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.)

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