Question

Evaluate the integral

Answer 1

`\displaystyle\int_(u=0)^(u=1)u\arcsin u^2\,\mathrm du`

Take
`t=u^2\implies\mathrm dt=2u\,\mathrm du`, so the integral becomes


`\displaystyle\frac12\int_(t=0)^(t=1)\arcsin t\,\mathrm dt`

Now integrate by parts, taking


`f=\arcsin t\implies\mathrm df=(\mathrm dt)/(√(1-t^2))`

`\mathrm dg=\mathrm dt\implies g=t`

so that the integral becomes


`\displaystyle\frac12\left(t\arcsin t\bigg|_(t=0)^(t=1)-\int_(t=0)^(t=1)\frac t{√(1-t^2)}\,\mathrm dt\right)`

Substitute
`s=1-t^2\implies\mathrm ds=-2t\,\mathrm dt` to get


`\displaystyle\frac12\left(t\arcsin t\bigg|_(t=0)^(t=1)+\frac12\int_(s=1)^(s=0)(\mathrm ds)/(\sqrt s)\right)`

`=\displaystyle\frac12\left(t\arcsin t\bigg|_(t=0)^(t=1)-\frac12\int_(s=0)^(s=1)s^(-1/2)\,\mathrm ds\right)`

`=\frac12(\arcsin1-0)-\frac14(2s^(1/2))\bigg|_(s=0)^(s=1)`

`=\frac\pi4-\frac12(\sqrt1-\sqrt0)`

`=\frac\pi4-\frac12`