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? 

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 $t=\pi$.
