To integrate $\int_0^{2\pi}\sqrt{\theta^2+1}\ d\theta$, why choose the change of variables $u=\theta+\sqrt{\theta^2+1}$?

Question

In order to find the length of the curve $r=\theta,\ \theta\in[0, 2\pi]$, the integral that must be solved is $$\int_0^{2\pi}\sqrt{\theta^2+1}\ d\theta$$ For which my proffesor opted to use the following change of variable: $u=\theta+\sqrt{\theta^2+1}$, since $$\theta=\frac{u^2-1}{2u}$$ and $$d\theta=\frac{u^2+1}{2u^2}du$$ And from there he wrote directly the solution $$L=\frac{1}{2}[\theta\sqrt{\theta^2+1}+log|\theta+\sqrt{\theta^2+1}|]^{2\pi}_0$$However, I do not undersand why is this the change of variable he decided to use.

Answer 1

There is an intermediate substitution you can make use of: $\theta = \sinh t$, which gives

$$I = \int_0^{\sinh^{-1}(2\pi)}\cosh^2t\:dt$$

From here, you can either use the $\cosh t$ double angle identity to solve the integral

$$I = \int_0^{\sinh^{-1}(2\pi)}\frac{1}{2}+\frac{1}{2}\cosh2t\:dt = \frac{t}{2}+\frac{1}{4}\sinh2t\Biggr|_0^{\sinh^{-1}(2\pi)}$$

$$ = \frac{1}{2}\sinh^{-1}(2\pi)+\frac{1}{2}\sinh\left(\sinh^{-1}(2\pi)\right)\cosh\left(\sinh^{-1}(2\pi)\right) = \boxed{\frac{1}{2}\sinh^{-1}(2\pi) + \pi\sqrt{4\pi^2+1}}$$

or you can rewrite the integral in terms $u = e^t = \sinh t + \cosh t = \theta + \sqrt{\theta^2+1}$ to get

$$I = \int_1^{2\pi+\sqrt{4\pi^2+1}}\left(\frac{u+u^{-1}}{2}\right)^2\frac{du}{u} = \frac{1}{4}\int_1^{2\pi+\sqrt{4\pi^2+1}}u+\frac{2}{u}+\frac{1}{u^3}\:du$$

and you can solve that integral to get the equivalent, but in my opinion less elegant, form of the answer.

Answer 2

Ninad Munshi’s answer is good - but I’m not sure it gets at the question “why use this particular change of variables?”

It is possible to solve the integral via the given substitution, after a fair bit of lengthy algebra, that turns out to be “nice” due to a few lucky coincidences. I’ll give an outline of how this may be done.

Say $$I = \int_0^{2 \pi} \sqrt{\theta^2+1} d \theta.$$

If we write $u=\theta + \sqrt{\theta^2+1}$, then

$$\frac{du}{d \theta} = 1 + \frac{\theta}{\sqrt{\theta^2+1}},$$

So $\sqrt{\theta^2+1} \frac{du}{d \theta} = \theta + \sqrt{\theta^2+1}= u$, i.e. $(u-\theta) du= u d \theta$. Note here that our ability to write $\frac{du}{d \theta}$ in terms of $u$ here in a “nice” way is somewhat lucky and may not happen in general.

We can substitute this into the integral to get

$$I = \int_{\alpha}^{\beta} (u-\theta)^2 u d u,$$

For some $\alpha$, $\beta$, whatever the appropriate new limits of the integral are under this change of variables.

Now, we can rewrite $\theta$ in terms of $u$. $(u - \theta)^2 = \theta^2+1$, so $u^2 - 2 u \theta - 1 = 0$, so $\theta = \frac{1 - u^2}{2u}$.

Substituting this into the integral, we get

$$I = \int_{\alpha}^{\beta} \left(u-\frac{1-u^2}{2u}\right)^2 u du$$

Expanding this out, we’ll get a rational function, that is, a ratio of two polynomials, for which there are algorithms that can be used to calculate the antiderivative (presumably how your teacher came up with the $L$ given in your question.)

Compare this instead to a substitution using $\theta = \sinh t$ and the use of trigonometric identities - much nicer.

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Also, try solving, say

$$\int_0^{2 \pi} \sqrt{\theta^2 + \theta + 1} d\theta,$$

using $u = \theta + \sqrt{\theta^2 + \theta + 1}$, or something similar like that. What goes wrong? You’ll see that the given substitution working in the original integral is more a result of luck.

Answer 3

This is another approach. (maybe a bit more complicated or "advanced") But at least in my mind , doing it this way makes a lot more sense and every step is more straight-forward.

