1. ## Arc Length

I could use some help determining the arc length determined by the function:

f(x) = sin x over the interval $[0, \pi]$

I know that I need to use the formula $s = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx$

which leads to evaluating: $s = \int_{0}^{\pi} \sqrt{1 + cos^{2}x} dx$,

which looks like it should be easy to do, but I'm just missing it. I've tried Pythagorean Indentity and power reduction formulas to replace $cos^{2}x$ but to no avail. As always, your help would be greatly appreciated.

2. Originally Posted by kaiser0792
I could use some help determining the arc length determined by the function:

f(x) = sin x over the interval $[0, \pi]$

I know that I need to use the formula $s = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx$

which leads to evaluating: $s = \int_{0}^{\pi} \sqrt{1 + cos^{2}x} dx$,

which looks like it should be easy to do, but I'm just missing it. I've tried Pythagorean Indentity and power reduction formulas to replace $cos^{2}x$ but to no avail. As always, your help would be greatly appreciated.
If we have the form

$\sqrt{a^2 + x^2}$ we let $x=atan \theta$

In this case

$cos \theta = tan \theta$

So

$\int \sqrt { 1 + cos^2 \theta } dx = \int \sqrt { 1 + tan^2 \theta } dx$

Of course, then we have

$\int \sqrt { 1 + tan^2 \theta } dx = \int \sqrt { sec^2 x } dx = \int sec x dx$

Change your bounds, and evaluate the integral. No problem!

3. Thanks Allan.

So you're essentially using trig substitution? I considered it, but it threw me using tan x in terms of cos x. I appreciate your help.

4. Originally Posted by kaiser0792
Thanks Allan.

So you're essentially using trig substitution? I considered it, but it threw me using tan x in terms of cos x. I appreciate your help.
Absolutely, but if you recall if we let

$cos \theta = tan \theta$

This is the same thing as saying

$cos \theta = \frac { sin \theta }{ cos \theta }$

$cos^2 \theta = sin \theta$

which is the same thing as saying

$1 - sin^2 \theta = sin \theta$

Of course this becomes the quadratic equation

$0 = sin^2 \theta + sin \theta - 1$

This has a value, and is therefore no problemmmmo!

In other words, what i'm trying to say is that

$cos \theta$ will be equal to $tan \theta$ for some value of theta. We need to find that out!

5. Excellent points and explanation. I am very grateful.

6. Originally Posted by AllanCuz
If we have the form

$\sqrt{a^2 + x^2}$ we let $x=atan \theta$

In this case

$cos \theta = tan \theta$
No, $cos(\theta)\ne tan(\theta)$! You mean $cos(x)= tan(\theta)$.

So

$\int \sqrt { 1 + cos^2 \theta } dx = \int \sqrt { 1 + tan^2 \theta } dx$

Of course, then we have

$\int \sqrt { 1 + tan^2 \theta } dx = \int \sqrt { sec^2 x } dx = \int sec x dx$

Change your bounds, and evaluate the integral. No problem!
No, because you can't have "dx" in your $\theta$ integral. If $cos(x)= tan(\theta)$ then $-sin(x)dx= sec^2(\theta)d\theta$. Since $sin(x)= \sqrt{1- cos^2(\theta)}$, the integral becomes $\int \frac{sec^4(\theta)}{\sqrt{1- tan^2(\theta)}} d\theta$ and I don't think you are any better off.

7. Originally Posted by HallsofIvy
No, $cos(\theta)\ne tan(\theta)$! You mean $cos(x)= tan(\theta)$.

No, because you can't have "dx" in your $\theta$ integral. If $cos(x)= tan(\theta)$ then $-sin(x)dx= sec^2(\theta)d\theta$. Since $sin(x)= \sqrt{1- cos^2(\theta)}$, the integral becomes $\int \frac{sec^4(\theta)}{\sqrt{1- tan^2(\theta)}} d\theta$ and I don't think you are any better off.
Yeah...that's right, i was drunk last night lol definately my bad on that one