How do you integrate #int (x^2 ) / sqrt(9 - x^2) dx# using trigonometric substitution?

Let

#x=3sintheta#

#x^2=9sin^2theta#

#dx=3costhetad theta#

We then have

#int(27sin^2thetacostheta)/(sqrt(9(1-sin^2theta))d theta#

Recall that #1-sin^2theta=cos^2theta#. Apply the identity:

#9int(sin^2thetacostheta)/sqrt(cos^2theta)d theta#

#9int(sin^2thetacancelcostheta)/(cancelcostheta)d theta#

#9intsin^2thetad theta#

Recall the identity #sin^2theta=1/2(1-cos2theta)#:

#9/2int(1-cos2theta)d theta#

Integrate:

#9/2int(1-cos2theta)d theta=9/2theta-9/4sin2theta+C#

We need to revert back to #x.# Recalling that #x=3sintheta, sintheta=x/3, theta=arcsin(x/3)#

#sin2theta# is still needed. Recalling that #sin2theta=2sinthetacostheta, sin^2theta+cos^2theta=1:#

#x^2/9+cos^2theta=9/9#
#cos^2theta=(9-x^2)/9#
#costheta=sqrt(9-x^2)/3#

Then #sin2theta=2sinthetacostheta=2(x/3)(sqrt(9-x^2)/3)=(2xsqrt(9-x^2))/9#

Then,

#int(x^2/sqrt(9-x^2))dx=9/2arcsin(x/3)-1/2xsqrt(9-x^2)#

#intx^2/(sqrt(9-x^2)) dx= x^2/(sqrt(3^2-x^2))dx# using our trig identities for integrals let us substitute #3sinphi = x# and #dx = 3cosphi dphi#

#therefore# #3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9-(9sin^2phi)) #

#3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9(1-(sin^2phi))#
where #1-sin^2phi = cos^2phi#

#3int(9sin^2phi)(cosphi)(dphi)/(sqrt(9(cos^2phi))#

#3int(9sin^2phi)(cancelcosphi)(dphi)/(3(cancelcosphi))#

#cancel3int(9sin^2phi)(dphi)/(cancel3)#

#int(9sin^2phi)(dphi) = 9int(sin^2phi)(dphi)#

Use the identity from trig where #cos(2phi) = 1-2sin^2phi# rearranging to solve such that #(cos(2phi) -1)/(-2)= sin^2phi# or #(1-cos(2phi))/(2)= sin^2phi#

#9int(1-cos(2phi))/(2)(dphi) = 9/2int(1-cos(2phi))(dphi)# use subtraction rules of the integral

#9/2int(dphi)-9/2intcos(2phi)dphi#

#9/2int(dphi) = (9/2)phi#

#-9/2intcos(2phi)dphi = -9/4sin(2phi)#

Now piece together and substitute in

#(9/2)phi-9/4sin(2phi)+C#

#phi = arcsin(x/3)#

#(9/2)phi-9/4sin(2phi)+C#

#(9/2)(arcsin(x/3)-1/2sin(2arcsin(x/3))+C#