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Question Number 211184 by nothing48 last updated on 30/Aug/24

Answered by mm1342 last updated on 30/Aug/24

$$x=sinu$$$$\Rightarrow \int {sin}^{3}udu=\int (1-{cos}^{2}u)sinudu$$$$=-cosu+\frac{1}{3}{cos}^{3}u+c$$$$=-\frac{1}{3}(3-{cos}^{2}u)cosu+c$$$$=-\frac{1}{3}(2+x^2)\sqrt{1-x^2}+c✓$$$$$$

Answered by Frix last updated on 30/Aug/24

$$\int \frac{x^3}{\sqrt{1-x^2}}dx\stackrel{t=\sqrt{1-x^2}}{=}\int (t^2-1)dt=\frac{t^3}{3}-t=$$$$=(t^2-3)\frac{t}{3}=-(x^2+2)\frac{\sqrt{1-x^2}}{3}+C$$

Commented by mm1342 last updated on 30/Aug/24

$$\underbrace{⌢}$$

Commented by Frix last updated on 30/Aug/24

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Answered by MathematicalUser2357 last updated on 27/Sep/24

$$\int \frac{x^3}{\sqrt{1-x^2}}dx\underset{du=-2xdx}{\stackrel{u=1-x^2}{\Rightarrow }}-\int \frac{u-1}{2u}du=-\frac{1}{2}\int (1-\frac{1}{u})du=-\frac{1}{2}(u-\ln u)+\mathrm{C}$$$$=\frac{1}{2}(x^2+\ln (1-x^2))+\mathrm{C}$$

Commented by MathematicalUser2357 last updated on 27/Sep/24

$$\mathrm{I}\mathrm{can}\mathrm{use}\mathrm{the}\mathrm{chain}\mathrm{rule}\frac{d}{dx}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)g^{\prime }\left(x\right)\mathrm{for}\mathrm{quick}\mathrm{integration}$$$$\divideontimes \int f^{\prime }\left(g\left(x\right)\right)g^{\prime }\left(x\right)dx=f\left(g\left(x\right)\right)+C$$

Commented by Frix last updated on 27/Sep/24

$$\int \frac{x^3}{\sqrt{1-x^2}}dx\stackrel{[u=1-x^2]}{=}\frac{1}{2}\int \frac{u-1}{\sqrt{u}}du=$$$$=\frac{1}{2}\int (u^{\frac{1}{2}}-u^{-\frac{1}{2}})du=\frac{u^{\frac{3}{2}}}{3}-u^{\frac{1}{2}}=\frac{(u-3)\sqrt{u}}{3}=$$$$=-\frac{(x^2+2)\sqrt{1-x^2}}{3}+C$$

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