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# ch06_04 - 530 CHAPTER 6 Integration Techniques 6-22 how...

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530 CHAPTER 6 . . Integration Techniques 6-22 how this works, suppose a second station broadcasts the sig- nal 3 sin t at frequency 32. The combined signal that the radio receives is 2 sin t sin 16 t + 3 sin t sin 32 t . We will decompose this signal. The first step is to rewrite the signal using the identity sin A sin B = 1 2 cos( B A ) 1 2 cos( B + A ) . The signal then equals f ( t ) = cos 15 t cos 17 t + 3 2 cos 31 t 3 2 cos 33 t . If the radio “knows” that the signal has the form c sin t , for some constant c , it can determine the constant c at frequency 16 by computing the integral π π f ( t ) cos 15 t dt and multi- plying by 2 . Show that π π f ( t ) cos 15 t dt = π , so that the correct constant is c = π (2 ) = 2. The signal is then 2 sin t . To recover the signal sent out by the second station, compute π π f ( t ) cos 31 t dt and multiply by 2 . Show that you cor- rectly recover the signal 3 sin t . 2. In this exercise, we derive an important result called Wallis’ product. Define the integral I n = π/ 2 0 sin n xdx for a pos- itive integer n . (a) Show that I n = n n 1 I n 2 . (b) Show that I 2 n + 1 I 2 n = 2 2 4 2 ··· (2 n ) 2 2 3 2 5 2 ··· (2 n 1) 2 (2 n + 1) π . (c) Conclude that π 2 = lim n →∞ 2 2 4 2 ··· (2 n ) 2 3 2 5 2 ··· (2 n 1) 2 (2 n + 1) . 6.4 INTEGRATION OF RATIONAL FUNCTIONS USING PARTIAL FRACTIONS In this section we introduce a method for rewriting certain rational functions that is very useful in integration as well as in other applications. We begin with a simple observation. Note that 3 x + 2 2 x 5 = 3( x 5) 2( x + 2) ( x + 2)( x 5) = x 19 x 2 3 x 10 . (4.1) So, suppose that you wanted to evaluate the integral of the function on the right-hand side of (4.1). While it’s not clear how to evaluate this integral, the integral of the (equivalent) function on the left-hand side of (4.1) is easy to evaluate. From (4.1), we now have x 19 x 2 3 x 10 dx = 3 x + 2 2 x 5 dx = 3 ln | x + 2 | − 2 ln | x 5 | + c . The second integrand, 3 x + 2 2 x 5 is called a partial fractions decomposition of the first integrand. More generally, if the three factors a 1 x + b 1 , a 2 x + b 2 and a 3 x + b 3 are all distinct (i.e., none is a constant multiple of another), then we can write a 1 x + b 1 ( a 2 x + b 2 )( a 3 x + b 3 ) = A a 2 x + b 2 + B a 3 x + b 3 , for some choice of constants A and B to be determined. Notice that if you wanted to integrate this expression, the partial fractions on the right-hand side are very easy to integrate, just as they were in the introductory example just presented. EXAMPLE 4.1 Partial Fractions: Distinct Linear Factors Evaluate 1 x 2 + x 2 dx .

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6-23 SECTION 6.4 . . Integration of Rational Functions Using Partial Fractions 531 Solution First, note that you can’t evaluate this as it stands and all of our earlier methods fail to help. (Consider each of these for this problem.) However, we can make a partial fractions decomposition, as follows.
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