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Thread: algebraic/trigonometric substitution

  1. #1
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    algebraic/trigonometric substitution

    $\displaystyle \int\frac{e^-x}{(9e^-2x+1)^3/2}\, dx$

    $\displaystyle \int\frac{dz}{(z^2-6z+18)^3/2}$

    $\displaystyle \int\frac{ln^3w}{w\sqrt{ln^2w-4}}\,dw$
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  2. #2
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    Krizalid's Avatar
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    Quote Originally Posted by cazimi View Post
    $\displaystyle \int\frac{e^{-x}}{(9e^{-2x}+1)^{3/2}}\, dx$
    Substitute $\displaystyle u=e^{-x},$

    $\displaystyle \int {\frac{{e^{ - x} }}
    {{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = - \int {\frac{{du}}
    {{\left( {9u^2 + 1} \right)\sqrt {9u^2 + 1} }}} .$

    Make another substitution defined by $\displaystyle u=\frac1\tau,$

    $\displaystyle \int {\frac{{e^{ - x} }}
    {{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = \int {\frac{\tau }
    {{\left( {9 + \tau ^2 } \right)\sqrt {9 + \tau ^2 } }}\,d\tau } .$

    One more substitution according to $\displaystyle \varphi = \sqrt {9 + \tau ^2 } ,$

    $\displaystyle \int {\frac{{e^{ - x} }}
    {{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = \int {\varphi ^{ - 2} \,d\varphi } = - \frac{1}
    {\varphi } + k.$

    Back substitute:

    $\displaystyle \begin{aligned}
    \varphi &= \sqrt {9 + \frac{1}
    {{u^2 }}}\\
    &= \sqrt {9 + \frac{1}
    {{e^{ - 2x} }}}\\
    &= \frac{{\sqrt {9e^{ - 2x} + 1} }}
    {{e^{ - x} }}.
    \end{aligned}$

    And we happily get $\displaystyle \int {\frac{{e^{ - x} }}
    {{\left( {9e^{ - 2x} + 1} \right)^{3/2} }}\,dx} = - \frac{{e^{ - x} }}
    {{\sqrt {9e^{ - 2x} + 1} }} + k.$
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  3. #3
    Senior Member
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    Second one

    On the second integral,

    First, I think you meant
    $\displaystyle \int\frac{dz}{(z^2-6z+18)^{3/2}} $, not $\displaystyle \int\frac{dz}{(z^2-6z+18)^3/2}$ (you forgot to bracket the exponent).

    First, we want to complete the square in the denominator: $\displaystyle z^2-6z+18=z^2-6z+9+9=(z-3)^2+9$

    So using $\displaystyle u=z-3$, we get $\displaystyle \int\frac{dz}{(z^2-6z+18)^{3/2}} = \int\frac{du}{(u^2+3^2)^{3/2}}$

    From here, you can use the trigonometric substitution $\displaystyle u=3 \tan \theta$, which gives $\displaystyle u^2+3^2 = 3^2 (\tan^2 \theta +1) = (3 \sec \theta)^2$ and $\displaystyle du = 3 \sec^2 \theta d\theta$
    And thus
    $\displaystyle \int\frac{dz}{(z^2-6z+18)^{3/2}} = \int\frac{3 \sec^2 \theta d\theta}{(3 \sec \theta)^3} = \int\frac{d\theta}{9 \sec \theta} = \frac{1}{9} \int \cos \theta d\theta$

    You can take it from there.

    --Kevin C.
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  4. #4
    Super Member PaulRS's Avatar
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    Well, the third

    Let $\displaystyle u=\ln(w)$ thus: $\displaystyle \frac{du}{dw}=\frac{1}{w}$

    Then $\displaystyle \int\frac{ln^3w}{w\sqrt{ln^2w-4}}dw=\int\frac{u^3}{\sqrt{u^2-4}}du$

    Now $\displaystyle z=\sqrt{u^2-4}$ so: $\displaystyle \frac{dz}{du}=\frac{u}{\sqrt{u^2-4}}$

    Therefore: $\displaystyle \int\frac{u^3}{\sqrt{u^2-4}}du=\int(z^2+4)dz$
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