Integration of a hyperbolic function

[penroseandpaper]

The integral of cothx is ln|sinhx|+C.

Does this mean the integral of coth²x is ln|sinh²x|+C?

If not, does anyone have a link to a page on how it is achieved - I'm trying to compile a list of all common hyperbolic function derivatives and integrals. However, I can't find anything to confirm if this is the right assumption.

My thought process went:
∫(cosh²x)/(sinh²x)

=∫1/u

=ln|u|+c,

where u=sinh²x

Thanks

 

[romsofia]

Well, before you work on the common hyperbolic intergrals, you should get all the derivatives down, and their general rules. You missed a chain rule! ##\frac {d}{dx} \sinh^2x \neq \cosh^2x##. Do the chain rule.

Now, onto your integral! Just like normal trig functions, the way to find the integrals for ##\int \cot^2x dx## is to use the Pythagorean identity ##1+\cot^2x = \csc^2x## . So, in this case use: ##\coth^2x - \csc h^2 x = 1## Once you make the substitution, you'll see it plays out like the ##\int \cot^2x dx## case.

 

[Mark44]

The integral of cothx is ln|sinhx|+C.
Does this mean the integral of coth²x is ln|sinh²x|+C?

If not, does anyone have a link to a page on how it is achieved - I'm trying to compile a list of all common hyperbolic function derivatives and integrals. However, I can't find anything to confirm if this is the right assumption.

Before you tackle this work, I would advise you to brush up on differentiation (particularly the chain rule), common trig identities (both circular and hyperbolic), and on the integrals of the circular trig functions.

This wiki page lists many integrals of hyperbolic trig functions - https://en.wikipedia.org/wiki/List_of_integrals_of_hyperbolic_functions

 

[penroseandpaper]

Well, before you work on the common hyperbolic intergrals, you should get all the derivatives down, and their general rules. You missed a chain rule! ##\frac {d}{dx} \sinh^2x \neq \cosh^2x##. Do the chain rule.

Now, onto your integral! Just like normal trig functions, the way to find the integrals for ##\int \cot^2x dx## is to use the Pythagorean identity ##1+\cot^2x = \csc^2x## . So, in this case use: ##\coth^2x - \csc h^2 x = 1## Once you make the substitution, you'll see it plays out like the ##\int \cot^2x dx## case.

Thanks for those pointers.
I was thinking about it in bed, and I thought I'd got it wrong because reversing my answer gets you something like (2cosh²x)/sinh²x.

I get a result of -cothx+x+C - using the fact that the derivative of cothx is -cosec2x; did I get there?

 

[romsofia]

Thanks for those pointers.
I was thinking about it in bed, and I thought I'd got it wrong because reversing my answer gets you something like (2cosh²x)/sinh²x.

I get a result of -cothx+x+C - using the fact that the derivative of cothx is -cosec2x; did I get there?

Yup, that's right!

 

[Mark44]

using the fact that the derivative of cothx is -cosec2x

I know what you're trying to say, but that's not what you actually wrote.

Here's the corrected version, using LaTeX:
##\frac d {dx}(\coth (x)) = -csch^2(x)##

What you wrote omits the 'h' for this hyperbolic function. I wrote this as csch, but it can also be written as written as cosech.

Also, cosec2x would be interpreted as cosec(2x) instead of what you intended.
To indicate an exponent, at least use the ^ character, as in x^2. What you wrote looks like -csc(2x).

What I wrote was using LaTeX - we have a pretty nice tuturial whose link is in the lower left corner, where it says LaTeX Guide.