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.