Elliptic Integral

As shown in the second link provided above, using the Maclaurin series expansion for $(1-x)^{\frac {-1}{2}},$

$\int_{0}^{\frac {\pi}{2}} \frac {d\mu}{\sqrt {1-k^2\sin^2(\mu)}} = \sum_{n=0}^\infty \bigg(\frac {(2n-1)!!}{n!}\bigg) \cdot \frac {k^{2n}\sqrt {\pi}\Gamma (n+\frac {1}{2})}{2^{n+1}\Gamma(n+1)}$

Observe that -

$\Gamma (0+\frac {1}{2}) = \sqrt {\pi}, \Gamma (1+\frac {1}{2}) = \frac {\sqrt {\pi}}{2}, \Gamma (2+\frac {1}{2}) = \frac {3\cdot1\sqrt {\pi}}{2^2} , \Gamma (3+\frac {1}{2}) = \frac {5\cdot3\cdot1\sqrt {\pi}}{2^3}$

$\therefore \Gamma \left(n+\frac {1}{2}\right) = \frac {(2n-1)!!\sqrt {\pi}}{2^n}$

by induction. Substituting into the first equation we get

$\sum_{n=0}^\infty \bigg(\frac {(2n-1)!!}{n!}\bigg)^2 \cdot \frac {k^{2n}\pi}{2^{2n+1}} = \int_{0}^{\frac {\pi}{2}} \frac {d\mu}{\sqrt {1-k^2\sin^2(\mu)}}$

It is clear from the above that $k^2=256$ hence $\boxed {\sqrt k=4}.$