# Integral using hyperbolic substitution

## Homework Statement

$$\int\left(1+x^{2}\right)^{\frac{3}{2}}dx$$

## Homework Equations

The hyperbolic functions.

## 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.

## Answers and Replies

SammyS
Staff Emeritus
Homework Helper
Gold Member
sinh2(u) + 1 = cosh2(u)

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

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

Your tex didn't work because you used \ instead of /

Omg i helped someone!!!

SammyS
Staff Emeritus
Homework Helper
Gold Member
$$\int\left(1+x^{2}\right)^{\frac{3}{2}}dx = \int\left(1+\sinh^{2}\left(u\right)\right)^{\frac{3}{2}}\cosh\left(u\right)\, du$$
$$=\int\sqrt{\left(\cosh^{2}\left(u\right)\right)^{3}}\cosh\left(u\right)\, du$$
$$=\int\sqrt{\cosh^{6}\left(u\right)} \cosh\left(u\right)\, du$$
$$=\int\cosh^{4}\left(u\right)\, du$$
Changed '\' to '/' in the /tex tag so we can read your LaTeX.

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
sinh2(u) + 1 = cosh2(u)

So, I would try letting x = sinh(u) → dx = cosh(u) du
By The Way: $u = \sinh^{-1}(x) = \ln \left( x+\sqrt{x^{2}+1} \right)$

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

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

Thanks very much for your help!