1. The problem statement, all variables and given/known data int sqrt 8x-x^2 2. Relevant equations trig sub 3. The attempt at a solution complete the square integral becomes int sqrt 16-(x-4)^2 let x-4= 16sin(Q) sqrt 16-(x-4)^2 =sqrt 16-256sin^2(Q) dx= 16cos(Q)dQ = 16cos(Q) int 16cos(Q)16cos(Q) dQ=256 int cos^2(Q)dQ double angle formula 128 int 1+cos(2Q) dQ let u=2Q 1/2du=dQ This is the part where i get messed up... 128 int 1 dQ + 128 int cos(u)du So the first part is just 128Q and the integral of 128intcos(u)du is 128sin(u) So do I fill 2Q back in for u? Then if I fill it back in do I use Q=[sqrt16-(x-4)^2] / 16 ??