7.1 Integration by Parts

There's still a few back in chapter 6 I didn't get to, but with the psychotic weather we've been having and the subsequent schedule adjustment, we've been skipping around and passing on entire sections. I'll go back and do the other parts eventually, but for now I have to press on.

First off, we learn Integration by Parts.
Since the fundamental theorem linked together integration and differentiation, each rule we've learned for differentiation must also have a corresponding rule of integration. The Product Rule of differention (d/dx(f(x)g(x) = f(x)g'(x)+g(x)f'(x)) is mirrored in Integration by the rule of integration by parts.

The above formula for the Product rule, when substituted for integration, can be stated a few different ways:
⌠[f(x)g'(x)+f'(x)g(x)]dx = f(x)g(x)
or, since the integral of products is the product of the integrals,
⌠[(f(x)g'(x)]dx + ⌠[f'(x)g(x)]dx = f(x)g(x)
and we can re-arrange this a little further to say that:
⌠[f(x)g'(x)dx = f(x)g(x) - ⌠[f'(x)g(x)dx]

That rearranged version of it is the formula for integration by parts. Since that's quite a few letters in there, we can make a little prettier by substituting it for other terms. So we'll say f(x) = u, g(x) = v, and thus f'(x) = du and g'(x) = dv.
So we can restate our formula in the easier to read
⌠u(dv) = uv - ⌠v du

I think that people did it like this just so it would be annoying to write for people whose handwriting is not precise enough to readily tell the difference between u and v.

In general, when you get an ugly-looking integral, pick one of the parts of it to be u and another part to be dv. Find your du and v, then plug it into the above formula and you'll be able to integrate. You may have to loop this several times to end up with your answer, but it works. There are more details but I'm kinda pressed for time so I have to move on 7.2.