How to find $\int_{-1}^1 \frac{\cos x}{a^x+1}\mathrm dx$ 
Evaluate $$\int_{-1}^1 \frac{\cos x}{a^x+1}\mathrm dx$$ where $a$ is a real parameter $a\geq1$.

I can easily find the definite integral for $a=1$. It is $\sin(1)$.
In wolframalpha.com when I put $\displaystyle\int_{-1}^1 \frac{\cos x}{a^x+1}\text{d}x$ it shows me a very complicated formula with complex numbers and functions I didn't study but it says that definite integral is $= 0.841471\ldots$.
How can I find that integral?
 A: One of the first observations we can make is that you're integrating a function over an interval that is symmetric around $0$, i.e. it can be written as $[-c,c]$. 
I have a trick for you, whenever you find similar definite integrals start by writing the function you're integrating as a sum of an even function and an odd function. In fact, it can be proven that for every function $f$, there exists two functions such that $f=f_{\mathrm e}+f_{\mathrm o}$ and $f_{\mathrm e}$ is even and $f_{\mathrm o}$ is odd. Specifically $$f_{\mathrm e}(x)=\dfrac{f(x)+f(-x)}{2}\qquad f_{\mathrm o}(x)=\dfrac{f(x)-f(-x)}{2}.$$
You may ask yourself: “Why bother? It seems we're just making a complicated expression to look more intimidating.” But just wait, you'll see the usefulness of this process in a bit. Let's get back to our integral, applying the property we've just discussed for the function $f\colon x\mapsto\tfrac{\cos (x)}{a^x+1}$
$$\eqalign{
\int_{-1}^1 \frac{\cos x}{a^x+1}\mathrm dx&=\int_{-1}^1 \big[f_{\mathrm o}(x)+f_{\mathrm e}(x)\big]\mathrm dx\\
&=\int_{-1}^1 f_{\mathrm o}(x)\,\mathrm dx+\int_{-1}^1 f_{\mathrm e}(x)\,\mathrm dx\\
&=\int_{-1}^1f_{\mathrm e}(x)\,\mathrm dx.}
$$ 
Since 
$$\int_{-c}^c(\text{odd function})(x)\,\mathrm dx=0.\tag{$\forall c\in\Bbb R$}$$ 
Now if you calculate $f_{\mathrm e}$ you'll find that it has the ridiculously simple form: 
$$\require{cancel}\eqalign{f_{\mathrm e}(x)&=\dfrac{f(x)+f(-x)}{2}\\&=\dfrac{\frac{\cos (x)}{a^x+1}+\frac{\cos(-x)}{a^{-x}+1}}{2}\\&=\dfrac{\frac{\cos (x)}{a^x+1}+\frac{\cos(x)}{a^{-x}+1}}{2}\\&=\dfrac{\cos x}{2}\left(\dfrac{1}{a^x+1}+\dfrac1{a^{-x}+1}\right)\\&=\dfrac{\cos x}2{\left(\dfrac{2+a^{-x}+a^x}{a^{-x}a^x+a^x+a^{-x}+1}\right)}\\&=\dfrac{\cos x}{2},
}$$
 which is way simpler to integrate than the original function. We thus get $$\int_{-1}^1 \frac{\cos x}{a^x+1}\mathrm dx=\int_{-1}^1\frac{\cos x}2\mathrm dx=\int_0^1\cos x\,\mathrm dx=\sin1.\tag{$\forall a\geqslant1$}$$
We can even give the following generalization:
$$\int_{-c}^c\frac{\cos x}{a^x+1}\mathrm dx=\int_{-c}^c\frac{\cos x}2\mathrm dx=\int_0^c\cos x\,\mathrm dx=\sin c.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\color{#66f}{\large%
\int_{-1}^{1}{\cos\pars{x} \over a^{x} + 1}\,\dd x}=
\int_{0}^{1}\cos\pars{x}\pars{{1 \over a^{x} + 1} + {1 \over a^{-x} + 1}}\,\dd x
\\[5mm]&=\int_{0}^{1}\cos\pars{x}
\pars{{1 \over a^{x} + 1} + {a^{x} \over 1 + a^{x}}}\,\dd x
=\int_{0}^{1}\cos\pars{x}\,\dd x
=\color{#66f}{\large\sin\pars{1}}\approx{\tt 0.8415}
\end{align}
A: Let $$I=\int_{-1}^1 \frac{\cos (x)}{a^x+1}\mathrm dx\tag1$$
Using identity $$\int_a^bf(x)\;\mathrm dx=\int_a^bf(a+b-x)\;\mathrm dx$$
we get
$$I=\int_{-1}^1 \frac{\cos (-x)}{a^{-x}+1}\mathrm dx=\int_{-1}^1 \frac{a^{x}\cos (x)}{1+a^{x}}\mathrm dx\tag2$$
Adding $(1)$ and $(2)$, we get
$$2I=\int_{-1}^1\cos x\;\mathrm dx=\sin x\,\Bigg|_{-1}^1=2\sin 1\qquad\implies\qquad I=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\sin 1}}$$
Note that, $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$. You might be interested in seeing even and odd functions.
