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Thread: 14 int u subst trig

  1. #1
    Super Member bigwave's Avatar
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    14 int u subst trig

    $\text{Evaluate the Integral:}$
    \begin{align*}\displaystyle
    I_{14}&=\int \frac{12\tan^2x \sec^2 x}{(4+\tan^3x)^2} \, dx \\
    \text{Use U substitution}&\\
    u&=4+\tan^3x\\
    \, \therefore dx& =\dfrac{1}{3\sec^2\left(x\right)\tan^2\left(x\righ t)}\,du\\
    &=4 \int\frac{1}{u^2}\,du\\
    &=4\left[-\dfrac{1}{u} \right]\\
    \text{Back substitute $u=4+\tan^3x$}\\
    I_{14}&=-\frac{4}{4+\tan^3x}+C
    \end{align*}

    Why won't this render????
    Last edited by bigwave; Nov 9th 2017 at 01:56 PM. Reason: latex won't render
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Re: 14 int u subst trig

    Why won't this render????
    Don't know ... is this what you want?

    $\displaystyle \int \frac{12\tan^2x \sec^2 x}{(4+\tan^3x)^2} \, dx$

    $u = 4+\tan^3{x}$

    $du = 3\tan^2{x}\sec^2{x} \, dx$

    $\displaystyle 4 \int \frac{du}{u^2}$

    $-\dfrac{4}{u} + C$

    $-\dfrac{4}{4+\tan^3{x}} + C$
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