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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 ⇒∫sin^3 udu=∫(1−cos^2 u)sinudu =−cosu+(1/3)cos^3 u+c =−(1/3)(3−cos^2 u)cosu+c =−(1/3)(2+x^2 )(√(1−x^2 ))+c ✓

$${x}={sinu} \\ $$$$\Rightarrow\int{sin}^{\mathrm{3}} {udu}=\int\left(\mathrm{1}−{cos}^{\mathrm{2}} {u}\right){sinudu} \\ $$$$=−{cosu}+\frac{\mathrm{1}}{\mathrm{3}}{cos}^{\mathrm{3}} {u}+{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3}−{cos}^{\mathrm{2}} {u}\right){cosu}+{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{c}\:\:\checkmark \\ $$$$ \\ $$

Answered by Frix last updated on 30/Aug/24

∫(x^3 /( (√(1−x^2 ))))dx =^(t=(√(1−x^2 ))) ∫(t^2 −1)dt=(t^3 /3)−t= =(t^2 −3)(t/3)=−(x^2 +2)((√(1−x^2 ))/3)+C

$$\int\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\:\overset{{t}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {=}\:\int\left({t}^{\mathrm{2}} −\mathrm{1}\right){dt}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−{t}= \\ $$$$=\left({t}^{\mathrm{2}} −\mathrm{3}\right)\frac{{t}}{\mathrm{3}}=−\left({x}^{\mathrm{2}} +\mathrm{2}\right)\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{3}}+{C} \\ $$

Commented by mm1342 last updated on 30/Aug/24

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$$\:\underbrace{\frown} \\ $$

Commented by Frix last updated on 30/Aug/24

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Answered by MathematicalUser2357 last updated on 27/Sep/24

∫(x^3 /( (√(1−x^2 ))))dx⇒_(du=−2xdx) ^(u=1−x^2 ) −∫((u−1)/(2u))du=−(1/2)∫(1−(1/u))du=−(1/2)(u−ln u)+C =(1/2)(x^2 +ln(1−x^2 ))+C

$$\int\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\underset{{du}=−\mathrm{2}{xdx}} {\overset{{u}=\mathrm{1}−{x}^{\mathrm{2}} } {\Rightarrow}}−\int\frac{{u}−\mathrm{1}}{\mathrm{2}{u}}{du}=−\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}−\frac{\mathrm{1}}{{u}}\right){du}=−\frac{\mathrm{1}}{\mathrm{2}}\left({u}−\mathrm{ln}\:{u}\right)+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\right)+\mathrm{C} \\ $$

Commented by MathematicalUser2357 last updated on 27/Sep/24

I can use the chain rule (d/dx)f(g(x))=f′(g(x))g′(x) for quick integration ⋇ ∫f′(g(x))g′(x)dx=f(g(x))+C

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

Commented by Frix last updated on 27/Sep/24

∫(x^3 /( (√(1−x^2 ))))dx =^([u=1−x^2 ]) (1/2)∫((u−1)/( (√u)))du= =(1/2)∫(u^(1/2) −u^(−(1/2)) )du=(u^(3/2) /3)−u^(1/2) =(((u−3)(√u))/3)= =−(((x^2 +2)(√(1−x^2 )))/3)+C

$$\int\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\:\overset{\left[{u}=\mathrm{1}−{x}^{\mathrm{2}} \right]} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{u}−\mathrm{1}}{\:\sqrt{{u}}}{du}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left({u}^{\frac{\mathrm{1}}{\mathrm{2}}} −{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right){du}=\frac{{u}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}−{u}^{\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\left({u}−\mathrm{3}\right)\sqrt{{u}}}{\mathrm{3}}= \\ $$$$=−\frac{\left({x}^{\mathrm{2}} +\mathrm{2}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{3}}+{C} \\ $$

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