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Question Number 162117 by CM last updated on 27/Dec/21

Commented by cortano last updated on 27/Dec/21

$$\int x^2\sqrt{4-{(1-x)}^2}dx=?$$$$[1-x=2\sin t\Rightarrow dx=-2\cos tdt]$$$$\int {(1-2\sin t)}^2(-2\cos {}^{2}t)dt$$$$=-\int (3-4\sin t-2\cos 2t)(1+\cos 2t)dt$$$$=-\int (2+\cos 2t-4\sin t-4\sin t\cos 2t-\cos 4t)dt$$$$=-2t-\frac{\sin 2t}{2}-4\cos t+\frac{\sin 4t}{4}+\int 4\sin t\cos 2tdt$$$$=-2\arcsin \left(\frac{1-x}{2}\right)-2\sqrt{3+2x-x^2}-\frac{(1-x)\sqrt{3+2x-x^2}}{4}$$$$+4\int (2\cos {}^{2}t-1)d\left(\cos t\right)$$$$=-2\arcsin \left(\frac{1-x}{2}\right)-\left(\frac{9-x}{4}\right)\sqrt{3+2x-x^2}$$$$+\frac{8}{3}\cos {}^{3}t-4\cos t+c$$

Answered by FongXD last updated on 27/Dec/21

$$\mathrm{I}=\int {\mathrm{x}}^2\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}\mathrm{dx}=\int {\mathrm{x}}^2\sqrt{4-{(\mathrm{x}-1)}^2}\mathrm{dx}$$$$\mathrm{let}\mathrm{x}-1=2\sin \theta ,\Rightarrow \mathrm{dx}=2\cos \theta \mathrm{d}\theta$$$$\mathrm{I}=\int {(2\sin \theta +1)}^2\sqrt{4-{4\sin }^2\theta }\times 2\cos \theta \mathrm{d}\theta$$$$\mathrm{I}=\int ({4\sin }^2\theta +4\sin \theta +1)\times {4\cos }^2\theta \mathrm{d}\theta$$$$\mathrm{I}=\int ({4\sin }^{2}2\theta +{4\cos }^2\theta )\mathrm{d}\theta +16\int {\cos }^2\theta \sin \theta \mathrm{d}\theta$$$$\mathrm{I}=\int (2-2\cos 4\theta +2+2\cos 2\theta )\mathrm{d}\theta -\frac{16}{3}{\cos }^3\theta$$$$\mathrm{I}=4\theta -\frac{1}{2}\sin 4\theta +\sin 2\theta -\frac{16}{3}{\cos }^3\theta +\mathrm{c}(\ast )$$$$\mathrm{but}\mathrm{x}-1=2\sin \theta ,\Rightarrow \sin \theta =\frac{\mathrm{x}-1}{2},\Rightarrow \theta =\arcsin \left(\frac{\mathrm{x}-1}{2}\right)$$$$\bullet {\cos }^3\theta =(1-{\sin }^2\theta )\sqrt{1-{\sin }^2\theta }=[1-{\left(\frac{\mathrm{x}-1}{2}\right)}^2]\sqrt{1-{\left(\frac{\mathrm{x}-1}{2}\right)}^2}=\frac{1}{8}(3+2\mathrm{x}-{\mathrm{x}}^2)\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}$$$$\bullet \sin 2\theta =2\sin \theta \sqrt{1-{\sin }^2\theta }=2\left(\frac{\mathrm{x}-1}{2}\right)\sqrt{1-{\left(\frac{\mathrm{x}-1}{2}\right)}^2}=\frac{1}{2}(\mathrm{x}-1)\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}$$$$\bullet \sin 4\theta =2\sin 2\theta \cos 2\theta =2\sin 2\theta (1-{2\sin }^2\theta )$$$$\iff \sin 4\theta =(\mathrm{x}-1)\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}[1-2{\left(\frac{\mathrm{x}-1}{2}\right)}^2]$$$$\Rightarrow \sin 4\theta =\frac{1}{2}(\mathrm{x}-1)(1+2\mathrm{x}-{\mathrm{x}}^2)\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}$$$$(\ast ):\mathrm{I}=4\arcsin \left(\frac{\mathrm{x}-1}{2}\right)-\frac{1}{4}(\mathrm{x}-1)(1+2\mathrm{x}-{\mathrm{x}}^2)\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}+\frac{1}{2}(\mathrm{x}-1)\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}-\frac{2}{3}(3+2\mathrm{x}-{\mathrm{x}}^2)\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}+\mathrm{c}$$$$\mathrm{I}=4\arcsin \left(\frac{\mathrm{x}-1}{2}\right)+\frac{1}{4}(\mathrm{x}-1)({\mathrm{x}}^2-2\mathrm{x}+1)\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}-\frac{2}{3}(3+2\mathrm{x}-{\mathrm{x}}^2)\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}+\mathrm{c}$$$$\mathrm{I}=4\arcsin \left(\frac{\mathrm{x}-1}{2}\right)+\frac{1}{12}[3({\mathrm{x}}^3-{3\mathrm{x}}^2+3\mathrm{x}-1)-8(3+2\mathrm{x}-{\mathrm{x}}^2)]\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}+\mathrm{c}$$$$\Rightarrow \mathrm{I}=4\arcsin \left(\frac{\mathrm{x}-1}{2}\right)+\frac{1}{12}({3\mathrm{x}}^3-{\mathrm{x}}^2-7\mathrm{x}-27)\sqrt{3+2\mathrm{x}-{\mathrm{x}}^2}+\mathrm{c}$$

