[tex]\displaystyle\int5x\sqrt{1-x^4}\,\mathrm dx[/tex]
Take [tex]x=\sqrt y[/tex], so that [tex]y^2=x^4[/tex] and [tex]\dfrac{\mathrm dy}{2\sqrt y}=\mathrm dx[/tex]. Note that this assumes [tex]x\ge0[/tex] and so requires [tex]y\ge0[/tex].
[tex]\displaystyle5\int\sqrt y\sqrt{1-y^2}\dfrac{\mathrm dy}{2\sqrt y}=\frac52\int\sqrt{1-y^2}\,\mathrm dy[/tex]
Now take [tex]y=\sin z[/tex], so that [tex]z=\arcsin y[/tex]. Note that for this invertibility condition to hold, we require that [tex]-\dfrac\pi2\le z\le\dfrac\pi2[/tex], which means [tex]-1\le y\le1[/tex]. But since we already fixed [tex]y\ge0[/tex] with the previous substitution, we thus have [tex]0\le y\le1[/tex], which in turn restricts us to [tex]0\le z\le\dfrac\pi2[/tex]. So with this substitution, we have [tex]\mathrm dy=\cos z\,\mathrm dz[/tex], which gives
[tex]\displaystyle\frac52\int\sqrt{1-\sin^2z}\cos z\,\mathrm dz[/tex]
[tex]=\displaystyle\frac52\int\sqrt{\cos^2z}\cos z\,\mathrm dz[/tex]
[tex]=\displaystyle\frac52\int|\cos z|\cos z\,\mathrm dz[/tex]
Now over the interval [tex]0<z<\dfrac\pi2[/tex], we have [tex]\cos z>0[/tex], which means [tex]|\cos z|=\cos z[/tex], and so the integral is equivalent to
[tex]\displaystyle\frac52\int\cos^2z\,\mathrm dz=\frac54\int(1+\cos2z)\,\mathrm dz[/tex]
[tex]=\dfrac54z+\dfrac58\sin2z+C[/tex]
[tex]=\dfrac54z+\dfrac54\sin z\cos z+C[/tex]
[tex]=\dfrac54\arcsin y+\dfrac54\sin(\arcsin y)\cos(\arcsin y)+C[/tex]
[tex]=\dfrac54\arcsin y+\dfrac54y\sqrt{1-y^2}+C[/tex]
[tex]=\dfrac54\arcsin(x^2)+\dfrac54x^2\sqrt{1-x^4}+C[/tex]
