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How do you integrate $$\int \frac{1}{a + \cos x} dx$$ Is it solvable by elementary methods?

I was trying to do it while incorrectly solving a homework problem. But, I couldn't find the answer.

Thanks!

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    $\begingroup$ The Weierstrass substitution $u=\tan(x/2)$ (Wikipedia) turns it into an integrating a rational function problem. $\endgroup$ Apr 20, 2012 at 23:22
  • $\begingroup$ Wolfram|Alpha gives a surprisingly complicated antiderivative. If you click on "Show Steps", you can see how it's derived using the Weierstraß substitution that André suggested. $\endgroup$
    – joriki
    Apr 20, 2012 at 23:25
  • $\begingroup$ @joriki: To remove your surprise, see my answer :-) $\endgroup$
    – Aryabhata
    Apr 20, 2012 at 23:42
  • $\begingroup$ @Aryabhata: I'm not sure why that should lessen my surprise :-) I can see that it's complicated, either from your answer or from Wolfram|Alpha's; what I meant was that I wouldn't have expected it to be that complicated just from the integrand, which seems relatively tame at first sight. $\endgroup$
    – joriki
    Apr 20, 2012 at 23:50
  • $\begingroup$ @joriki: I see, I should have chosen the word 'lessen' instead of 'remove' :-) $\endgroup$
    – Aryabhata
    Apr 20, 2012 at 23:52

6 Answers 6

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This doesn't help you to evaluate the indefinite integral, but I though I would add that the definite integral $\int_0^{2 \pi } \frac{1}{a + \cos x} \ dx$ can also be evaluated using methods from complex analysis. Let us make the simplifying assumption that $a > 1$ to avoid a blow up in the integral. Other values of $a$ (including complex ones!) can be made to work too.

We have \begin{align*} \int_0^{2 \pi } \frac{dx}{a + \cos x} &= \int_0^{2 \pi} \frac{dx}{a + \frac{e^{ix} + e^{-ix}}{2}} \\ &= 2\int_0^{2 \pi} \frac{e^{ix} \ dx}{2ae^{ix} + e^{2ix} + 1} && \text{Let } z=e^{ix}, \text{ so } dz = ie^{ix} \ dx. \\ &= \frac{2}{i} \int_{|z|=1} \frac{dz}{z^2 + 2az + 1} \\ &= \frac{2}{i} \int_{|z|=1} \frac{dz}{(z-z_1)(z-z_2)} \end{align*} where the circle $|z|=1$ is parametrized counterclockwise and \begin{align*} z_1 = -a - \sqrt{a^2-1} && z_2 = -a + \sqrt{a^2-1}. \end{align*} Clearly $z_1 < -a < -1$, so $z_1$ is outside of the unit circle. As for $z_2$, it is convenient to notice that \begin{align*} z_1 z_2 = 1. \end{align*} Thus, since $z_1$ is outside the unit circle, its inverse $z_2$ is inside the unit circle. So, applying the residue theorem, we have

\begin{align*} \frac{2}{i} \int_{|z|=1} \frac{dz}{(z-z_1)(z-z_2)} &= \frac{2}{i} \ 2 \pi i \ \mathrm{Res}\left( \frac{1}{(z-z_1)(z-z_2)}, z_2\right) \\ &= 4 \pi \frac{1}{z_2 - z_1} \\ &= 4 \pi \frac{1}{2 \sqrt{a^2 -1}} \\ &= \frac{2 \pi}{\sqrt{a^2-1}}. \end{align*}

One can check this result against the one obtained from Arybatha's antiderivative, although a bit of care needs to be taken as an improper integral crops up while making the various substitutions. \begin{align*} \int_0^{2 \pi } \frac{dx}{a + \cos x} &= \int_0^\pi \frac{2 \ dy}{a+1 + (a-1) \tan^2(y)} && y= \frac{x}{2}\\ &= \int_0^\infty \frac{ 2 dt}{ a + 1 + (a-1)t^2} +\int_{-\infty}^0 \frac{ 2 dt}{ a + 1 + (a-1)t^2} && t = \tan(y) \\ &= \frac{2}{\sqrt{a^2-1}} \arctan\left( \sqrt{ \frac{a-1}{a+1}} t \right) \big|_0^\infty + \frac{2}{\sqrt{a^2-1}} \arctan\left( \sqrt{ \frac{a-1}{a+1}} t \right) \big|_{-\infty}^0 \\ &= \frac{2 \pi}{\sqrt{a^2-1}} \end{align*}

