Here’s our second important identity which is a generalization of the first one.

$(2) \displaystyle \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$

Proof: Let $t = a+b-x$. Then, $dt = - dx$. Therefore,

$\displaystyle \int_a^b f(a+b-x) \, dx = - \int_b^a f(t) \, dt = \int_a^b f(t) \, dt$. And, we are done.

Let us now look at an integral that is quite easy to solve, though it looks quite formidable at first glance.

Problem 1. (Putnam 1987/B1) Evaluate $\displaystyle \int_2^4 \frac{\sqrt{\ln (9-x)}}{\sqrt{\ln (9-x)} + \sqrt{\ln (3+x)}} \, dx$.

Solution. Let us denote the given integral by $I$. Then applying identity (2) to $I$, we obtain

$\displaystyle I = \int_2^4 \frac{\sqrt{\ln (9-(6-x))}}{\sqrt{\ln (9-(6-x))} + \sqrt{\ln (3+(6-x))}} \, dx$, which implies

$\displaystyle I = \int_2^4 \frac{\sqrt{\ln (3+x)}}{\sqrt{\ln (3+x)} + \sqrt{\ln (9-x)}} \, dx$, which implies

$\displaystyle I + I = \int_2^4 \frac{\sqrt{\ln (9-x)} + \sqrt{\ln (3+x)}}{\sqrt{\ln (3+x)} + \sqrt{\ln (9-x)}}\, dx = \int_2^4 \, dx = 2$.

Hence, $I = 1$. Now, that was very easy!

(I might post a few more problems later, but you get the idea now.)