One of my Ph.D. students, Victor Zankoni is reading the paper Relativistically corrected Schrödinger equation with Coulomb interaction by J. L. Friar and E. L. Tomusiak, Physical Review C, Volume 29, Number 4, 1537. Following their calculations, he stumbled upon an integral of the form
$$\int\frac{dx}{a+b\cos x}$$
where $b^2>a^2$. Thinking it may not be easy to calculate, I suggested him to look up the well-known book Tables of Integrals, Series, and Products by Gradshteyn and Ryzhik. He said he couldn’t find a relevant formula (actually there is and he somehow missed it) and instead he decided to calculate it. He used a clever substitution and it worked beautifully, so here I am introducing his calculation. First note the identity
$$\cos x=\frac{1-\tan^2\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)}$$
Here, I assume $a>0$ and $b>0$ but similar calculations can be carried out for other cases as well. With the above substitution, we have
\begin{align*} \int\frac{dx}{a+b\cos x}&=\int\frac{\sec^2\left(\frac{x}{2}\right)}{b+a-(b-a)\tan^2\left(\frac{x}{2}\right)}dx\\ &=\frac{2}{b-a}\int\frac{du}{\left(\sqrt{\frac{b+a}{b-a}}\right)^2-u^2}\\ &=\frac{1}{\sqrt{b^2-a^2}}\left[\int\frac{du}{\sqrt{\frac{b+a}{b-a}}+u}+\int\frac{du}{\sqrt{\frac{b+a}{b-a}}-u}\right]\\ &=\frac{1}{\sqrt{b^2-a^2}}\ln\frac{u+\sqrt{\frac{b+a}{b-a}}}{u-\sqrt{\frac{b+a}{b-a}}}\\ &=\frac{1}{\sqrt{b^2-a^2}}\ln\frac{\sqrt{b^2-a^2}u+a+b}{\sqrt{b^2-a^2}u-(a+b)}\\ &=\frac{1}{\sqrt{b^2-a^2}}\ln\frac{\sqrt{b^2-a^2}\tan\left(\frac{x}{2}\right)+a+b}{\sqrt{b^2-a^2}\tan\left(\frac{x}{2}\right)-(a+b)} \end{align*}
The last expression coincides with 2.553 #3 of Gradshteyn and Ryzhik, 8th Edition on page 172. Gradshteyn and Ryzhik 7th Edition does not contain this expression but one can find a supposedly equivalent expression in 2.553 #3 of the 7th Edition but unfortunately, it is stated incorrectly. It has been corrected in the 8th Edition and is listed as 2.553 #4:
$$\int\frac{dx}{a+b\cos x}=\frac{2}{\sqrt{b^2-a^2}}\ln\left|\frac{(b-a)\tan\left(\frac{x}{2}\right)+\sqrt{b^2-a^2}}{(b-a)\tan\left(\frac{x}{2}\right)-\sqrt{b^2-a^2}}\right|\ [b^2>a^2]$$
There is another incorrect formula that caught my eyes:
\begin{align*} \int\frac{dx}{a+b\cos x}&=\frac{2}{\sqrt{b^2-a^2}}\mathrm{arctanh}\left(\frac{(a-b)\tan\left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)\ \left[b^2>a^2,\ \left|(b-a)\tan\left(\frac{x}{2}\right)\right|<\sqrt{b^2-a^2}\right]\\ &=\frac{2}{\sqrt{b^2-a^2}}\mathrm{arccoth}\left(\frac{(a-b)\tan\left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)\ \left[b^2>a^2,\ \left|(b-a)\tan\left(\frac{x}{2}\right)\right|>\sqrt{b^2-a^2}\right] \end{align*}
in both 7th Edition and the 8th Edition. It should be corrected to
\begin{align*} \int\frac{dx}{a+b\cos x}&=\frac{2}{\sqrt{b^2-a^2}}\mathrm{arctanh}\left(\frac{(b-a)\tan\left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)\ \left[b^2>a^2,\ \left|(b-a)\tan\left(\frac{x}{2}\right)\right|<\sqrt{b^2-a^2}\right]\\ &=\frac{2}{\sqrt{b^2-a^2}}\mathrm{arccoth}\left(\frac{(b-a)\tan\left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)\ \left[b^2>a^2,\ \left|(b-a)\tan\left(\frac{x}{2}\right)\right|>\sqrt{b^2-a^2}\right] \end{align*}
I don’t use Gradshteyn and Ryzhik’s book much but I have heard that it contains numerous typos and mistakes. It appears to be true. So, please use it with caution when you do.