arc length of $\sqrt{x}$ from $0$ to $a$.

Question

Find the arc length of $\sqrt{x}$ from $0$ to $a\ (a\in \mathbb{R})$.

I found that the arc length above is the same arc length as $x^2$ from $0$ to $\sqrt{a}$. Then I used the formula to calculate arc length, I obtained that $$s(x)=\int_{0}^{\sqrt a} \sqrt{1+4x^2}\ dx$$ and made the substitute $2x= \tan(t)$ but then I got stuck.

Any help would be great.

Answer 1

$$2x = \tan \theta$$ $$dx = \frac12 \sec^2 \theta \: d\theta$$ $$\displaystyle\int_0^{\sqrt{a}} \sqrt{1 + 4x^2} \: dx = \frac12 \int_0^{\arctan \sqrt{a}} \sec^3 \theta \: d\theta$$ Use integration by parts with $u = \sec \theta$ and $dv = \sec^2 \theta \: d\theta$. Then $du = \sec \theta \tan \theta \: d \theta$ and $v = \tan \theta$. This leads to $$\frac12 \int_0^{\arctan \sqrt{a}} \sec^3 \theta \: d\theta = \frac12 \sec \theta \tan \theta \biggr|_0^{\arctan \sqrt a} - \int_0^{\arctan \sqrt{a}} \tan \theta \sec^2 \theta \: d\theta$$ You have to apply the identity $\sec^2 \theta = 1 + \tan^2 \theta$. You will end up with the original integral on the RHS, which you then add to both sides, then divide by 2.

Answer 2

Hint:

Rescale $x$ to get rid of the $4$, for convenience.

Then, by parts,

$$I=\int\sqrt{1+x^2}dx=x\sqrt{1+x^2}-\int\frac{x^2}{\sqrt{1+x^2}}dx.$$

$$\int\frac{x^2}{\sqrt{1+x^2}}dx=\int\frac{1+x^2-1}{\sqrt{1+x^2}}dx=I-\int\frac{dx}{\sqrt{1+x^2}}.$$

The last integrand is known to be the derivative of the inverse hyperbolic sine.

Or if you prefer, set $x=\sinh(t)$, and

$$\int\frac{dx}{\sqrt{1+x^2}}=\int\frac{\cosh(t)}{\cosh(t)}dt.$$