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Math Help - [SOLVED] Solving This Indefinite Integral With A Trig Substitution

  1. #1
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    [SOLVED] Solving This Indefinite Integral With A Trig Substitution

    The problem is the integral of x/(4-x^2), and I'm asked to solve it using a trig substitution despite there being no square root present in the problem, how can I go about doing this?
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    Hello, fattydq!

    \int\frac{x}{4-x^2}\,dx . . . . using a trig substitution.

    Let x \:=\:2\sin\theta \quad\Rightarrow\quad dx \:=\:2\cos\theta\,d\theta

    . . and: . 4 - x^2 \:=\:4 - (2\sin\theta)^2 \:=\:4-4\sin^2\!x \:=\:4(1-\sin^2\!x) \:=\:4\cos^2\!x


    Substitute: . \int\frac{2\sin\theta}{4\cos^2\!\theta}\,(2\cos\th  eta\,d\theta) \;=\;\int\frac{\sin\theta}{\cos\theta}\,d\theta \;=\;\int\tan\theta\,d\theta \;=\;-\ln(\cos\theta) + C  .[1]


    Bacl-substitute: . 2\sin\theta \:=\:x \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{2} \:=\:\frac{opp}{hyp}

    We have: .  opp = x,\;hyp = 2\quad\Rightarrow\quad adj = \sqrt{4-x^2}

    . . Hence: . \cos\theta \:=\:\frac{adj}{hyp} \:=\:\frac{\sqrt{4-x^2}}{2}


    Then [1] becomes: . -\ln\left(\frac{\sqrt{4-x^2}}{2}\right)  + C \;=\;-\bigg[\ln(\sqrt{4-x^2}) - \ln(2)\bigg] + C

    . . . . . . = \;-\ln\left(\sqrt{4-x^2}\right) + \underbrace{\ln(2) + C}_{\text{This is a constant}} \;=\; -\ln\left(\sqrt{4-x^2}\right) + C

    . . . . . . = \;-\ln\left(4-x^2\right)^{\frac{1}{2}} + C \;=\;-\tfrac{1}{2}\ln\left(4-x^2\right) + C

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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fattydq View Post
    The problem is the integral of x/(4-x^2), and I'm asked to solve it using a trig substitution despite there being no square root present in the problem, how can I go about doing this?
    for the record, if the method of integration was not specified, a regular u-substitution of u = 4 - x^2 would take care of this guy with much less difficulty.
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    Quote Originally Posted by Jhevon View Post
    for the record, if the method of integration was not specified, a regular u-substitution of u = 4 - x^2 would take care of this guy with much less difficulty.
    Yep, I actually had to solve it that way first, and then in part 2 I was asked to solve it with a trig substitution, which did indeed prove to be much more painful
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