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U sub involving inverse trig

  1. Mar 7, 2009 #1
    I need help with this problem.


    uit.jpg

    As of now I have absolutely no idea how to do this. I know that I should use inverse tan, but no further than that. What should u equal and how could I get the numerator to 1?

    Please explain so I can do similar problems later on.
     
  2. jcsd
  3. Mar 7, 2009 #2

    lanedance

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    there are a few ways to do this

    if you want to use arctan, have a look at the derivative of arctan...

    otherwise you might be able to use trigonometric substitution

    teh main way to do substitition is write

    u = u(x)
    then
    du = u'(x)dx for the integral
     
  4. Mar 7, 2009 #3
    The teacher asks to choose u, find u. My textbook doesn't show any example of problems involving inverse trig so I'm stuck trying to figure out how to make the denominator look similar to the derivative of arctan.


    Can you give me more hints?
     
  5. Mar 7, 2009 #4
    The derivative of arctan is [tex]\frac{1}{u^2 + 1}[/tex]. so what substitution should u have so that [tex]\frac{1}{10 + 3x^2} = \frac{1}{1 + u^2} [/tex]
     
  6. Mar 8, 2009 #5

    lanedance

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    well before you even substitute, why not manipulate the expression so the constant on the denominator is 1 (instead of 10), this should help you pick u(x)
     
  7. Mar 8, 2009 #6
    I can't find anyway to do that. :(

    Would I divide everything by 10?
     
  8. Mar 8, 2009 #7
    Yes! :smile:

    You get [tex]\frac{1/10}{1+(3/10)x^2}[/tex].

    Do you see a good choice for [tex]u[/tex] now? Hint: Take the constant factor 1/10 out of the integral.
     
  9. Mar 8, 2009 #8

    tiny-tim

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    Hi musicfairy! :smile:
    uhhh? :confused: why is everyone being so obscure?

    make u the tan of something (so that the bottom of the fraction becomes the sec of something :wink:)​
     
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