# Integral using hyperbolic substitution

1. Aug 14, 2011

### Malby

1. The problem statement, all variables and given/known data
$$\int\left(1+x^{2}\right)^{\frac{3}{2}}dx$$

2. Relevant equations
The hyperbolic functions.

3. The attempt at a solution
We've been going over hyperbolic substitutions in class so I assume I'm meant to use one of those, but I'm just not sure how to choose which one. Any help greatly appreciated.

2. Aug 14, 2011

### SammyS

Staff Emeritus
sinh2(u) + 1 = cosh2(u)

So, I would try letting x = sinh(u) → dx = cosh(u) du

3. Aug 15, 2011

$$\begin{eqnarray*} \int\left(1+x^{2}\right)^{\frac{3}{2}}dx & = & \int\left(1+\sinh^{2}\left(u\right)\right)^{\frac{3}{2}}\cosh\left(u\right)\, du\\ & = & \int\sqrt{\left(\cosh^{2}\left(u\right)\right)^{3}}\cosh\left(u\right)\, du\\ & = & \int\sqrt{\cosh^{6}\left(u\right)}\cosh\left(u\right)\, du\\ & = & \int\cosh^{4}\left(u\right)\, du \end{eqnarray*}[\tex] Last edited: Aug 15, 2011 4. Aug 15, 2011 ### Malby Oops, that was meant to be nice neat LaTeX... Well, I tried doing that but ended up with int(cosh^4(u))du... not sure if there is a nice little trick for solving that or if I've ended up with a harder problem than I started with. Also, why didn't that tex code just work? 5. Aug 15, 2011 ### flyingpig Use the identity [tex]cosh^2 (t) = \frac{1 + cosh(2t)}{2}$$

Omg i helped someone!!!

6. Aug 15, 2011

### SammyS

Staff Emeritus
Changed '\' to '/' in the /tex tag so we can read your LaTeX.

7. Aug 15, 2011

### GreenPrint

You could evaluate this integral without using the hyperbolic trig functions and just using the regular ones.

8. Aug 15, 2011

### SammyS

Staff Emeritus
By The Way: $u = \sinh^{-1}(x) = \ln \left( x+\sqrt{x^{2}+1} \right)$

9. Aug 15, 2011

### Malby

Aha! I was wondering how it got to that step.

Thanks very much for your help!