39. (a) Use trigonometric substitution to verify that. 40. The parabola y 2x divides the disk into two

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1 35. Prove the formula A r for the area of a sector of a circle with radius r and central angle. [Hint: Assume 0 and place the center of the circle at the origin so it has the equation. Then is the sum of the area of the triangle POQ A and r the area of the region PQR in the figure.] P SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS (a) Use trigonometric substitution to verif that sa t dt 0 a sin a sa (b) Use the figure to give trigonometric interpretations of both terms on the right side of the equation in part (a). a =œ a@-t@ ; 36. Evaluate the integral d 4 s Graph the integrand and its indefinite integral on the same screen and check that our answer is reasonable. ; 37. Use a graph to approimate the roots of the equation. Then approimate the area bounded b the s4 curve s4 and the line. 38. A charged rod of length L produces an electric field at point Pa, b given b EP La b a 40 b 3 d where is the charge densit per unit length on the rod and 0 is the free space permittivit (see the figure). Evaluate the integral to determine an epression for the electric field EP. O P(a,b) Q 0 L R The parabola divides the disk into two parts. Find the areas of both parts Find the area of the crescent-shaped region (called a lune) bounded b arcs of circles with radii r and R. (See the figure.) r R 4. A water storage tank has the shape of a clinder with diameter 0 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft, what percentage of the total capacit is being used? 43. A torus is generated b rotating the circle about the -ais. Find the volume enclosed R b the r torus. t 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS In this section we show how to integrate an rational function (a ratio of polnomials) b epressing it as a sum of simpler fractions, called partial fractions, that we alread know how to integrate. To illustrate the method, observe that b taking the fractions and to a common denominator we obtain 5 If we now reverse the procedure, we see how to integrate the function on the right side of

2 474 CHAPTER 7 TECHNIQUES OF INTEGRATION ++ -) this equation: d ln ln C To see how the method of partial fractions works in general, let s consider a rational function where P and Q are polnomials. It s possible to epress f as sum of simpler fractions provided that the degree of Pis less than the degree of Q. Such a rational function is called proper. Recall that if P an n an n a a0 where an 0, then the degree of P is n and we write degp n. If f is improper, that is, degp degq, then we must take the preliminar step of dividing Q into P (b long division) until a remainder R is obtained such that degr degq. The division statement is 5 d f P Q f P R S Q Q where S and R are also polnomials. As the following eample illustrates, sometimes this preliminar step is all that is required. EXAMPLE Find 3. d V SOLUTION Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write 3 d d 3 M 3 ln C The net step is to factor the denominator Q as far as possible. It can be shown that an polnomial Q can be factored as a product of linear factors (of the form a ) and irreducible quadratic factors (of the form a, where b ). For instance, if Q 4 6, we could factor it as b c b 4ac 0 Q The third step is to epress the proper rational function RQ (from Equation ) as a sum of partial fractions of the form A A or B a b i a b c j

3 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 475 A theorem in algebra guarantees that it is alwas possible to do this. We eplain the details for the four cases that occur. CASE I N The denominator Q() is a product of distinct linear factors. This means that we can write Q a ba b ak bk where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there eist constants A, A,..., Ak such that A B N Another method for finding,, and is given in the note after this eample. C R Q A a b Ak a b ak bk These constants can be determined as in the following eample. EXAMPLE Evaluate 3. 3 d V SOLUTION Since the degree of the numerator is less than the degree of the denominator, we don t need to divide. We factor the denominator as Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand () has the form A B C 3 To determine the values of A, B, and C, we multipl both sides of this equation b the product of the denominators,, obtaining A 4 A B C Epanding the right side of Equation 4 and writing it in the standard form for polnomials, we get 5 A B C 3A B C A The polnomials in Equation 5 are identical, so their coefficients must be equal. The coefficient of on the right side,, must equal the coefficient of on the left side namel, A B C. Likewise, the coefficients of are equal and the constant terms are equal. This gives the following sstem of equations for A, B, and C: A B C 3A B C A B C