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  • $\begingroup$ For $z_1=z_2=-1$, can we still use the residue theorem and use Riemman's integral theorem to expand the unit disk to something that includes $-1$? And in this case, will the integral just be undefined? Thanks! $\endgroup$
    – Korn
    Aug 12, 2022 at 20:30
  • $\begingroup$ What if a < 1 ? Then z1 and z2 are both on the $|z|=1$ contour $\endgroup$ Nov 25, 2023 at 14:08
  • $\begingroup$ @anoffercan'trefuse: well for $a<-1$, one again has one root inside the circle and one root outside. I guess a similar argument should work for those cases. For $-1\leq a \leq 1$, you are right that both roots are on the circle. But in that case, the integrand $\frac{1}{a+\cos x}$ has a blowup, so one has to decide what kind of improper/prinicipal value integration one is doing no? Honestly I haven't thought about it. $\endgroup$
    – Mike F
    Nov 27, 2023 at 4:13
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Let $ y = \frac{x}{2}$.

$$\frac{1}{a + \cos 2y} = \frac{1}{a -1 + 2\cos ^2 y} = \frac{\sec^2 y}{(a-1)\sec^2 y + 2} = \frac{\sec^2 y}{a + 1 + (a-1)\tan^2 y} $$

Thus

$$\int \frac{1}{a + \cos x} \text{d}x = \int \frac{2}{a + \cos 2y} \text{d}y $$

$$ = \int \frac{ 2\sec^2 y}{ a + 1 + (a-1)\tan^2 y} \text{d} y$$

Now make the subsitution $t = \tan y$.

I remember having used the same trick before: Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$

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  • $\begingroup$ Clever! Thanks for the solution :) $\endgroup$
    – badatmath
    Apr 23, 2012 at 4:13
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Generalization:

Let's consider $\cos x = \frac{1-t^2}{1+t^2}; t = \tan\frac{x}{2}; dx=\frac{2}{1+t^2} dt.$ Then we get that our integral becomes:

$$J = \int \frac{2dt}{(a+b)+(a-b) t^2}$$

I. For the case $a>b$, consider $a+b=u^2$ and $a-b=v^2$, and obtain that: $$J = 2\int \frac{dt}{u^2+v^2 t^2}=\frac{2}{uv} \arctan\frac{vt}{u} +C.$$ Turning back to our notation we get: $$I=\frac{2}{\sqrt{a^2-b^2}} \arctan\left(\sqrt{\frac{a-b}{a+b}} \tan\frac{x}{2} \right) + C.$$

II. For the case $a<b$, consider $a+b=u^2$ and $a-b=-v^2$, and obtain that: $$J = 2\int \frac{dt}{u^2-v^2 t^2}=\frac{1}{uv}\ln\frac{u+vt}{u-vt} \ +C.$$

Turning back again to out initial notation and have that:

$$I=\frac{2}{\sqrt{b^2-a^2}} \ln\frac{b+a \cos x + \sqrt{b^2-a^2} \sin x}{a+b \cos x} + C.$$ Also, note that $x$ must be different from ${+}/{-}\arccos(-\frac{a}{b})+2k\pi$ if $|\frac{a}{b}|\leq1$.

Q.E.D.

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Also a generalised solution, borrowing from and expanding upon user1357113's answer,

I. For the case $|a| > |b|$, note that the substitution $t=\tan \left( \frac{x}{2} \right)$ is not injective. So to retrieve a continuous antiderivative, we have as a complete answer

$$ \int \frac{1}{a+b\cos(x)} {\rm d}x = \frac{2}{\sqrt{a^2-b^2}} \arctan \left( \sqrt{\frac{a-b}{a+b}}\tan \left( \frac{x}{2} \right) \right) + \frac{2\pi}{\sqrt{a^2-b^2}} \left\lfloor \frac{x+\pi}{2\pi} \right\rfloor +C $$

However, we can simplify the ugly floor function in terms of arctangents and tangents. First, we have

$$ \pi \left\lfloor \frac{x+\pi}{2\pi} \right\rfloor = \frac{x}{2} - \arctan \left( \tan \left( \frac{x}{2} \right) \right) $$

