TITLE: Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ d}x$
QUESTION [7 upvotes]: I'm learning integral. Here is my homework: 
$$\int_0^1 \sqrt{1+x^2}\;dx$$
I think this problem solve by change $x$ to other variable. Can you tell me how please. (just direction how to solve)
thanks :)

REPLY [5 votes]: Before we proceed let us recall some basics on the hyperbolic functions $$\sinh x=\frac{e^x-e^{-x}}{2}\qquad \text{and }\qquad \cosh x=\frac{e^x+e^{-x}}{2}$$
Below are some relations that are easy to prove
$$\cosh^2 x - \sinh^2x = 1\\
\cosh2x=\cosh^2 x + \sinh^2 x\qquad \sinh2x=2\cosh x\sinh x\\
(\cosh x)' =\sinh x\qquad\qquad (\sinh x)' =\cosh x$$ 

Next, we turn to the integral of the OP. 
Let us make the substitution $x=\sinh t$, then
$$\int_0^1\sqrt{1+x^2}dx=\int_0^{\rm{arcsinh 1}}\sqrt{1+\sinh^2t}\cosh t\,dt\\
\qquad =\int_0^{\rm{arcsinh 1 }}\cosh^2 t\,dt =\frac12 \int_0^{\rm{arcsinh 1 }}1+\cosh 2 t\,dt\\ =\frac12\left({\rm{arcsinh}}1 + \frac{\sinh2({\rm{arcsinh}} 1)}{2}\right)$$
Now, a calculation based on the definition shows that ${\rm{arcsinh}}x=\log(\sqrt{1+x^2}+x)$, and hence, after using the definition of $\sinh$ once more we arrive at
$$\int_0^1\sqrt{1+x^2}dx=\frac{\log(\sqrt{2}+1)}{2} + \frac{\sinh(2\log(\sqrt{2}+1))}{4}\\ = \frac{\log(\sqrt{2}+1)}{2} + \frac{(\sqrt{2}+1)^2 -\frac{1}{(\sqrt{2}+1)^2}}{8} = \frac{\log(\sqrt{2}+1)}{2} + \frac{\sqrt{2}}{2}\\ $$