\(\displaystyle\int_0^{1/16}\frac{\arcsin8x}{\sqrt{1-64x^2}}\,\mathrm dx\)
First let \(y=8x\), so that \(\mathrm dx=\dfrac{\mathrm dy}8\) to write the integral as
\(\displaystyle\frac18\int_0^{1/2}\frac{\arcsin y}{\sqrt{1-y^2}}\,\mathrm dy\)
Now recall that \((\arcsin y)'=\dfrac1{\sqrt{1-y^2}}\), so substituting \(z=\arcsin y\) should do the trick. The integral then becomes
\(\displaystyle\frac18\int_0^{\pi/6}z\,\mathrm dz=\frac1{16}z^2\bigg|_{z=0}^{z=\pi/6}=\frac{\pi^2}{576}\)
