Integrate using trig substitution

1. Mar 27, 2008

n77ler

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 ??

2. Mar 27, 2008

rocomath

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

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

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

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

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

3. Mar 27, 2008

n77ler

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?

4. Mar 27, 2008

n77ler

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

5. Mar 27, 2008

rocomath

You want it in terms of x?

6. Mar 27, 2008

n77ler

hmm Im not familiar with doing it like that

7. Mar 27, 2008

rocomath

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

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

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

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

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

Last edited: Mar 27, 2008