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Integrate using trig substitution

  • Thread starter n77ler
  • Start date
89
0
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 ??
 

Answers and Replies

1,750
1
[tex]\int\sqrt{8x-x^2}dx[/tex]

[tex]\int\sqrt{4^2-(x-4)^2}dx[/tex]

*[tex]x=a\sin\theta \rightarrow a=4[/tex]

[tex]x-4=4\sin\theta[/tex]
[tex]dx=4\cos\theta d\theta[/tex]

[tex]\int\sqrt{4^2-4^2\cos^2\theta}d\theta[/tex]

Re-check your substitutions!
 
89
0
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?
 
89
0
I end up with 8[Q]+8[sin(2Q)] But im not sure how to put a value back in for Q
 
1,750
1
You want it in terms of x?

Use your substitutions ... cosine = adj/hyp and sine = opp/hyp.
 
89
0
hmm Im not familiar with doing it like that
 
1,750
1
Draw a triangle and you will know what I mean ...

[tex]x-4=4\sin\theta[/tex]

[tex]\sin\theta=\frac{x-4}{4}[/tex]

[tex](x-4)^2+\cos^2\theta=4^2[/tex]

[tex]\cos\theta=\frac{\sqrt{4^2-(x-4)^2}}{4}[/tex]
 
Last edited:

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