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Math Help - trig substitution, where did i go wrong?

  1. #1
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    trig substitution, where did i go wrong?

    problem: 1/(4-x^2)

    (ill be using o to represent theta)
    x=sino
    dx=2coso do
    sin=x/2


    integrate 1/2coso * 2coso do = integrate do

    Looking at the triangle I draw I can tell that i screwed up somewhere but the form a-x is sin

    the answer I end up with is arcsin(x/2) but I know it should be arctan(x/2) i just cant make it work.
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  2. #2
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    Re: trig substitution, where did i go wrong?

    Quote Originally Posted by NecroWinter View Post
    problem: 1/(4-x^2)

    (ill be using o to represent theta)
    x=sino
    dx=2coso do
    sin=x/2


    integrate 1/2coso * 2coso do = integrate do

    Looking at the triangle I draw I can tell that i screwed up somewhere but the form a-x is sin

    the answer I end up with is arcsin(x/2) but I know it should be arctan(x/2) i just cant make it work.
    Hyperbolic substitution would work better here, since \displaystyle \begin{align*} 1 - \textrm{tanh}^2{t} \equiv \textrm{sech}^2{t} \end{align*}, so make the substitution \displaystyle \begin{align*} x = 2\,\textrm{tanh}\,{t} \implies dx = 2\,\textrm{sech}^2{t}\,dt \end{align*}.
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  3. #3
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    Re: trig substitution, where did i go wrong?

    Ive never heard of that, this question is homework that I got wrong and im just wondering what I did wrong. weve only covered basic trig substitution
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  4. #4
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    Re: trig substitution, where did i go wrong?

    You made a mistake working out 4-x^2
    If we sub x=2sino 4-x^2 becomes 4cos^2o and you get an integral you can't do
    Do you know about partial fractions. If you partial fraction the question you get an integral you can do (if you have learnt about differentiating logs)
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