Section 8.2 Reading Assignment: Integration by Parts

Answer Only Exercise 1, 2, and 3 by using a screenshot provided Calculus Pearson textbook. Make sure you read these three questions very carefully and see on what it is asking for.

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Section 8.2 Reading Assignment: Integration by Parts Instructions. Read through this assignment and complete the three exercises below by reading the a ppropriate passages of the textbook. The integration by parts form ula co mes fro m reversing the prod uct rule for derivatives. Typically, integration by parts applies when you have a product [i.e. multiplication} of functions, but not in a situation where u-substitution applies. A mnemonic that might help is LIATE, which stands for: logarithms, inverse functions, algebraic functions, trigonometric functions, and exponentials. The order indicates the preference for which function should be placed in the u position when doing integration by parts. So, when integrating a logarithm times anything, the u function should be that logarithm. Dr if doing an algebraic function [such as a polynomial} times a trigonometric function, then u should be the algebraic function and the 1;\" [also called by]1 should be the trigonometric function. Exercise 1. Read Example 1 ip. 45? 453}. Describe the choice of parts for this example and ex plain why the choice of parts helps evaluate the integral. Using LIATE [described above]: can be used here. when writing out these computations, it is an important part of showing your process to include what these various parts are. So be sure to include the u{x}| and v'lxi that you're using. Then differentiate to get u'{x} and integrate to get vix]. All four of these functions are crucial to include with your worlr. when using integration by parts. Exercise 2. Read Example 1 [p. 458}. In contrast to what I wrote above, the integrand is not the product [i.e. multiplication} of functions. Despite this, integration by parts still applies. Explain how this is done in this example by describing the choice of parts. Example 3 (p. 459} is an example of repeated use of integration by parts, which is an important strategy. Be careful about the minus signs when doing this type of computation. I recommend immediately cancelling two minus signs to a plus rather than letting the minus signs pile up. The other minus sign issue here is that the minus sign when applying integration by parts the first time applies to the entire integral, which means it must be distributed.I Exercise 3. Read Example 4 (p. 459 no}. Explain how this integral is computed. Be careful with this one, a bit of algebraic trickery is involved here.a We will avoid this type of problem for this class and focus on more direct integration by parts problems, but these integrals may come up, for example, in a differential equations course. So, it may be a good idea to keep this as refemnce. 1 Sometimes integration by parts is written with 111' and do instead of f and u'. This is the classical way of writing this formular but the-5r refer to the same thing. 3 Another way of doing this Is using a table known as the DI method or the tabular method. A 'll'l'dED of this is available In the Section 3.2 video page. This method makes this repeated process a lot easier to manage. 3 Notice that the choice of pars is reverse from LIATE; this Is not the relevant part of this oomputatlon. Using LIATE would have worked equally well and would been better for this computation In my opinion. Chapter 8 Techniques of Integration 8.2 Integration by Parts 467 In terms of indefinite integrals, this equation becomes DT Rearranging the terms of this last equation, we get leading to the integration by parts formula Integration by Parts Formula (1) This formula allows us to exchange the problem of computing the integral J uxv'(x) dx with the problem of computing a different integral, / w(x) #'(x) dx. In many cases, we can choose the functions a and a so that the second integral is easier to compute than the first. There can be many choices for # and v, and it is not always clear which choice works best, so sometimes we need to try several. The formula is often given in differential form. With v'(x) dx = du and a'(x) dx = du. the integration by parts formula becomesIntegration by Parts Formula-Differential Version Jude = w - fudu (2) The next examples illustrate the technique. EXAMPLE 1 Find & cos x dx. Solution There is no obvious antiderivative of a cos.x, so we use the integration by parts formula to change this expression to one that is easier to integrate. We first decide how to choose the functions u(x) and w(x). In this case we factor the expression x cos r into u(x) = x and v'(x) = cos x. Next we differentiate a(x) and find an antiderivative of v'(t), w'(x) = 1 and v(x) = sin.r.Chapter 8 Techniques of Integration 468 Chapter 8 Techniques of Integration When finding an antiderivative for v'(x) we have a choice of how to pick a constant of integration C. We choose the constant C = 0, since that makes this antiderivative as sim- ple as possible. We now apply the integration by parts formula: x cos x dx = x sinx - / six (1) ex Integration by parts formula m(x) u'(x) mix) u(x) = x sin x + cos x + c Integrate and simplify. and we have found the integral of the original function. There are four apparent choices available for a(x) and u'(x) in Example 1: 1. Let u(x) = 1 and v'(x) = xcosx. 