So, implementing this and then asking What is $\arctan(x) + \arctan(y)$? We get

$$ \begin{align} \int \frac{1}{a+b\cos(x)} {\rm d}x & = \frac{2}{\sqrt{a^2-b^2}} \arctan \left( \sqrt{\frac{a-b}{a+b}}\tan \left( \frac{x}{2} \right) \right) + \frac{2}{\sqrt{a^2-b^2}} \left( \frac{x}{2} - \arctan \left( \tan \left( \frac{x}{2} \right) \right) \right) +C \\ & = \frac{1}{\sqrt{a^2-b^2}} \left( x + 2\arctan \left( \sqrt{\frac{a-b}{a+b}}\tan \left( \frac{x}{2} \right) \right) - 2\arctan \left( \tan \left( \frac{x}{2} \right) \right) \right) + C \\ & = \frac{1}{\sqrt{a^2-b^2}} \left( x + 2\arctan \left( \frac{ \sqrt{\frac{a-b}{a+b}}\tan \left(\frac{x}{2}\right) - \tan \left(\frac{x}{2}\right) }{1 + \sqrt{\frac{a-b}{a+b}} \tan^2 \left(\frac{x}{2}\right) } \right) \right) + C \\ \end{align} $$

Which ultimately simplifies to a satisfying compact

$$ \int \frac{1}{a+b\cos(x)} {\rm d}x = \frac{1}{\sqrt{a^2-b^2}} \left( x - 2\arctan \left( \frac{ (\sqrt{a+b}-\sqrt{a-b})\tan \left(\frac{x}{2}\right) }{\sqrt{a+b} + \sqrt{a-b} \tan^2 \left(\frac{x}{2}\right) } \right) \right) + C $$

II. And for the case $|a|<|b|$, we will have, from user1357113's partial fraction decomposition,

$$ \int \frac{1}{a+b\cos(x)} {\rm d}x = \frac{1}{\sqrt{b^2-a^2}}\ln\left(\left|\frac{b+a\cos\left(x\right)+\sqrt{b^2-a^2}\sin\left(x\right)}{a+b\cos\left(x\right)}\right|\right) +C $$

III a. And for the case $a = b$, we will have $$ \begin{align} \int \frac{1}{a+b\cos(x)} {\rm d}x & = \frac1a \int \frac{1}{1+\cos(x)} {\rm d}x \\ & = \frac1a \int \frac{1}{1+2\cos^2(\frac{x}{2})-1}{\rm d}x \\ & = \frac1{2a} \int \sec^2 \left(\frac{x}{2}\right) {\rm d}x \\ & = \frac1a \tan\left(\frac{x}{2}\right) + C \end{align} $$

III b. Finally, for the case $a = -b$, we will have $$ \begin{align} \int \frac{1}{a+b\cos(x)} {\rm d}x & = \frac1a \int \frac{1}{1-\cos(x)} {\rm d}x \\ & = \frac1a \int \frac{1}{1 - 1 + 2\sin^2(\frac{x}{2})}{\rm d}x \\ & = \frac1{2a} \int \csc^2 \left(\frac{x}{2}\right) {\rm d}x \\ & = \frac1b \cot\left(\frac{x}{2}\right) + C \end{align} $$

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Expanding André's comment

Say we have an integral of the form

$$\int R(\sin x,\cos x) dx$$

Then the substitution

$$t= \tan\frac x 2 $$

will change the integral into a rational function of

$$\sin x = \frac{2t}{1+t^2}$$

$$\cos x = \frac{1-t^2}{1+t^2}$$

and of course

$$dx = \frac{2 dt}{1+t^2}$$

Would you like to try solve it that way or want a full solution?

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The substitution $t= a\cot x+\csc x $ is more desirable than the half-angle one \begin{align}\int \frac{dx}{a+\cos x} =\int\frac {dt}{1-a^2-t^2}= \begin{cases} -\frac1{\sqrt{a^2-1}}\tan^{-1}\frac{t}{\sqrt{a^2-1}}&a^2>1\\ \frac1 {t}& a^2=1\\ \frac1{\sqrt{1-a^2}}\coth^{-1}\frac{t}{\sqrt{1-a^2}}&a^2<1\\ \end{cases} \end{align}

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