<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] $$I=\int_{0}^{1} \frac{r^3 dr}{\sqrt{1-r^2}}$$
Let $r=\sin t \implies dr= \cos t dt$, then
$$I=\int_{0}^{\pi/2} \sin^3 t dt=\int_{0}^{\pi/2} \sin t(1-\cos^2 t) dt$$
$$I=\int_{0}^{\pi/2} (\sin t -\cos^2 t \sin t) dt$$
$$ I=-\cos t+\frac{\cos^3 t}{3}|_{0}^{\pi/2}=-(-1+1/3)=2/3.$$
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