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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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