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Are hyperbolic substitutions absolutely necessary?

  1. Sep 6, 2015 #1
    I am familiar with both trigonometric (circular) and hyperbolic substitutions, and I have solved several integrals using both substitutions.
    I feel like trigonometric substitutions are a lot simpler, however. Even in cases where the substitution yields an integral of secant raised to an odd power. I feel like it's a lot easier to apply the reduction formula for secant than to memorize and apply hyperbolic identities.
    Granted, hyperbolic identities are not that different from circular identities, but oftentimes I forget the logarithmic form of inverse hyperbolic functions.
    So what my question boils down to is:
    Are there any cases where trigonometric substitution fails?
  2. jcsd
  3. Sep 10, 2015 #2
    Since both substitutions are suitable for dealing with square roots inside integrals if you can do a problem with normal trig you can do it with hyperbolic trig. In the end it boils down to how well you can manipulate trigonometric identities and integrals vs hyperbolic ones. Even if you are not that good with trigonometry using the exponential expressions for the trigonometric and hyperbolic functions gets the job done, so basically both substitutions are a clever way to use exponentials to simplify the problem and thus are equivalent.
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