Integral using hyperbolic substitution

[Malby]

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.

 

[SammyS]

sinh²(u) + 1 = cosh²(u)

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

 

[Malby]

[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]

 

[Malby]

[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?

 

[flyingpig]

Use the identity

$$cosh^2 (t) = \frac{1 + cosh(2t)}{2}$$Your tex didn't work because you used \ instead of /

Omg i helped someone!

 

[SammyS]

$$\int\left(1+x^{2}\right)^{\frac{3}{2}}dx = \int\left(1+\sinh^{2}\left(u\right)\right)^{\frac{3}{2}}\cosh\left(u\right)\, du$$
$$=\int\sqrt{\left(\cosh^{2}\left(u\right)\right)^{3}}\cosh\left(u\right)\, du$$
$$=\int\sqrt{\cosh^{6}\left(u\right)} \cosh\left(u\right)\, du$$
$$=\int\cosh^{4}\left(u\right)\, du$$

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

 

[GreenPrint]

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

 

[SammyS]

sinh²(u) + 1 = cosh²(u)

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

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

 

[Malby]

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!