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Integration with inverse trigonometric substitution

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data

    The problem is: integrate: (1/(4x2+4x+5)2)dx

    2. Relevant equations



    3. The attempt at a solution
    4x2+4x+5=(2x+1)2+4 gives:

    ∫dx/((2x+1)2+4)2

    use regular substitution:
    u=2x+1
    du=2dx
    dx=1/2du
    gives: 1/2∫du/(u2+4)2)

    trigonometric substitution:
    u=tan(z)
    du=1/(cos(z))2dz
    gives: 1/2∫dz/((sin2(z)/cos2(z)) +4)2

    I have tried most trigonometric tricks I can think off to make that last integral into something that I know how to integrate, but it has usually just gotten much worse, and certainly not better. I think it will be helpful to show my different attempts beyond this point.
     
  2. jcsd
  3. Nov 10, 2009 #2

    tiny-tim

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    Homework Helper

    Hi Toutatis! :smile:

    Try u = 2tan(z). :wink:

    By Toutatis! :biggrin:
     
  4. Nov 11, 2009 #3
    of course!

    Thank you!
     
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