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