Finding the arc length of the parabola $y=x^2 \; from \; (0,0)\;to\;(1,1)$ As the title says, I need to find the arc length of that. This is what I have so far (I'm mostly stuck on the integration part): $${dy\over dx}=2x \Rightarrow L=\int_0^1 \sqrt{1+(2x)^2}dx$$
Substitute $$x=\tan\theta, \qquad dx=\sec^2\theta\,d\theta ,$$
giving $$\int_0^1 \sqrt{1+(2\tan\theta)^2}\sec^2\theta\,d\theta=\int_0^1 \sqrt{1+4\tan^2\theta}\sec^2\theta\,d\theta$$
That is where I'm stuck. Any help is appreciated, thank you.
Let $2x = \tan\theta$ instead. Then, the integral becomes $\displaystyle \int_0^{\arctan 2} \sqrt{1+\tan^2 \theta} \cdot \dfrac14\sec^2\theta \ \mathrm d\theta$ which is equal to $\displaystyle \frac14 \int_0^{\arctan2} \sec^3\theta \ \mathrm d\theta$.
You can make the substitution $2x=sh(t)$ it is simple ot obtain the result.