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Question Number 107212 by bemath last updated on 09/Aug/20

$$⊚bemath⊚$$$$\int x^6\sqrt{1-x^2}dx?$$

Answered by Ar Brandon last updated on 09/Aug/20

$$\mathcal{I}=\int {\mathrm{x}}^6\sqrt{1-{\mathrm{x}}^2}\mathrm{dx},\mathrm{x}=\sin \theta$$$$\mathcal{I}=\int {\sin }^6\theta {\cos }^2\theta \mathrm{d}\theta =\int ({\sin }^6\theta -{\sin }^8\theta )\mathrm{d}\theta$$$${(\mathrm{z}-\frac{1}{\mathrm{z}})}^6={\left(2\mathrm{isin}\theta \right)}^6,\mathrm{z}=\cos \theta +\mathrm{isin}\theta$$$${\mathrm{z}}^6-{6\mathrm{z}}^5\left(\frac{1}{\mathrm{z}}\right)+{15\mathrm{z}}^4\left(\frac{1}{{\mathrm{z}}^2}\right)-{20\mathrm{z}}^3\left(\frac{1}{{\mathrm{z}}^3}\right)+{15\mathrm{z}}^2\left(\frac{1}{{\mathrm{z}}^4}\right)-6\mathrm{z}\left(\frac{1}{{\mathrm{z}}^5}\right)+\frac{1}{{\mathrm{z}}^6}=-{64\sin }^6\theta$$$${\mathrm{z}}^6+\frac{1}{{\mathrm{z}}^6}-6({\mathrm{z}}^4+\frac{1}{{\mathrm{z}}^4})+15({\mathrm{z}}^2+\frac{1}{{\mathrm{z}}^2})-20=-{64\sin }^6\theta$$$$2\cos 6\theta -12\cos 4\theta +30\cos 2\theta -20=-{64\sin }^6\theta$$$${(\mathrm{z}-\frac{1}{\mathrm{z}})}^8={\left(2\mathrm{isin}\theta \right)}^8$$$$17213535217118285670562881$$$${\mathrm{z}}^8+\frac{1}{{\mathrm{z}}^8}-8({\mathrm{z}}^6+\frac{1}{{\mathrm{z}}^6})+28({\mathrm{z}}^4+\frac{1}{{\mathrm{z}}^4})-56({\mathrm{z}}^2+\frac{1}{{\mathrm{z}}^2})+70={256\sin }^8\theta$$$$2\cos 8\theta -16\cos 6\theta +56\cos 4\theta -112\cos 2\theta +70={256\sin }^8\theta$$$$\Rightarrow \mathcal{I}=-\frac{1}{64}[\frac{2}{6}\sin 6\theta -\frac{12}{4}\sin 4\theta +\frac{30}{2}\sin 2\theta -20\theta ]$$$$-\frac{1}{256}[\frac{2}{8}\sin 8\theta -\frac{16}{6}\sin 6\theta +\frac{56}{4}\sin 4\theta -\frac{112}{2}\sin 2\theta +70\theta ]+\mathcal{C}$$$$\mathcal{I}=\frac{1}{256}[10\theta -4\sin 2\theta -2\sin 4\theta +\frac{4}{3}\sin 6\theta -\frac{1}{4}\sin 8\theta ]+\mathcal{C}$$

Answered by bobhans last updated on 09/Aug/20

$${\mathrm{I}}_1=\int -\frac{1}{2}{\mathrm{x}}^5(-2\mathrm{x}\sqrt{1-{\mathrm{x}}^2})\mathrm{dx}=\int -\frac{1}{2}{\mathrm{x}}^5\sqrt{1-{\mathrm{x}}^2}\mathrm{d}(1-{\mathrm{x}}^2)$$$${\mathrm{I}}_1=-\frac{1}{2}{\mathrm{x}}^5\left(\frac{2}{3}{(1-{\mathrm{x}}^2)}^{3/2}\right)+\int \frac{5}{3}{\mathrm{x}}^4{(1-{\mathrm{x}}^2)}^{3/2}\mathrm{dx}$$$${\mathrm{I}}_2=\int -\frac{5}{6}{\mathrm{x}}^3(-2\mathrm{x}{(1-{\mathrm{x}}^2)}^{3/2})\mathrm{dx}$$$${\mathrm{I}}_2=\int -\frac{5}{6}{\mathrm{x}}^3{(1-{\mathrm{x}}^2)}^{3/2}\mathrm{d}(1-{\mathrm{x}}^2)$$$${\mathrm{I}}_2=-\frac{1}{3}{\mathrm{x}}^3{(1-{\mathrm{x}}^2)}^{5/2}+\int {\mathrm{x}}^2{(1-{\mathrm{x}}^2)}^{5/2}\mathrm{dx}$$$${\mathrm{I}}_3=\int -\frac{1}{2}\mathrm{x}(-2\mathrm{x}{(1-{\mathrm{x}}^2)}^{5/2})\mathrm{dx}$$$${\mathrm{I}}_3=-\frac{1}{2}\mathrm{x}.\frac{2}{7}{(1-{\mathrm{x}}^2)}^{7/2}+\int \frac{1}{7}{(1-{\mathrm{x}}^2)}^{7/2}$$$${\mathrm{I}}_3=-\frac{1}{7}\mathrm{x}{(1-{\mathrm{x}}^2)}^{7/2}+\int \frac{1}{7}{(1-{\mathrm{x}}^2)}^{7/2}\mathrm{dx}$$

