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Homework Help: Hints for integral

  1. Oct 10, 2005 #1
    Ok so after a substituion I have this: (I dont know how to proceed)

    "integral of" (u-2) / sqrt(u^2 + 1) du

    and by the way, can someone tell me what cot^3(tan^-1(x)) equals?

  2. jcsd
  3. Oct 10, 2005 #2


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    sqrt(u^2+1) just screams for a trig substitution.
  4. Oct 10, 2005 #3
    Bah, I see it. I easily get confused when having to substitution i several steps :P

  5. Oct 10, 2005 #4


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    Well, it may be a bit easier if you split at first.

    [tex]\int {\frac{{u - 2}}{{\sqrt {u^2 + 1} }}du} = \int {\frac{u}{{\sqrt {u^2 + 1} }}du} - \int {\frac{2}{{\sqrt {u^2 + 1} }}du} [/tex]

    So the first one gives

    [tex]\int {\frac{u}{{\sqrt {u^2 + 1} }}du} = \frac{1}{2}\int {\frac{{d\left( {u^2 + 1} \right)}}{{\sqrt {u^2 + 1} }}} [/tex]

    Which is easy to integrate, without a trig subst.
    Then of course, still the second part :smile:
  6. Oct 10, 2005 #5


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    You might want to break it up into 2 integrals. The first one is very straightforward, the second one is easy with a trig substitution as tmc suggested.

    EDIT: near simultaneous post w/ TD
    Last edited: Oct 10, 2005
  7. Oct 10, 2005 #6
    Ok, thanks. I have that one now.

    Can you please hint me with one more:
    I have the integral of (sqrt(1-x^2)) / x^4 dx

    Which with trig.subs leads to:
    integral of cot^2(y) *csc^2(y) dy

    Again I'm stuck :P
  8. Oct 10, 2005 #7


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    Correct, but remember that

    [tex]\left( {\cot x} \right)^\prime = - \csc ^2 x[/tex]

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