Integration of hyperbolic functions

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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
 
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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