<s>[INST] A calculation problem about differential equation 
Let $n>2$ and $n\in \mathbb{N}_+$, $\lambda  \ge  - \frac{{{{\left( {n - 1} \right)}^2}}}{4}$.
  For $x>0$, we have $$f\left( x \right) = {\left( {\sinh x} \right)^{2 - n}}\int_0^\pi  {{{\left( {\cosh x + \cos t} \right)}^\eta }{{\left( {\sin t} \right)}^{2\sqrt {\lambda  + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} .$$
  Solving the value of $\eta$ such that function $f(x)$ satisfies differential equation $$f''\left( x \right) + \left( {n - 1} \right)\frac{{\cosh x}}{{\sinh x}}f'\left( x \right) - \lambda f\left( x \right) = 0.$$

Adding: This is a difficult problem! I have get 
\begin{align*}&f'\left( x \right) = \left( {2 - n} \right){\left( {\sinh x} \right)^{1 - n}}\cosh x\int_0^\pi  {{{\left( {\cosh x + \cos t} \right)}^\eta }{{\left( {\sin t} \right)}^{2\sqrt {\lambda  + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} \\+ &\eta {\left( {\sinh x} \right)^{3 - n}}\int_0^\pi  {{{\left( {\cosh x + \cos t} \right)}^{\eta  - 1}}{{\left( {\sin t} \right)}^{2\sqrt {\lambda  + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} .\end{align*}
and 
\begin{align*}&f''\left( x \right) = \left( {2 - n} \right)\left[ {\left( {1 - n} \right){{\left( {\sinh x} \right)}^{ - n}}{{\left( {\cosh x} \right)}^2} + {{\left( {\sinh x} \right)}^{2 - n}}} \right]\int_0^\pi  {{{\left( {\cosh x + \cos t} \right)}^\eta }{{\left( {\sin t} \right)}^{2\sqrt {\lambda  + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} \\+ &\left( {5 - 2n} \right)\eta {\left( {\sinh x} \right)^{2 - n}}\cosh x\int_0^\pi  {{{\left( {\cosh x + \cos t} \right)}^{\eta  - 1}}{{\left( {\sin t} \right)}^{2\sqrt {\lambda  + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} \\+ &\eta \left( {\eta  - 1} \right){\left( {\sinh x} \right)^{4 - n}}\int_0^\pi  {{{\left( {\cosh x + \cos t} \right)}^{\eta  - 2}}{{\left( {\sin t} \right)}^{2\sqrt {\lambda  + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} .\end{align*}
Note that $$f''\left( x \right) + \left( {n - 1} \right)\frac{{\cosh x}}{{\sinh x}}f'\left( x \right) - \lambda f\left( x \right) = 0.$$
So we get \begin{align*}&\left( {2 - n-\lambda} \right)\int_0^\pi  {{{\left( {\cosh x + \cos t} \right)}^\eta }{{\left( {\sin t} \right)}^{2\sqrt {\lambda  + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} \\+& \left( {4 - n} \right)\eta \cosh x\int_0^\pi  {{{\left( {\cosh x + \cos t} \right)}^{\eta  - 1}}{{\left( {\sin t} \right)}^{2\sqrt {\lambda  + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} \\+& \eta \left( {\eta  - 1} \right){\left( {\sinh x} \right)^2}\int_0^\pi  {{{\left( {\cosh x + \cos t} \right)}^{\eta  - 2}}{{\left( {\sin t} \right)}^{2\sqrt {\lambda  + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt}  = 0.\end{align*}
But I have no way to continue, thanks for your help!
 [/INST] This is not a complete answer, but rather another way of attacking the problem, that might be more fruitful. It is too long to put in a comment.
First, we write the differential equation as
$$
\bigl(f'(x)\sinh^{n-1}x\bigr)'=\lambda f(x)\sinh^{n-1}x.\tag{1}
$$
Differentiating the integral that should equal $f$, and multiplying with $\sinh^{n-1}x$, we find that
$$
f'(x)\sinh^{n-1}x=\int_0^\pi \bigl[(2-n)\cosh x+\frac{\eta\sinh x}{\cosh x+\cos t}\bigr](\cosh x+\cos t)^{\eta}(\sin t)^{2\sqrt{\lambda+(n-1)^2/4}}\,dt\tag{2}
$$
The right-hand side of $(1)$ equals
$$
\lambda f(x)\sinh^{n-1}x=\lambda\int_0^{\pi}\sinh x(\cosh x+\cos t)^{\eta}(\sin t)^{2\sqrt{\lambda+(n-1)^2/4}}\,dt\tag{3}
$$
Now, instead of differentiating $(2)$, it looks easier to integrate $(3)$,
$$
\int\lambda f(x)\sinh^{n-1}x\,dx
=\frac{\lambda}{1+\eta}\int_0^\pi(\cosh x+\cos t)^{\eta+1}(\sin t)^{2\sqrt{\lambda+(n-1)^2/4}}\,dt\tag{4}
$$
Maybe you can find $\eta$ so that $(2)$ and $(4)$ differs by a constant?
</s>