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Integration using method of substitution

  1. Jun 16, 2008 #1
    Problem:
    (tdt)/(4-t^4)^(1/2)
    Attempt:
    I want the derivate of whatever i make u equal to, to equal something outside of u therefore I will factorize the denominator to equal -(-2+t^2)(2+t^2) and make u equal to (2+t^2) so that du=2tdt
    Balance the equation so that one side is equal to the numerator
    (1/2)du=tdt
    Now I have
    1/2[tex]\int[/tex] = 1/(-(-2+t^2)u)^1/2
    How do I apply this elementary integral in this situation?:
    [tex]\int[/tex] 1/(a^2-x^2)^1/2 dx = sin^-1(x/a) + C
    I don't know what to do with the u in
    1/(-(-2+t^2)u)^1/2

    If it helps the final answer is supposed to be:
    (1/2)sin^-1(t^2/2)
     
  2. jcsd
  3. Jun 16, 2008 #2

    Dick

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    Put u=t^2 first. Then you can apply the arcsin formula.
     
  4. Jun 17, 2008 #3
    Could you elaborate? Thanks
     
  5. Jun 17, 2008 #4

    Dick

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    Uh, u=t^2. du=2*t*dt. There's your t for the numerator. What does the denominator become?
     
  6. Jun 17, 2008 #5
    How can u be made to equal t^2? Can't you only make u equal to a whole polynomial, not just part of it?
     
  7. Jun 17, 2008 #6

    Dick

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    You can make u whatever you want. What does the denominator become in terms of u?
     
  8. Jun 17, 2008 #7
    If u = t^2
    then
    (-(-2+u)(2+u))^(1/2)?
     
  9. Jun 17, 2008 #8

    Dick

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    Yes. But no need to factor it. So now you have (1/2)*du/(4-u^2)^(1/2). Now apply the arcsin formula.
     
  10. Jun 17, 2008 #9
    Can the arsin formula only be applied if the variable in the denominator has an exponent of 2?
     
  11. Jun 17, 2008 #10

    Dick

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    Yes. But if you do the substitution u=t^2 then the denominator variable does have an exponent of 2. That was the whole point.
     
  12. Jun 17, 2008 #11
    Can you tell me how I would go about solving
    (sin(t))^4 x (cos(t))^5) dt
    if i make u = sin t
    du=cost dt
    but that doesnt work because cos(t) has an exponent of 5
     
  13. Jun 17, 2008 #12

    Defennder

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    [tex] sin^4t cos t = (sin^2 t)^2 cos t = \frac{1}{2}^2 (1-cos2t)^2 cos t [/tex]

    Expand that out. Change the square of the trigo term into a non-square trigo function of a different angle each time you get a sin or cos squared. (What the heck is that trigo identity called?)

    Then make use of some trigo identities to express the final integrand into something easily integrable.
     
  14. Jun 17, 2008 #13

    Dick

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    Use cos^2(t)=(1-sin^2(t)).
     
  15. Jun 17, 2008 #14
    double angle formula?
     
  16. Jun 17, 2008 #15
    How did you get
    [tex]
    (cos t)
    [/tex]

    from

    [tex]
    (cos t)^5
    [/tex]
     
  17. Jun 17, 2008 #16

    Defennder

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    Ok, my mistake it should be [tex] (sintcost)^4 cos t = \frac{1}{2}^4 (sin(2t))^4 cos t = \frac{1}{2}^4 \frac{1}{2}^2(1-cos 4t)^2cos t[/tex]

    Just continue from here.
     
  18. Jun 17, 2008 #17
    I dont understand how you get that at all. Could you please explain it to me?
     
  19. Jun 17, 2008 #18

    Defennder

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    Use these formulae:

    [tex]sin(t) cos(t) = \frac{1}{2}sin(2t)[/tex]
    [tex]sin^2(t) = \frac{1}{2}(1-cos(2t))[/tex]
     
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