Reconcile two anti-derivatives of $\int \frac{dx}{A + \cos x} $

Question

I see in a calculus textbook that, for $|A| \neq 1$, \begin{align} f(x) &\equiv \int \frac{1}{A + \cos x} \,\text{d}x \\&=\frac{1}{ \sqrt{A^2 - 1} } \left( x - 2 \tan^{-1} \frac{ \sin x }{ A + \sqrt{A^2 - 1} + \cos x } \right) \end{align} which I have verified using Mathematica

$$\frac{\text{d}\, f(x)}{\text{d}\, x} = \frac{1}{A + \cos x}$$

However, I have failed to show the equivalence (up to a constant) of $f(x)$ to the more sensible form obtained by Mathematica (as explained here): $$g(x) \equiv \int \frac{1}{A + \cos x} \,\text{d}x = \frac{-2}{\sqrt{1-A^2}} \tanh ^{-1}\frac{(A-1) \tan \frac{x}{2}}{\sqrt{1-A^2}}$$ With the half angle substitution, one can replace $\tan \frac x2$ in above result. However, what really stumps me is how to get the linear term and arctangent in $f(x)$. It seems reasonable to try some identities involving inverses like $$\tanh^{-1}(\sin x) = \sinh^{-1}(\tan x) $$ or equivalently $$ \sin^{-1}( \tanh x ) = \tan^{-1}( \sinh x )$$ together with some common ways to combine $\sin x$ and $\cos x$. But, so far, I have gotten nowhere.

Answer 1

There are several ways to compute $$I=\int \frac{dx}{A + \cos x} $$ Using the tangent half-angle substitution $t=\tan \left(\frac{x}{2}\right)$, it reduces to $$I=2\int \frac{dt}{(A-1) t^2+(A+1)}=\frac 2{A-1}\int\frac{dt}{t^2+\frac{A+1}{A-1}}$$ So, now, using $$t=\sqrt{\frac{A+1}{A-1}}u\implies I=\frac{2 }{\sqrt{A-1} \sqrt{A+1}}\tan ^{-1}\left(\frac{\sqrt{A-1} }{\sqrt{A+1}}t\right)$$ but,as you can see, there could be some problem depending on the value of $A$. If $A >1$, I think that this is the best form to use.

If $A<1$, the argument of the tangent becomes an imaginary number but we can use the identity $\tan(i \theta)=i\tanh(\theta)$ and the the second formula.

Answer 2

The equivalence of $g(x)$ and $f(x)$ can indeed be established, as shown below.

\begin{align} g(x) &= \frac{-2}{\sqrt{1-A^2}} \tanh ^{-1}\frac{(A-1) \tan \frac{x}{2}}{\sqrt{1-A^2}}\\ &= \frac{2}{\sqrt{A^2-1}} \tan^{-1}\frac{(A-1) \tan \frac{x}{2}}{\sqrt{A^2-1}}\\ &= \frac{2}{\sqrt{A^2-1}} \bigg(\frac x2 -\tan^{-1}\tan\frac x2+\tan^{-1}\frac{(A-1) \tan \frac{x}{2}}{\sqrt{A^2-1}}\bigg)\\ &= \frac{1}{\sqrt{A^2-1}} \bigg(x-2\tan^{-1}\frac{\sqrt{A^2-1} -A+1}{\sqrt{A^2-1}\cot\frac x2+(A-1)\tan\frac x2}\bigg)\\ \end{align} Next, utilize $\tan\frac x2=\frac{1-\cos x}{\sin x}$ and $\cot\frac x2=\frac{1+\cos x}{\sin x}$, as well as $$\frac{\sqrt{A^2-1}+A-1}{\sqrt{A^2-1}-A+1}= A+ \sqrt{A^2-1}$$ to arrive at \begin{align} g(x) &= \frac{1}{\sqrt{A^2-1}} \bigg(x-2\tan^{-1}\frac{\sin x}{A+\sqrt{A^2-1}+\cos x}\bigg)=f(x) \end{align}