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Math Help - Trig Substitutions Need Help

  1. #1
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    Trig Substitutions Need Help

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

    \int \frac{(4+(4tan^2\theta)^2)^2}{x^3} dx

    \int \frac{(4+(4tan^2\theta)^2)^2}{4tan^2\theta} d\theta

    \int \frac{4+4tan^8\theta}{4tan^2\theta} d\theta

    \int \frac{4}{4tan^2\theta} d\theta + \int \frac{4tan^8\theta}{4tan^2\theta} d\theta

    \int \frac{1}{tan^2\theta} d\theta + \int tan^6\theta d\theta

    I'm stuck at here...
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by circuscircus View Post
    \int \frac{(4+x^2)^2}{x^3} dx
    Try x = 2tan(\theta) instead.

    -Dan
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  3. #3
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    \int \frac{(4+x^2)^2}{x^3} dx

    \int \frac{(4+(2tan^2\theta)^2)^2}{2\tan^6\theta} dx

    \int \frac{4+2tan^8\theta}{2tan^6\theta} dx

    \int \frac{2+tan^8\theta}{tan^6\theta} dx

    What to do next?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    Try x = 2tan(\theta) instead.

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

    \int \frac{(4+(2tan^2\theta)^2)^2}{2\tan^6\theta} dx

    \int \frac{4+2tan^8\theta}{2tan^6\theta} dx

    \int \frac{2+tan^8\theta}{tan^6\theta} dx

    What to do next?
    Not quite what I was thinking:
    \int \frac{(4+x^2)^2}{x^3} dx

    Let x = 2tan{\theta} \implies dx = 2sec^2(\theta) d \theta

    \int \frac{(4+x^2)^2}{x^3} dx = \int \frac{(4 + 4tan^2{\theta})^2}{8tan^3(\theta)} \cdot 2sec^2(\theta) d\theta

    = 4 \int \frac{(1 + tan^2{\theta})^2}{tan^3(\theta)} \cdot sec^2(\theta) d\theta

    = 4 \int \frac{sec^4{\theta}}{tan^3(\theta)} \cdot sec^2(\theta) d\theta

    = 4 \int \frac{sec^6{\theta}}{tan^3(\theta)} d\theta

    And you can proceed from there.

    However upon reflection:
    \int \frac{(4+x^2)^2}{x^3} dx = \int \frac{x^4 + 8x^2 + 16}{x^3}

    = \int \left ( x + \frac{8}{x} + \frac{16}{x^3} \right ) dx
    is a much simpler method.

    -Dan
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  5. #5
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    Hello, circuscircus!

    You seem to have missed the point of Trig Substitution . . .


    \int \frac{(4+x^2)^2}{x^3}\,dx
    Let: x \:=\:2\tan\theta \quad\Rightarrow\quad dx \:=\:2\sec^2\!\theta\,d\theta

    Also: 4 + x^2 \:=\:4 + 4\tan^2\!\theta \:=\:4(1 + \tan^2\!\theta) \:=\:4\sec^2\!\theta


    Substitute: . \int\frac{(4\sec^2\!\theta)^2}{(2\tan\theta)^3}\,(  2\sec^2\!\theta\,d\theta) \;=\;4\int\frac{\sec^4\!\theta}{\tan^3\!\theta}(\s  ec^2\!\theta\,d\theta)  \;= \;4\int\frac{(\sec^2\!\theta)^2}{\tan^3\!\theta}(\  sec^2\!\theta\,d\theta)

    . . . . =\;4\int\frac{(\tan^2\!\theta + 1)^2}{\tan^3\!\theta}(\sec^2\!\theta\,d\theta) \;=\;4\int\frac{\tan^4\theta + 2\tan^2\theta + 1}{\tan^3\theta}(\sec^2\!\theta\,d\theta)

    . . . . = \;4\int\left(\tan\theta + \frac{2}{\tan\theta} + (\tan\theta)^{-3}\right)(\sec^2\!\theta\,d\theta)


    Let: u \,=\,\tan\theta\quad\Rightarrow\quad du \,=\,\sec^2\!\theta\,d\theta

    Substitute: . 4\int\left(u + \frac{2}{u} + u^{-3}\right)\,du


    Can you finish it now?

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  6. #6
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    <br />
\int\frac{(4\sec^2\!\theta)^2}{(2\tan\theta)^3}\,(  2\sec^2\!\theta\,d\theta) \;=\;4\int\frac{\sec^4\!\theta}{\tan^3\!\theta}(\s  ec^2\!\theta\,d\theta) \;= \;4\int\frac{(\sec^2\!\theta)^2}{\tan^3\!\theta}(\  sec^2\!\theta\,d\theta)<br />

    Ok gotcha, but I'm confused why (4sec0^2)^2 / (2tan0)^3 is 4... shouldn't it be 4^2 = 16 and 2^3=8, so 16/8 = 2?
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