Evaluating $\int_{-20}^{20}\sqrt{2+t^2}\,dt$ I have this integral:
$$\int_{-20}^{20}\sqrt{2+t^2}\,dt$$
I tried solving it many times but without success.
The end result is this:
$$2\left( 10\sqrt{402}+\mathop{\mathrm{arcsinh}}(10\sqrt{2})\right).$$
I can't seem to get this end result. I got a few wrong ones but cant find this one. Perhaps it's wrong? Could anyone confirm it?
I tried Sage, and it calculates it correctly, but not with steps.

The typical way to start dealing with an integral that has $\sqrt{a^2+t^2}$ (in your case, $a=\sqrt{2}$) is to use a trigonometric substitution or a hyperbolic substitution.
If we use hyperbolic substitution, we want to use the fact that
$$1 + \sinh^2 z = \cosh^2 z.$$
Set $t=\sqrt{2}\sinh z$. Then
$$\sqrt{2 + t^2} = \sqrt{2 + 2\sinh^2z} = \sqrt{2(1+\sinh^2 z)} = \sqrt{2\cosh^2 z} = \sqrt{2}\cosh z.$$
Also, if $t=\sqrt{2}\sinh z$, then $dt =\sqrt{2}\cosh z\,dz$. Therefore,
$$\int\sqrt{2+t^2}\,dt = \int \sqrt{2}\cosh z \sqrt{2}\cosh z\,dz = 2\int\cosh^2 z\,dz.$$
Now we need to solve the integral of $\cosh^2 z$. We can do this by parts, similarly to how we solve the integral of $\cos^2t$. Let $u=\cosh z$, $dv=\cosh z\,dz$. Then $du=\sinh z\,dz$, $v=\sinh z$, so
$$\begin{align*}
\int\cosh^2 z\,dz &= \cosh z\sinh z - \int \sinh^2z\,dz\\
&= \cosh z \sinh z - \int(cosh^2z - 1)\,dz\\
&= \cosh z\sinh z +z - \int\cosh^2z\,dz.
\end{align*}$$
Hence
$$\begin{align*}
\int\cosh^2z\,dz &= z + \cosh z\sinh z - \int\cosh^2z\,dz\\
2\int\cosh^2z\,dz &= z+ \cosh z\sinh z + C\\
\int\cosh^2 z\,dz &= \frac{1}{2}z + \frac{1}{2}\cosh z\sinh z + C.
\end{align*}$$
Then to get it back into a function of $t$ we remember that $t=\sqrt{2}\sinh z$, so $\sinh z = \frac{\sqrt{2}}{2}t$. Then $z= \mathop{\mathrm{arcsinh}}\left(\frac{\sqrt{2}}{2}t\right)$, and
$$\sinh z\cosh z = \sinh z \sqrt{1+\sinh^2z} = \frac{\sqrt{2}}{2}t\sqrt{1 + \frac{1}{2}t^2} = \frac{1}{2}\sqrt{2+t^2}.$$
Therefore,
$$\int\sqrt{2+t^2}\,dt = 2\int\cosh^2z\,dz = 2\mathop{\mathrm{arcsinh}}\left(\frac{\sqrt{2}}{2}t\right) + \sqrt{2+t^2}+C.$$
Now plugging into the definite integral gives you a solution.

If we use trigonometric substitutions, we want to use the fact that
$$1 + \tan^2 \theta = \sec^2\theta$$
Set $t=\sqrt{2}\tan\theta$, with $-\frac{\pi}{2} \lt \theta\lt\frac{\pi}{2}$. Then
$$\sqrt{2+t^2} = \sqrt{2+2\tan^2\theta} = \sqrt{2}\sqrt{1+\tan^2\theta} = \sqrt{2}\sqrt{\sec^2\theta} = \sqrt{2}|\sec\theta|.$$
Since $\sec\theta\gt 0$ on $-\frac{\pi}{2}\lt \theta\lt \frac{\pi}{2}$, we get that $\sqrt{2+t^2} = \sqrt{2}\sec\theta$.
Also, if $t=\sqrt{2}\tan\theta$, then $dt = \sqrt{2}\sec^2\theta\,d\theta$. Therefore,
$$\int\sqrt{2+t^2}\,dt = \int \sqrt{2}\sec\theta\sqrt{2}\sec^2\theta\,d\theta = 2\int\sec^3\theta\,d\theta.$$
So we need to find $\int\sec^3\theta\,d\theta = \int\frac{1}{\cos^3\theta}\,d\theta$.  
This can be done any number of ways. Using integration by parts, we get
$$\int\frac{d\theta}{\cos^3\theta} = \frac{1}{2}\tan\theta + \frac{1}{2}\int\frac{d\theta}{\cos\theta}.$$
And
$$\int\frac{d\theta}{\cos\theta} = \int\sec\theta\,d\theta = \ln|\sec\theta + \tan\theta|+C.$$
Thus, we get that
$$\int\sqrt{2+t^2}\,dt = \tan\theta + \ln|\sec\theta+\tan\theta|+C.$$
To change it into a formula using $t$, we use the fact that $t=\sqrt{2}\tan\theta$. Therefore, $\tan\theta = \frac{\sqrt{2}}{2}t$; and 
$$\sec\theta = \sqrt{1 + \tan^2\theta} = \sqrt{1 + \frac{t^2}{2}} = \frac{\sqrt{2}}{2}\sqrt{2+t^2}.$$
Plugging in and evaluating gives the desired result.
