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Math Help - Improper Intergrals...

  1. #1
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    Improper Intergrals...

    evaluate:

    \int_2^{4} \frac{dt}{t\sqrt{t^2-4}}


    after a trig sub I end up here:

    \int_2^{4} \frac{d\theta}{tan\theta}

    I'm not sure where to go from here?

    This is what I was thinking:

    \lim_{b\rightarrow 2} \left[ \int \frac{d\theta}{tan\theta} \right]_{b}^{4}

    but not sure. Any help in finishing this is greatly appreciated!
    Last edited by Got5onIt; November 6th 2007 at 09:26 PM.
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  2. #2
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    Hello, Got5onIt!

    Evaluate: . \int_2^{4} \frac{dt}{t\sqrt{t^2-4}}

    After a trig sub I end up here: . \int_2^{4} \frac{d\theta}{\tan\theta} . . . . no

    This is a standard form: . \int\frac{dx}{x\sqrt{x^2-a^2}} \;=\;\frac{1}{a}\text{ arcsec}\,\frac{x}{a} + C


    So we have: . \frac{1}{2}\text{ arcsec}\,\frac{t}{2}\,\bigg]^4_2 \;=\;\frac{1}{2}\bigg[\text{arcsec}\,2 - \text{arcsec}\,1\bigg] \;=\;\frac{1}{2}\bigg[\frac{\pi}{3} - 0\bigg] \;=\;\frac{\pi}{6}

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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    if you don't want to use the standard form for the answer, you could have used the trig substitution \frac t2 = \sec \theta (after some algebraic manipulation), and you would arrive at the answer

    (by the way, when you change the variable, the limits usually changes with it, in this case, when you went from t to \theta, the limits would not be 2 to 4 anymore)
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  4. #4
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    Thanks for the help!

    I used the trig sub  t= 2sec\theta

    and dt= 2\sec\theta tan\theta d\theta

    and \sqrt{t^2-4} = \sqrt{4(sec^2\theta-1} = 2tan\theta

    Looks like I was all wrong or going the "unnecessary" route.
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  5. #5
    Math Engineering Student
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    Quote Originally Posted by Got5onIt View Post
    \int_2^{4} \frac{dt}{t\sqrt{t^2-4}}
    When the discontinuity is visible (of course, if it's in the upper or lower integration limit, not on the interval [a,b]), you can avoid the limit procedure, then evaluate. The rest is routine.

    --

    Why do we have to apply trig. sub.? Before that, I suggest to set u=\sqrt{t^2-4}\implies du=\frac t{\sqrt{t^2-4}}\,dt.

    Since u^2+4=t^2, you'll have an easy integral to evaluate, and the conclusion follows.
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