Let $θ=\ \tan(u)$ and $dθ=\sec^2(u)$ then $\sqrt{θ^2+1} = \sqrt{\tan^2(u)+1}=\sqrt{\ sec^2(u)}$

This substitution is invertible over $0<u<\tan^{-1}(2π) $ with inverse $u=\tan^{-1}(θ)$.

This gives a new lower bound $u_1=\tan^{-1} (0)=0$ and upper bound $u_2=\tan^{-1}(2π)$

$$I=\int_0^{\tan^{-1}(2π)}\sec^2(u)^{3/2}du = \int_0^{\tan^{-1}(2π)}\sec^3(u)du$$

Then we can use the formula $\int \sec^m(u)du=\frac{\sin(u) \sec^{m-1}(u) }{m-1}+\frac{m-2}{m-1}\int \sec^{-2+m}(u)du$, where $m=3$

$$I=\frac{1}{2}\tan(u)\sec(u)|^ {\tan^{-1}(2π)}_0 + \frac{1}{2} \int_0^{\tan^{-1}(2π)}\sec(u)du$$

Evaluating the antiderivative at the limits and subtracting we get

$$I=π \sqrt{1+4π^2}+ \frac{1}{2}\int_0^{\tan^{-1}(2π)}\sec(u)du$$

Multiplying and dividing by $\sec(u)+\tan(u)$

$$I=π \sqrt{1+4π^2}+ \frac{1}{2} \int_0^{\tan^{-1}(2π)}\frac{\sec^2(u)+\sec(u)\tan(u)} {\sec(u)+\tan(u)}du$$

Then substituting $s=\tan(u)+\sec(u)$ and $ds=(\sec^2(u)+\ tan(u)\sec(u))du$ this gives a lower bound $s_1=\tan(0)+\sec(0)=1$ and upper bound $s_2=\tan(\tan^{-1}(2π))+\sec(\tan^{-1}(2π))=\sqrt{1+4π^2}+2π$

$$I=π\sqrt{1+4π^2}+\frac{1}{2}\int_1^{\sqrt{1+4π^2}+2π}\frac{1}{s}ds=π\sqrt{1+4π^2}+\frac{\log(s)}{2}|^{{\sqrt{1+4π^2}+2π}}_1$$

$$=π\sqrt{1+4π^2}+\frac{1}{2}\log(2π+\sqrt{1+4π^2})$$

Which is equal to :

$$=π\sqrt{1+4π^2}+\frac{1}{2}\sinh^{-1}(2π)$$

Answer 4

We first use integration by parts. $$ \begin{aligned} J & =\int \sqrt{\theta^2+1} d \theta \\ & =\theta \sqrt{\theta^2+1}-\int \frac{\theta^2}{\sqrt{\theta^2+1}} d \theta \\ & =\theta \sqrt{\theta^2+1}-\int\left(\sqrt{\theta^2+1}-\frac{1}{\sqrt{\theta^2+1}}\right) d \theta \end{aligned} $$ Rearranging gives $$ J=\frac{1}{2}\left(\theta \sqrt{\theta^2+1}+\int \frac{d \theta}{\sqrt{\theta^2+1}}\right) $$ For the last integral, let $\theta=\sinh x$, we have $$ \begin{aligned} \int_0^{2 \pi} \frac{d \theta}{\sqrt{\theta^2+1}} & =\int_0^{\sinh ^{-1}(2 \pi)} \frac{\cosh x}{\sqrt{\sinh ^2 x+1}} d x \\ & =\int_0^{\sinh ^{-1}(2 \pi)} 1 d x \\ & =\sinh ^{-1}(2 \pi) \end{aligned} $$ Conclusively, we have $$ \begin{aligned} \int_0^{2 \pi} \sqrt{\theta^2+1} d \theta & =\frac{1}{2}\left[\theta\sqrt{\theta^2+1}\right]_0^{2 \pi}+\sinh ^{-1}(2 \pi) \\ & =\pi \sqrt{4 \pi^2+1}+\frac{1}{2} \sinh ^{-1}(2 \pi) \end{aligned} $$ Wish it helps!

Answer 5

When the expression $\sqrt{\theta^2+1}$ shows up it is very tempting to substiute $\theta: =\sinh u.$ In this way $$e^u=\sinh u+\cosh u=\theta+\sqrt{\theta^2 +1}$$ $u=\log(\theta+\sqrt{\theta^2+1})$ and $du={1\over \sqrt{\theta^2+1}}.$ Therefore the substitions $u= \log(\theta+\sqrt{\theta^2+1})$ or $v=\theta+\sqrt{\theta^2+1}$ are pretty common.