$\newcommand{\cosech}{\operatorname{cosech}}$
I ran into this integral when trying to solve a problem, and I could not get my head around it.
$$\int \cosech^4(x)dx$$
I tried to split into two $\text{cosech}^2(x)$ terms, and then substitute in the identity $\cosech^2(x)=\coth^2(x)-1$ to get:
$$ \begin{align} & =\int \cosech^2(x)(\coth^2(x)-1)dx \\[8pt] & =\int \left(\cosech^2(x)\coth^2(x) - \cosech^2(x) \right) \, dx \\[8pt] & =\int \cosech^2(x)\coth^2(x)dx - \int\cosech^2(x) \, dx \end{align} $$
From here I was okay to integrate the second part, since its a standard integral, but I was really not sure what to do with the first part. I tried to use integrating by parts because that is what we've been doing in class, but it didn't get me anywhere.
So if anyone could give me some advice as to what do from here, or whether my whole method is kind of faulty, I would really appreciate it.
No integration by parts necessary:
$$\int dx \, \text{csch}^4{x} = -\int d(\coth{x}) (\coth^2{x}-1) = -\frac13 \coth^3{x} - \coth{x}+C$$