How do you integrate #int x^3/sqrt(144-x^2)dx# using trigonometric substitution?

Use the substitution #x=12sintheta#. From here we see that #dx=12costhetad theta#. Substituting:

#intx^3/sqrt(144-x^2)dx=int(1728sin^3theta(12costheta))/sqrt(144-144sin^2theta)d theta#

Factoring #sqrt(1/144)=1/12# from the square root:

#=int(1728sin^3theta(12costheta))/(12sqrt(1-sin^2theta))d theta#

Canceling the #12#s and recalling that since #sin^2theta+cos^2theta=1#, we know that #costheta=sqrt(1-sin^2theta)#:

#=1728int(sin^3thetacostheta)/costhetad theta=1728intsin^3thetad theta#

We will use here #sin^2theta=1-cos^2theta#:

#=1728intsin^2thetasinthetad theta=1728int(1-cos^2theta)sinthetad theta#

Now, using the substitution #u=costheta#, so that #du=-sinthetad theta#:

#=-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 #u=costheta#:

#=576cos^3theta-1728costheta#

Because our substitution is #sintheta=x/12#, we should write this in terms of sine functions. That is, #costheta=sqrt(1-sin^2theta)# and #cos^3theta=(1-sin^2theta)^(3/2)#:

#=576(1-sin^2theta)^(3/2)-1728sqrt(1-sin^2theta)#

Factoring:

#=sqrt(1-sin^2theta)(576(1-sin^2theta)-1728)#

Now using #sintheta=x/12#:

#=sqrt(1-x^2/144)(576(1-x^2/144)-1728)#

#=sqrt((144-x^2)/144)(576-4x^2-1728)#

Factoring #sqrt(1/144)=1/12# from the square root and #-4# from the parentheses:

#=-1/3sqrt(144-x^2)(x^2+288)+C#