(adsbygoogle = window.adsbygoogle || []).push({}); 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 ??

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# Homework Help: Integrate using trig substitution

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