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Okay, I thought earlier that part 3 of the above series of posts would be my last one. For some reason, this series has turned out to be a somewhat popular one when considering the fact that a big chunk of blog visitors visit it. Probably, this is due to the fact that the 2008 MIT Integration Bee is going to be held sometime soon – I have no idea exactly when – or perhaps, there are other Integration Bees that are going to be held in other colleges/universities sometime soon. If I get some more feedback/interest, then I will consider posting more results/identities/tricks on this same topic.

As you might have noticed from the title of the post, this identity isn’t related to definite integrals; it is related to indefinite integrals. Of course, since definite integrals form a “subset” of indefinite integrals, we can apply this identity to either one of them.

Let me begin by posing two problems, which I ask you to solve in your head. If you are able to do so, then you probably know the trick that is stated below, and hence you may stop reading this post; else, continue reading.

Problem 1: Evaluate $\displaystyle \int e^x (\frac{x-2}{x^3}) \, dx$.

Problem 2: Evaluate $\displaystyle \int e^x (\ln x + \frac1{x}) \, dx$.

And, here’s our identity.

$(4) \displaystyle \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C$,

where $C$ is the constant of integration.

Proof: We use integration by parts. Recall, $\displaystyle \int u\, dv = uv - \int v\, du$, where $u \equiv f(x)$ and $v \equiv g(x)$. So, if we let $\displaystyle v = e^x$, then we have $\displaystyle \int e^x f(x) \, dx = \int f(x) \, d(e^x) = f(x) e^x - \int e^x d(f'(x))$, which leads us to our identity.

Now, you should be able to solve the above problems in your head in just a couple of seconds if not less.

Solution 1: $\displaystyle e^x/x^2 + C$.

Solution 2: $e^x \ln x + C$.

If you have found this particular series of posts useful, drop me a comment. Doing so will provide me the motivation to post more stuff on this topic in the near future.

Okay, this is the final part in the above series of posts on some identities related to definite integrals (before I get too lazy and forget to post the same).

So, what is the magic identity? Here it is.

$\displaystyle (3) \int_{-a}^{a} f(x) \, dx = \int_0^a \left( f(x) + f(-x)\right) \, dx$

Proof: Let $t = -x$ in the second integral on the right hand side. Then, we have $\displaystyle \int_0^a f(-x) \, dx = - \int_0^{-a} f(t) \, dt = \int_{-a}^0 f(x) \, dx$, and combining this with the first integral on the right hand side yields the desired result.

Now, apply the above identity to the “difficult” integral in problem $(6)$ from the Integration Bee, Challenging Integrals post to evaluate the integral. The solution turns out to be an easy one. The answer is $\pi /4$, just in case you need to verify.

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.)

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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