# Trigonometric Substitution - Arc Length

Good day to everyone,

I'm not so good in maths as I wish. But, I'm doing my job to improve it. So, be patience and thanks in advance for any detailed clue.

Description of the problem:

The arc is described by this parabola: $\ x=y^2$. Having the integral of the length of the arc, in Leibniz notation. The term "$\ dx/dy$" is $\ 2y$. (in this case $\ x=g(y)$ )

In the integral, there is a step I cannot understand. They make the following substitution: "$\ y=(1/2)*Tan(\theta)$ "

Question: Why $\ Tan(\theta)$ ?

I still cannot understand how to choose "the right trigonometric substitution".

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Fabian's post seems fine, but to clear up your confusion about which trig sub to use, I'll mention three cases.

When you see integrals of the form $$\sqrt{a^2+x^2}$$

where $a \in \mathbb R$, the substitution $x = a \tan \theta$ usually cleans up the integral quite nicely and makes it easier to work with. Note that the new integral may require another common integration method (such as integration by parts.)

Similarly, if you see integrals of the form $$\sqrt{a^2-x^2}$$

try the substitution $x = a \sin \theta$. For integrals of the form $$\sqrt{x^2-a^2}$$ try the substitution $x = a \sec \theta$.

Addendum Based on Stewart's text, the problem is to find the length of the arc of the parabola $y^2 = x$ from $(0,0)$ to $(1,1)$.

Arc length is given by $$L = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \ dy$$

We are given $$x = y^2 \implies \frac{dx}{dy} = 2y$$

$$L = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \ dy = \int_0^1 \sqrt{1 + (2y)^2} \ dy = \int_0^1 \sqrt{1 + 4y^2} \ dy$$

$$\int_0^1 \sqrt{1 + 4y^2} \ dy = \int_0^1 \sqrt{4\left(\frac{1}{4} + y^2\right)}\ dy = \int_0^1 2\sqrt{\left(\frac{1}{4} + y^2\right)} \ dy$$

The key to this is to note that the part involving $x^2$ cannot have a constant out front - so we need to factor and be clever to get in in a form that matches our three templates we know.

So, in general, if you see $$\sqrt{a^2 + kx^2} \quad a,k \in \mathbb R$$

Factor out $k$ to be left with $\sqrt{k\left(\frac{a^2}{k} + x^2 \right)}.$ Most of the time in nice examples (I imagine most of those picked out by Stewart are), $k$ will be a perfect square, which means we can factor it out of the square root as I did in your example. This helps clean our integral up nicely!

Now since $\left(\frac{1}{2}\right)^2 = \frac{1}4$, it should make sense that this follows the $a^2 + u^2$ template above. Carry out the substitution of $y = \frac{ \tan \theta}{2} \implies dy = \frac{\sec^2 \theta}{2} \ d \theta$. You end up needing to deal with

$$\frac{1}{2}\int_{y = 0}^{y=1} \sec^3 \theta \ d\theta$$

This is a common integration by parts example done in a Calc 2 course, in which many proofs of it can be found online. I've answered a question involving this integral here.

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Thanks Joe!!! You clear me up in another way, thanks for that. But, in the book, they use that substitution, literaly $\ y=1/2*Tan(\theta)$. And I don't understand where or how that comes or what was behind that substitution. Thanks so much for any clue. – Another.Chemist Jan 22 '13 at 23:40
What book are you using? This seems to be something that would be easier if you posted all of the details in the problem/proof, or gave me an idea of where to look in a book if I can find it online, so I can better help you. I assume the $tan \theta$ substitution comes from the definition of the arc length involving an integral as Fabian pointed out. Are you sure the $\frac{1}{2}$ is not to be confused with the differential $dx$ in the integral? – Joe Jan 22 '13 at 23:42
ok, Thanks so much for being so nice. The book is James Stewart Calculus 7e, if you want to find it to your personal use, you can find it by this web address gen.lib.rus.ec. The page of the book is: 564. Thanks so much in advance. – Another.Chemist Jan 22 '13 at 23:51
I actually have the book on hand - it's what I used for my calculus courses. I'll give it a look in one minute. – Joe Jan 22 '13 at 23:55
I think the addendum should help clear up your confusion. Let me know if something is unclear. – Joe Jan 23 '13 at 0:24

I guess you have misunderstood something. The arclength is given by $$s=\int_0^{x_0} \sqrt{1+ \left( \frac{dy}{dx}\right)^{2}} dx$$ and not by $\int (dy/dx) dx$ as I read from your post. This last expression equals $$s=\int_0^{x_0} \sqrt{1+ 4 x^2} dx$$ for the parabola. Substitution $x=\tfrac12 \tan\theta$ yields $$s= \int_0^{\arctan(2x_0)} \frac{\sqrt{1+\tan^2\theta}}{2\cos^2\theta} d\theta = \int_0^{\arctan(2x_0)} \frac{d\theta}{2\cos^3 \theta}.$$

An easier substitution is $x=\tfrac12\sinh\theta$ which yields $$s=\frac12\int_0^{\text{asinh}( 2 x_0)} \cosh^2 \theta \,d\theta.$$

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Thanks for answering so fast, by I'm using the integral of arclenght in this case: $\ x=g(y)$. Maybe, I didn't myself clear. But, I still cannot understand how to choose the "right trigonometric substitution". – Another.Chemist Jan 22 '13 at 23:27