4 476 CHAPTER 7 TECHNIQUES OF INTEGRATION N We could check our work b taking the terms to a common denominator and adding them. N Figure shows the graphs of the integrand in Eample and its indefinite integral (with K 0). Which is which? _3 FIGURE _ 3 Solving, we get A, B 5, and C 0, and so 3 3 d 5 0 d ln 0ln 0ln K In integrating the middle term we have made the mental substitution u, which gives du d and d du. M NOTE We can use an alternative method to find the coefficients A, B, and C in Eample. Equation 4 is an identit; it is true for ever value of. Let s choose values of that simplif the equation. If we put 0in Equation 4,then the second and third terms on the right side vanish and the equation then becomes A, or. Likewise, gives and gives, so and. (You ma object that 5B4 Equation 3 is not B valid for 0,,or,so wh should A C Equation 0C be valid for those values? In fact,equation 4 is true for all values of,even, 0,and. See Eercise69 for the reason.) EXAMPLE 3 Find d, where a 0. a SOLUTION The method of partial fractions gives a a a A a B a and therefore A a B a Using the method of the preceding note, we put a in this equation and get Aa, so A a. If we put a, we get Ba, so B a. Thus a a d a (ln a ln a ) C Since ln ln ln, we can write the integral as d a a ln 6 a C See Eercises for was of using Formula 6. CASE N Q() d a a is a product of linear factors, some of which are repeated. a b r Q Aa b Suppose the first linear factor is repeated times; that is, a b occurs in the factorization of. Then instead of the single term in Equation r, we M

5 N Another method for finding the coefficients: Put in (8): B. Put : C. Put :. 0 A B C SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 477 would use a b Ar 7 a b B wa of illustration, we could write 3 A 3 B C D but we prefer to work out in detail a simpler eample. EXAMPLE 4 Find d A SOLUTION The first step is to divide. The result of long division is The second step is to factor the denominator Q. Since Q 0, we know that is a factor and we obtain 3 3 Since the linear factor occurs twice, the partial fraction decomposition is 4 A B C Multipling b the least common denominator,, we get 8 4 A B C A C B C A B C Now we equate coefficients: A B C 0 A B C 4 A B C 0 Solving, we obtain A, B, and C, so d ln d ln K A a b r E 3 ln K M

6 478 CHAPTER 7 TECHNIQUES OF INTEGRATION CASE III N Q() contains irreducible quadratic factors, none of which is repeated. If Q has the factor a, where, then, in addition to the partial fractions in Equations b c b and 7, the epression 4ac 0 for RQ will have a term of the form A B 9 a b c where A and B are constants to be determined. For instance, the function given b f 4 has a partial fraction decomposition of the form A B 4 C D E 4 The term given in (9) can be integrated b completing the square and using the formula 0 d a a tan a C EXAMPLE 5 Evaluate 4 V 3. 4 d SOLUTION Since can t be factored further, we write 4 A B C 4 4 Multipling b 4, we have 4 A 4 B C A B C 4A Equating coefficients, we obtain A B C 4A 4 Thus A, B, and C and so d 4 d In order to integrate the second term we split it into two parts: 4 d 4 d 4 d We make the substitution u in the first of these integrals so that du d. We evaluate the second integral 4 b means of Formula 0 with a : 4 d 4 d 4 d 4 d ln ln 4 tan K M

7 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 479 EXAMPLE 6 Evaluate d SOLUTION Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain Notice that the quadratic 4 is irreducible because its discriminant is b. This means it can t be factored, so we don t need to use the partial 4ac fraction technique. To integrate the given function we complete the square in the denominator: This suggests that we make the substitution u. Then, and du d u, so d d u u du 4 u u du u 4 u du 4 u du 8lnu 4 s u tan sc 8ln4 4 3 M 4 s tan s C NOTE Eample 6 illustrates the general procedure for integrating a partial fraction of the form A B where b 4ac a 0 b c We complete the square in the denominator and then make a substitution that brings the integral into the form Cu D u a du u C u a du D u a du Then the first integral is a logarithm and the second is epressed in terms of tan. contains a repeated irreducible quadratic factor. CASE IV N Q() If Q has the factor a, where, then instead of the single partial fraction (9), the sum b c r b 4ac 0 A B A a b c B Ar a Br b c a b c r

