Difficult integral evaluation I'm working through Vector Calculus by Marsden and Tromba to review for my GRE (and because it has really interesting historical snippets) and I ran into a wall on a problem where I have to evaluate the arc length of the vector valued function $<t, tsint, tcost>$ for $0 \leq t \leq \pi$. Here is what I have so far:
$$\int_{0}^{\pi} \sqrt{1^2 + (\sin(t) + t\cos(t))^2 + (\cos(t) - t \sin(t))^2} dt\rightarrow\int_{0}^{\pi} \sqrt{1 + (\sin(t))^2 + (\cos(t))^2 + t^2((\cos(t))^2 + (\sin(t))^2)} dt\\\rightarrow \int_{0}^{\pi} \sqrt{2+t^2} dt$$
But once I get there I cant figure out how to evaluate the integral. I plugged it into wolfram alpha in hopes that by seeing what they get I would have some insight on how to proceed but I have no idea (they had a term with $\sinh^{-1}(t)$).

You can use integration by parts, or you can use the hyperbolic substitution $t = \sqrt{2}\sinh(u)$. I'll use the latter. 
With $t = \sqrt{2}\sinh(u)$, we have $\sqrt{2 + t^2} = \sqrt{2}\cosh(u)$ and $dt = \sqrt{2}\cosh(u)\, du$, so
$$\int_0^\pi \sqrt{2+t^2}\, dt = \int_0^{\sin^{-1}(\pi/\sqrt{2})} 2\cosh^2(u)\, du.$$
Using the identity $2\cosh^2(u) = 1 + \cosh(2u)$, we get
\begin{align}\int_0^{\sinh^{-1}(\pi/\sqrt{2})} 2\cosh^2(u)\, du &= \int_0^{\sinh^{-1}(\pi/\sqrt{2})} (1 + \cosh(2u))\, du \\
&= u + \frac{\sinh(2u)}{2}\bigg|_{u = 0}^{u = \sinh^{-1}(\pi/\sqrt{2})}\\
& = \sinh^{-1}\left(\frac{\pi}{\sqrt{2}}\right)+ \frac{\sinh(2\sinh^{-1}(\pi/\sqrt{2}))}{2}.
\end{align}
From the identity $\sinh(2x) = 2\sinh(x)\, \cosh(x)$ and the computations $\sinh(\sinh^{-1}(\pi/\sqrt{2})) = \pi/\sqrt{2}$ and $\cosh(\sinh^{-1}(\pi/\sqrt{2})) = \sqrt{1 + (\pi/\sqrt{2})^2} = \sqrt{(2 + \pi^2)/2}$, we find
$$\sinh(2\sinh^{-1}(\pi/\sqrt{2})) =2\cdot \frac{\pi}{\sqrt{2}}\cdot \sqrt{\frac{2 + \pi^2}{2}} = \pi\sqrt{2 + \pi^2}.$$
Hence,
$$\int_0^{\pi} \sqrt{2 + t^2}\, dt = \sinh^{-1}\left(\frac{\pi}{\sqrt{2}}\right) + \frac{1}{2}\pi\sqrt{2 + \pi^2}.$$
