Help please with this integral involving an inverse trig function

  • Thread starter madafo3435
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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} ##
 
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Answers and Replies

  • #2
BvU
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I'd like to see which ones you tried. PF guidelines require you post your attempt(s)
 
  • #3
etotheipi
Use Weierstrass, ##t = \tan{\frac{x}{2}}##. It is quite a useful substitution for fractions with stubborn trig functions.
 
  • #4
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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} ##
 
  • #5
55
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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.
 
  • #6
BvU
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You want to use it anyway :smile: maybe it doesn't crash...
What is ##dx## in terms of ##t## and ##dt## ?
 
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  • #7
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You want to use it anyway :smile: 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?
 
  • #8
55
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You want to use it anyway :smile: 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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  • #9
PeroK
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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.
 
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  • #10
George Jones
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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
$$\int_\pi ^ {2\pi} \frac{dx}{3 + \cos x} = \frac{1}{\sqrt{2}} \tan^{-1} u|_{-\infty}^0 = \frac{1}{\sqrt{2}} \left[0 - \left( -\frac{\pi}{2} \right)\right]$$
 
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  • #11
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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
 
  • #12
55
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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
 
  • #13
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Thank you all for your support, I am satisfied!
 
  • #14
George Jones
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One way to avoid this is to note (by, e.g., looking at the graph) that the integrand is even about ##\pi##. It is easily shown algebraically that the integrand evaluated at ##\pi - x## is the same as the integrand evaluated at ##\pi + x##. Consequently, ##\int_0^{2\pi} = 2 \int_0^{\pi}##.
 
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  • #15
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One way to avoid this is to note (by, e.g., looking at the graph) that the integrand is even about ##\pi##. It is easily shown algebraically that the integrand evaluated at ##\pi - x## is the same as the integrand evaluated at ##\pi + x##. Consequently, ##\int_0^{2\pi} = 2 \int_0^{\pi}##.
I will keep it in mind, thank you very much for your consideration
 

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