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Help with integration

  1. Jul 16, 2006 #1
    integra of (x^2 + y^2)^(3/2) dy is it equals to 2[(x^2 + y^2)^(5/2)] / 10y
    ??

    pls let me know
    thanx
     
  2. jcsd
  3. Jul 16, 2006 #2

    Hootenanny

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    Nope, that is not correct. Perhaps if you showed your steps....
     
  4. Jul 16, 2006 #3
    i just integrate as (x^2 + y^2)^(5/2) / (5/2)(2y) .may i know how to do??
     
  5. Jul 16, 2006 #4
    How to do what ?
     
  6. Jul 16, 2006 #5
    pls show me how to integrate the eqn above
     
  7. Jul 16, 2006 #6
    try to substitute u = x^2 + y^2

    Substitutions are a handy trick when the denominator is the derivative of the numerator.
     
    Last edited: Jul 16, 2006
  8. Jul 16, 2006 #7

    Hootenanny

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    HINT: You may have to substitute more than once.
     
  9. Jul 16, 2006 #8
    pls show,how to substitute more than once

    thanx
     
  10. Jul 16, 2006 #9
    Perseverance, my friend, is the key to math. enlightment. In the worst case scenario, erase and start over. Try to figure it out by yourself, and if you really can't get to the answer, show what went wrong and then maybe....
     
  11. Jul 16, 2006 #10
    i subs u=x^2 + y^2 and differentiate then get du=2x dx
    then integra sqrt[u/(u-y^2)] du and fromhere onwards i don't know how to integra wif respect to u

    pls help

    thanx
     
  12. Jul 16, 2006 #11
    This is where the wisdom of hootnanny comes in play. Once you find du , you must find an expression with du where dy/5y can be "eliminated". You know that du = 2ydy. Therefore du/(10y^2) = dy/5y .

    From that, substitue y^2 again ....

    EDIT : Yo uare integrating with respect to what variable ??
    Plz specify ....I don't think I got your question right when I saw your last post.
     
    Last edited: Jul 16, 2006
  13. Jul 16, 2006 #12
    If you are trying to integrate wrt x, try trig substitutions. If you are trying to integrate wrt y, try the substitution i mentionned earlier .
     
  14. Jul 16, 2006 #13

    Hootenanny

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    If you are trying to solve;

    [tex]\int (x^{2} + y^{2})^{\frac{3}{2}} \; dy[/tex]

    The after your double substitution you should end up with something like this (unless I have made a mistake somewhere);

    [tex]\int \frac{U^{\frac{3}{2}}}{2(U - x^{2})^{\frac{1}{2}}} \; du[/tex]
     
  15. Jul 16, 2006 #14
    ya,i got stuck here also,how to proceed??
     
  16. Jul 17, 2006 #15

    VietDao29

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    Let a be some constant.
    In an integral, when you meet the form of a2 + x2, then you may use the trig substitution [tex]x = a \tan \theta[/tex]
    If you meet the form of a2 - x2, then you can to use a trig substitution [tex]x = a \sin \theta[/tex]
    If you meet the form of x2 - a2, then you can to use a trig substitution [tex]x = a \sec \theta[/tex].
    You can take a look here.
    So in this problem, what you should do is to use the trig substitution [tex]y = x \tan \theta[/tex].
    [tex]\Rightarrow dy = \frac{x d ( \theta )}{\cos ^ 2 \theta}[/tex]
    Ok, let's see if you can go from here. :)
    Hint: You should use the identity 1 + tan2x = sec2x.
    By the way, I advise you to re-read the book. Read again from the very first lesson of integral, if possible, and make sure you understand everything. It won't do you any harm, I promise. Just for your own sake. :)
     
  17. Jul 17, 2006 #16
    thanx........
     
  18. Jul 17, 2006 #17

    arildno

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    An even simpler choice would be to set [itex]y=x\sinh(u)[/tex]
     
  19. Jul 17, 2006 #18
    may i know how to integrate sec^5(tita) d(tita) pls
     
  20. Jul 17, 2006 #19
    i got x^4 cosh^4(u) du but how can i integra from here??
    what shoulc i do with cosh^4(u) du??

    pls help
     
  21. Jul 17, 2006 #20

    VietDao29

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    Integrating odd power of secant, and cosecant is a bit tricky. You can use Integrate by Parts to solve this. Say you want to integrate:
    [tex]I_n = \int \sec ^ n \theta d( \theta )[/tex].
    Let [tex]u = \sec ^ {n - 2} \theta \quad \mbox{and} \quad dv = \sec ^ 2 \theta d ( \theta )[/tex]
    [tex]\Rightarrow du = (n - 2) \sec ^ {n - 2} n \theta \tan \theta d( \theta ) \quad \mbox{and} \quad v = \tan \theta[/tex]
    So, we have:
    [tex]I_n = \int \sec ^ n \theta d( \theta ) = \sec ^ {n - 2} \theta \tan \theta - (n - 2) \int \sec ^ {n - 2} \tan ^ 2 \theta d( \theta ) = \sec ^ {n - 2} \theta \tan \theta - (n - 2) \int \sec ^ {n - 2} \left( \sec ^ 2 - 1 \right) d( \theta )[/tex]
    [tex]= \sec ^ {n - 2} \theta \tan \theta - (n - 2) \int \sec ^ n \theta d( \theta ) + (n - 2) \int \sec ^ {n - 2} \theta d( \theta ) = \sec ^ {n - 2} \theta \tan \theta - (n - 2) I_n + (n - 2) I_{n - 2}[/tex]
    Rearranging it gives:
    [tex](n - 1) I_n = \sec ^ {n - 2} \theta \tan \theta + (n - 2) I_{n - 2}[/tex] or:
    [tex]I_n = \frac{1}{n - 1} \left( \sec ^ {n - 2} \theta \tan \theta + (n - 2) I_{n - 2} \right)[/tex].
    Appling it here, we have:
    [tex]I_5 = \frac{1}{4} \left( \sec ^ {3} \theta \tan \theta + (3) I_{3} \right)[/tex].
    Can you go from here?
    Hint: You may need to apply it once more, and use the fact that:
    [tex]\int \sec \theta d( \theta ) = \ln | \sec \theta + \tan \theta | + C[/tex]
    --------------------
    You can also use the u substitution:
    [tex]u = \sin \theta[/tex] to integrate [tex]\int \sec ^ 5 \theta d( \theta ) = \int \frac{1}{\cos ^ 5 \theta} d( \theta )[/tex], since the power of cos is odd. And then use Partial fraction. Something like this:
    [tex]\int \sec ^ 5 \theta d( \theta ) = \int \frac{1}{\cos ^ 5 \theta} d( \theta ) = \frac{\cos \theta}{\cos ^ 6 \theta} d( \theta ) = \frac{1}{(1 - u ^ 2) ^ 3} du = ...[/tex]
    Can you go from here? :)
     
    Last edited: Jul 17, 2006
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