# Thread: Simple integral with hyperbolic functions

1. ## 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.

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

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

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

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

3. 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 ...

4. Thanks Soroban, I knew it wasn't that complicated I just failed to see it :P