2. Let u(x) = x and v'(x) = cos x. 3. Let u(x) = x cos x and v'(x) = 1. 4. Let u(x) = cos x and v'(x) = x. Choice 2 was used in Example 1. The other three choices lead to integrals we don't know how to integrate. For instance, Choice 3. with #'(x) = cos x - & sinx. leads to the integral (x cos.x - 8 sin .x ) dx. The goal of integration by parts is to go from an integral / w(x)w'(x) dx that we don't see how to evaluate to an integral / wxju'(x) dx that we can evaluate. Generally, you choose v'(x) first to be as much of the integrand as we can readily integrate; u(x) is the leftover part. When finding v(x) from v'(x), any antiderivative will work, and we usually pick the simplest one: no arbitrary constant of integration is needed in v(x) because it would simply cancel out of the right-hand side of Equation (2).EXAMPLE 2 Find / In x dx. Solution We have not yet seen how to find an antiderivative for In x. If we set "(x) = In x, then u'(x) is the simpler function 1/x. It may not appear that a second func- tion v'(x) is multiplying In x, but we can choose v'(x) to be the constant function v'(x) = 1. We use the integration by parts formula Equation (1) with u( x) = In x and v'(x) = 1. We differentiate w(x) and find an antiderivative of v'(x), u'(x) = and v(x) = x. Then In x. I dx = (Inx) x - Integration by parts formula Mix) u'(r) = xInx - x+c Simplify and integrate. In the following examples we use the differential form to indicate the process of inte- gration by parts. The computations are the same, with du and du providing shorter expres- sions for a'(x) dx and v'(x) dx. Sometimes we have to use integration by parts more than once, as in the next example.Chapter 8 Techniques of Integration 8.2 Integration by Parts 469 EXAMPLE 3 Evaluate xhe dx. Solution We use the integration by parts formula Equation (1) with u(x) - x and v'(x) = 8. We differentiate w(x) and find an antiderivative of v'(x). w'(x) = 2x and v(x) = e. We summarize this choice by setting du = u'(x) dx and du = v'(x) dx, so du = 2x dx and du = edx. We then have redx = xe' - | ezxax. Integration by parts formula The new integral is less complicated than the original because the exponent on x is reduced by one. To evaluate the integral on the right. we integrate by parts again with a = x du = edx. Then du = dx. v = e, and Integration by parts Equation (!) xedx = xe' - edx - xe' - e' + C . " = x, du = edx v= , du = dx w du Using this last evaluation, we then obtain xedx = re' - 2 xe' dx - re - 2x' + 2* + C.where the constant of integration is renamed after substituting for the integral on the right. The technique of Example 3 works for any integral / re dx in which n is a positive integer, because differentiating " will eventually lead to zero and integrating of is casy. Integrals like the one in the next example occur in electrical engineering. Their evalu- ation requires two integrations by parts, followed by solving for the unknown integral. EXAMPLE 4 Evaluate cos x dx. Solution Let u = e' and du = cos x dx. Then du = o' dx. v = sin x, and cos xdx = e sinx - / e sin x dx. M(x) = wr) = sinx The second integral is like the first except that it has sin x in place of cos x. To evaluate it, we use integration by parts with du = sin x ax, du = edr.Chapter 8 Techniques of Integration 470 Chapter 8 Techniques of Integration Then cos x dx = et sinx - (-e cos x - / (-cos met x) ) "(x) = 1x) =-cos x = el sin x + el cos x - / e cos x x. The unknown integral now appears on both sides of the equation, but with opposite signs. Adding the integral to both sides and adding the constant of integration give 2 ex cos x dx = e sin x + e'cos x + C. Dividing by 2 and renaming the constant of integration give e' cos x dx = el sin x + el cos * + C. 2 EXAMPLE 5 Obtain a formula that expresses the integral cost x dx in terms of an integral of a lower power of cos x. Solution We may think of cos" x as cos"- x . cos x. Then we let u = cos"-Ix and du = cos x dx, so that du = (n - 1) cos"-2x (-sin x dr) and 1 = sin T.Integration by parts then gives cos" x dx = cos" x sinx + (n - 1)/ sin' x cos"-2x dx = cost-xsinx + (x - 1)/ (1 - cos x) cog"-2 x dx = cos"x sinx + (n - 1)/ cos"-3xdx - (a - 1)/ cos" xax. If we add (n - 1)/ cos" x dix to both sides of this equation, we obtain "/ cos" x dx = cos" -1x sinx + (n - 1)/ cos"-2x x. We then divide through by n, and the final result is cos" rar = cos" x sin x _ " 1/ cox"- 2x dr. The formula found in Example 5 is called a reduction formula because it replaces an integral containing some power of a function with an integral of the same form having the