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!
 A: 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
$$ 
\begin{align}
\int \frac{1}{a+b\cos(x)} {\rm d}x
& = \frac1a \int \frac{1}{1-\cos(x)} {\rm d}x \\
& = \frac1a \int \frac{1}{1 - 1 + 2\sin^2(\frac{x}{2})}{\rm d}x \\
& = \frac1{2a} \int \csc^2 \left(\frac{x}{2}\right) {\rm d}x \\
& = \frac1b \cot\left(\frac{x}{2}\right) + C
\end{align}
$$
A: 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?
A: 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*} 
A: 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}$
A: 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.
