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.
While I was writting this answer, Artem posted his. My answer is essentially his 1st one.
Let $f(t)=\sqrt{2+t^{2}}$. Since $f(-t)=$ $f(t)$, we have
$$\begin{equation*} I=\int_{-20}^{20}\sqrt{2+t^{2}}dt=2\int_{0}^{20}\sqrt{2+t^{2}}dt \end{equation*}$$
As the integrand $f(t)$ is a quadratic irrational we can also use the following Euler substitution $$\begin{eqnarray*} x &=&\sqrt{2+t^{2}}-t\Leftrightarrow t=\frac{2-x^{2}}{2x}\Leftrightarrow \sqrt{2+t^{2}}=\frac{2+x^{2}}{2x} \\ dt &=&-\frac{2+x^{2}}{2x^{2}}dx, \end{eqnarray*}$$ which transforms the integrand into a rational fraction in $x$ $$\begin{eqnarray*} \int \sqrt{2+t^{2}}dt &=&\int \frac{2+x^{2}}{2x}\left( -\frac{2+x^{2}}{2x^{2} }\right) dx \\ &=&-\int \frac{1}{x}+\frac{1}{x^{3}}+\frac{x}{4}dx \\ &=&-\left( \ln \left\vert x\right\vert -\frac{1}{2x^{2}}+\frac{x^{2}}{8} \right) +C. \end{eqnarray*}$$ The limits of integration of $I/2$ are $$\begin{eqnarray*} t &=&0,x=\sqrt{2} \\ t &=&20,x=\sqrt{2+20^{2}}-20=\sqrt{402}-20. \end{eqnarray*}$$ Thus $$\begin{eqnarray*} I &=&\left. -2\left( \ln \left\vert x\right\vert -\frac{1}{2x^{2}}+\frac{ x^{2}}{8}\right) \right\vert _{\sqrt{2}}^{\sqrt{402}-20} \\ &=&2\ln \left( \frac{\sqrt{2}}{\sqrt{402}-20}\right) +\frac{1}{\left( \sqrt{ 402}-20\right) ^{2}}-\frac{\left( \sqrt{402}-20\right) ^{2}}{4}+\frac{1}{2}. \end{eqnarray*}$$
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.
Here are two more ways to calculate $\int \sqrt{a^2+x^2}\ dx$.
Another way integrate $\sqrt{2 + t^2}$ is to use the Euler substitution $u = \sqrt{2+t^2} + t$. Subtracting $t$ and squaring gives $t = \frac{1}{2}(u - \frac{2}{u})$. This implies
$$\sqrt{2+t^2} = u - t = \frac{1}{2}(u + \frac{2}{u})$$
and
$$dt = \frac{1}{2}(1 + \frac{2}{u^2})\,du$$
Thus \begin{align*} \int \sqrt{2+t^2}\,dt &= \int \frac{1}{4}(u + \frac{4}{u} + \frac{4}{u^3})\,du \\ &= \frac{1}{8}u^2 - \frac{1}{2u^2} + \ln{|u|} + C \\ &= \frac{1}{8}(u - \frac{2}{u})(u + \frac{2}{u}) + \ln{|u|} + C \\ &= \frac{1}{2}t\sqrt{2 + t^2} + \ln(t + \sqrt{2 + t^2}) + C \end{align*}
Applying this antiderivative to your integral, you get $$20\sqrt{402} + \ln(20 + \sqrt{402}) - \ln(-20 + \sqrt{402}) = 2(10\sqrt{402} + \ln(10\sqrt{2} + \sqrt{201}))$$
You can check that $\ln(10\sqrt{2} + \sqrt{201}) = \operatorname{arcsinh}(10\sqrt{2})$ so that this solution is the same as the one in your question.
