# ch06_01 - 510 CHAPTER 6 . . Integration Techniques 6-2 In...

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Unformatted text preview: 510 CHAPTER 6 . . Integration Techniques 6-2 In section 6.2, we introduce a powerful technique called integration by parts that can be used to find antiderivatives of many such functions. The new techniques of integration introduced in this chapter provide us with a broad range of tools used to solve countless problems of interest to engineers, mathematicians and scientists. 6.1 REVIEW OF FORMULAS AND TECHNIQUES In this brief section, we draw together all of the integration formulas and the one integration technique (integration by substitution) that we have developed so far. We use these to develop some more general formulas, as well as to solve more complicated integration problems. First, look over the following table of the basic integration formulas developed in Chapter 4. x r dx = x r + 1 r + 1 + c , for r = 1 (power rule) 1 x dx = ln | x | + c , for x = sin x dx = cos x + c cos x dx = sin x + c sec 2 x dx = tan x + c sec x tan x dx = sec x + c csc 2 x dx = cot x + c csc x cot x dx = csc x + c e x dx = e x + c e x dx = e x + c tan x dx = ln | cos x | + c 1 1 x 2 dx = sin 1 x + c 1 1 + x 2 dx = tan 1 x + c 1 | x | x 2 1 dx = sec 1 x + c Recall that each of these follows from a corresponding differentiation rule. So far, we have expanded this list slightly by using the method of substitution, as in example 1.1. EXAMPLE 1.1 A Simple Substitution Evaluate sin( ax ) dx , for a = 0. Solution The obvious choice here is to let u = ax , so that du = a dx . This gives us sin( ax ) dx = 1 a sin( ax ) sin u a dx du = 1 a sin u du = 1 a cos u + c = 1 a cos( ax ) + c . There is no need to memorize general rules like the ones given in examples 1.1 and 1.2, although it is often convenient to do so. You can reproduce such general rules any time you need them using substitution. 6-3 SECTION 6.1 . . Review of Formulas and Techniques 511 EXAMPLE 1.2 Generalizing a Basic Integration Rule Evaluate 1 a 2 + x 2 dx , for a = 0. Solution Notice that this is nearly the same as 1 1 + x 2 dx and we can write 1 a 2 + x 2 dx = 1 a 2 1 1 + ( x a ) 2 dx . Now, letting u = x a , we have du = 1 a dx and so, 1 a 2 + x 2 dx = 1 a 2 1 1 + ( x a ) 2 dx = 1 a 1 1 + ( x a ) 2 1 + u 2 1 a dx du = 1 a 1 1 + u 2 du = 1 a tan 1 u + c = 1 a tan 1 x a + c ....
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## This note was uploaded on 10/03/2010 for the course CHE 2C CHE 2C taught by Professor Atsumi during the Spring '10 term at UC Davis.

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ch06_01 - 510 CHAPTER 6 . . Integration Techniques 6-2 In...

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