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.
\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}