Doing integrals are often hard things. Beyond the traditional bag of tricks that include standard change of variables, trigonometric substitutions, and such, there are often integrals here and there that requires some ingenuity.

The very first such integral most students encounter is The integral defies most attempts at finding substitution and in fact has no easily recognizable analytic solution. The standard trick is to notice two trigonometric identities: Together they imply

The second one that a person might see later on in a statistics class or a class on Fourier transformations, is the evaluation of the Gaussian integral This one also defies normal changes of variables and any attempts to integrate by parts. The trick here is actually to double up the integral Then implicitly using Fubini's theorem, the product under the integral can be evaluated as Which we can evaluate to be and we have that the integral of the Gaussian is the square root of π.

A third one I am going to give here is a cute question that I saw the other day on Yahoo Answers. It took me a while to find the solution. We are asked to evaluate (allegedly this problem and its original solution was due to Feynman. He solved it by embedding this integrand into a 1-parameter family that satisfy a simple ordinary differential equation, with the parameter s = 0 being the standard Gaussian, and the parameter s = 1 being the desired exponential. Then he just need to integrate the ordinary differential equation from 0 to 1 to find the answer. Here I give a different proof.)

The two main ideas are, firstly, the integrand is invariant under change of coordinates y = x^-1, while the volume form changes to -y^-2dy = dx: and secondly, that we can define the variable z = x - x^-1, so dz = dx + x^-2dx (notice that dz is never 0, so z is a good coordinate). Now so Combining everything we get