<s>[INST] Evaluate $\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$ Evaluate
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$$
My attempt : I put  $t= \sqrt{1-r^2}$ now  $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$
$$= 2\int_{0}^{1} r^2 dt$$
$t= \sqrt{1-r^2}\implies r^2= t^2-1$
$$2\int_{0}^{1} t^2-1 dt= 2[\frac{t^3}{3} -t]_{0}^{1}==-4/3$$
 [/INST] Or just integration by parts:
\begin{eqnarray*}\int_0^1 r^2 \frac{r}{\sqrt{1-r^2}}dr
& = & \left.-\sqrt{1-r^2}\cdot r^2\right|_0^1 + \int_0^1 (2r)\sqrt{1-r^2}\;dr \\
& = & \left.-\frac 23\left(1-r^2\right)^{\frac 32}\right|_0^1 \\
& = & \frac 23
\end{eqnarray*}
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