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Homework Help: Evaluating these integrals

  1. Oct 2, 2005 #1
    I'm having a bit of trouble evaluating this integral: (This is not a homework problem btw)
    Integral between Pi and 0 (x sin x)/(1 + cos^2 x) d x
    I don't even know where to begin with the substitutions or anything...... I was thinking the denominator could be sin^2 x..... only when I realized this is trig and not hyperbolic. :(

    Also, something in my book:
    Integral (2t^2 + 3t*t^3) d(t^2)
    This was after the substitution of dy to d(t^2) when dealing with a line integral; I don't understand... how do you integrate this?

    Thanks for any ideas.
  2. jcsd
  3. Oct 2, 2005 #2


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    I don't think you're going to find a closed form for your integral. If it's any help, the numerical value of your integral is approximately -2.4674011...

    For the second integral, replace [itex]d (t^2)[/itex] with [itex]2 t dt[/itex].
  4. Oct 2, 2005 #3
    Thanks for the reply,
    My book shows something weird with the integral...
    Here's the work.... (Assuming everything I have below has the limits pi and 0)
    Let x = pi y, then:
    integral (xsin x)/(1 + cos^2 x) dx = ---> integ (pi-y)(sin y)/(1 + cos^2 y) dy = pi * integ (sin y)/(1 + cos^2 y) dy - integ (y sin y)/(1 + cos^2 y) dy = -pi * integ d(cos y)/(1 + cos^2 y) - I = -pi*arctan(cos y) - I = Pi^2 /2 - I
    And as it says:
    i.e. I = Pi^2 / 2 - I or I = Pi^2 / 4
    I got lost at the point where I showed the arrow (And I'm confused about why they chose that particular substitution)...
    Also, what's up with this - I thing? Can somebody explain it?
    Thanks loads for any replies.
    Last edited: Oct 2, 2005
  5. Oct 2, 2005 #4


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    Oh, I like that! Basically, it's a simple transformation of the integral with the result that the original integral = some other integral - the original integral and, luckily "some other integral" can be evaluated!
  6. Oct 3, 2005 #5


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    U can part integrate

    [tex] \int_{\pi}^{0} x \frac{-d(\cos x)}{1+\cos^{2}x} =...[/tex]

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