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I need Help with integrals

  1. Oct 8, 2005 #1
    I am stuck in this integral please could i get some help or sugestions.

    [tex] \int \frac {x+5} {\sqrt{9-(x-3)^2}} * dx [/tex]

    If I set:

    a= 3
    u=(x-3)
    [tex] \frac {du} {dx} = 1 [/tex]
    du=dx

    ,and then I don't know what to do.
     
    Last edited: Oct 8, 2005
  2. jcsd
  3. Oct 8, 2005 #2

    hotvette

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    You are on the right track. Just keep going. Make substitutions back into the original integral to get rid of x and have the integral be only in terms of u and du. Hint: you'll end up with 2 integrals.
     
  4. Oct 8, 2005 #3

    ok i have [tex] \int \frac {x} {\sqrt{9-(x-3)^2}} * dx + \int \frac {5} {\sqrt{9-(x-3)^2}} * dx [/tex]

    Is this correct, i mean is this legal in math. I only broke down the integral into two separte integrals. Im still working on it!!!
     
    Last edited: Oct 8, 2005
  5. Oct 8, 2005 #4

    TD

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    That won't help very much, I'm affraid.

    Manipulating the denominator may help you. Factor out the 9, get it out of the square root to get:

    [tex]\int {\frac{x}{{\sqrt {9 - \left( {x - 3} \right)^2 } }}dx} = \frac{1}{3}\int {\frac{x}{{\sqrt {1 - \left( {\frac{{x - 3}}{3}} \right)^2 } }}dx} [/tex]

    That should smell arcsin-ish, but don't forget the x in your nominator :smile:
     
  6. Oct 8, 2005 #5

    hotvette

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    Ok, that works. Now use the substitution you already stated (i.e. u = x - 3) and that should get you integrals that are readily solved.
     
    Last edited: Oct 8, 2005
  7. Oct 8, 2005 #6

    GCT

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    [tex] \int \frac {x+5} {\sqrt{9-(x-3)^2}} * dx [/tex]

    the function within the square root simplifies to [tex]-x^{2}+6x[/tex]
    The derivative of this is [tex]-2x+6[/tex]

    [tex] \int \frac{-2(x+5)+16~dx}{\sqrt{9-(x-3)^2}} [/tex]

    ring a bell?
     
  8. Oct 8, 2005 #7

    hotvette

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    I believe either approach will obtain the same result
     
  9. Oct 8, 2005 #8

    TD

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    Hopefully :biggrin:
     
  10. Oct 8, 2005 #9

    ok how did he get this: [tex] \int \frac{-2(x+5)+16~dx}{\sqrt{9-(x-3)^2}} [/tex]

    that confused me even more.
     
  11. Oct 8, 2005 #10

    what No help??
     
  12. Oct 8, 2005 #11

    hotvette

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    I believe he used a little trick to get the upper part to look like the derivative of what is inside the radical. Consider the following example (nothing to do with this problem, but shows the idea):

    x + 5 = x + 5 + 3 - 3 = (x + 8) - 3
     
  13. Oct 8, 2005 #12

    Is it going to help me out, Ive been looking at the problem for hours but i really dont get it, when i take the derivative of u it the x of the upper part doesnt cancel so thats bothering me a bit. I need more explanaition please.
     
  14. Oct 8, 2005 #13

    hotvette

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    Here is my suggestion. Take what you originally did (i.e. make the substitution u = x-3). The net result is an integrand that you can break apart: Yes, what you did in your 2nd post is valid. Simple example:

    [tex]\frac{5 + 2}{3} = \frac{5}{3} + \frac{2}{3}[/tex]

    The 2nd integral should be readily solveable (Hint: TD mentioned something about arcsin). The first is actually much easier to solve (an additional substitution will make it easier to see).

    GCT has a valid approach, but the one you were on originally is valid as well.
     
    Last edited: Oct 8, 2005
  15. Oct 8, 2005 #14

    mezarashi

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    My approach would be to make the substitution

    u = x-3, which also means that x = u + 3

    then differentiating
    du = dx

    Your original integral will now reduce to

    [tex] \int \frac {u+8} {\sqrt{9-u^2}} du [/tex]

    This integral has two parts and can be written as:

    [tex] \int \frac {u} {\sqrt{9-u^2}} du + \int \frac {8} {\sqrt{9-u^2}} du [/tex]

    The first part can be solved using the chain rule as you know, since the numerator term is the derivative of the term inside the square root.

    The second part is the harder one. The solution is (so you can check your work). Edit: It is solved though the substitution sin w = x/a.

    [tex] \int \frac {dx} {\sqrt{a^2 - x^2}} dx = arcsin(x/a)[/tex]

    Note: Of course, it is only valid for absolute value x less than or equal to a. Or else you would have an imaginary.
     
    Last edited: Oct 8, 2005
  16. Oct 8, 2005 #15
    thanks for all the help. I think i got it
     
  17. Oct 8, 2005 #16

    mezarashi

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    =D great news ^^
     
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