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Math Help - More trig substitution

  1. #1
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    More trig substitution

     \int \frac{x^2}{(3+4x-4x^2)^{3/2}}

    I'm pretty sure I need to complete the square...but when I start that process the problem gets really nasty.

    A push por favor?
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  2. #2
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    A push
    A push down a cliff?

    Quote Originally Posted by silencecloak View Post
     \int \frac{x^2}{(3+4x-4x^2)^{3/2}}

    I'm pretty sure I need to complete the square...but when I start that process the problem gets really nasty.
    3+4x-4x^2 = - (4x^2 -4x - 3) = - (4x^2 - 4x + 1 - 4).
    Therefore, -((2x-1)^2-4) = 4 - (2x-1)^2.
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  3. #3
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    This problem is super confusing to me
    \int \frac{x^2}{(3+4x-4x^2)^{3/2}}

    =

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

    Would I set  u = 2sec\theta

    as my next step? And after that I have no clue
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  4. #4
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    Quote Originally Posted by silencecloak View Post
    This problem is super confusing to me
    \int \frac{x^2}{(3+4x-4x^2)^{3/2}}

    =

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

    Would I set  u = 2sec\theta

    as my next step? And after that I have no clue
    First t=2x-1.
    Second use a trignometric substitution.
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  5. #5
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    Hello, silencecloak!

    This is a nasty problem . . .


     \int \frac{x^2\,dx}{(3+4x-4x^2)^{3/2}}
    Yes, we must complete the square . . .

    We have: . 3 + 4x - 4x^2 \;=\;4 - 1 + 4x - 4x^2\;=\; 4 - (1 - 4x + 4x^2) \;=\;1 - (2x - 1)^2

    The integral becomes: . \int\frac{x^2\,dx}{\left[1 - (2x-1)^2\right]^{\frac{3}{2}}}


    Let: 2x-1 \:=\:\sin\theta \quad\Rightarrow\quad 2\,dx \:=\:\cos\theta\,d\theta \quad\Rightarrow\quad dx \:=\:\tfrac{1}{2}\cos\theta\,d\theta

    . . and: . x \:=\:\frac{\sin\theta + 1}{2}


    Substitute: . \int\frac{\left(\frac{\sin\theta+1}{2}\right)^2} {(\cos\theta)^3}\left(\tfrac{1}{2}\cos\theta\,d\th  eta\right) \;=\; \frac{1}{8}\int \frac{\sin^2\!\theta + 2\sin\theta + 1}{\cos^2\theta}\,d\theta

    . . . = \;\frac{1}{8}\int\bigg[\frac{\sin^2\!\theta}{\cos^2\!\theta} + \frac{2\sin\theta}{\cos^2\!\theta} + \frac{1}{\cos^2\!\theta}\bigg]\,d\theta

    . . . =\;\frac{1}{8}\int\bigg[\left(\frac{\sin\theta}{\cos\theta}\right)^2 + 2\left(\frac{1}{\cos\theta}\right)\left(\frac{\sin  \theta}{\cos\theta}\right) + \left(\frac{1}{\cos\theta}\right)^2 \bigg]\,d\theta

    . . . = \;\frac{1}{8} \int\bigg[\tan^2\!\theta + 2\sec\theta\tan\theta + \sec^2\!\theta\bigg]\,d\theta


    Can you finish it?

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  6. #6
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    Well now I'm just more confused

    Two different answers from completing the square

    @Soroban

    you lost me at

    2x-1 \:=\:\sin\theta \quad\Rightarrow\quad 2\,dx<br />
\:=\:\cos\theta\,d\theta \quad\Rightarrow\quad dx<br />
\:=\:\tfrac{1}{2}\cos\theta\,d\theta<br />

    Edit: this is the first time I have done a trig sub problem like this...
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