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A integrate question

  1. Apr 17, 2007 #1
    question in the attachment...

    integrate this function from 1 to 2.

    i can't find a way to do this quesiton
    it's from previous year final exam so there's no answer key to it..

    hope someone would teach me..

    Attached Files:

  2. jcsd
  3. Apr 17, 2007 #2


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    Have you learned trigonometric substitution? Let [itex]x = \sqrt(5)\sin\theta[/itex], and use the pythagorean identity to write [itex]5(1-\sin^2\theta) = 5\cos^2\theta[/itex].

    Under this change of variables, [itex]dx = \sqrt(5)\cos\theta d\theta[/itex]. Your integral then becomes an integral over a constant times [itex]1/\cos^2\theta = sec^2\theta[/itex]. If you know the integral of [itex]sec^2\theta[/itex], you're pretty much done then.
  4. Apr 17, 2007 #3
    but where did the 3/2 go..
  5. Apr 17, 2007 #4
    right now i have...

    4/5 integral sec^2 theta..

    if i'm right at this point
    integral of sec^2 is tan x
    but i set x = root5 sin theta earlier..
    how do i finish this question..
    Last edited: Apr 18, 2007
  6. Apr 17, 2007 #5

    Gib Z

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    The Exponents canceled each other out.

    [tex]\int \frac{4}{(5-x^2)^{3/2}} dx[/tex].

    Now We do the substitution Mute said. [itex]dx = \sqrt(5)\cos\theta d\theta[/itex]

    So now the integral is [tex]\int {4\sqrt{5} \cos \theta}{5^{3/2}\cdot (\cos^2 \theta)^{3/2}} d\theta[/tex]

    We can take all constants out. When we have a power to a power, we multiply the powers. 3/2 times 2 is just 3.

    [tex]\frac{4\sqrt{5}}{5^{3/2}} \int \frac{\cos \theta}{\cos^3 \theta} d\theta = \frac{4\sqrt{5}}{5^{3/2}} \int \frac{1}{\cos^2 \theta} d\theta[/tex]. As Mute said, 1/cos^2 theta is sec^2 theta. And the integral of that is easy, you should know that.
  7. Apr 17, 2007 #6

    Gib Z

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    O and the constant on the outside simplifies to 4/5.
  8. Apr 18, 2007 #7
    that's exactly what i said ..
    but if i take the integral from 1 to 2..
    i dont' get the answer 6/5

    cuz i think if you set x to some value.
    the upper and lower bound needs to change..
    but i dont' remember if there's such thing..
  9. Apr 18, 2007 #8

    Gib Z

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    Well The integral is (4/5) tan theta.

    x = sqrt{5} sin theta.
    sin theta = x/(sqrt5)
    theta = arcsin (x/sqrt5)
    The integral is therefore 4/5 tan (arcsin x/sqrt5). Plug in x values and subtract.
  10. Apr 18, 2007 #9
    my x values are 1 to 2..
    if i sub into arcsinx it's undefined.
  11. Apr 18, 2007 #10


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    That would be a problem, except that you want to sub into arcsin(x/sqrt5), not arcsinx.
  12. Apr 18, 2007 #11
    got it...
    thanks mute and GibZ..
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