calculus - Evaluating the integral $int_0^{pi/4}sqrt{1-16sin^2(x)}mathop{}!mathrm dx$

How can we evaluate this integral?
$$\int_\limits{0}^{\pi/4}\sqrt{1-16\sin^2(x)}\mathop{}\!\mathrm dx$$
I tried a substitution
$$u=4\sin x,\quad \mathrm dx=\frac{\mathrm du}{\sqrt{16-u^2}}$$
hence the integration will be

$$\int_\limits{u=0}^{u=2\sqrt{2}}\frac{\sqrt{1-u^2}}{\sqrt{16-u^2}}\mathop{}\!\mathrm du$$
But I could not complete the solution using this substitution.

Answer

\begin{align}
\int_{0}^{\pi/4}\,\sqrt{\,{1 - 16\sin^{2}\left(\, x\,\right)}\,}\,\,\mathrm{d}x & =
\int_{0}^{\pi/4}\,\sqrt{\,{1 - 4^{2}\sin^{2}\left(\, x\,\right)}\,}\,\,\mathrm{d}x
\\[5mm] & =
\bbox[10px,#ffd,border:1px groove navy]{\mathrm{E}\left(\,{{\pi \over 4},4}\,\right)}
\end{align}

$\displaystyle\mathrm{E}$ is a
Legendre Integral.