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x-3-1-x-2-dx-




Question Number 96257 by bobhans last updated on 31/May/20
∫ x^3  (√(1−x^2 )) dx ?
$$\int\:{x}^{\mathrm{3}} \:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx}\:?\: \\ $$
Answered by john santu last updated on 31/May/20
∫ x^2  (x(√(1−x^2 )) ) dx = J  set (√(1−x^2 )) = z ⇒x^2 =1−z^2   x dx = −zdz   J = ∫ (1−z^2 )z (−zdz)   J= ∫ (z^4 −z^2 ) dz   J= (1/5)z^5 −(1/3)z^3  + c   J= (1/(15))z^3 (3z^2 −5) + c   J= (((−3x^2 −2)(√((1−x^2 )^3 )))/(15))+ c
$$\int\:{x}^{\mathrm{2}} \:\left({x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\right)\:{dx}\:=\:{J} \\ $$$${set}\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:=\:{z}\:\Rightarrow{x}^{\mathrm{2}} =\mathrm{1}−{z}^{\mathrm{2}} \\ $$$${x}\:{dx}\:=\:−{zdz}\: \\ $$$${J}\:=\:\int\:\left(\mathrm{1}−{z}^{\mathrm{2}} \right){z}\:\left(−{zdz}\right)\: \\ $$$${J}=\:\int\:\left({z}^{\mathrm{4}} −{z}^{\mathrm{2}} \right)\:{dz}\: \\ $$$${J}=\:\frac{\mathrm{1}}{\mathrm{5}}{z}^{\mathrm{5}} −\frac{\mathrm{1}}{\mathrm{3}}{z}^{\mathrm{3}} \:+\:{c}\: \\ $$$${J}=\:\frac{\mathrm{1}}{\mathrm{15}}{z}^{\mathrm{3}} \left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{5}\right)\:+\:{c}\: \\ $$$${J}=\:\frac{\left(−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\right)\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} }}{\mathrm{15}}+\:{c}\: \\ $$
Commented by bobhans last updated on 31/May/20
nice solution
$$\mathrm{nice}\:\mathrm{solution}\: \\ $$

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