Overachiever's Edition of Integration by Parts

For those who bored with current studies in calculus, I will provide some notes for the general purpose of anyone. I don't actually know how to integrate by parts, but I'm going to look into the textbook right now just for you guys. And for as Kenny's comment in calculus, "Why do you learn the new stuff if you know that we're just going to go over it in class and be bored?" Well, even though I'm bored as shit, there are still things that I pick up from the second lesson that I wouldn't get the first time. If I'm learning something for the first time in a classroom, I never completely get it. Whereas, if I walk into the classroom knowing something about the topic, I'm just sharpening my skills.

As Mr. Cole has stated, there is somewhat of a product rule for integrals, but its method is not mechanical in any sense. The method is called integration by parts, which is an expansion of the derivative product rule discovered by Leibniz. I've removed the x from the functions because it makes it way easier to read.

d/dx [ f*g ] = f*g' + g*f'

If we take the integral of this equation, we get:


f*g = ∫f*g' dx + ∫f'*g dx


If we let u= f(x) and v=g(x), which assumes du=f'(x) and dv=g'(x), we get:


uv= ∫u dv + ∫v du


However, the standard notation has it written as:


∫u dv = uv - ∫v du


I'm sure that this formula doesn't mean a heck of a lot to anyone, but here's how you use it. Let's use an example integration.

∫x*sinx dx

First, create a u and v substitution. Notice the integral ∫u dv. This is the formula that your integral should resemble, so you must find a way to substitute ∫x*sinx dx in a way such that it equals ∫u dv.

Well, u appears in the integral as plain old u, so assign one of the factors the u. The v, however, has been differentiated. Therefore, in order to correctly substitute into v, you must set the second factor equal to the derivative of v.

The corresponding step taken in the example is as follows:

let u=x
let dv=sinx

Therefore:

∫u dv = ∫x*sinx dx


So now all that's left to do is plug into the rest of the values:

∫u dv = uv - ∫v du

[to get the value of v, integrate dv]
[∫sinx dx= -cosx ]

∫x*sinx dx = -xcosx - ∫(-cosx) dx

Now evaluate the integral:

∫x*sinx dx = -xcosx + sinx + C


And voilà, there you have the Integral. That was easy, right?
Well the fun part is realizing that this only works for a select group of functions where the new integral formed is a product that can be determined by other methods. Have a great weekend guys.