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2 integral questions

  1. Apr 20, 2007 #1
    i added a file with the the two integral and the method that i tried to solve them

    i could'nt solve them plz help
     

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    Last edited: Apr 20, 2007
  2. jcsd
  3. Apr 20, 2007 #2

    Gib Z

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    Your substitutions seem to be the way to go for the first one. I hope you know those expressions for sin and cos actually come about from t= tan(x/2)?

    Therefore by the chain rule, [itex]dt=\frac{1}{2} \sec^2 (x/2) dx[/itex].
    Then we use the Pythatgorean Identities to reduce sec^2 to tan^2 +1.

    [tex]dt = \frac{1}{2} (\tan^2 (x/2) +1 ) dx = \frac{1}{2} (t^2 +1) dx[/tex]
    Getting dx alone yields [tex]\frac{2 dt}{t^2+1} = dx[/tex] which unfortunately is not what you had. So try it with that correction.

    For the second one, in partial fractions the numerator always has to be 1 degree less than the denominator. I sure you knew that because for the first partial fraction you put Ax+b over 1+x^2. The second partial fraction has degree 4, so instead of putting cx+d, try cx^3+dx^2+e^x+f .
     
  4. Apr 20, 2007 #3
    For the second one, instead of splitting it into partial fractions try the substitution:
    [tex]x = tan(u)[/tex]

    There's still a fair bit of work to do after that, but it should get you started.
     
  5. Apr 20, 2007 #4

    Gib Z

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    Actually, for the second one there is a much easier way that I just saw!

    x= tan z
    dx = sec^2 z dz
    [tex]\int \frac{1}{(1+x^2)^2} dx = \int \frac{\sec^2 z}{(1+\tan^2 z)^2} dz=\int \frac{\sec^2 z}{\sec^4 z} dz =\int \frac{1}{\sec^2 z}dz = \int \cos^2 z dz[/tex]

    Now finish that off with [itex]\cos^2 z = \frac{1}{2} (\cos 2z +1)[/itex]

    EDIT: DAMN IT
     
  6. Apr 21, 2007 #5
    thank you very much
     
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