# Primitive of $\frac{1-a\cosh(x)}{(\cosh(x)-a)^2}$

I would like to solve the following primitive:

$$\int\frac{1-a\cosh(x)}{(\cosh(x)-a)^2}\,dx$$

where $$a$$ is a constant, $$0\leq a\leq1$$.

I really don't know how to start. I can't relate the numerator with the denominator. There's no $$\sinh(x)$$ in the integrand, and I can't see any similarity with the well known hyperbolic functions derivatives. How would you do it?

• What does $$[…]$$ mean? – Dr. Sonnhard Graubner Feb 21 at 12:22
• That's the same as $(...)$. It's there because I already used $(...)$ inside of it. I can change it to avoid confusion. – Élio Pereira Feb 21 at 12:47
• It would be interesting to know if there is a way of evaluating this integral without "guessing" or "by inspection" arriving at the right answer first. – James Arathoon Feb 21 at 14:08

Here is a slightly more structured approach to arriving at an answer which is not simply based, as you say, on a "pure guess."

Let $$J = \int \frac{1 - a \cosh x}{(\cosh x - a)^2} \, dx.$$

To find the integral $$J$$, the appearance of the all squared term in the denominator of the integrand suggests it may be possible to make use of the so-called reverse quotient rule (For another example using this method see here).

Recall that if $$u$$ and $$v$$ are differentiable functions, from the quotient rule $$\left (\frac{u}{v} \right )' = \frac{u' v - v' u}{v^2},$$ and it is immediate that $$\int \frac{u' v - v' u}{v^2} \, dx = \int \left (\frac{u}{v} \right )' \, dx = \frac{u}{v} + C. \tag1$$

For the integral $$J$$ we see that $$v = \cosh x - a$$. So $$v' = \sinh x$$. Now for the (generally) harder bit but in this particular case it is quite easy. We need to find a function $$u(x)$$ such that $$u' v - v' u = u'(\cosh x - a) - u \sinh x = 1 - a \cosh x.$$ After a little trial and error it is quite easy to see that if $$u = \sinh x,$$ as $$u' = \cosh x,$$ this gives $$u' v - v' u = \cosh^2 x - a \cosh x - \sinh^2 x = 1 - a \cosh x,$$ where we have used $$\cosh^2 x - \sinh^2 x = 1$$, as required.

Our integral can now be readily found. The result is: \begin{align} J &= \int \left (\frac{\sinh x}{\cosh x - a} \right )' \, dx = \frac{\sinh x}{\cosh x - a} + C. \end{align} as expected.

Let's start by conjecturing an antiderivative of the form $$\frac{f(x)}{\cosh x-a}$$, so $$f^\prime(x)(\cosh x-a)-f(x)\sinh x=1-a\cosh x=\cosh^2x-a\cosh x-\sinh^2x.$$By inspection, one solution is $$f(x)=\sinh x.$$

Hint: Differentiate $$\frac{\sinh(x)}{\cosh(x)-a}+C$$ with respect to $$x$$

• I see that this is indeed the primitive of $\frac{1-a\cosh(x)}{(\cosh(x)-a)^2}$. The function $\frac{1-a\cosh(x)}{(\cosh(x)-a)^2}$ can be achieved by differentiating $\frac{\sinh(x)}{\cosh(x)-a}+C$ and using the identity $\cosh^2(x)-\sinh^2(x)=1$. Did you get this result just by pure guessing? – Élio Pereira Feb 21 at 15:21