Integration Techniques: Substitution and Integration by Parts, Study notes of Differential and Integral Calculus

Download Integration Techniques: Substitution and Integration by Parts and more Study notes Differential and Integral Calculus in PDF only on Docsity! 4 Integration techniques We are now out of Part I of the course, where everything goes back to number sense, and into a segment of the course that involves learning a skill. It’s a high level skill, but you’re good at that kind of thing or you wouldn’t be here. So relax and enjoy some clean and satisfying computation. The material corresponds to Sections 5.5, 5.6 and 8.2 of the textbook. The first method ought to be review, and the second ought to be new, though due to your varied backgrounds some might find both to be review or both to be new. Curricular note: in Math 104 they spend a whole lot of time on integration techniques, nearly half the course (rmember the slide I showed you of the Math 104 final?). These days you can get your computer (or even your Wolfram Alpha iPhone app) to do this for you so there isn’t as great a need. But you need some familiarity in order to make sense of things, and learning the two most pervasive techniques strikes a reasonable balance. 4.1 Substitution The most common way of doing a integral by substitution, and the only way for indefinite integrals, is as follows. 1. Change variables from x to u (hence the common name “u-substitution”) 2. Keep track of the relation between dx and du 3. If you chose correctly you can now do the u-integral 4. When you’re done, substitute back for x The most common substitution is when you let u = h(x) for some function h. Then du = h￿(x) dx. Usually you don’t do this kind of substitution unless there will be an h￿(x) dx term waiting which you can then turn into du. Also, you don’t do this unless the rest of the occurences of x can also be turned into u. If h has an inverse function, you can do this by substituting h−1(u) for x everywhere. Now when you reach the fourth step, it’s easier because you can just plug in u = h(x) to get things back in terms of x. Please read the examples in Section 5.5 – there are a ton. I will give just one. 34 Example: Compute ￿ sinn x cos x dx. Solution: substitute u = sin x and du = cosx dx. This turns the integral into ￿ un du which is easily valuated as un+1/(n+1). Now plug back in u = sin x and you get the answer sinn+1 x n+ 1 . You might think to worry whether the substitution had the right domain and range, was one to one, etc., but you don’t need to. When computing an indefinite integral you are computing an anti-derivative and the proof of correctness is whether the derivative is what you started with. You can easily check that the derivative of sinn+1 x/(n+ 1) is sinn x cos x. There are a zillion examples of this in Section 5.5. When evaluating a definite integral you can compute the indefinite integral as above and then evaluate. A second option is to change variables, including the limit of integration, and then never change back. Example: Compute ￿ 2 1 x x2 + 1 dx. If we let u = x2 + 1 then du = 2x dx, so the integrand becomes (1/2) du/u. If x goes from 1 to 2 then u goes from 2 to 5, thus the integral becomes ￿ 5 2 1 2 du u = 1 2 (ln 5− ln 2) . Of course you can get the same answer in the usual way: the indefinite integral is (1/2) ln u; we substitute back and get (1/2) ln(x2 + 1). Now we evaluate at 2 and 1 instead of 5 and 2, but the result is the same: (1/2)(ln 5− ln 2). Some useful derivatives A large part of exact integration is recognizing when something is a derivative of something familiar. Here is a list of functions whose derivatives you should stare at long enough to recognize if they come up. (Yes, you can put them on a cheatsheet when exam time comes.) 35