# Integrate using trig substitution

1. Homework Statement

int sqrt 8x-x^2

2. Homework 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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$$\int\sqrt{8x-x^2}dx$$

$$\int\sqrt{4^2-(x-4)^2}dx$$

*$$x=a\sin\theta \rightarrow a=4$$

$$x-4=4\sin\theta$$
$$dx=4\cos\theta d\theta$$

$$\int\sqrt{4^2-4^2\cos^2\theta}d\theta$$

Ok so other than that does it look right? It is a definate integral so in the end i just plug my value in correct?

I end up with 8[Q]+8[sin(2Q)] But im not sure how to put a value back in for Q

You want it in terms of x?

hmm Im not familiar with doing it like that

Draw a triangle and you will know what I mean ...

$$x-4=4\sin\theta$$

$$\sin\theta=\frac{x-4}{4}$$

$$(x-4)^2+\cos^2\theta=4^2$$

$$\cos\theta=\frac{\sqrt{4^2-(x-4)^2}}{4}$$

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