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Please help with t-substitution

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data

    [itex]\int_{0}^{2\pi} \frac{\cos^2\alpha}{A + \sin^2\alpha}d\alpha[/itex]

    3. The attempt at a solution

    I believe when the trig functions all appear squared one may use the substitution [itex] t =\tan \alpha[/itex]. Then

    [itex]
    t^2 + 1 = \sec^2\alpha \implies \cos^2\alpha = \frac{1}{t^2 +1}
    [/itex]
    [itex]
    \sin^2\alpha = t^2\cos^2\alpha = \frac{t^2}{t^2+1}
    [/itex]
    [itex]
    dt = \sec^2\alpha d\alpha \implies d\alpha = \frac{dt}{1 + t^2}
    [/itex]

    Now use

    [itex]\int_{0}^{2\pi} = \int_0^\pi + \int_\pi^{2\pi} [/itex].

    [itex]\int_{0}^{\pi} \frac{\cos^2\alpha}{A + \sin^2\alpha}d\alpha = \int_{0}^\infty\frac{dt}{A(1+t^2)^2 +t^2(1+t^2)}[/itex]
    [itex]= \int_{0}^\infty\frac{dt}{(1+t^2)(A+t^2+At^2)}[/itex]

    [itex]
    1 = C_1(A+ (1+A)t^2) + C_2 (1+t^2) \implies C_1 = -1,C_2 = -\frac{1+A}{A}
    [\latex]

    [itex]I = \int_{0}^\infty \frac{-1}{1+t^2}- \frac{(1+A)/A}{A+(1+A)t^2}[/itex]
    [itex]I = - \frac{\pi}{2}- (1+A)/A\int_{0}^\infty\frac{dt}{A+(1+A)t^2}[/itex]
    [itex]I = - \frac{\pi}{2}- \frac{(1+A)/A}{\sqrt{1+A}}\frac{1}{\sqrt{A}}\left.\tan^{-1}\frac{t}{\sqrt{A}}\right|_{0}^\inft[/itex]

    [itex]I = - \frac{\pi}{2}- \frac{(1+A)/A}{\sqrt{1+A}}\frac{1}{\sqrt{A}}\frac{\pi}{2}[/itex]

    Thanks.
     
    Last edited: Oct 19, 2007
  2. jcsd
  3. Oct 20, 2007 #2

    dynamicsolo

    User Avatar
    Homework Helper

    I'm with you up to the point where you derive the constants for the integration by partial fractions. (And I dropped the quotation because there's a hidden character that's really messing everything up here...)

    For the [tex]t^{2}[/tex] terms, you have

    [tex]C_{1}(1+A) + C_{2} = 0[/tex]

    and for the constant terms,

    [tex]C_{1}A + C_{2} = 1 [/tex].

    So [tex]C_{2} = 1 - AC_{1} [/tex],

    giving [tex] C_{1} + AC_{1} + ( 1 - AC_{1} ) = 0 [/tex].

    Sure, you get [tex]C_{1} = -1[/tex], but wouldn't that lead to just

    [tex] C_{2} = 1 - A(-1) = 1 + A [/tex]?

    As another suggestion: upon plotting this function, I find that the area from x = 0 to x = [tex]2\pi[/tex] is four times the area from x = 0 to x = [tex]\frac{\pi}{2}[/tex].
     
    Last edited: Oct 20, 2007
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