How do i integrate this? Hyperbolic substitution? I'm trying to integrate 
$\int_0^\pi \sqrt{2+t^2}\; dt$
I know there it involves with something along the lines of either hyperbolic substitution or trigonometric. 
I tried $t= \sqrt {2}\tan(u)$ and $u=\arctan(\frac{x}{\sqrt{2}})$. But then, I get lost and confused. Any guidance please? 
 A: Try out $t=\sqrt{2} \sinh(z)$. This leads to an easier calculation compared to the gonimetric substitution. More precisely, $t=\sqrt{2} \sinh(z)$, $dt=\sqrt{2} \cosh(z) dz$ and our integral becomes:
$$ I= \sqrt{2} \int \sqrt{1+\sinh^2(z)} \cosh(z) \, dz = \sqrt{2} \int \cosh^2(z) \, dz = \sqrt{2} \int \frac{1+\cosh(2z)}{2} \, dz = \frac{\sqrt{2}}{2} z + \frac{\sqrt{2}}{4} \sinh(2z) = \frac{1}{2} t \sqrt{2+t^2} + \sinh^{-1}\left(\frac{\sqrt{2}}{2} t \right). $$
Now keep in mind that $\sinh^{-1} (x)= \log(x + \sqrt{1+x^2})$. Finally, of course, evaluate the primitive in $t=0$ and $t=\pi$.
A: Using your substitution $\text{t}=\sqrt{2}\tan\left(\text{u}\right)$ and $\text{d}\text{t}=\sqrt{2}\sec^2\left(\text{u}\right)\space\text{d}\text{u}$:
$$\text{I}=\int_0^\pi\sqrt{2+\text{t}^2}\space\text{d}\text{t}=2\int_0^{\arctan\left(\frac{\pi}{\sqrt{2}}\right)}\sec^3\left(\text{u}\right)\space\text{d}\text{u}$$
Using integration by parts:
$$\mathcal{I}\left(\text{u}\right)=\int\sec^3\left(\text{u}\right)\space\text{d}\text{u}=\sec\left(\text{u}\right)\tan\left(\text{u}\right)-\int\sec^3\left(\text{u}\right)\space\text{d}\text{u}+\int\sec\left(\text{u}\right)\space\text{d}\text{u}=$$
$$\sec\left(\text{u}\right)\tan\left(\text{u}\right)-\mathcal{I}\left(\text{u}\right)+\int\sec\left(\text{u}\right)\space\text{d}\text{u}$$
Solving, $\mathcal{I}\left(\text{u}\right)$:
$$\mathcal{I}\left(\text{u}\right)=\frac{1}{2}\left\{\sec\left(\text{u}\right)\tan\left(\text{u}\right)+\int\sec\left(\text{u}\right)\space\text{d}\text{u}\right\}$$
And:
$$\int\sec\left(\text{u}\right)\space\text{d}\text{u}=\ln\left|\tan\left(\text{u}\right)+\sec\left(\text{u}\right)\right|+\text{C}$$
