# Integration of hyperbolic functions

## Homework Statement

$$\int cosh(2x)sinh^{2}(2x)dx$$

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:

$$\int cosh(2x)sinh^{2}(2x)dx =$$ $$\frac{1}{2}$$$$\int sinh^{2}(2x)2cosh(2x) dx$$

= $$\frac{sinh^{3}2x}{6} + C$$

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

Joe

Last edited:

Mark44
Mentor

## Homework Statement

$$\int cosh(2x)sinh^{2}(2x)dx$$

## Homework Equations

Not sure
Hyperbolic trig identities would be very relevant.

## 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.
$$\int cosh(2x)sinh^{2}(2x)dx = \frac{1}{2}\int sinh^{2}(2x)2cosh(2x) dx = \frac{1}{2}\frac{sinh^{3}2x}{3} + C$$

My issue with this problem is I don't understand what happened to the $$2cosh(2x)$$. It relates to $$sinh^{2}(x)+cosh^{2}(x)$$ 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