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)(\arcsin8x)/(√(1-64x^2))\,\mathrm dx`

First let
`y=8x`, so that
`\mathrm dx=\frac{\mathrm dy}8` to write the integral as


`\displaystyle\frac18\int_0^(1/2)(\arcsin y)/(√(1-y^2))\,\mathrm dy`

Now recall that
`(\arcsin y)'=\frac1{√(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)=(\pi^2)/(576)`