Calculating the integral with an undefined function f(x) I am having a problem with an excersice of calculating the integral
I= $\int_{0}^{a}\frac{1}{1+f(x)}\,dx$
and we know that $f(x)f(a-x)=1$ , $a>0$  and $f(x)$ is continous and positive on the interval $[0,a]$
i've tried manipulating the expression with the fact that $f(x)= \ \frac{1}{f(a-x)} $ in order to find something to cancel or something that makes me solve the integral "nicely"
However after 20 minutes i have no idea what to do, or what i am missing
so if anyone could give me a tip or point me in the right direction i would appreciate it
 A: If the answer is independent of $f(x)$, then you can choose one consistent with the stated conditions, e.g., $f(x)=1$, in which case we find:
$$\int\limits_{x=0}^a \frac{1}{1 + 1} dx = \frac{a}{2}.$$
A: Same answer, different approach.  By symmetry,
$$
\begin{align}
\int_0^a\frac1{1+f(x)}\,\mathrm{d}x
&=\frac12\left(\int_0^a\frac1{1+f(x)}\,\mathrm{d}x+\int_0^a\frac1{1+f(a-x)}\,\mathrm{d}x\right)\tag1\\
&=\frac12\int_0^a\left(\frac1{1+f(x)}+\frac1{1+f(a-x)}\right)\,\mathrm{d}x\tag2\\
&=\frac12\int_0^a\frac{\color{#C00}{1}+f(x)+\color{#C00}{1}+f(a-x)}{\color{#090}{1}+f(x)+f(a-x)+\color{#090}{f(x)f(a-x)}}\,\mathrm{d}x\tag3\\
&=\frac12\int_0^a\frac{\color{#C00}{2}+f(x)+f(a-x)}{\color{#090}{2}+f(x)+f(a-x)}\,\mathrm{d}x\tag4\\[3pt]
&=\frac a2\tag5
\end{align}
$$
Explanation:
$(1)$: by substituting $x\mapsto a-x$, we get $\int_0^a\frac1{1+f(x)}\,\mathrm{d}x=\int_0^a\frac1{1+f(a-x)}\,\mathrm{d}x$
$\phantom{\text{(1):}}$ average the two
$(2)$: sum of integrals is the integral of the sum
$(3)$: $\frac1u+\frac1v=\frac{u+v}{uv}$
$(4)$: the sum of the integrands is $1$ since $f(x)f(a-x)=1$
$(5)$: integrate
A: $$
\int_0^a\frac{1}{1+f(x)}\,dx = \int_0^a\frac{1}{1+f(a-x)}\,dx = \int_0^a\frac{1}{1+\frac 1{f(x)}}\,dx = \int_0^a\frac{f(x)}{1+f(x)}\,dx,
$$
where in the first transformation we changed variables $x = a-y$. Hence,
$$
2I = I + I = \int_0^a\frac{1}{1+f(x)}\,dx + \int_0^a\frac{f(x)}{1+f(x)}\,dx = \int_0^a1\,dx = a.
$$
Thus, $I = \frac a2$.
