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Question Number 133591 by benjo_mathlover last updated on 23/Feb/21

$$\underset{0}{\stackrel{1}{\int }}{\mathrm{x}}^2\sqrt{1-{\mathrm{x}}^2}\mathrm{dx}?$$

Answered by EDWIN88 last updated on 23/Feb/21

$$\mathrm{by}\mathrm{Ostrogradsky}\mathrm{method}$$$$\int x^2\sqrt{1-x^2}dx=({ax}^3+{bx}^2+cx+d)\sqrt{1-x^2}+\int \frac{\mathrm{e}}{\sqrt{1-x^2}}dx$$$$\mathrm{differentiating}\mathrm{both}\mathrm{sides}\mathrm{give}$$$$x^2\sqrt{1-x^2}=\frac{d}{dx}({ax}^3+{bx}^2+cx+d)\sqrt{1-x^2}+\frac{e}{\sqrt{1-x^2}}$$$$x^2\sqrt{1-x^2}=(3{ax}^2+2bx+c)\sqrt{1-x^2}-\frac{x({ax}^3+{bx}^2+cx+x)}{\sqrt{1-x^2}}+\frac{e}{\sqrt{1-x^2}}$$$$\mathrm{after}\mathrm{solving}\mathrm{for}\mathrm{coefficient}\mathrm{we}\mathrm{get}$$$$a=\frac{1}{4};\mathrm{b}=0;c=-\frac{1}{8};\mathrm{d}=0;\mathrm{e}=\frac{1}{8}$$$$\mathrm{then}\int x^2\sqrt{1-x^2}dx=(\frac{1}{4}{x}^3-\frac{1}{8}x)\sqrt{1-x^2}+\int \frac{1}{8\sqrt{1-x^2}}dx$$$$or{\int }_0^1{x}^2\sqrt{1-x^2}dx={[(\frac{1}{4}{x}^3-\frac{1}{8}x)\sqrt{1-x^2}+\frac{1}{8}\arcsin \left(x\right)]}_0^1$$$$=\frac{\pi }{16}$$

Commented by benjo_mathlover last updated on 23/Feb/21

$$\mathrm{amazing}$$

Answered by Ñï= last updated on 23/Feb/21

$${\int }_0^1{x}^2\sqrt{1-x^2}dx$$$$={\int }_0^{\pi /2}{sin}^2\theta {cos}^2\theta d\theta$$$$=\frac{{\mathrm{\Gamma }}^2\left(\frac{3}{2}\right)}{2\mathrm{\Gamma }\left(3\right)}=\frac{1}{16}\pi$$

Answered by SEKRET last updated on 23/Feb/21

$${\int }_0^1{\mathbf{x}}^2\cdot \sqrt{1-{\mathbf{x}}^2}\mathbf{dx}\left[\begin{array}{c}\mathbf{x}=\mathbf{sin}\left(\mathbf{t}\right){/}_0^1\to {/}_0^{\frac{\mathit{\pi }}{2}}\\ \mathbf{dx}=\mathbf{cos}\left(\mathbf{t}\right)\mathbf{dt}\end{array}\right]$$$${\int }_0^{\frac{\mathit{\pi }}{2}}{\mathbf{sin}}^2\left(\mathbf{x}\right)\cdot {\mathbf{cos}}^2\left(\mathbf{t}\right)\mathbf{dt}=\frac{1}{4}{\int }_0^{\frac{\mathit{\pi }}{2}}{\mathbf{sin}}^2\left(2\mathbf{t}\right)\mathbf{dt}$$$$\frac{1}{4}(\frac{1}{2}\mathbf{t}-\frac{\mathbf{sin}\left(4\mathbf{t}\right)}{8}){/}_0^{\frac{\mathit{\pi }}{2}}=\frac{\mathit{\pi }}{16}$$$$\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{L}\mathit{A}\mathit{Z}\mathit{I}\mathit{Z}\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{V}\mathit{A}\mathit{L}\mathit{I}\mathit{Y}\mathit{E}\mathit{V}$$

Answered by benjo_mathlover last updated on 23/Feb/21

Answered by Dwaipayan Shikari last updated on 23/Feb/21

$$\frac{1}{2}{\int }_0^1{u}^{\frac{1}{2}}{(1-u)}^{\frac{1}{2}}du{x}^2=u$$$$=\frac{{\mathrm{\Gamma }}^2\left(\frac{3}{2}\right)}{2\mathrm{\Gamma }\left(3\right)}=\frac{\pi }{16}$$

Answered by mathmax by abdo last updated on 24/Feb/21

$$\mathrm{I}={\int }_0^1{\mathrm{x}}^2\sqrt{1-{\mathrm{x}}^2}\mathrm{dx}\mathrm{changement}\mathrm{x}=\mathrm{sint}\mathrm{give}$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}{\sin }^2\mathrm{t}.\mathrm{cost}.\mathrm{cost}\mathrm{dt}={\int }_0^{\frac{\pi }{2}}{\left(\mathrm{sint}\mathrm{cost}\right)}^2\mathrm{dt}=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}{\sin }^2\left(2\mathrm{t}\right)\mathrm{dt}$$$$=\frac{1}{8}{\int }_0^{\frac{\pi }{2}}(1-\cos \left(4\mathrm{t}\right))\mathrm{dt}=\frac{\pi }{16}-\frac{1}{32}{\left[\sin \left(4\mathrm{t}\right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{16}-0\Rightarrow \mathrm{I}=\frac{\pi }{16}$$

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