Integration by hyperbolic substitution

**mathieumg** · June 4th 2009, 10:44 AM

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

I am asked to integrate:

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

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...

I am asked to integrate:

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

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:

$x = \sinh u$

$dx = \cosh u \ du$

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

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

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

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

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

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

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.

The good answer is:

$\frac{x \sqrt{1+x^2}}{2} - \frac{\text{argsinh} x}{2} + C$

Thank you very much in advance.

**red_dog** · June 4th 2009, 11:08 AM

$\sinh 2u=2\sinh u\cosh u$

$\sinh u=x, \ \cosh u=\sqrt{1+\sinh^2u}=\sqrt{1+x^2}$

**mathieumg** · June 4th 2009, 11:30 AM

Thanks a lot red_dog, didn't think of using that identity at all!

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