Integral Problem Homework: Solving \int\frac{dx}{cos(x)-1}

[just_see]

Homework Statement

$$\int\frac{dx}{cos(x)-1}$$

Homework Equations

None

The Attempt at a Solution

I've had extreme problems in even attempting to begin this problem. I initially tried to multiply by the conjugate to get int[(cosx+1)/(cos^2(x)-1)], but this doesn't seem to be leading me anywhere.

Since then, I've tried looking for a possible substitution to make, but have been coming up blank. Using the search function on physics forums I found a similar problem in which the substitution u=tan(x/2) was suggested. However I have not been able to figure out if (or how) this same substitution would apply in this case.

 

[d_leet]

Can you simplify cos²(x)-1 at all?

 

[rocomath]

let u = cosx - 1

now tell me what you get when you substitute

 

[just_see]

um... cos^2(x)-1 = -sin^2(x)...

int[cotxcscxdx] - int[csc^2(x)]

-cscx + cotx?

Or did I drop something somewhere? (which is quite possible)

 

[bob1182006]

How to Exit Trigonometry

the substitution u=tan(x/2) is a great suggestion, I was about to recommend it!

I don't know the proof in it's entirety but here are the formula's:

This can only be used if you have ONLY trig functions along with *, /, +, - a constant, if you have a ln (x), e^x, etc... you have to find another way.

$$tan(\frac{x}{2}) = y$$ then:

$$sin(x) = \frac{2y}{1+y^2}$$

$$cos(x) = \frac{1-y^2}{1+y^2}$$

$$dx = \frac{2dy}{1+y^2}$$

This method to "Exit Trigonometry" is unbelievably helpful I think.

These formula's are nicer, but there's a catch. They only work on EVEN powers with respect to sin and cos.

$$tan (x) = y$$

$$sin^2 (x) = \frac{y^2}{1+y^2}$$

$$cos^2 (x) = \frac{1}{1+y^2}$$

$$cos(x)sin(x) = \frac{y}{1+y^2}$$

$$dx = \frac{dy}{1+y^2}$$

 

[d_leet]

I think you missed a minus sign. You should have the integral of (cos(x)+1)/(-sin²(x))=-(csc(x)cot(x)+csc²(x)).

 

[rocomath]

ima go ahead and show some of my work (practicing latex) so ur in luck

$$\int\frac{dx}{\cos{x}-1} \ \ \ \ \ u=\cos{x}-1 \ \ \ \ \ -\csc{x}du=dx$$

$$-\csc{x}\int\frac{du}{u}$$

integrating

$$-\csc{x}\ln{|\cos{x}-1|}+C$$

it can also be written as (power rule)

$$\ln{|\cos{x}-1|}^{-\csc{x}}+C$$

differentiating to check work

$$\frac{-\csc{x}(-\sin{x})}{\cos{x}-1}$$

 

[d_leet]

ima go ahead and show some of my work (practicing latex) so ur in luck

$$\int\frac{dx}{\cos{x}-1} \ \ \ \ \ u=\cos{x}-1 \ \ \ \ \ -\csc{x}du=dx$$

$$\-csc{x}\int\frac{du}{u}$$

integrating

You can not pull functions of x outside of an integral like that. So from the very start this is the wrong way to a solution.

integrating

$$-\csc{x}\ln{|\cos{x}-1|}+C$$

it can also be written as (power rule)

$$\ln{|\cos{x}-1|}^{-\csc{x}}+C$$

Your integral is correct, but since it was based on a false premise of being able to pull out the cosecant function it will not give the correct answer to this problem.

differentiating to check work

$$\frac{-\csc{x}(-\sin{x})}{\cos{x}-1}$$

How did you differentiate this? You either didn't use the chain rule correctly, or did not use the product rule correctly because that is not the derivative you should end up with.

 

[rocomath]

lol ... my baddd, it's been 2 days since I've gotten back into Calculus ... damn

edit: ok i re-did it, the last part i understand why it was incorrect, the check i forgot the product rule, the first part i don't get.

 

[bob1182006]

hehe at least you didn't do that on a test :s

@just_see try using the substitutions I posted. of the top formulas all you need is cos (x) and dx, well tan x/2 = y also to go back to x when you're done integrating.

 

[rocomath]

is this a Cal 2 problem? i do not remember learning any of those things you posted bob.

 

[d_leet]

lol ... my baddd, it's been 2 days since I've gotten back into Calculus ... damn

edit: ok i re-did it, the last part i understand why it was incorrect, the check i forgot the product rule, the first part i don't get.

You can't pull a function of x outside of an integral with respect to x. dx is part of that integral, and after you made your substitution you factored part of it outside of the integral, that isn't allowed.

du=-sin(x)dx, however you don't have a sin(x) in the original integral. The goal is to do some work on the original expression so that you CAN transform dx into du.

 

[rocomath]

so the only things i can pull outside of an integral are constants, gotcha.

 

[bob1182006]

Those formula's aren't usually taught nowadays. I heard years ago they were stressed in high school but not now :/.
In my Calc book (Stewart) they're not shown explicitly, the derivation is given as one of the problems in some chapter so nobody really pays attention.

My Calc 2/3 teacher uses those formula's all the time. If there's no formula to reduce to a nice trig function then just "Exit Trigonometry".

 

[Gib Z]

the substitution u=tan(x/2) is a great suggestion, I was about to recommend it!

I don't know the proof in it's entirety but here are the formula's:

This can only be used if you have ONLY trig functions along with *, /, +, - a constant, if you have a ln (x), e^x, etc... you have to find another way.

$$tan(\frac{x}{2}) = y$$ then:

$$sin(x) = \frac{2y}{1+y^2}$$

$$cos(x) = \frac{1-y^2}{1+y^2}$$

$$dx = \frac{2dy}{1+y^2}$$

We know that $$\tan (2x) =\frac{2 \tan x}{1- \tan^2 x}$$
Letting x=y/2, $$\tan y = \frac{2\tan (y/2)}{1-\tan^2 (y/2)}$$.

And again, let t= tan (y/2)
$$\tan y = \frac{ 2t}{1-t^2}$$.

To get the expressions for sine and cosine of y, we draw A right angled triangle, with one of the acute angles being y. Then we know the opposite side is 2t, and the adjacent side is 1-t^2. By Pythagoras, the hypotenuse is 1+t^2. Its simple to see what sin and cos y are from that.

As for the derivative:
Using the chain rule and Trig Pythagorean identities;
$$\frac{d}{dx} \tan (x/2) = \frac{1}{2} \sec^2 (x/2) = \frac{1}{2} (\tan^2 (x/2) + 1) = \frac{t^2+1}{2}$$ so the expression for dx follows.