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Inverse trig fuction

  1. Apr 27, 2004 #1
    This is a physics problem, but I have some trouble finishing it up. I evaluated an integral, which gave me

    inv tan of (1-E)tan(theta/2)/(sqrt (1-E^2)

    evaluated from 0 to 2pi. I changed the limits to 0 to pi, and multipied by 2, because tan x doesn't exist at 2pi.
    I know that if you take the inverse tangent of a tangent, you'll just get the angle. But because the tan has this ugly coefficient, I can't simply get the angle, right?
    The answer is pi/2 so I'm inclined to just ignore it...but I know you can't do that...?

    Thanks a lot!
     
    Last edited: Apr 27, 2004
  2. jcsd
  3. Apr 27, 2004 #2
    tan (x) is defined at 2pi. tan(x) = sin(x)/cos(x), and cos(2pi) = 1. Might want to rethink that. I'm just wondering, what's the E? that's where i'm having trouble! After all the dirty work (not really i guess) i get tan^-1(E-1)/sqrt(1-E^2). I can only guess that E is 0 or something , which would mean that you'd get tan^-1(-1) which is only defined in the second and fourth quadrants.

    Maybe your integral is wrong? It sounds to me by your language that you used a computer ("it gave me") or other device to get this integral. You might want to go back and type it in correctly. That's all I have to say.
     
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