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Question Number 142116 by Eric002 last updated on 26/May/21
use trigonometric substitution to solve ∫(x^3 /( (√(9−x^2 ))))dx
$${use}\:{trigonometric}\:{substitution}\:{to}\:{solve} \\ $$$$\int\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{dx} \\ $$
Answered by ZiYangLee last updated on 27/May/21
∫(x^3 /( (√(9−x^2 )) )) dx let x=3sin θ, dx=3cos θ dθ =∫((27 sin^3 θ)/( (√(9−9 sin^2 θ)) ))(3 cos θ dθ) =∫ ((9sin^3 θ)/(cos θ))(3 cos θ dθ) =27∫ sin^3 θ dθ =27∫ sin θ sin^2 θ dθ =27∫ sin θ (1−cos^2 θ) dθ =27∫ sin θ dθ−27∫ sin θ cos^2 θ dθ let u=cos θ⇒du=−sin θ dθ =27(−cos θ)−27∫ u^2 (−du) =27(−cos θ)+27∫ u^2 du =−27cos θ+9u^3 +C =−27cos θ+9 cos^3 θ+C =−27(((√(9−x^2 ))/3))+9(((√(9−x^2 ))/3))^3 +C =−9(√(9−x^2 ))+(((9−x^2 )^(3/2) )/3)+C #
$$\int\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }\:}\:{dx} \\ $$$${let}\:{x}=\mathrm{3sin}\:\theta,\:{dx}=\mathrm{3cos}\:\theta\:{d}\theta \\ $$$$=\int\frac{\mathrm{27}\:\mathrm{sin}^{\mathrm{3}} \theta}{\:\sqrt{\mathrm{9}−\mathrm{9}\:\mathrm{sin}^{\mathrm{2}} \theta}\:}\left(\mathrm{3}\:\mathrm{cos}\:\theta\:{d}\theta\right) \\ $$$$=\int\:\frac{\mathrm{9sin}^{\mathrm{3}} \theta}{\mathrm{cos}\:\theta}\left(\mathrm{3}\:\mathrm{cos}\:\theta\:{d}\theta\right) \\ $$$$=\mathrm{27}\int\:\mathrm{sin}^{\mathrm{3}} \theta\:{d}\theta \\ $$$$=\mathrm{27}\int\:\mathrm{sin}\:\theta\:\mathrm{sin}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$=\mathrm{27}\int\:\mathrm{sin}\:\theta\:\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta\right)\:{d}\theta \\ $$$$=\mathrm{27}\int\:\mathrm{sin}\:\theta\:{d}\theta−\mathrm{27}\int\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{let}\:{u}=\mathrm{cos}\:\theta\Rightarrow{du}=−\mathrm{sin}\:\theta\:{d}\theta \\ $$$$=\mathrm{27}\left(−\mathrm{cos}\:\theta\right)−\mathrm{27}\int\:{u}^{\mathrm{2}} \:\left(−{du}\right) \\ $$$$=\mathrm{27}\left(−\mathrm{cos}\:\theta\right)+\mathrm{27}\int\:{u}^{\mathrm{2}} \:{du} \\ $$$$=−\mathrm{27cos}\:\theta+\mathrm{9}{u}^{\mathrm{3}} +{C} \\ $$$$=−\mathrm{27cos}\:\theta+\mathrm{9}\:\mathrm{cos}^{\mathrm{3}} \theta+{C} \\ $$$$=−\mathrm{27}\left(\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{\mathrm{3}}\right)+\mathrm{9}\left(\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{\mathrm{3}}\right)^{\mathrm{3}} +{C} \\ $$$$=−\mathrm{9}\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }+\frac{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}+{C}\:# \\ $$
Answered by mathmax by abdo last updated on 27/May/21
Φ=∫ (x^3 /( (√(9−x^2 ))))dx changement x=3sint give Φ=∫ ((27sin^3 t)/(3 cost))(3cost)dt =27 ∫ sin^3 t dt =27 ∫ sint(1−cos^2 t)dt =27∫ sint dt−27∫ cos^2 t sintdt =−27 cost +9 cos^3 t +C =−27(√(1−sin^2 t))+9((√(1−sin^2 t)))^3 )+C =−27(√(1−(x^2 /9)))+9((√(1−(x^2 /9))))^3 +C
$$\Phi=\int\:\frac{\mathrm{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{9}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}\:\mathrm{changement}\:\mathrm{x}=\mathrm{3sint}\:\mathrm{give} \\ $$$$\Phi=\int\:\:\frac{\mathrm{27sin}^{\mathrm{3}} \mathrm{t}}{\mathrm{3}\:\mathrm{cost}}\left(\mathrm{3cost}\right)\mathrm{dt}\:=\mathrm{27}\:\int\:\:\mathrm{sin}^{\mathrm{3}} \mathrm{t}\:\mathrm{dt} \\ $$$$=\mathrm{27}\:\int\:\mathrm{sint}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{t}\right)\mathrm{dt}\:=\mathrm{27}\int\:\mathrm{sint}\:\mathrm{dt}−\mathrm{27}\int\:\mathrm{cos}^{\mathrm{2}} \mathrm{t}\:\mathrm{sintdt} \\ $$$$=−\mathrm{27}\:\mathrm{cost}\:+\mathrm{9}\:\mathrm{cos}^{\mathrm{3}} \mathrm{t}\:\:+\mathrm{C} \\ $$$$\left.=−\mathrm{27}\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{t}}+\mathrm{9}\left(\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{t}}\right)^{\mathrm{3}} \:\right)+\mathrm{C} \\ $$$$=−\mathrm{27}\sqrt{\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{9}}}+\mathrm{9}\left(\sqrt{\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{9}}}\right)^{\mathrm{3}} \:+\mathrm{C} \\ $$$$ \\ $$

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