Different ways to solve $\int_0^1 \sqrt{1+x^2} dx $ So I have the task to solve this integral in $4$ different ways, but have solved it only with substitution. ($x=tg(t)$, $dx=\sec^2(t)dt$) and so on. Any advice on the other $3$ ways ? Thank you :)
 A: IBP? $u = \sqrt{1+x^2}$ and $\mathrm{d}v = 1$ so $v = x$. Hence $$I = \int_0^1 \sqrt{1+x^2} \, \mathrm{d}x = \bigg[x\sqrt{1+x^2}\bigg]_0^1 - \int_0^1 \frac{x^2}{\sqrt{1+x^2}} \, \mathrm{d}x$$
Then $\int_0^1 \frac{x^2}{\sqrt{1+x^2}} \, \mathrm{d}x = \int_0^1 \frac{1+x^2}{\sqrt{1+x^2}} \, \mathrm{d}x - \int_0^1 \frac{1}{\sqrt{1+x^2}} \, \mathrm{d}x$. So $$I = \sqrt{2} - I + \int_0^1 \frac{1}{\sqrt{1+x^2}} \, \mathrm{d}x$$ or equivalently $$I = \frac{\sqrt{2}}{2} + \frac{1}{2}\text{arsinh} \,1$$

Alternatively, if you accept other substitutions $\sqrt{x^2 + 1} + x = t$ should work; this is known as Euler's substitution. 
A: One way is to use the Euler substitutions
Euler substitution 1 
Let $\sqrt{1+x^2} = x + t$, then $x = \cfrac{1 - t^2}{2t}$
Euler substitution 2
Let $\sqrt{1+x^2} = xt + 1$. Then $x = \cfrac{2t}{1-t^2}$
Hyperbolic trigonometric functions
We define $$\sinh x = \frac{e^{x} - e^{-x}}{2} \qquad \text{and} \qquad \cosh x = \frac{e^{x} + e^{-x}}{2}.$$ With the definitions above it can be shown with relative ease that 
$$1+\sinh^2x = \cosh^2x \qquad \text{and} \qquad \frac{\mathrm{d}}{\mathrm{d}x}\sinh x = \cosh x.$$
Thus, we can solve our integral with the hyperbolic substitution $x \mapsto \sinh t$ and $\mathrm{d}x = \cosh t \,\mathrm{d}t$,
$$
\int \sqrt{1+x^2} \,\mathrm{d}x = \int \sqrt{1+\sinh^2t} \cosh t \,\mathrm{d}t = \int \cosh^2t \,\mathrm{d}x\,.
$$
Where the last integral can be solved in a number of ways, perhaps the simplest is to use the definition of $\cosh t$ and expand. Another is to use the familiar looking identity $\cosh^2t = (1+\cosh 2t)/2$ and $\int \cosh t\,\mathrm{d}t = \sinh t + C$.
A: Substitute $x=\sinh(t/2)$, so $\sqrt{1+x^2}=\cosh(t/2)$ and $dx=\frac{1}{2}\cosh(t/2)\,dt$, so the integral becomes
$$
\frac{1}{2}\int\cosh^2(t/2)\,dt=\frac{1}{4}\int(\cosh t-1)\,dt
$$