Answered by 1549442205PVT last updated on 10/Aug/20

$$\mathrm{Set}\frac{1}{{\mathrm{x}}^2}-1={\mathrm{u}}^2\Rightarrow 2\mathrm{udu}=-\frac{2}{{\mathrm{x}}^3}\mathrm{dx}$$$$\Rightarrow \mathrm{dx}=-{\mathrm{ux}}^3\mathrm{du},\sqrt{1-{\mathrm{x}}^2}=\mathrm{ux},{\mathrm{x}}^2=\frac{1}{{\mathrm{u}}^2+1}$$$$\mathrm{F}=-\int {\mathrm{x}}^6.\left(\mathrm{ux}\right)\left({\mathrm{ux}}^3\right)\mathrm{du}=-\int {\mathrm{x}}^{10}{\mathrm{u}}^2\mathrm{du}$$$$=-\int \frac{{\mathrm{u}}^2\mathrm{du}}{{(1+{\mathrm{u}}^2)}^5}=-\int \frac{{\mathrm{u}}^2+1-1}{{({\mathrm{u}}^2+1)}^5}\mathrm{du}$$$$=-\int \frac{\mathrm{du}}{{(1+{\mathrm{u}}^2)}^4}+\int \frac{\mathrm{du}}{{(1+{\mathrm{u}}^2)}^5}={\mathrm{I}}_5-{\mathrm{I}}_4$$$$$$$$\mathrm{Apply}\mathrm{the}\mathrm{current}\mathrm{formula}:$$$${\mathrm{I}}_{\mathrm{n}}=\frac{1}{{2\mathrm{a}}^2(\mathrm{n}-1)}.\frac{\mathrm{t}}{{({\mathrm{t}}^2+{\mathrm{a}}^2)}^{\mathrm{n}-1}}+\frac{1}{{\mathrm{a}}^2}.\frac{2\mathrm{n}-3}{2\mathrm{n}-2}.{\mathrm{I}}_{\mathrm{n}-1}$$$${\mathrm{I}}_4=\frac{1}{6}.\frac{\mathrm{u}}{{({\mathrm{u}}^2+1)}^3}+\frac{5}{6}{\mathrm{I}}_3=$$$$\frac{\mathrm{u}}{6{({\mathrm{u}}^2+1)}^3}+\frac{5}{6}[\frac{1}{4}.\frac{\mathrm{u}}{{({\mathrm{u}}^2+1)}^2}+\frac{3}{4}.{\mathrm{I}}_2]$$$$=\frac{\mathrm{u}}{6{({\mathrm{u}}^2+1)}^3}+\frac{5\mathrm{u}}{24{({\mathrm{u}}^2+1)}^2}+\frac{5}{8}{\mathrm{I}}_2$$$$=\frac{\mathrm{u}}{6{({\mathrm{u}}^2+1)}^3}+\frac{5\mathrm{u}}{24{({\mathrm{u}}^2+1)}^2}+\frac{5}{8}[\frac{1}{2}.\frac{\mathrm{u}}{({\mathrm{u}}^2+1)}+\frac{1}{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2+1}]$$$$=\frac{\mathrm{u}}{6{({\mathrm{u}}^2+1)}^3}+\frac{5\mathrm{u}}{24{({\mathrm{u}}^2+1)}^2}+\frac{5\mathrm{u}}{16({\mathrm{u}}^2+1)}+\frac{5}{16}{\tan }^{-1}\left(\mathrm{u}\right)$$$${\mathrm{I}}_5=\frac{1}{8}.\frac{\mathrm{u}}{{({\mathrm{u}}^2+1)}^4}+\frac{7}{8}.{\mathrm{I}}_4.\mathrm{Therefore},$$$$\mathrm{F}={\mathrm{I}}_5-{\mathrm{I}}_4=\frac{1}{8}.\frac{\mathrm{u}}{{({\mathrm{u}}^2+1)}^4}-\frac{1}{8}{\mathrm{I}}_4$$$$=\frac{\mathrm{u}}{8{({\mathrm{u}}^2+1)}^4}-\frac{1}{8}[\frac{\mathrm{u}}{6{({\mathrm{u}}^2+1)}^3}+\frac{5\mathrm{u}}{24{({\mathrm{u}}^2+1)}^2}+\frac{5\mathrm{u}}{16({\mathrm{u}}^2+1)}+\frac{5}{16}{\tan }^{-1}\left(\mathrm{u}\right)]+\mathrm{C}$$$$=\frac{\mathbf{u}}{8{({\mathbf{u}}^2+1)}^4}-\frac{\mathbf{u}}{48{({\mathbf{u}}^2+1)}^3}-\frac{5\mathbf{u}}{192{({\mathbf{u}}^2+1)}^2}$$$$-\frac{5\mathbf{u}}{128({\mathbf{u}}^2+1)}-\frac{5}{128}{\mathbf{tan}}^{-1}\left(\mathbf{u}\right)+\mathrm{C}$$$$\mathrm{Substituting}{\mathrm{u}}^2+1=1/{\mathrm{x}}^2\mathrm{and}\mathrm{u}=\frac{\sqrt{1-{\mathrm{x}}^2}}{\mathrm{x}}$$$$\mathrm{we}\mathrm{get}:$$$$\mathbf{F}=\frac{{\mathbf{x}}^7\sqrt{1-{\mathbf{x}}^2}}{8}-\frac{{\mathbf{x}}^5\sqrt{1-{\mathbf{x}}^2}}{48}-\frac{5{\mathbf{x}}^3\sqrt{1-{\mathbf{x}}^2}}{192}$$$$-\frac{5\mathbf{x}\sqrt{1-{\mathbf{x}}^2}}{128}-\frac{5}{128}{\mathbf{tan}}^{-1}\left(\frac{\sqrt{1-{\mathbf{x}}^2}}{\mathbf{x}}\right)+\mathrm{C}$$

