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Trig Substitution (Integrals)

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data integral of
    dx/((9-(x^2))^(3/2)) A = 0, B = 3/2



    2. Relevant equations Trigonometry Substitutions



    3. The attempt at a solution: I am stuck with this question. So far, I got
    (1/9)integral of (1/cos^2(θ)) dθ
     
  2. jcsd
  3. Feb 21, 2013 #2

    haruspex

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    Do you know a trig function with derivative sec-squared?
     
  4. Feb 21, 2013 #3
    That's my problem. My trigonometry is very limited. I do not know of a Anti-derivative of sec^2θ
     
  5. Feb 21, 2013 #4

    Dick

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    What is the derivative of tan(θ)?
     
  6. Feb 22, 2013 #5
    the derivative of tan(∅) is sec^2(∅). I get 1/9(tan((sin(∅))) | , Could someone solve the entire problem for me ? and reply with the steps to get there.
    This would be a lot of help!
     
  7. Feb 22, 2013 #6

    Dick

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    That's not right. It's just a little confused. What you need to do is show how you got there and then people can tell you where you are going wrong.
     
  8. Feb 22, 2013 #7
    A = 0 , B = 3/2 ∫ dx/((9-x2)3/2)

    First I substituted , x = 3sin∅ , dx = 3cos∅d∅

    ∫ 3cos∅d∅/(32-(3sin∅)2)3/2

    Then I factored.

    ∫ 3cos∅d∅/(32(1-sin2∅))3/2

    Then I used the trig identity to convert 1-sin2∅ to cos2θ

    ∫ 3cos∅d∅/(32cos2∅)3/2

    Then I simplified

    (1/9) ∫ (1/cos2θ) d∅

    Then I anti-derived that into..

    (1/9) tanθ | [ I do not know what to do after this, Since it is respect to ∅]

    To conclude, I cannot use A and B, Since those are respect to x. I am stuck on this part.
     
  9. Feb 22, 2013 #8
    differential equation

    Can you tell me how to crack the following problems?:

    1) y=2x(dy/dx)-y(dy/dx)2

    2) (d2y/dx2)+y=2 cos2(x)+sin(3x)
     
  10. Feb 22, 2013 #9

    Dick

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    Pretty good. You've got both thetas and phis floating around there. Intended to mean the same thing, I hope. One way to do it is the find the theta limits, since you substituted x=3*sin(θ) you just need to solve 3/2=3*sin(θ) and 0=3*sin(θ). If you want to eliminate the trig functions altogether you can solve for theta using the arcsin function. θ=arcsin(x/3). There's also a way to simplify tan(arcsin(x/3)). Any of this sounding familiar?
     
  11. Feb 22, 2013 #10

    Dick

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    Re: differential equation

    That has nothing to do with this thread. You really need to start a new one.
     
  12. Feb 22, 2013 #11
    Ok thats what I was going for.

    So tan(sin-1(x/3))

    (sin-1((3/2)/3)) is ∏/6 right?


    1/9 (Tan (∏/6) - Tan(0)) ? Am I on the right track?
     
  13. Feb 22, 2013 #12

    Dick

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    That's exactly right. It's also good practice to think about how you might express tan(θ) in terms of x. Since sin(θ)=x/3 and sine=opposite/hypotenuse draw a right triangle with opposite side length x and hypotenuse length 3. Then figure out the length of the adjacent side. Now tan(θ)=opposite/adjacent.
     
  14. Feb 22, 2013 #13
    Yeah thanks for the tip.

    So the answer is 1/(9√3)? Am I correct?
     
  15. Feb 22, 2013 #14

    Dick

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    Yes, it is.
     
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