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

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}$

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*}

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.