Answered by mathmax by abdo last updated on 10/Aug/20

$$\mathrm{A}=\int {\mathrm{x}}^6\sqrt{1-{\mathrm{x}}^2}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\mathrm{sint}\Rightarrow$$$$\mathrm{A}=\int {\sin }^6\mathrm{t}{\cos }^2\mathrm{t}\mathrm{dt}=\int {\left({\sin }^2\mathrm{t}\right)}^3\left(\frac{1+\cos \left(2\mathrm{t}\right)}{2}\right)\mathrm{dt}$$$$=\frac{1}{2}\int {\left(\frac{1-\cos \left(2\mathrm{t}\right)}{2}\right)}^3(1+\cos \left(2\mathrm{t}\right))\mathrm{dt}$$$$=\frac{1}{16}\int {(1-\cos \left(2\mathrm{t}\right))}^2(1-{\cos }^2\left(2\mathrm{t}\right))\mathrm{dt}$$$$=\frac{1}{16}\int (1-2\cos \left(2\mathrm{t}\right)+{\cos }^2\left(2\mathrm{t}\right))(1-\frac{1+\cos \left(4\mathrm{t}\right)}{2})\mathrm{dt}$$$$=\frac{1}{32}\int (\frac{1+\cos \left(4\mathrm{t}\right)}{2}+1-2\cos \left(2\mathrm{t}\right))(1-\cos \left(4\mathrm{t}\right))\mathrm{dt}$$$$=\frac{1}{64}\int (3+\cos \left(4\mathrm{t}\right)-4\cos \left(2\mathrm{t}\right))(1-\cos \left(4\mathrm{t}\right))\mathrm{dt}$$$$=\frac{1}{64}\int (3-3\cos \left(4\mathrm{t}\right)+\cos \left(4\mathrm{t}\right)-{\cos }^2\left(4\mathrm{t}\right)-4\cos \left(2\mathrm{t}\right)+4\cos \left(2\mathrm{t}\right)\cos \left(4\mathrm{t}\right))\mathrm{dt}$$$$=\frac{1}{64}\int (3-2\cos \left(4\mathrm{t}\right)-\frac{1+\cos \left(8\mathrm{t}\right)}{2}-4\cos \left(2\mathrm{t}\right)+2\{\cos \left(6\mathrm{t}\right)+\cos \left(2\mathrm{t}\right)\})\mathrm{dt}$$$$=\frac{1}{128}\int \{6-4\cos \left(4\mathrm{t}\right)-1-\cos \left(8\mathrm{t}\right)-8\cos \left(2\mathrm{t}\right)+4\cos \left(6\mathrm{t}\right)+4\cos \left(2\mathrm{t}\right)\}\mathrm{dt}$$$$=\frac{1}{128}\int \{5-4\cos \left(4\mathrm{t}\right)-\cos \left(8\mathrm{t}\right)-4\cos \left(2\mathrm{t}\right)+4\cos \left(6\mathrm{t}\right)\}\mathrm{dt}$$$$=\frac{5}{128}\mathrm{t}-\frac{4}{128}\int \cos \left(4\mathrm{t}\right)\mathrm{dt}-\frac{1}{128}\int \cos \left(8\mathrm{t}\right)\mathrm{dt}-\frac{4}{128}\int \cos \left(2\mathrm{t}\right)\mathrm{dt}$$$$+\frac{4}{128}\int \cos \left(6\mathrm{t}\right)\mathrm{dt}$$$$=\frac{5}{128}\mathrm{t}-\frac{1}{128}\sin \left(4\mathrm{t}\right)-\frac{1}{8.128}\sin \left(8\mathrm{t}\right)-\frac{1}{64}\sin \left(2\mathrm{t}\right)+\frac{4}{6.128}\sin \left(6\mathrm{t}\right)+\mathrm{C}$$$$\mathrm{t}=\mathrm{arcsinx}$$

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