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Math Help - Simple integral with hyperbolic functions

  1. #1
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    Simple integral with hyperbolic functions

    I have looked every way in this, tried some identities, substitutions etc. and always came close to something but never found a winning combination. I am sure it is a pretty simple substitution, I just don't see it.

    \int x^2 \cdot \text{tanh}^2 \ (x^3) \cdot dx

    I thought of setting u = \text{cosh} \ (x^3) but it would give me 2 \cdot \int \frac{\text{sinh} \ (x^3)}{u^2} \cdot du which I am still unable to resolve. (I also managed to get something similar but divided by u instead of u^2 with identities, but it is worthless since I cannot resolve it either.)

    Like I said, I know it is really simple and I just don't see it. Thanks for helping me out.
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  2. #2
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    Hello, mathieumg!

    \int x^2 \tanh^2(x^3)\,dx

    We have: . \int \tanh^2(x^3)\,(x^2\,dx)

    Let u = x^3\quad\Rightarrow\quad du = 3x^2\,dx \quad\Rightarrow\quad x^2\,dx \:=\:\tfrac{1}{3}du

    Substitute: . \int \tanh^2\!u\left(\tfrac{1}{3}du\right)

    . . . . . . =\;\tfrac{1}{3}\int \underbrace{\tanh^2\!u}\,du

    . . . . = \;\tfrac{1}{3}\int\overbrace{(1 - \text{sech}^2u)}\,du

    . . . . =\; \tfrac{1}{3}(u - \tanh u) + C \quad\hdots \text{etc.}

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  3. #3
    MHF Contributor Amer's Avatar
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    let
    u=e^{x^3}

    du=x^2e^{x^3}dx

    du=x^2udx

    dx=\frac{du}{x^2u}

    \frac{1}{u}=e^{-x^3}

    \int x^2tanh^2(x^3).dx

    since
    tanh^2(x^3)=\frac{e^{x^3}-e^{-x^3}}{e^{x^3}+e^{-x^3}}

    \int \frac{u-\frac{1}{u}}{u+\frac{1}{u}}x^2\left(\frac{du}{x^2u  }\right)

    the rest for you ...
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  4. #4
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    Thanks Soroban, I knew it wasn't that complicated I just failed to see it :P
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