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Question Number 142116 by Eric002 last updated on 26/May/21

$$usetrigonometricsubstitutiontosolve$$$$\int \frac{x^3}{\sqrt{9-x^2}}dx$$

Answered by ZiYangLee last updated on 27/May/21

$$\int \frac{x^3}{\sqrt{9-x^2}}dx$$$$letx=3\sin \theta ,dx=3\cos \theta d\theta$$$$=\int \frac{27{\sin }^3\theta }{\sqrt{9-9{\sin }^2\theta }}\left(3\cos \theta d\theta \right)$$$$=\int \frac{{9\sin }^3\theta }{\cos \theta }\left(3\cos \theta d\theta \right)$$$$=27\int {\sin }^3\theta d\theta$$$$=27\int \sin \theta {\sin }^2\theta d\theta$$$$=27\int \sin \theta (1-{\cos }^2\theta )d\theta$$$$=27\int \sin \theta d\theta -27\int \sin \theta {\cos }^2\theta d\theta$$$$letu=\cos \theta \Rightarrow du=-\sin \theta d\theta$$$$=27(-\cos \theta )-27\int u^2(-du)$$$$=27(-\cos \theta )+27\int u^{2}du$$$$=-27\cos \theta +9u^3+C$$$$=-27\cos \theta +9{\cos }^3\theta +C$$$$=-27\left(\frac{\sqrt{9-x^2}}{3}\right)+9{\left(\frac{\sqrt{9-x^2}}{3}\right)}^3+C$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

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

$$\mathrm{\Phi }=\int \frac{{\mathrm{x}}^3}{\sqrt{9-{\mathrm{x}}^2}}\mathrm{dx}\mathrm{changement}\mathrm{x}=3\mathrm{sint}\mathrm{give}$$$$\mathrm{\Phi }=\int \frac{{27\sin }^3\mathrm{t}}{3\mathrm{cost}}\left(3\mathrm{cost}\right)\mathrm{dt}=27\int {\sin }^3\mathrm{t}\mathrm{dt}$$$$=27\int \mathrm{sint}(1-{\cos }^2\mathrm{t})\mathrm{dt}=27\int \mathrm{sint}\mathrm{dt}-27\int {\cos }^2\mathrm{t}\mathrm{sintdt}$$$$=-27\mathrm{cost}+9{\cos }^3\mathrm{t}+\mathrm{C}$$$$=-27\sqrt{1-{\sin }^2\mathrm{t}}+9{\left(\sqrt{1-{\sin }^2\mathrm{t}}\right)}^3)+\mathrm{C}$$$$=-27\sqrt{1-\frac{{\mathrm{x}}^2}{9}}+9{\left(\sqrt{1-\frac{{\mathrm{x}}^2}{9}}\right)}^3+\mathrm{C}$$$$$$

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