How do you find the integral #intx^3/(sqrt(16-x^2))dx# ?
1 Answer
Aug 27, 2014
#=(16-x^2)^(3/2)/3-16sqrt(16-x^2)+c# , where#c# is a constantSolution
#=intx^3/sqrt(16-x^2)dx=int(x^2*x)/sqrt(16-x^2)dx# Using Integration by Substitution
let's assume
#(16-x^2)=t^2# ,#=>-2xdx=2tdt#
#xdx=-tdt#
#=int(-(16-t^2)t)/tdt#
#=int(t^2-16)dt#
#=intt^2dt-16intdt#
#=t^3/3-16t+c# , where#c# is a constant
#=(16-x^2)^(3/2)/3-16sqrt(16-x^2)+c# , where#c# is a constant
