Hyperbolic substitution vs Trigonometric Substitution

A

Al-Layth

Junior Member
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Dec 28, 2021
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it seems they are used in the exact same circumstances:

Component​
Substitution Comparison​
[math]\sqrt{a^2 - x^2 }[/math][math]x = a\tanh{\theta}[/math][math]x = a\sin{\theta}[/math]
[math]\sqrt{x^2 - a^2 }[/math][math]x = a\sec{\theta}[/math][math]x = a\cosh{\theta}[/math]
[math]\sqrt{x^2 + a^2 }[/math][math]x = a\sinh{\theta}[/math][math]x = a\tan{\theta}[/math]

So my question is, is there ever any significant difference in the effect of hyperbolic vs trig substitutions to simplify an integrands?

So far I have considered:
(1) Some of these are easier to differentiate and integrate than others.
Sin(x) is "easier" than Tanh(x)
cosh is "easier" than sec
sinh is "easier" than tan
Based on that, Is it a good idea to simply ignore the other substitutions since these will always be simpler?
Although for me, I've already memorised the differentials and integrals for all base functions, their reciprocals and inverses so I'm not sure if this is useful to me.

(2) Hyperbolic terms can be easily substituted into their exponential form, which is easy to integrate. You can do the same for Trig though, but with imaginary numbers. I don't know if that's a big deal though, it seems easy enough to use euler's identity to cancel the imaginary numbers in the end. (assuming the antiderivative is real)

(3) Hyperbolic subs are more "exotic" and may help me seduce my maths teacher

So is there ever a situation where hyperbolic will be far more effective than trig subs or vice versa?
thanks
 
Work through an integral table like I suggested and then you'll stop thinking about these questions. Why? Because you'll have your own opinion on which method is better to use than others.
Your questions are all excellent questions. The problem is that all the answers are the same--it comes from experience.

I once was extremely good at doing integrals. Why was that? I solved integrals all the time and deciding which technique to use was becoming easier each day.

You want the tutors here to answer your questions but without you experiencing things for yourself our answer will mean nothing to you.

Just practice solving integrals. It is just like solving a puzzle and getting better comes with practice.

I doubt that I will respond to anymore of your questions of this type as my answer will always be the same: Practice solving integrals!

I would start off by 1st proving that \(\displaystyle \int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx\) for any real number c and \(\displaystyle \int_a^b f(x)dx = -\int_b^a f(x)dx\)
 
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