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Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 8, Section 2 1 Integration by Parts Integration by Parts is a technique of integration that is useful when the integrand involves a product of an algebraic with a transcendental expression, such as … x cos( x) dx See Exercise 1. x 3 e ln( x) dx See Exercise 2. x sin( x) dx See Exercise 5. Integration by Parts is based on the Product Rule for taking derivatives. Product Rule: (u v) = where u and v are functions of x If u and v are continuous, then you can integrate both sides of the Product Rule equation. (u v) dx This produces: = = u vdx + v u dx where dv = and du = Subtracting v du from both sides of the previous equation, we obtain the following: *** *** This last equation that we just created is the Integration by Parts formula. u dv is the “original integral,” the one that we are trying to integrate. Notice that it is written in terms of another integral, v du . Choosing u and dv wisely will make it easier to determine the integral v du than it would be to determine the original integral, u dv . Let’s look at some guidelines for choosing u and dv well. (“You have chosen…wisely.” This quote is from what movie?) Guidelines for Choosing u and dv 1. Try letting dv be the most “complicated” portion of the integrand that fits an elementary integration rule. Then u will be the remaining factor(s) of the integrand. 2. Try letting u be the portion of the integrand whose derivative is a function that is “simpler” than u. The dv will be the remaining factor(s) of the integrand. Exercise 1: Determine x cos( x) dx . Let u = Then du = Thus, x cos( x) dx = and let dv = and v = dv = Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 8, Section 2 2 Exercise 2: Determine x 3 ln( x) dx . Fruitless First Attempt at Exercise 2 Hmmm. If we let u x 3 , then we are forced to let dv ln( x) dx . Okay, du will be easy to compute. But then v = dv = ln( x) dx = ??? I hope and pray…that you did not say…that ln( x) dx equals 1 is the derivative of ln(x). x 1 … for you’d be astray! x 1 is NOT an anti-derivative of ln(x). x Fruitful Second Attempt at Exercise 2 Let u = Then du = Thus, x 3 and let dv = and v = dv = ln( x) dx = An Integrand with a Single Factor A surprising use of Integration by Parts involves some problems in which the integrand consists of a single factor. Bonusly (a new adverb…and a near-antonym of “bogusly”), we’ll also solve the problem raised in the Fruitless First Attempt at Exercise 2. Exercise 3: Determine ln( x) dx . In these sorts of cases, let dv = Let u = and let dv = Then du = and v = dv = Thus, ln( x) dx = Check this answer to see that you are right by derivating our anti-derivative answer. x ln( x) x = x ln( x) - x = . Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 8, Section 2 3 Repeated Use of Integration by Parts Some integrals require repeated use of the Integration by Parts formula, such as… Exercise 4: Determine x 2 cos( x) dx . becomes “simpler,” whereas the derivative of The derivative of Therefore, let u = and let dv = Then du = and v = Thus, x 2 (Compare with Exercise 1.) cos( x) dx = x 2 sin( x) - does not. sin( x) 2 xdx . We now have to integrate sin( x) 2 xdx by parts, with a new u and a new dv. Let new u = and let new dv = Then new du = and new v = Thus, sin( x) 2 xdx = Finally, x 2 cos( x) dx = x 2 sin( x) - = Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 8, Section 2 A Clever Trick Exercise 5: Determine e x sin( x) dx . 4 Let u = du = Let dv = v= Thus, e x sin( x) dx = Then, e x sin( x) dx = e x sin( x) dx = Let new u = new du = Let new dv = new v = Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 8, Section 2 5 Tabular Method For problems involving repeated applications of Integration by Parts, a tabular method can help to organize the work. This method works well for integrands with a factor that is a power of x, such as… x n cos( ax) dx x n sin( ax) dx x n e ax dx Exercise 6: Determine x 5 cos( 2 x) dx . Use the tabular method. Let u = and let dv = Alternating Signs u and its derivatives dv = v and its antiderivatives + x5 cos(2x) - + - Thus, x 5 cos( 2 x) dx =