Question

Find the definite integral from 0 to 1/16 of arcsin(8x)/sqrt(1-64x^2)dx

Answer 1

$\displaystyle\int_0^{1/16}\frac{\arcsin8x}{\sqrt{1-64x^2}}\,\mathrm dx$

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

$\displaystyle\frac18\int_0^{\pi/6}z\,\mathrm dz=\frac1{16}z^2\bigg|_{z=0}^{z=\pi/6}=\frac{\pi^2}{576}$