Integration of a second degree polynomial underneath a radical. I have a question about how to apply the substitution rule in an arc length problem. The problem asks for the arc length of $x^2$ on the interval $[-1,2]$.
Here is what I have come up with:
$$y=x^2
\frac {dy}{dx} = 2x
(\frac {dy}{dx}^2 = 4x^2$$
So, $$L = \int_{-1}^2 \sqrt {1+ \frac {dy}{dx}^2}dx
= \int_{-1}^2 \sqrt {1+4x^2}dx$$
So the question is, how do I solve the last step, i.e. $\int_{-1}^2 \sqrt {1+4x^2}dx$?
(If the mathjax is bad, please forgive me, I am just learning)
 A: As you wrote, for your problem, the arc length is given by $$L = \int_{-1}^2 \sqrt {1+ \Big(\frac {dy}{dx}\Big)^2}dx
= \int_{-1}^2 \sqrt {1+4x^2}dx$$ For the calculation of the antiderivative, the change of variable $x=\frac{\sinh (y)}{2}$ seems to be a simple one and then $$\int \sqrt {1+4x^2}dx=\frac{1}{2}\int \cosh ^2(y) dy=\frac{1}{4}\int \Big(1+\cosh(2y)\Big)dy=\frac{y}{4}+\frac{1}{8} \sinh (2 y)$$ For $y$, the integration bounds are $-\sinh ^{-1}(2)$ and $\sinh ^{-1}(4)$.
I am sure that you can take from here.
A: I shall give you the indefinite integral using IBP:
$$I=\int \sqrt{4x^2+1}\ dx=x\sqrt{4x^2+1}-\int \frac {(4x^2+1)-1}{\sqrt{4x^2+1}}\ dx$$
This means, $$2I=x\sqrt {4x^2+1}+\frac 12 \int \frac {dx}{\sqrt{x^2+(\frac 12)^2}}$$
Now, $$\int \frac {dx}{\sqrt {x^2+a^2}}=\int \frac {1+\frac {x}{\sqrt {x^2+a^2}}}{x+\sqrt {x^2+a^2}}\ dx=\ln|x+\sqrt {x^2+a^2}|+C$$
Hence we get:
$$I=\frac {x\sqrt {4x^2+1}}{2}+\frac 14 \ln|x+\sqrt {x^2+\frac 14}|+C$$
Now you can put the required limits to obtain the answer.
A: This is a good problem. It demonstrates a lot of what is learned in integral calculus.
$$\int_a^b \sqrt{1+ (\frac{dy}{dx})^2} dx$$
Since $y=x^2$, $y'=2x$.
So the arc length formula becomes:
$$\int_a^b \sqrt{1+ 4x^2} dx$$
Trig sub gives you :
$$\sqrt{1+4x^2}=\sec{\theta}$$
$$\frac{1}{2}\tan{\theta}=x$$
$$\frac{1}{2}\sec^2{\theta}d\theta=dx$$
So:
$$\int \sqrt{1+4x^2}dx=\frac{1}{2}\int \sec^3{\theta}d\theta=\frac{1}{2}\int \sec{\theta}+\sec{\theta}\tan^2{\theta} \ d\theta$$
$$\int \sec{\theta} \ d\theta = \ln{|\sec{\theta}+\tan{\theta}|}+C$$
Use Integration by Parts to integrate the second term.
$$u=\tan{\theta}$$
$$dv=\sec{\theta}\tan{\theta}$$
$$du=\sec^2{\theta}$$
$$v=\sec{\theta}$$
$$\int \sec{\theta}\tan^2{\theta} \ d\theta=\sec{\theta}\tan{\theta}-\int \sec^3{\theta} \ d\theta$$
$$\int \sec^3{\theta} \ d\theta = \frac{1}{2}\ln{|\sec{\theta}+\tan{\theta}|}+\frac{1}{2}\sec{\theta}\tan{\theta}+C$$
So finally:
$$\int \sqrt{1+4x^2} \ dx = \frac{1}{4}\ln{|2x +\sqrt{1+4x^2}|}+\frac{1}{4}2x\sqrt{1+4x^2}+C$$
Plug in the end points to get the length.
A related problem, find the ratio of the  arclength of the curve $y=x^2$ to the length of the latis rectum between the points of intersection of the latis rectum to the parabola. This essentially establishes an analog to $\pi$ for parabolas, i.e. all parabolas are similar with a constant ratio between two characteristic lengths.
