How do you integrate #int (6-x^2)/sqrt(9-x^2) dx# using trigonometric substitution?

#I=int(6-x^2)/sqrt(9-x^2)dx#

Rewrite the integral for simplification:

#I=int(-3+9-x^2)/sqrt(9-x^2)dx=-3intdx/sqrt(9-x^2)+int(9-x^2)/sqrt(9-x^2)dx#

#color(white)I=-3intdx/sqrt(9-x^2)+intsqrt(9-x^2)dx#

First focusing on #J=intdx/sqrt(9-x^2)#. For this, let #x=3sintheta#. This implies that #dx=3costhetad theta#.

#J=intdx/sqrt(9-x^2)=int(3costhetad theta)/sqrt(9-9sin^2theta)=intcostheta/sqrt(1-sin^2theta)d theta=intd theta#

#color(white)J=theta=arcsin(x/3)#

Recall that the #theta=arcsin(x/3)# relationship comes from our earlier substitution #x=3sintheta#.

Now we can work on the next integral #K=intsqrt(9-x^2)dx#. We will use the same substitution as before, #x=3sinphi#, implying that #dx=3cosphidphi#.

#K=intsqrt(9-x^2)dx=intsqrt(9-9sin^2phi)(3cosphidphi)#

#color(white)K=9intsqrt(1-sin^2phi)(cosphi)dphi=9intcos^2phidphi#

Rewrite by solving for #cos^2x# in the identity #cos2x=2cos^2x-1#.

#K=9int(cos2phi+1)/2dphi=9/2intcos2phidphi+9/2intdphi#

For the first integral, we can use the substitution #u=2phi# so #du=2dphi# and #dphi=1/2du#. The second integral is a fundamental one:

#K=9/4intcosudu+9/2phi=9/4sinu+9/2phi=9/4sin2phi+9/2phi#

Using #sin2x=2sinxcosx#:

#K=9/2sinphicosphi+9/2phi#

Now to undo these substitutions, recall that #x=3sinphi#. Thus, #phi=arcsin(x/3)# and #sinphi=x/3#. Thus #cosphi=sqrt(1-sin^2phi)=sqrt(1-x^2/9)=1/3sqrt(9-x^2)#. So:

#K=9/2(x/3)(1/3sqrt(9-x^2))+9/2arcsin(x/3)#

#color(white)K=1/2xsqrt(9-x^2)+9/2arcsin(x/3)#

Returning to the original integral:

#I=-3J+K#

#color(white)I=-3arcsin(x/3)+[1/2xsqrt(9-x^2)+9/2arcsin(x/3)]#

#color(white)I=1/2xsqrt(9-x^2)+3/2arcsin(x/3)+C#