How do you integrate #int x^3/sqrt(144-x^2)dx# using trigonometric substitution?
1 Answer
Explanation:
Use the substitution
#intx^3/sqrt(144-x^2)dx=int(1728sin^3theta(12costheta))/sqrt(144-144sin^2theta)d theta#
Factoring
#=int(1728sin^3theta(12costheta))/(12sqrt(1-sin^2theta))d theta#
Canceling the
#=1728int(sin^3thetacostheta)/costhetad theta=1728intsin^3thetad theta#
We will use here
#=1728intsin^2thetasinthetad theta=1728int(1-cos^2theta)sinthetad theta#
Now, using the substitution
#=-1728int(1-u^2)du=1728intu^2du-1728intdu#
Integrating using the power rule for integration:
#=1728(u^3/3)-1728u=576u^3-1728u#
Reverse substituting with
#=576cos^3theta-1728costheta#
Because our substitution is
#=576(1-sin^2theta)^(3/2)-1728sqrt(1-sin^2theta)#
Factoring:
#=sqrt(1-sin^2theta)(576(1-sin^2theta)-1728)#
Now using
#=sqrt(1-x^2/144)(576(1-x^2/144)-1728)#
#=sqrt((144-x^2)/144)(576-4x^2-1728)#
Factoring
#=-1/3sqrt(144-x^2)(x^2+288)+C#