[tex]\displaystyle\int_{u=0}^{u=1}u\arcsin u^2\,\mathrm du[/tex]
Take [tex]t=u^2\implies\mathrm dt=2u\,\mathrm du[/tex], so the integral becomes
[tex]\displaystyle\frac12\int_{t=0}^{t=1}\arcsin t\,\mathrm dt[/tex]
Now integrate by parts, taking
[tex]f=\arcsin t\implies\mathrm df=\dfrac{\mathrm dt}{\sqrt{1-t^2}}[/tex]
[tex]\mathrm dg=\mathrm dt\implies g=t[/tex]
so that the integral becomes
[tex]\displaystyle\frac12\left(t\arcsin t\bigg|_{t=0}^{t=1}-\int_{t=0}^{t=1}\frac t{\sqrt{1-t^2}}\,\mathrm dt\right)[/tex]
Substitute [tex]s=1-t^2\implies\mathrm ds=-2t\,\mathrm dt[/tex] to get
[tex]\displaystyle\frac12\left(t\arcsin t\bigg|_{t=0}^{t=1}+\frac12\int_{s=1}^{s=0}\frac{\mathrm ds}{\sqrt s}\right)[/tex]
[tex]=\displaystyle\frac12\left(t\arcsin t\bigg|_{t=0}^{t=1}-\frac12\int_{s=0}^{s=1}s^{-1/2}\,\mathrm ds\right)[/tex]
[tex]=\dfrac12(\arcsin1-0)-\dfrac14(2s^{1/2})\bigg|_{s=0}^{s=1}[/tex]
[tex]=\dfrac\pi4-\dfrac12(\sqrt1-\sqrt0)[/tex]
[tex]=\dfrac\pi4-\dfrac12[/tex]
