I will discuss a list of some “identities” that one may employ in evaluating certain types of definite integrals. Without knowing them, it may virtually be impossible to integrate certain functions. The knowledge of such identities greatly enhances one’s ability to integrate!

(Below, a and b are real numbers and f is some “suitable” integrable function in the Riemannian sense.)

Here’s our first one.

(1) \displaystyle \int_0^a f(x)\, dx = \int_0^a f(a-x)\, dx.

Proof: Let t = a-x. Then, dt = - dx. Therefore, \displaystyle \int_0^a f(a-x)\, dx = - \int_a^0 f(t)\, dt = \int_0^a f(t)\, dt. And, we are done.

Let us now solve the following integral.

Problem 1. Evaluate \displaystyle \int_0^{\pi /2} \frac{\sin x}{\sin x + \cos x} \, dx.

Solution. Let \displaystyle I = \int_0^{\pi /2} \frac{\sin x}{\sin x + \cos x} \, dx. Then, applying identity (1) to the above integral, we obtain

\displaystyle I = \int_0^{\pi /2} \frac{\sin (\pi /2 - x)}{\sin (\pi /2 - x) + \cos (\pi /2 - x)} \, dx = \int_0^{\pi /2} \frac{\cos x}{\cos x + \sin x} \, dx.

Therefore, \displaystyle I + I = \int_0^{\pi /2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx = \int_0^{\pi /2}  \, dx = \pi /2.

Hence, I = \pi /4, which is our answer.

Okay, let us now evaluate a more difficult integral that appeared on the Putnam contest in 2005.

Problem 2. (Putnam 2005/A5) Evaluate \displaystyle \int_0^1\frac {\ln(x + 1)}{x^2 + 1}\, dx.

Solution. There are several ways of solving this problem, but the easiest way is the one that employs identity (1).

First, we use a “natural” trigonometric substitution, viz. x = \tan t. Then, dx = \sec^2 t \, dt. Denoting the given integral by I, we thus have

\displaystyle I = \int_0^{\pi / 4} \ln (1 + \tan t) \, dt = \int_0^{\pi / 4} \ln (1 + \tan (\pi /4 - t)) \, dt

\ldots (applying identity (1) to I)

Now, using the identity \displaystyle \tan (a-b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}, we get

\displaystyle I = \int_0^{\pi /4} \ln (\frac{2}{1 + \tan t}) \, dx = \int_0^{\pi /4} \ln 2 \, dx- I, which implies

\displaystyle 2I = \ln 2 \int_0^{\pi /4}\, dx. Hence, \displaystyle I = \frac{\pi}{8} \ln 2.

The second identity is a generalization of the first one, and I will discuss it (along with some sample problems) in my next post.

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