How do you integrate $x^3/sqrt(1-x^2) dx$ ?

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Answer 1 · 2016-08-05T10:57:29

$= -1/3 (x^2 + 2 ) sqrt(1-x^2) \ + C$

$int \ x^3/sqrt(1-x^2) \ dx$

$=int \ x^2 d/dx (-sqrt(1-x^2)) \ dx$

$=-x^2 sqrt(1-x^2) + int \d/dx( x^2) sqrt(1-x^2) \ dx$

$=-x^2 sqrt(1-x^2) + int \ 2x sqrt(1-x^2) \ dx$

$=-x^2 sqrt(1-x^2) + int \ d/dx(-2/3 (1-x^2)^(3/2) )\ dx$

$=-x^2 sqrt(1-x^2) -2/3 (1-x^2)^(3/2) + C$

$= sqrt(1-x^2) (-x^2 -2/3 (1-x^2) ) + C$

$= sqrt(1-x^2) (-x^2/3 -2/3 ) + C$

$= -1/3 (x^2 + 2 ) sqrt(1-x^2) \ + C$

How do you integrate x^3/sqrt(1-x^2) dxx^3/sqrt(1-x^2) dx?