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Thread: 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.

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

    I thought of setting $\displaystyle u = \text{cosh} \ (x^3)$ but it would give me $\displaystyle 2 \cdot \int \frac{\text{sinh} \ (x^3)}{u^2} \cdot du$ which I am still unable to resolve. (I also managed to get something similar but divided by $\displaystyle u$ instead of $\displaystyle u^2$ 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!

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

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

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

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

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

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

    . . . . $\displaystyle =\; \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
    $\displaystyle u=e^{x^3}$

    $\displaystyle du=x^2e^{x^3}dx$

    $\displaystyle du=x^2udx$

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

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

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

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

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

    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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