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Trig integral

  1. Apr 10, 2007 #1
    1. The problem statement, all variables and given/known data

    how would one calculate [tex] 4 \int_0^{\frac{\pi}{2}} \frac{\cos^2 \theta}{(1 + \cos^2 \theta)^2} d \theta [/tex]?

    3. The attempt at a solution

    someone suggested a [tex]u = \tan \theta[/tex] substitution, but i don't understand why and how this would help me. couldn't i just use [tex]u = \cos t[/tex]?
     
  2. jcsd
  3. Apr 10, 2007 #2

    AlephZero

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    Divide top and bottom by [tex]\cos^4\theta[/tex].

    The function becomes [tex]\frac{\sec^2\theta}{(\sec^2\theta+1)^2}[/tex]

    Does that make [tex]u=\tan\theta[/tex] looks like a good move?

    u = cos t is probably a bad move, because du = -sin t dt and the sin t doesn't cancel with anything.
     
    Last edited: Apr 10, 2007
  4. Apr 10, 2007 #3
    the only trig functions we learn here are tan, cos and sin, so i'm not really sure... :)

    wouldn't i get du = -sin t dt = -sqrt(1 - cos^2 t) = -sqrt(1 - u^2) dt?

    because then the integral should translate to [tex]-4 \int_1^0 \frac{u^2}{ (u^2+1)(1-u^2)^{1/2} } du[/tex], or am i wrong?
     
  5. Apr 10, 2007 #4
    :surprised

    I agree with what you had in the last post, except, wouldn't it be (1+u^2)^2 in the denominator? Anyway, now what?
     
    Last edited: Apr 10, 2007
  6. Apr 10, 2007 #5
    yeah, i've never understood why that is so...
     
  7. Apr 10, 2007 #6

    mjsd

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    what is the derivative of [tex]\tan \theta[/tex]?
     
  8. Apr 10, 2007 #7

    mjsd

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    also do you know [tex]1+\tan^2\theta =\sec^2\theta[/tex]?
     
  9. Apr 10, 2007 #8
    I edited my post just a little too late... isn't it [tex](u^2 + 1)^2[/tex] in the denominator?
     
  10. Apr 10, 2007 #9
    oops. yeah, that's right.
     
  11. Apr 10, 2007 #10
    no, i didn't know that. :)

    so i would get [tex]4 \int_0^{\pi/2} \frac{\sec^2 \theta}{(\sec^2 \theta + 1)^2} d \theta = 4 \int_0^{\pi/2} \frac{\tan^2 \theta + 1}{(\tan^2 \theta + 2)^2} d \theta = 4 \int_0^{\Omega} \frac{u^2 + 1}{(u^2 + 2)^2(u^2 + 1)} du = 4 \int_0^{\Omega} \frac{du}{(u^2 + 2)^2}[/tex]

    as [tex]u = \tan \theta[/tex] and [tex]du = (1 + \tan^2 \theta) d \theta = (1 + u^2) d \theta[/tex].

    does that seem correct? i'm unsure on what [tex]\Omega[/tex] would be though, as [tex]\tan \pi/2 = \infty[/tex]
     
    Last edited: Apr 10, 2007
  12. Apr 11, 2007 #11

    Gib Z

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    Ahh Nope, [tex]\int \frac{\sec^2\theta}{(\sec^2\theta+1)^2} d\theta[/tex] becomes [tex]\int \frac{1}{u^2} du[/tex] when u = tan theta, because the derivative of tan theta is (sec theta)^2.

    In case you don't know Sec x = 1/(cos x) by definition.
     
  13. Apr 11, 2007 #12

    AlephZero

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    Oops.... you mean [tex]\int \frac{1}{(u^2+2)^2} du[/tex]

    1 + tan^2 x = sec^2 x

    NOT 1 + sec^x = tan^2 x !!!

    (And even if that was true, it would have been u^4 not u^2)
     
  14. Apr 12, 2007 #13

    Gib Z

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    Sorry about that, your right of course AlephZero.
     
  15. Apr 13, 2007 #14

    dextercioby

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    [tex]\int_{0}^{\frac{\pi }{2}}\frac{\cos ^{2}x}{\left( 1+\cos ^{2}x\right) ^{2}}dx=\allowbreak \int_{0}^{\infty }\frac{1}{\left( 1+p^{2}\right) ^{2}\left( 1+\frac{1}{1+p^{2}}\right) ^{2}}\,dp=\int_{0}^{\infty }\frac{1}{\left( 2+p^{2}\right) ^{2}}\,dp\allowbreak [/tex]

    [tex] =\allowbreak \int_{0}^{\frac{1}{2}\pi }\frac{\sqrt{2}}{4+4\tan ^{2}q}\,dq=\frac{\sqrt{2}}{4}\int_{0}^{\frac{1}{2}\pi }\cos ^{2}q \ {} \ dq=\frac{\sqrt{2}}{4}\frac{\pi }{4}=\frac{\pi\sqrt{2}}{16}[/tex]
     
  16. Apr 13, 2007 #15

    Gib Z

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    Would I be the only person to be a tiny bit confused as to what exactly dexter did ...
     
    Last edited: Apr 13, 2007
  17. Apr 13, 2007 #16

    dextercioby

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    Would it help if you knew that [itex] \tan x=p [/itex] and [itex] p=\sqrt{2}\tan q [/itex] ?
     
  18. Apr 13, 2007 #17

    Gib Z

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    Ahh yup it did :D
     
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