I've tried substituting $u=\cosh(t)$ whence
$$\int\nolimits_{\cosh^{-1}(0)}^{\cosh^{-1}(3)} \sqrt{\cosh^2(t)-\cosh{t}}\,dt$$ becomes
$$ \tag{1} \int _0^3 \sqrt{\frac{u}{u+1}}\,du $$ since $\sinh(\cosh^{-1}(u))=\sqrt{\dfrac{u^2-1}{u+1}}$ according to Wolfram Alpha. Equation (1) doesn't look that hard, and Wolfram Alpha gave the answer of $\sqrt{3} - \frac{1}{2}\sinh^{-1}(\sqrt{3})$, but I'd be much obliged if someone can give me some pointers as to how to evaluate this integral analytically/manually. I have a feeling it might be another substitution.
Thanks in advance.
One method is to make the substitution $$ u = \frac{t^2}{1-t^2} \iff t = \sqrt{\frac{u}{u+1}}. $$ However, this approach needs a bit of a struggle to complete.
A more clever approach (inspired by the answer given by Wolfram|Alpha) is to substitute $u = \sinh^2 t$. Then $$ \sqrt{\frac{u}{u+1}} = \sqrt{\frac{\sinh^2 t}{\cosh^2 t}} = \frac{\sinh t}{\cosh t}, $$ and $$ du = 2 \sinh t \cosh t. $$ Therefore, the integrand becomes: $$ \int_0^{\sinh^{-1}(\sqrt{3})} 2 \sinh^2 t dt = \frac 12 \int_0^{\sinh^{-1}(\sqrt{3})} (e^t - e^{-t})^2 dt = \frac 12 \int_0^{\sinh^{-1}(\sqrt{3})} (e^{2t} + e^{-2t} - 2) dt, $$ which can be easily evaluated.