MHB When to use a hyperbolic trig substitution in integration problems?

[Dethrone]

I read somewhere that:
sqrt(a^2-x^2), you can use x = asinx, acosx
sqrt(a^2+x^2), you can use x = atanx (or acotx), asinhx
sqrt(x^2-a^2), you can use x = asecx (or a cscx), acoshx

When would it be beneficial to use a hyperbolic trig substitution as oppose to the regular trig substitutions (sin, tan sec)?

 

[Prove It]

It all comes down to the following identities:

$\displaystyle \begin{align*} \sin^2{ \left( \theta \right) } + \cos^2{ \left( \theta \right) } &\equiv 1 \\ \\ 1 + \tan^2{ \left( \theta \right) } &\equiv \sec^2{ \left( \theta \right) } \\ \\ \cosh^2{ (t) } - \sinh^2{(t)} &\equiv 1 \end{align*}$

and the fact that the derivatives of trigonometric and hyperbolic functions end up being very similar to some of these identities:

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}\theta} \left[ \sin{ \left( \theta \right) } \right] &= \cos{ \left( \theta \right) } \\ \\ \frac{ \mathrm{d}}{\mathrm{d}\theta} \left[ \tan{ \left( \theta \right) } \right] &= \sec^2{ \left( \theta \right) } \\ \\ \frac{\mathrm{d}}{\mathrm{d}t} \left[ \sinh{(t)} \right] &= \cosh{(t)} \end{align*}$

With whatever situation you are given, you need to look at your integrand and think about which substitution might end up simplifying using one of these trigonometric or hyperbolic identities, to something that could cancel with one of these derivatives. It takes a bit of experience, and there's not really a definitive answer as to when to use one over another (for example, the substitution $\displaystyle \begin{align*} x = \tan{(\theta)} \end{align*}$ often does the same job as the substitution $\displaystyle \begin{align*} x = \sinh{(t)} \end{align*}$ (can you see why)? It just depends on personal preference and experience.