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.

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    $\begingroup$ Accordingly, this cannot be solved in terms of elementary functions. In other words, there is no closed form solution and you must resort to numerical integration over a given interval. $\endgroup$ – Decaf-Math Dec 13 '18 at 20:08
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    $\begingroup$ You have an elliptic integral. Furthermore, at $x = \frac {\pi}{4}$ the integrand is complex. $\endgroup$ – Doug M Dec 13 '18 at 20:22
  • $\begingroup$ u = 2.sin x is a nice substitution but results in complex numbers. $\endgroup$ – William Elliot Dec 13 '18 at 20:30
  • $\begingroup$ You're taking the square root of negative numbers in the integral. Is that what you want? $\endgroup$ – zhw. Dec 13 '18 at 21:44

\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.

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    $\begingroup$ @Clayton I don't think so. Maybe, that's ONE of the reasons people define a new "special function". $\endgroup$ – Felix Marin Dec 13 '18 at 20:46
  • $\begingroup$ Thanks, Felix!${}$ $\endgroup$ – Clayton Dec 13 '18 at 20:51

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