Integration of hyperbolic functions

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SUMMARY

The integration of the hyperbolic function \(\int cosh(2x)sinh^{2}(2x)dx\) can be solved using substitution. By letting \(u = sinh(2x)\), the differential \(du\) becomes \(2cosh(2x)dx\). This leads to the transformation of the integral into \(\frac{1}{2}\int sinh^{2}(2x)2cosh(2x) dx\), ultimately yielding the result \(\frac{sinh^{3}(2x)}{6} + C\). Understanding hyperbolic identities is crucial for this integration process.

PREREQUISITES
  • Hyperbolic function identities
  • Integration techniques, specifically substitution
  • Understanding of LaTeX formatting for mathematical expressions
  • Basic calculus knowledge
NEXT STEPS
  • Study hyperbolic function identities in detail
  • Practice integration by substitution with various functions
  • Learn to format mathematical expressions using LaTeX
  • Explore advanced integration techniques, such as integration by parts
USEFUL FOR

Students studying calculus, mathematics educators, and anyone looking to deepen their understanding of hyperbolic functions and integration techniques.

Agent M27
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Homework Statement



[tex]\int cosh(2x)sinh^{2}(2x)dx[/tex]

Homework Equations



Not sure

The Attempt at a Solution



This was an example problem in the book and was curious how they got to the following answer:

[tex]\int cosh(2x)sinh^{2}(2x)dx =[/tex] [tex]\frac{1}{2}[/tex][tex]\int sinh^{2}(2x)2cosh(2x) dx[/tex]

= [tex]\frac{sinh^{3}2x}{6} + C[/tex]

My issue with this problem is I don't understand what happened to the [tex]2cosh(2x)[/tex]. It relates to [tex]sinh^{2}(x)+cosh^{2}(x)[/tex] but that only equals 1 in normal trig, not hyperbolic. Thanks in advance.

Joe
 
Last edited:
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Agent M27 said:

Homework Statement



[tex]\int cosh(2x)sinh^{2}(2x)dx[/tex]

Homework Equations



Not sure
Hyperbolic trig identities would be very relevant.
Agent M27 said:

The Attempt at a Solution



This was an example problem in the book and was curious how they got to the following answer:
For some reason, your LaTeX wasn't showing up correctly. I fixed it by removing several pairs of [ tex] and [ /tex] tags.
Tip: Use one pair of these tags per block.
Agent M27 said:
[tex]\int cosh(2x)sinh^{2}(2x)dx = \frac{1}{2}\int sinh^{2}(2x)2cosh(2x) dx<br /> <br /> = \frac{1}{2}\frac{sinh^{3}2x}{3} + C[/tex]

My issue with this problem is I don't understand what happened to the [tex]2cosh(2x)[/tex]. It relates to [tex]sinh^{2}(x)+cosh^{2}(x)[/tex] but that only equals 1 in normal trig, not hyperbolic. Thanks in advance.

Joe

The integration was done using an ordinary substitution, u = sinh(2x).
 
Ya I just realized that if I set u= sinh(2x) then du=2cosh(2x) dx then substitute from there. Thanks

Joe
 

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