Problem :
if f is continuous on [0,a] and $\displaystyle f(x) \cdot f(a-x) = 1$ , Evaluate :
$\displaystyle \int_0^a \dfrac{dx}{1+f(x)}$.
Final answer should be $\displaystyle \dfrac{a}{2}$
Let $\displaystyle I = \int_0^a \dfrac{dx}{1+f(x)}$
Make the change of variable $\displaystyle x = a - u$ so your integral becomes
$\displaystyle I = - \int_a^0 \dfrac{du}{1+f(a-u)}$
Now your your $\displaystyle f(x) \cdot f(a-x) = 1$ relation
so
$\displaystyle I = \int_0^a \dfrac{du}{1 + 1/f(u)} = \int_0^a \dfrac{f(u)}{1 + f(u)}du$
Now replace the $\displaystyle u$ with $\displaystyle x$ and add the two integrals
$\displaystyle 2I = \int_0^a \dfrac{dx}{1+f(x)} + \int_0^a \dfrac{f(x)dx}{1+f(x)} = \int_0^a dx$
and solve for I.