# Help please with this integral involving an inverse trig function

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Homework Statement:
I have problems evaluating this integral because every substitution I use turns out to have a discontinuous derivative, or the composition is not well defined.

I write the integral below
Relevant Equations:
I can't advance much in the integral because of what I mention below
## \int_0 ^ {2 \pi} \frac {dx} {3 + cos (x)} ##

las únicas formas que probé fueron, multiplicar por ## \frac{3-cos (x)}{3-cos (x)} ## pero no me gusta esto porque obtengo una expresión muy complicada. También recurrí a la sustitución ## t = tan (\frac {x} {2}) ## que me gusta bastante, pero encuentro que esta función es discontinua en ## \frac {\pi } {2} ## así que no he podido escribir mucho. También probé la sustitución ## cos (x) = \frac {1-t ^ 2} {1 + t ^ 2} ##, pero obtengo el mismo problema con ## \frac{\pi }{2} ##

Last edited:

BvU
Homework Helper
I'd like to see which ones you tried. PF guidelines require you post your attempt(s)

etotheipi
Use Weierstrass, ##t = \tan{\frac{x}{2}}##. It is quite a useful substitution for fractions with stubborn trig functions.

PeroK
I'd like to see which ones you tried. PF guidelines require you post your attempt(s)
sorry, it was my mistake not to indicate my attempts. However, the only ways I tried were, multiply by ##\frac {3-cos(x)}{3-cos(x)}## but I don't like this because I get a very complicated expression. I also resorted to the substitution recommended by etotheipi, using the substitution ##t=tan(\frac {x}{2})## which I quite like, but I find that this function is discontinuous in ## \frac {\pi}{2}## so I haven't been able to write much. Also i tried the substitution ##cos(x)=\frac {1-t^2}{1+t^2} ##, but i get the same problem whit ##\frac {\pi}{2} ##

Use Weierstrass, ##t = \tan{\frac{x}{2}}##. It is quite a useful substitution for fractions with stubborn trig functions.
i tried that substitution, but i have problem whit ##\frac {\pi}{2} ## because this function it not continous at this point.

BvU
Homework Helper
You want to use it anyway maybe it doesn't crash...
What is ##dx## in terms of ##t## and ##dt## ?

You want to use it anyway maybe it doesn't crash...
What is ##dx## in terms of ##t## and ##dt## ?
i find this: ##dx=2cos^2(\frac{x}{2})dt ##. But, I understand that any substitution must be to belong to ##C^1## or is it not necessary?

You want to use it anyway maybe it doesn't crash...
What is ##dx## in terms of ##t## and ##dt## ?
please observe this attempted solution. By applying the weistrass substitution carelessly, I obtain that the value of the proposed integral gives 0, which is contradictory, since this integral exists and must be positive.

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PeroK
Homework Helper
Gold Member
2020 Award
i find this: ##dx=2cos^2(\frac{x}{2})dt ##. But, I understand that any substitution must be to belong to ##C^1## or is it not necessary?
As a first step, I looked at the graph of ##\cos x## and reduced the integral to:
$$\int_0^{2\pi} \frac {dx} {3 + cos x} = 2\int_0^{\pi/2} \frac 1 {3 + \cos x} + 2\int_0^{\pi/2} \frac 1 {3 - \cos x} dx = 12\int_0^{\pi/2} \frac {dx} {9 - \cos^2 x}$$
That gets rid of all the problems with discontinuities.

Now, think ##\sec^2x## and ##\tan^2 x## perhaps.

Last edited:
George Jones
Staff Emeritus
Gold Member
please observe this attempted solution. By applying the weistrass substitution carelessly, I obtain that the value of the proposed integral gives 0, which is contradictory, since this integral exists and must be positive.

I haven't lookd carefully at this, but doesn't
$$\int_\pi ^ {2\pi} \frac{dx}{3 + \cos x} = \frac{1}{\sqrt{2}} \tan^{-1} u|_{-\infty}^0 = \frac{1}{\sqrt{2}} \left[0 - \left( -\frac{\pi}{2} \right)\right]$$

I haven't lookd carefully at this, but doesn't
$$\int_\pi ^ {2\pi} \frac{dx}{3 + \cos x} = \frac{1}{\sqrt{2}} \tan^{-1} u|_{-\infty}^0 = \frac{1}{\sqrt{2}} \left[0 - \left( -\frac{\pi}{2} \right)\right]$$
you're right, the integration limits confused me. If so, I am satisfied with a solution. thank you

As a first step, I looked at the graph of ##\cos x## and reduced the integral to:
$$\int_0^{2\pi} \frac {dx} {3 + cos x} = 2\int_0^{\pi/2} \frac 1 {3 + \cos x} + 2\int_0^{\pi/2} \frac 1 {3 - \cos x} dx = 12\int_0^{\pi/2} \frac {dx} {9 - \cos^2 x}$$
That gets rid of all the problems with discontinuities.

Now, think ##\sec^2x## and ##\tan^2 x## perhaps.
interesting your idea, I like it a lot, I will try to solve it by this method. thanks for your idea

Lnewqban
Thank you all for your support, I am satisfied!

BvU
George Jones
Staff Emeritus