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Question Number 131453 by Engr_Jidda last updated on 04/Feb/21
∫(x^3 /( (√(16−x^2 ))))dx
$$\int\frac{{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }}{dx} \\ $$
Answered by bramlexs22 last updated on 04/Feb/21
(√(16−x^2 )) = l ⇒ x^2 =16−l^2 ; x dx = −l dl L=∫ (((16−l^2 )(−l dl))/l)=∫ (l^2 −16) dl L=(1/3)l^3 −16l + c = (1/3)l (l^2 −48)+c L=(1/3)(√(16−x^2 )) (−x^2 −32)+ c
$$\:\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }\:=\:{l}\:\Rightarrow\:{x}^{\mathrm{2}} =\mathrm{16}−{l}^{\mathrm{2}} \:;\:{x}\:{dx}\:=\:−{l}\:{dl} \\ $$$${L}=\int\:\frac{\left(\mathrm{16}−{l}^{\mathrm{2}} \right)\left(−{l}\:{dl}\right)}{{l}}=\int\:\left({l}^{\mathrm{2}} −\mathrm{16}\right)\:{dl} \\ $$$${L}=\frac{\mathrm{1}}{\mathrm{3}}{l}^{\mathrm{3}} −\mathrm{16}{l}\:+\:{c}\:=\:\frac{\mathrm{1}}{\mathrm{3}}{l}\:\left({l}^{\mathrm{2}} −\mathrm{48}\right)+{c} \\ $$$${L}=\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }\:\left(−{x}^{\mathrm{2}} −\mathrm{32}\right)+\:{c} \\ $$

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