Solving a problem by adding 0 and multiplying by 1

Has it ever occurred to you that when solving a problem, a non-trivial step might make the problem trivial?

Well, if you are a mathematician, it probably has. The only difference might be the number of steps needed to turn the problem into something easier.

Take for example the following problem which I solved yesterday in a forum on which I have the privilege to moderate.

Compute the following integral :


Two students are trying to solve it but they are getting nowhere, thus they ask for help. So, here am I, thinking about it. First, I get rid of the limits (temporarily).

\int\frac{e^x+cos(x)}{e^x+sin(x)+cos(x)}dx (1)

OK, cool. Then, I notice a nice pattern. If I had a -sin(x) in the numerator, I could just set u=e^x+sin(x)+cos(x) and problem solved. So, I add it and then remove it, to get.


So, one of the integrals is easy to solve, while the other is kind of harder. Wouldn’t it be nice if we had another pair of e^x,\ cos(x)? Then, we would have two easy integrals and problem solved.

OK, so we add those two and to keep the balance, we divide by 2.

\int\frac{1}{2}\frac{2e^x+2cos(x)+sin(x)-sin(x)}{e^x+sin(x)+cos(x)}dx (2)

And amazingly, while the only thing we did to get from (1) to (2) was to multiply by 1 and then add 0, the second problem is directly solvable while the first is not. How cool is that?

Answer : very!