How do you integrate #int x^2sqrt(16-x^2)# using trig substitutions?

Solving for:

#I=intx^2sqrt(16-x^2)dx#

Make the substitution #x=4sintheta# where #dx=4costhetad theta#:

#I=int16sin^2thetasqrt(16-16sin^2theta)4costhetad theta#

Bringing all the multiplicative constants out, including the #sqrt16=4#:

#I=256intsin^2thetacosthetasqrt(1-sin^2theta)d theta#

Since #sin^2theta+cos^2theta=1#, we see that #costheta=sqrt(1-sin^2theta)#:

#I=256intsin^2thetacos^2thetad theta#

Here, since #sin2theta=2sinthetacostheta#, we see that #sin^2 2theta=4sin^2thetacos^2theta#:

#I=64int4sin^2thetacos^2thetad theta#

#I=64intsin^2 2theta d theta#

Note that, according to the cosine double angle formula, #cos4theta=1-2sin^2 2theta#, so #sin^2 2theta=(1-cos4theta)/2#:

#I=32int(1-cos4theta)d theta#

#I=32intd theta-32intcos4thetad theta#

Integrating both (substitution can be used for the second integral, or just see it for the reverse chain rule of #sin4theta# that it is):

#I=32theta-8int4cos4thetad theta#

#I=32theta-8sin4theta#

Since #x=4sintheta#, solving for #theta# gives #theta=arcsin(x/4)# and #4theta=4arcsin(x/4)#. Thus:

#I=32arcsin(x/4)-8sin(4arcsin(x/4))+C#