Help please with this integral involving an inverse trig function

[madafo3435]

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 $\frac{\pi }{2}$

 

[BvU]

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.

 

[madafo3435]

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 $\frac {\pi}{2}$

 

[madafo3435]

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 continuous at this point.

 

[BvU]

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

 

[madafo3435]

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?

 

[madafo3435]

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.

 

[PeroK]

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.

 

[George Jones]

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
$$\int_\pi ^ {2\pi} \frac{dx}{3 + \cos x} = \frac{1}{\sqrt{2}} \tan^{-1} u|_{-\infty}^0 = \frac{1}{\sqrt{2}} \left[0 - \left( -\frac{\pi}{2} \right)\right]$$

 

[madafo3435]

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

 

[madafo3435]

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

 

[madafo3435]

Thank you all for your support, I am satisfied!

 

[George Jones]

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, $\int_0^{2\pi} = 2 \int_0^{\pi}$.

 

[madafo3435]

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