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# , where#c# 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# , where#c# is a constant
#=t^5/5-3t^3+c# , where#c# is a constantSubstituting
#t# back,
#=(9-x^2)^(5/2)/5-3(9-x^2)^(3/2)+c# , where#c# is a constant
