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

1 Answer
Aug 5, 2016

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

Explanation:

#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#