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Thread: Trig Substitution

  1. #1
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    Trig Substitution

    I need to integrate

    $\displaystyle \int 3/ (4-3x^2)^{1/2}$

    I know I'm supposed to use the substitution x= a sin theta but I don't know what happens to the coefficient 3...
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Hellreaver View Post
    I need to integrate

    $\displaystyle \int 3/ (4-3x^2)^{1/2}$

    I know I'm supposed to use the substitution x= a sin theta but I don't know what happens to the coefficient 3...
    Try using the substitution $\displaystyle \sqrt{3}x=2\sin\theta$. If you get stuck, post back.
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  3. #3
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    Hello, Hellreaver!

    $\displaystyle \int \frac{3}{\sqrt{4-3x^2}}\,dx$

    I know I'm supposed to use the substitution $\displaystyle x= a\sin\theta$
    Look ahead a step or two . . .


    We want that radical to change from $\displaystyle \sqrt{4-3x^2}$

    . . to: $\displaystyle \sqrt{4-4\sin^2\!\theta} \;=\;\sqrt{4(1-\sin^2\!\theta)} \;=\;\sqrt{4\cos^2\!\theta} \;=\;2\cos\theta$


    To do that, we let: .$\displaystyle 3x^2 \:=\:4\sin^2\!\theta \quad\Rightarrow\quad x^2 \:=\:\tfrac{4}{3}\sin^2\theta \quad \Rightarrow\quad x \:=\:\tfrac{2}{\sqrt{3}}\sin\theta$ .[1]

    . . Then: .$\displaystyle dx \:=\:\tfrac{2}{\sqrt{3}}\cos\theta\,d\theta $


    Substitute: .$\displaystyle \int\frac{3}{2\cos\theta}\left(\tfrac{2}{\sqrt{3}} \cos\theta\,d\theta\right) \;=\;\sqrt{3}\int d\theta \;=\;\sqrt{3}\,\theta + C$ .[2]


    From [1], we have: .$\displaystyle \sin\theta \:=\:\tfrac{\sqrt{3}}{2}\,x \quad\Rightarrow\quad \theta \:=\:\arcsin\left(\tfrac{\sqrt{3}}{2}x\right)$

    . . Substitute into [2]: .$\displaystyle \sqrt{3}\,\arcsin\left(\tfrac{\sqrt{3}}{2}\,x\righ t) + C $

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