ and on the lower bank it has the negative of this value. The upper bank runs from $1$ to $-1$, and the lower bank runs from $-1$ to $1$. The endpoint circles contribute $O(\sqrt\epsilon)$ and vanish. Hence
$$
\frac{3\pi i}{4}=2i\int_{-1}^1\frac{x^4}{\sqrt{1-x^2}}\,dx,\qquad
\boxed{\int_{-1}^1\frac{x^4}{\sqrt{1-x^2}}\,dx=\frac{3\pi}{8}.}
$$
The <change of variables> $x=\sin t$ independently reduces the result to the <integral> of $\sin^4t$ over $[-\pi/2,\pi/2]$.
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