Let $I={\int }_0^a\frac{1}{1+f(x)}dx$. Using the property ${\int }_0^{a}g(x)dx={\int }_0^{a}g(a-x)dx$: $I={\int }_0^a\frac{1}{1+f(a-x)}dx$.

Given $f(x)f(a-x)=1$, we imply $f(a-x)=1/f(x)$. Substitute this back: $I={\int }_0^a\frac{1}{1+1/f(x)}dx={\int }_0^a\frac{f(x)}{f(x)+1}dx$.

Now add the original $I$ and the new form: $2I={\int }_0^a\frac{1}{1+f(x)}dx+{\int }_0^a\frac{f(x)}{1+f(x)}dx$ $2I={\int }_0^a\frac{1+f(x)}{1+f(x)}dx={\int }_0^{a}1dx=a$. Thus, $I=a/2$.