Can someone help me with the intermediate steps here please, because they have been left out of the textbook and I am unsure.
so we have this equation:
$\sqrt{1-e}\tan\frac{\theta - \theta_{0}}{2} = \sqrt{1+e}\tan\frac{\eta}{2}$
and taking t=0 to occur at the pericentre passage $r_1 = a(1-e)$ and from $L=r^2 \dot{\theta}$ we have
$t = \int_{\theta_{0}}^{\theta} \frac{d\theta}{\dot{\theta}} = \int d\theta \frac{r^2}{L} = \frac{a^2}{L} \int_{0}^{\eta} d\eta \frac{d\theta}{d\eta} (1-e\cos\eta)^2$
evaluating $\frac{d\theta}{d\eta}$ from the first equation, integrating and using trigonometric identities to simplify the result, we obtain
$t = \frac{a^2}{L} \sqrt{1-e^2} (\eta - e\sin\eta) = \frac{T_{r}}{2\pi}(\eta - e\sin\eta)$
where the 2nd identity follows because the bracket on the right increases by $2\pi$ over an orbital period.
Any help would be great thanks.
Parts of this come from information you have not reproduced, so I can't really say for sure. It is evidently related to orbital equations, and I haven't looked at those in many years. I have no idea what "at the pericentre passage" even means. However, this is what I can figure out.
Obviously $$t = \int_0^tdt = \int_{\theta_0}^\theta \frac{dt}{d\theta}d\theta = \int_{\theta_0}^\theta \frac{d\theta}{\left(\frac{d\theta}{dt}\right)} = \int_{\theta_0}^\theta\frac{d\theta}{\dot\theta}$$ Now make the substitution $\dot \theta = L /r^2$ and then a second substitution for $r = a(1-e\cos\eta)$, an equation that does not follow from anything you've written here, but is evidently a formula for an elliptical orbit (or more general conic section). I'm not sure that $\eta$ is, but apparently, $\eta = 0$ at the "pericentre passage", since the lower limit of the integral (which still corresponds to $t = 0$) is $0$.
$$t = \int_{\theta_0}^\theta\frac{r^2}L\, d\theta=\frac{a^2}L\int_{0}^{\eta} (1-e\cos\eta)^2\frac{d\theta}{d\eta}\,d\eta$$
To determine $\frac{d\theta}{d\eta}$, we need to differentiate the formula you gave. To make it less messy, let $\alpha = \frac{\theta - \theta_0}2, \beta = \frac\eta2, A = \sqrt{1-e}, B = \sqrt{1+e}$. It is easy to confirm that $\frac{d\theta}{d\eta}=\frac{d\alpha}{d\beta}$. So $$A\tan\alpha = B\tan\beta\\A\left(\sec^2\alpha\right) \frac{d\alpha}{d\beta} = B\sec^2\beta\\A\left(1 + \tan^2\alpha\right)\frac{d\alpha}{d\beta} = B\left(1 + \tan^2\beta\right)\\\left(A^2 + A^2\tan^2\alpha\right) \frac{d\alpha}{d\beta}= AB\left(1+\tan^2\beta\right)\\\left(A^2 + B^2\tan^2\beta\right) \frac{d\alpha}{d\beta}= AB\left(1+\tan^2\beta\right)\\\frac{d\alpha}{d\beta} = \frac{AB\left(1+\tan^2\beta\right)}{A^2 + B^2\tan^2\beta}$$ Substitute back in the original values, we get $$\frac{d\theta}{d\eta} = \frac{\sqrt{1 - e^2}\left(1+\tan^2\frac\eta2\right)}{\left(1+\tan^2\frac\eta2\right) - e^2 + e^2\tan^2\frac\eta2}$$ So $$t = \frac{a^2}L\int_{0}^{\eta} (1-e\cos\eta)^2\frac{\sqrt{1 - e^2}\left(1+\tan^2\frac\eta2\right)}{\left(1+\tan^2\frac\eta2\right) - e^2 + e^2\tan^2\frac\eta2}\,d\eta$$
A horrid mess, but there are a couple of potential approaches: make one or both of the substitutions $$\tan\frac\eta2 = \frac{\sin \eta}{1 - \cos\eta} = \frac{1 + \cos\eta}{\sin\eta}$$ or make the substitution $$u = \tan\frac\eta2,\quad \cos\eta = \frac{1-u^2}{1+u^2}, \quad d\eta = \frac{2du}{1+u^2}$$
I've not explored which gives a more useful result yet, but I suspect that one or the other will produce something that can be integrated.