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# ch7 - CHAPTER 7 Techniques of Integration Integration...

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CHAPTER 7 Techniques of Integration 7.1. Substitution Integration, unlike differentiation, is more of an art-form than a collection of algorithms. Many problems in applied mathematics involve the integration of functions given by complicated formulae, and practi- tioners consult a Table of Integrals in order to complete the integration. There are certain methods of integration which are essential to be able to use the Tables effectively. These are: substitution, integration by parts and partial fractions. In this chapter we will survey these methods as well as some of the ideas which lead to the tables. After the examination on this material, students will be free to use the Tables to integrate. The idea of substitution was introduced in section 4.1 (recall Proposition 4.4). To integrate a differen- tial f x dx which is not in the table, we first seek a function u u x so that the given differential can be rewritten as a differential g u du which does appear in the table. Then, if g u du G u C , we know that f x dx G u x C . Finding and employing the function u often requires some experience and ingenuity as the following examples show. Example 7.1 x 2 x 1 dx ? Let u 2 x 1, so that du 2 dx and x u 1 2. Then x 2 x 1 dx u 1 2 u 1 2 du 2 1 4 u 3 2 u 1 2 du 1 4 2 5 u 5 2 2 3 u 3 2 C (7.1) 1 30 u 3 2 3 u 5 C 1 30 2 x 1 3 2 6 x 2 C 1 15 2 x 1 3 2 3 x 1 C (7.2) where at the end we have replaced u by 2 x 1. Example 7.2 tan xdx ? 107

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Chapter 7 Techniques of Integration 108 Since this isn’t on our tables, we revert to the definition of the tangent: tan x sin x cos x . Then, letting u cos x du sin xdx we obtain (7.3) tan xdx sin x cos x dx du u ln u C lncos x C lnsec x C Example 7.3 sec xdx ?. This is tricky, and there are several ways to find the integral. However, if we are guided by the principle of rewriting in terms of sines and cosines, we are led to the following: (7.4) sec x 1 cos x cos x cos 2 x cos x 1 sin 2 x Now we can try the substitution u sin x du cos xdx . Then (7.5) sec xdx du 1 u 2 This looks like a dead end, but a little algebra pulls us through. The identity (7.6) 1 1 u 2 1 2 1 1 u 1 1 u leads to (7.7) du 1 u 2 dx 1 2 1 1 u 1 1 u du 1 2 ln 1 u ln 1 u C Using u sin x , we finally end up with (7.8) sec xdx 1 2 ln 1 sin x ln 1 sin x C 1 2 ln 1 sin x 1 sin x C Example 7.4 As a circle rolls along a horizontal line, a point on the circle traverses a curve called the cycloid . A loop of the cycloid is the trajectory of a point as the circle goes through one full rotation. Let us find the length of one loop of the cycloid traversed by a circle of radius 1. Let the variable t represent the angle of rotation of the circle, in radians, and start (at t 0) with the point of intersection P of the circle and the line on which it is rolling. After the circle has rotated through t radians, the position of the point is as given as in figure 7.1. The point of contact of the circle with the line is now t units to the right of the original point of contact (assuming no slippage), so (7.9) x t t sin t y t 1 cos t To find arc length, we use ds 2 dx 2 dy 2 , where dx 1 cos t dt dy sin
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ch7 - CHAPTER 7 Techniques of Integration Integration...

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