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I a multiple answer math question and i have solved the problem, the answer exists, but it's not the right one. $$I =\int \:\frac{1}{a+\cos \left(x\right)}dx$$ Using the substitution $\tan(\frac{x}{2}) = u$ , I got

$$I_1=\int \:\frac{\frac{2}{1+u^2}}{\frac{a\left(1+u^2\right)+\left(1+u^2\right)}{1+u^2}}du=2\int \:\frac{du}{a\left(1+u^2\right)+\left(1-u^2\right)}=\frac{2}{a-1}\int \:\frac{du}{u^2+\sqrt{\frac{a+1}{a-1}}^2}\: => I=\frac{2}{\sqrt{a^2-1}}\arctan \left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{\frac{a+1}{a-1}}}\right)+C$$

The correct answer is: $\frac{1}{\sqrt{a^2-1}}\left(x-2\arctan \left(\frac{\sin \left(x\right)}{a+\sqrt{a^2-1}+\cos \left(x\right)}\right)\right)$

Am I missing something ? I really can't tell.

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    Have you checked both options by differentiating them? In particular with trigonometric functions, it often occurs that different ways to find a primitive give different expressions that however represent the same function (at least where both are defined). – 2017-02-18
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    @DanielFischer I tried, but the simplifications are a pretty hard thing to do so i get to that final form. Stiil, there is a single correct answer and it's not the one i got to, even though it is written as a possible answer. – 2017-02-18
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    If things are complicated, [you can ask a bot to do it](http://www.wolframalpha.com/input/?i=derivative+of+arctan%28\sqrt{%28a-1%29%2F%28a%2B1%29}\tan%28x%2F2%29%29+with+respect+to+x). Look at "Alternate form assuming $a$ and $x$ are positive", and note that I left off the constant factor. So you found a primitive. The problem with your form is that it's only valid for $\lvert x\rvert<\pi$, since $\tan(x/2)$ has poles at $(2k+1)\pi,\,k\in\mathbb{Z}$. One can piece things together by going from one branch of $\arctan$ to the next at poles, but the other form is (if $a>1$) valid for all $x$. – 2017-02-18

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int{\dd \theta \over a + \cos\pars{\theta}} & = \int{\dd \theta \over a + \bracks{2\cos^{2}\pars{\theta/2} - 1}} = \int{\sec^{2}\pars{\theta/2} \over \pars{a - 1}\sec^{2}\pars{\theta/2} + 2} \,\dd \theta \\[5mm] & = \int{\sec^{2}\pars{\theta/2} \over \pars{a - 1}\tan^{2}\pars{\theta/2} + a + 1} \,\dd \theta \\[5mm] & = {1 \over a + 1}\int{\sec^{2}\pars{\theta/2} \over \bracks{\pars{a - 1}/\pars{a + 1}}\tan^{2}\pars{\theta/2} + 1} \,\dd \theta \end{align}

With $\ds{t \equiv \root{a - 1 \over a + 1}\tan\pars{\theta \over 2}\quad}$ and $\quad\left\{\begin{array}{rcl} \ds{\theta} & \ds{=} & \ds{2\arctan\pars{\root{a + 1 \over a - 1}\,t}} \\[2mm] \ds{\quad\sec^{2}\pars{\theta \over 2}\,\dd\theta} & \ds{=} & \ds{2\root{a + 1 \over a - 1}\dd t} \end{array}\right. $:

\begin{align} \int{\dd \theta \over a + \cos\pars{\theta}} & = {1 \over a + 1}\bracks{2\root{a + 1 \over a - 1}}\int{\dd t \over t^{2} + 1} = {2 \over \root{a^{2} - 1}}\,\arctan\pars{t} \\[5mm] & = \bbx{\ds{{2 \over \root{a^{2} - 1}}\arctan\pars{\root{a - 1 \over a + 1} \tan\pars{\theta \over 2}} + \pars{~\mbox{a constant}~}}} \end{align}

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    Is it the correct answer i wrote in the post ? Because i have to get to that form and I don't know if i could get that $x - ...$. And why is my answer wrong ? – 2017-02-19
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    @Liviu Your answer is correct. In order to arrive to your cited result, you have to write $\tan\left(\theta/2\right)$ in terms of $\sin\left(\theta\right)$ and $\cos\left(\theta\right)$. – 2017-02-19