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

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.

• 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. – Decaf-Math Dec 13 '18 at 20:08
• You have an elliptic integral. Furthermore, at $x = \frac {\pi}{4}$ the integrand is complex. – Doug M Dec 13 '18 at 20:22
• u = 2.sin x is a nice substitution but results in complex numbers. – William Elliot Dec 13 '18 at 20:30
• You're taking the square root of negative numbers in the integral. Is that what you want? – 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.
• Thanks, Felix!${}$ – Clayton Dec 13 '18 at 20:51