<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] With $r=\sin t$ it's$$\int_0^{\pi/2}\sin^3tdt=\int_0^{\pi/2}(\sin t-\sin t\cos^2t)dt=[\tfrac13\cos^3t-\cos t]_0^{\pi/2}=\tfrac23.$$
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