8 480 CHAPTER 7 TECHNIQUES OF INTEGRATION N It would be etremel tedious to work out b hand the numerical values of the coefficients in Eample 7. Most computer algebra sstems, however, can find the numerical values ver quickl. For instance, the Maple command convertf, parfrac, or the Mathematica command Apart[f] gives the following values: A, B 8, C D, E 5 8, F 8, G H 3 4, I, J N In the second and fourth terms we made the mental substitution u. occurs in the partial fraction decomposition of integrated b first completing the square.. Each of the terms in () can be EXAMPLE 7 Write out the form of the partial fraction decomposition of the function 3 3 SOLUTION 3 3 A B C D RQ E F G H I J 3 EXAMPLE 8 Evaluate 3. d SOLUTION The form of the partial fraction decomposition is 3 A B C D E Multipling b, we have 3 A B C D E A 4 B 4 C 3 D E A B 4 C 3 A B D C E A If we equate coefficients, we get the sstem A B 0 C A B D C E A which has the solution A, B, C, D, and E 0. Thus 3 d d d d d d ln ln tan M K We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral d 3 M

9 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 48 could be evaluated b the method of Case III, it s much easier to observe that if u 3 3 3, then du 3 3 d and so d 3 3ln 3 3 C RATIONALIZING SUBSTITUTIONS Some nonrational functions can be changed into rational functions b means of appropriate substitutions. In particular, when an integrand contains an epression of the form s, then the substitution ma be effective. Other instances appear in the eercises. n t u s n t EXAMPLE 9 Evaluate s 4 d. SOLUTION Let u s 4. Then u, so and. Therefore 4 u 4 d u du s 4 d u u 4 u du u 4 u 4 du u 4 du We can evaluate this integral either b factoring u as u u and using partial fractions or b using Formula 6 with a 4 : s 4 d du du 8 u u 8 ln 4 u C s 4 ln s M 4 C 7.4 EXERCISES 6 Write out the form of the partial fraction decomposition of the function (as in Eample 7). Do not determine the numerical values of the coefficients.. (a) (b) (a) (b) 3. (a) 4 (b) (a) 3 (b) (a) (b) 6. (a) 3 (b) Evaluate the integral. 6 d 9 5 d t t t 4 t r r 4 dr t 4t dt

10 48 CHAPTER 7 TECHNIQUES OF INTEGRATION d a b d d 4 7 d 3 5 d d d 0 d 9 3 d 4 5 d 3 d d d d d 5 6 d ds s s d d d d d Make a substitution to epress the integrand as a rational function and then evaluate the integral. d s d s s 0 s 3 d 9 4 d 3 s d s 3 d [Hint: Substitute u s s 6.] s 3 d s s 46. d d a d b d 4 3 d 3 3 d d e e 3e d cos sin sin d sec t tan t 3 tan t dt e e e d 5 5 Use integration b parts, together with the techniques of this section, to evaluate the integral. 5. ln d 5. tan d ; 53. Use a graph of f to decide whether is positive or negative. Use the graph to give a rough estimate 0 f d 3 of the value of the integral and then use partial fractions to find the eact value. ; 54. Graph both and an antiderivative on the same screen Evaluate the integral b completing the square and using Formula 6. d d The German mathematician Karl Weierstrass (85 897) noticed that the substitution t tan will convert an rational function of sin and cos into an ordinar rational function of t. (a) If t tan,, sketch a right triangle or use trigonometric identities to show that cos s t and sin t (b) Show that cos (c) Show that t t and sin t t d t dt s t 58 6 Use the substitution in Eercise 57 to transform the integrand into a rational function of t and then evaluate the integral. d sin sin 3 sin cos d 4 cos d