<s>[INST] Calculating arc length $y=x^2$ I picked this example for practice and got stuck with it. Someone moderate me if I am in the right path.
I need to calculate the length of arc s, on the section of the curve $y=x^2$ with $0≤x≤1$
My Workings:
The formula is $s=\int^b_a\sqrt{1+[f'(x)]^2}dx$
I work out everything under the $\sqrt{}$
my $f(x) =x^2$. I need to get $f'(x)$, which =$2x$
So our function becomes $s=\int^1_0\sqrt{1+[2x]^2}dx$
Then I work out $1+[f'(x)]^2 = 1+4x^2 $
$s=\int^1_0\sqrt{1+[f'(x)]^2}dx =\int^1_0\sqrt{(1+4x^2)}dx$
$s=\int^1_0\sqrt{(1+4x^2)}dx =\int^1_0(1+4x^2)^\frac{1}{2}dx$
I reckon I have to integrate $\int^1_0(1+4x^2)^\frac{1}{2}dx$
The integration part:
I used u-substitution:
let $ u = 1+4x^2$ then $du = 8x dx$
hence $dx = \frac{du}{8x}$
But, now there is trouble with the $8x$ because if I plug in $dx$, I get:   
$\int^1_0(u)^\frac{1}{2}\frac{du}{8x} = \int^1_0\dfrac{(u)^\frac{1}{2}}{8x}du$
$= \dfrac{1}{8}\int^1_0\dfrac{1}{x}.{(u)^\frac{1}{2}}du$
Doesn't this give me $(\dfrac{1}{8}.ln|x|.\dfrac{2}{3}.u^\frac{3}{2})]^1_0$
...but... my answer doesn't look right......shouldnt' $x >0$? 
How do I end this?
 [/INST] Use the substitution $2x = \sinh u$. This gives you $\frac{dx}{du} = \frac{1}{2}\cosh u$.
So you get: $$ \int_0^1 (1+4x^2)^\frac{1}{2} dx= \int_0^{\sinh^{-1}2} (1+\sinh^{2} u)^\frac{1}{2} \frac{1}{2} \cosh udu = \frac{1}{2} \int_0^{\sinh^{-1}2}\cosh^{2} udu = \frac{1}{2} \int_0^{\sinh^{-1}2} \frac{1}{2} (\cosh2u +1)du = \frac{1}{4}[\frac{1}{2}\sinh 2u + u]_0^{\sinh^{-1}2}$$
This substitution is often very useful when you have an expression of the sort $(1+x^2)^{\frac{1}{2}}$ because of the hyperbolic identity $\cosh^{2} u = 1+\sinh^{2} u$.
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