So I have this integral:
$$I = \int \frac{\gamma^4v^2}{\pi\sinh^2(\sqrt{2}\gamma v/\sqrt{h})}dh $$
I used a $t$-substitution but only got as far as:
$$t = \frac{\sqrt{2}\gamma v}{\sqrt{h}} \Rightarrow da = \frac{-\sqrt{2}h^{3/2}}{\gamma v}$$
When the stated solution is
$$\int \frac{dt}{t^2\sinh^2{t}}$$
How is this done?
I use to work the other way
Let $$\frac{\sqrt{2} \gamma v}{\sqrt{h}}=t \implies h=\frac{2 \gamma ^2 v^2}{t^2}\implies dh=-\frac{4 \gamma ^2 v^2}{t^3}\,dt$$ this makes the integral to be $$-\frac{4 \gamma ^6 v^4}\pi \int\frac{\text{csch}^2(t)}{ t^3}\,dt$$
Probably typo's.