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Polar cords and integral help

  1. Feb 21, 2005 #1
    two questions...

    1) by a change of variable show the following...
    [tex] \int^{\infty} _{-\infty} \frac{dt}{(a^2 + t^2)^{3/2}} = \frac{2}{a^2}\int^{\pi/2} _0 cos \ t \ dt [/tex]

    i'm thinking about changing this to polar cords and see where that take me anyone?

    2) [tex] F(x) = \int^{cos \ x} _0 e^{xt^2} \ dt , \ G(x) = \int^{cos \ x} _0 t^2 e^{xt^2} \ dt, \ H(t) = G(x) - F'(x) [/tex] express H(x) in elementry functions.

    [tex] F'(x) = \frac{\partial cos \ x}{\partial x} e^{xt^2} - 0 + \int \frac{\partial}{\partial x} e^{xt^2} \ dt \\
    = -sin \ x \ e^{xt^2} + \int t^2e^{xt^2} \ dt [/tex]

    [tex]
    H(x) = \int^{cos \ x} _0 t^2 e^{xt^2} \ dt + sin \ x \ e^{xt^2} - \int t^2e^{xt^2} \ dt

    = sin \ x \ e^{xt^2} [/tex]

    Did i do the dirv correctly? and how do I show that [tex]H(\pi/4) = \frac{e^{\pi/8}}{\sqrt{2}}[/tex]
     
    Last edited: Feb 21, 2005
  2. jcsd
  3. Feb 21, 2005 #2

    Curious3141

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    Hint : try [tex]\frac{t}{a} = \tan \theta[/tex]
     
  4. Feb 21, 2005 #3

    dextercioby

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    For the second,u should plug [itex] t\rightarrow \cos x [/itex] and then it will be simple to find the final formula...

    Daniel.

    P.S.What polar coordinates...?It's not a double integral...
     
  5. Feb 21, 2005 #4
    why would i plug t in for cos x? do u mean t for x so its cos t? im confused why i would do that
     
  6. Feb 21, 2005 #5
    i have to take the function at the upper limit don't I? yep yea i do, ok thanks all
     
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