Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

$$\mathrm{nice}\:\mathrm{solution}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com