I hope you like substitutions, because we're about to do a lot of them. To start, let #u = x-3 implies du = dx#
Integral becomes:
#int (u+8)/(sqrt(9-u^2)) du#
Now we want to deal with the denominator. This type of function will often involve a trig substitution - if you learn/can derive the pythagorean identities they are very helpful in figuring out what will cancel out etc. Whatever we choose for #u# in this case, it should have a coefficient of 3 so that we can take the 9 outside the square root.
Let's go with #u = 3sintheta implies du = 3costheta d theta and theta = sin^(-1)(u/3)#
Integral becomes:
#3int (3sintheta + 8)/(3sqrt(1 - sin^2theta)) costhetad theta#
#1 - sin^2theta = cos^2theta# so:
#3int (3sintheta + 8)/(3sqrt(cos^2theta))cos thetad theta#
#=int(3sintheta+8)d theta#
Can split this up into two seperate integrals:
#=3int sintheta d theta + 8 int d theta#
# = -3costheta + 8theta + C#
Substitute back in that #theta = sin^(-1)(u/3)#
# = 8sin^(-1)(u/3) - 3cos(sin^(-1)(u/3)) + C#
We need to figure out what the general form of #cos(sin^(-1)(phi))# is:
Consider #y = sin^(-1)(phi)#
#implies phi = sin(y)#
#phi^2 = sin^2(y)#
#phi^2 = 1 - cos^2(y)#
#cos(y) = cos(sin^(-1)(phi)) = sqrt(1 - phi^2)#
Hence #cos(sin^(-1)(u/3)) = sqrt(1 - (u/3)^2) = 1/3sqrt(9 - u^2)#
Solution becomes:
#8sin^(-1)(u/3) - sqrt(9-u^2) + C#
Now back substitute #u = x-3#
Solution is:
#8sin^(-1)((x-3)/3) - sqrt(9-(x-3)^2) + C#