Answered by mathmax by abdo last updated on 27/Dec/21

$$\mathrm{\Phi }=\int {\mathrm{x}}^2\sqrt{-{\mathrm{x}}^2+2\mathrm{x}+3}\mathrm{dx}\Rightarrow \mathrm{\Phi }=\int {\mathrm{x}}^2\sqrt{-({\mathrm{x}}^2-2\mathrm{x}-3)}\mathrm{dx}$$$$=\int {\mathrm{x}}^2\sqrt{-({\mathrm{x}}^2-2\mathrm{x}+1-4)}\mathrm{dx}=\int {\mathrm{x}}^2\sqrt{4-{(\mathrm{x}-1)}^2}\mathrm{dx}$$$$\mathrm{changement}\mathrm{x}-1=2\sin \theta \mathrm{give}$$$$\mathrm{\Phi }=\int {(1+2\sin \theta )}^{2}2\cos \theta \left(2\cos \theta \right)\mathrm{d}\theta$$$$=4\int {(2\sin \theta +1)}^2{\cos }^2\theta \mathrm{d}\theta$$$$=4\int ({4\sin }^2\theta +4\sin \theta +1){\cos }^2\theta \mathrm{d}\theta$$$$=16\int {\sin }^2\theta .{\cos }^2\theta +16\int {\cos }^2\theta \sin \theta \mathrm{d}\theta +4\int {\cos }^2\theta \mathrm{d}\theta$$$$=4\int {\sin }^2\left(2\theta \right)\mathrm{d}\theta -\frac{16}{3}{\cos }^3\theta +2\int (1+\cos \left(2\theta \right))\mathrm{d}\theta$$$$=2\int (1-\cos \left(4\theta \right))\mathrm{d}\theta -\frac{16}{3}{\cos }^3\theta +2\theta +\sin \left(2\theta \right)+\mathrm{C}$$$$=4\theta -\frac{1}{2}\sin \left(2\theta \right)-\frac{16}{3}{\cos }^3\theta +\sin \left(2\theta \right)+\mathrm{C}=4\theta +\frac{1}{2}\sin \left(2\theta \right)-\frac{16}{3}{\cos }^3\theta +\mathrm{c}$$$$\mathrm{but}\theta =\arcsin \left(\frac{\mathrm{x}-1}{2}\right)\Rightarrow$$$$\mathrm{\Phi }=4\arcsin \left(\frac{\mathrm{x}-1}{2}\right)+\frac{1}{2}\sin \left(2\arcsin \left(\frac{\mathrm{x}-1}{2}\right)\right)$$$$-\frac{16}{3}{\left(\sqrt{1-{\left(\frac{\mathrm{x}-1}{2}\right)}^2}\right)}^3+\mathrm{C}$$

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