<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] First of all, since $t= \sqrt{1-r^2}$, we should have $dt/dr = \frac {-r}{\sqrt{1-r^2}}$ and $t^2 + r^2 = 1$.
Now
$$\int_0^1 \frac {r^3}{\sqrt{1-r^2}}dr = \int_{\color{red}1}^{\color{red}0} \frac {r^3}{\sqrt{1-r^2}}\frac {\sqrt{1-r^2}}{-r}dt = \int_0^1r^2dt = \int_0^1(\color{red}{1-t^2})dt = [t-\frac{t^3}3]_0^1 = \frac23$$
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