Prove that:
$$I_{n}=\displaystyle\int_{0}^{2\alpha}{x^{n}\sqrt{2\alpha{x}-x^2}\;dx}=\frac{\pi\,\alpha^{n+2}}{2}\mathop{\prod}\limits_{k=2}^{n}{\frac{2k+1}{k+2}}\,,\quad n\in\mathbb{N}\,,\;\alpha>0$$
Solution
We are invoking Beta and Gamma functions, hence:
$$\begin{align*} I_{n}&=\displaystyle\int_{0}^{2\alpha}{x^{n}\sqrt{2\alpha{x}-x^2}\;dx}=\int_{0}^{2\alpha}{x^{n+\frac{1}{2}}\,({2\alpha-x})^{\frac{1}{2}}\,dx}\\ &\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,\frac{x}{2\alpha}} \\ {2\alpha\,dt\,=\,dx} \end{subarray}}\,\int_{0}^{1}{({2\alpha})^{n+\frac{1}{2}}\,t^{n+\frac{1}{2}}\,({2\alpha-2\alpha{t}})^{\frac{1}{2}}\,2\alpha\,dt}=({2\alpha})^{n+2}\int_{0}^{1}{t^{n+\frac{1}{2}}\,({1-t})^{\frac{1}{2}}\,2\alpha\,dt}\\ &=({2\alpha})^{n+2}\int_{0}^{1}{t^{n+\frac{3}{2}-1}\,({1-t})^{\frac{3}{2}-1}\,dt}=({2\alpha})^{n+2}\,{\rm{B}}\bigl({n+\tfrac{3}{2},\tfrac{3}{2}}\bigr)\\ &\stackrel{(1)}{=}({2\alpha})^{n+2}\,\frac{\Gamma\bigl({n+\tfrac{3}{2}}\bigr)\,\Gamma\bigl({\tfrac{3}{2}}\bigr)}{\Gamma(n+3)}\stackrel{(2)}{=}2^{n+2}\alpha^{n+2}\,\frac{\frac{\sqrt{\pi}}{2^{n+1}}\prod\limits_{k=1}^{n}{(2k+1)}\,\frac{\sqrt{\pi}}{2}}{2\prod\limits_{k=1}^{n}{(k+2)}}\\ &=\frac{\pi\,\alpha^{n+2}}{2}\mathop{\prod}\limits_{k=1}^{n}{\frac{2k+1}{k+2}}=\frac{\pi\,\alpha^{n+2}}{2}\mathop{\prod}\limits_{k=2}^{n}{\frac{2k+1}{k+2}}\,,\;n\in\mathbb{N}\end{align*}$$
ending the calculations.
We used the formulae:
$$ \begin{align*} &{\rm{B}}({x,y})=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} \quad&(1)\\ &\left\{{\begin{array}{rl} \Gamma\bigl({n+\tfrac{3}{2}}\bigr)\!&=\displaystyle\frac{\sqrt{\pi}}{2^{n+1}}\mathop{\prod}\limits_{k=1}^{n}{(2k+1)}\\ \Gamma\bigl({\tfrac{3}{2}}\bigr)\!&=\dfrac{\sqrt{\pi}}{2}\\ \Gamma({n+3})\!&=\displaystyle2\mathop{\prod}\limits_{k=1}^{n}{(k+2)} \end{array}}\right\}\quad&(2) \end{align*}$$
The exercise can also be found in mathimatikoi.org
$$I_{n}=\displaystyle\int_{0}^{2\alpha}{x^{n}\sqrt{2\alpha{x}-x^2}\;dx}=\frac{\pi\,\alpha^{n+2}}{2}\mathop{\prod}\limits_{k=2}^{n}{\frac{2k+1}{k+2}}\,,\quad n\in\mathbb{N}\,,\;\alpha>0$$
Solution
We are invoking Beta and Gamma functions, hence:
$$\begin{align*} I_{n}&=\displaystyle\int_{0}^{2\alpha}{x^{n}\sqrt{2\alpha{x}-x^2}\;dx}=\int_{0}^{2\alpha}{x^{n+\frac{1}{2}}\,({2\alpha-x})^{\frac{1}{2}}\,dx}\\ &\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,\frac{x}{2\alpha}} \\ {2\alpha\,dt\,=\,dx} \end{subarray}}\,\int_{0}^{1}{({2\alpha})^{n+\frac{1}{2}}\,t^{n+\frac{1}{2}}\,({2\alpha-2\alpha{t}})^{\frac{1}{2}}\,2\alpha\,dt}=({2\alpha})^{n+2}\int_{0}^{1}{t^{n+\frac{1}{2}}\,({1-t})^{\frac{1}{2}}\,2\alpha\,dt}\\ &=({2\alpha})^{n+2}\int_{0}^{1}{t^{n+\frac{3}{2}-1}\,({1-t})^{\frac{3}{2}-1}\,dt}=({2\alpha})^{n+2}\,{\rm{B}}\bigl({n+\tfrac{3}{2},\tfrac{3}{2}}\bigr)\\ &\stackrel{(1)}{=}({2\alpha})^{n+2}\,\frac{\Gamma\bigl({n+\tfrac{3}{2}}\bigr)\,\Gamma\bigl({\tfrac{3}{2}}\bigr)}{\Gamma(n+3)}\stackrel{(2)}{=}2^{n+2}\alpha^{n+2}\,\frac{\frac{\sqrt{\pi}}{2^{n+1}}\prod\limits_{k=1}^{n}{(2k+1)}\,\frac{\sqrt{\pi}}{2}}{2\prod\limits_{k=1}^{n}{(k+2)}}\\ &=\frac{\pi\,\alpha^{n+2}}{2}\mathop{\prod}\limits_{k=1}^{n}{\frac{2k+1}{k+2}}=\frac{\pi\,\alpha^{n+2}}{2}\mathop{\prod}\limits_{k=2}^{n}{\frac{2k+1}{k+2}}\,,\;n\in\mathbb{N}\end{align*}$$
ending the calculations.
We used the formulae:
$$ \begin{align*} &{\rm{B}}({x,y})=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} \quad&(1)\\ &\left\{{\begin{array}{rl} \Gamma\bigl({n+\tfrac{3}{2}}\bigr)\!&=\displaystyle\frac{\sqrt{\pi}}{2^{n+1}}\mathop{\prod}\limits_{k=1}^{n}{(2k+1)}\\ \Gamma\bigl({\tfrac{3}{2}}\bigr)\!&=\dfrac{\sqrt{\pi}}{2}\\ \Gamma({n+3})\!&=\displaystyle2\mathop{\prod}\limits_{k=1}^{n}{(k+2)} \end{array}}\right\}\quad&(2) \end{align*}$$
The exercise can also be found in mathimatikoi.org
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