# Thread: Integration by hyperbolic substitution

1. ## Integration by hyperbolic substitution

Once again I feel I am not too far from the answer...

$\displaystyle \int \frac{x^2}{\sqrt{1+x^2}} \ dx$

using an hyperbolic substitution.

I tried helping myself with this: Integration by Hyperbolic Substitution | calculuspowerup.com but I have trouble converting the hyperbolic function with the triangle.

I did:

Once again I feel I am not too far from the answer...

$\displaystyle \int \frac{x^2}{\sqrt{1+x^2}} \ dx$

using an hyperbolic substitution.

I tried helping myself with this: Integration by Hyperbolic Substitution | calculuspowerup.com but I have trouble converting the hyperbolic function with the triangle.

I did:

$\displaystyle x = \sinh u$

$\displaystyle dx = \cosh u \ du$

$\displaystyle \int \frac{x^2}{\sqrt{1+x^2}} \ dx =$

$\displaystyle \int \frac{{\sinh}^2 u \cdot \cosh u}{\cosh u} \ du =$

$\displaystyle \int {\sinh}^2 u \ du =$

$\displaystyle \frac{1}{2} \int (\cosh 2u - 1) \ du =$

$\displaystyle \frac{\sinh 2u}{4} - \frac{u}{2} + C =$

$\displaystyle \frac{\sinh 2u}{4} - \frac{\text{argsinh} x}{2} + C$

That is where I'm not sure how to isolate u to get my x value and how to work with the triangle etc. I need help with that.

$\displaystyle \frac{x \sqrt{1+x^2}}{2} - \frac{\text{argsinh} x}{2} + C$
2. $\displaystyle \sinh 2u=2\sinh u\cosh u$
$\displaystyle \sinh u=x, \ \cosh u=\sqrt{1+\sinh^2u}=\sqrt{1+x^2}$