Next we look at an example of the actual application of 714 In the example

# Next we look at an example of the actual application

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Next we look at an example of the actual application of (7.14). In the example below, the arrangement of terms is as one would work the problem with pencil and paper, except for the implication arrows which we will omit in subsequent problems, and possibly the under-braces which become less and less necessary with practice. Example 7.2.2 Compute integraldisplay xe x dx . Solution : First recall our boxed formula (7.14). For this problem, we write u = x dv = e x dx du = dx v = e x integraldisplay x bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright u e x dx bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright dv = uv integraldisplay v du = ( x )( e x ) integraldisplay ( e x ) dx = xe x e x + C. It is interesting to note that we choose u and dv , and then compute du and v , with one qualifi- cation. That is that v is not unique; the computation from dv to v is of an antidifferentiation nature and so we really only know v up to an additive constant. In fact any v so that dv = e x dx (we took v = e x = dv = e x dx ) will work in (7.14). Any additive constant, while legitimate, will eventually cancel in the final computation, and so we usually omit it. For instance, if we had chosen v = e x + 100, we would have had uv integraldisplay v du = x ( e x + 100) integraldisplay ( e x + 100) dx = xe x + 100 x e x 100 x + C = xe x e x + C, with the same final answer as before. For most cases, we will just assume that the additive constant is zero and we will use the simplest antiderivative for v . We will also not continue to write the implication arrows as they are technical and perhaps confusing. Now we will revisit the first example we gave in Subsection 7.2.1, using what will be our basic style for this method. Example 7.2.3 Compute integraldisplay x sec 2 x dx . Solution : First, as is standard for these, we complete a chart like in the previous example. u = x dv = sec 2 x dx du = dx v = tan x
602 CHAPTER 7. ADVANCED INTEGRATION TECHNIQUES integraldisplay x sec 2 x dx = integraldisplay u dv = uv integraldisplay v du = x tan x integraldisplay tan x dx = x tan x ln | sec x | + C. Since this method is more complicated than substitution, there are more complicated consid- erations in how to apply it. First of course, one should attempt an earlier, simpler method. But if those fail, and integration by parts is to be attempted, 5 the following guidelines for choosing u and dv should be considered for our formula integraltext u dv = uv integraltext v du : 1. u and dv must account for all factors of the original integral, and no more. 1.5. Of course, dv must contain the differential term (for example, dx ) as a factor, but can contain more terms. 2. v = integraltext dv should be computable with relative ease. 3. du = u ( x ) dx (assuming the original integral was in x ) should not be overly complicated.
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