How do you find the integral intx^3*sqrt(9-x^2)dx ?
1 Answer
Aug 29, 2014
=(9-x^2)^(5/2)/5-3(9-x^2)^(3/2)+c , wherec is a constantExplanation :
=intx^3*sqrt(9-x^2)dx Using Integration by Substitution,
let's assume
9-x^2=t^2 , then
-2xdx=2tdt =>xdx=-tdt
=int-(9-t^2)t^2dt
=int(t^4-9t^2)dt
=intt^4dt-9intt^2dt
=t^5/5-9t^3/3+c , wherec is a constant
=t^5/5-3t^3+c , wherec is a constantSubstituting
t back,
=(9-x^2)^(5/2)/5-3(9-x^2)^(3/2)+c , wherec is a constant