# How do you integrate $\int \frac{1}{a + \cos x} dx$?

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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The Weierstrass substitution $u=\tan(x/2)$ (Wikipedia) turns it into an integrating a rational function problem. –  André Nicolas Apr 20 '12 at 23:22
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. –  joriki Apr 20 '12 at 23:25
@joriki: To remove your surprise, see my answer :-) –  Aryabhata Apr 20 '12 at 23:42
@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. –  joriki Apr 20 '12 at 23:50
@joriki: I see, I should have chosen the word 'lessen' instead of 'remove' :-) –  Aryabhata Apr 20 '12 at 23:52

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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Clever! Thanks for the solution :) –  badatmath Apr 23 '12 at 4:13

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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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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