y sec 3 x dx sec x tan x y sec x tan 2 x dx y sec 3 x dx 1 2(sec x tan x ln sec x tan x ) C

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1 SECTION 7. TRIGONOETRIC INTEGRLS 465 Then sec 3 sec tan sec tan sec tan sec sec sec tan sec 3 sec Using ormula and solving for the required integral, we get sec 3 (sec tan ln sec tan ) C Integrals such as the one in the preceding eample ma seem ver special but the occur frequentl in applications of integration, as we will see in Chapter 8. Integrals of the form cot m csc n can be found b similar methods because of the identit cot csc. inall, we can make use of another set of trigonometric identities: N These product identities are discussed in ppendi D. To evaluate the integrals (a) sin m cos n, sin m sin n, or (c) cos m cos n, use the corresponding identit: (a) sin cos B sin B sin B sin sin B cos B cos B (c) cos cos B cos B cos B EXPLE 9 Evaluate sin 4 cos 5. SOLUTION This integral could be evaluated using integration b parts, but it s easier to use the identit in Equation (a) as follows: sin 4 cos 5 sin sin 9 sin sin 9 (cos 9 cos 9 C 7. EXERCISES 49 Evaluate the integral.. sin 3 cos. sin 6 cos sin cos sin 5 cos 3 cos d cos 5 sin3 (s ) s sin d 9.. sin4 3t dt. cos d. cos 3. sin cos 5. cos5 6. cos cos 5 sin d ssin d 4. cos6 d sin t cos 4 t dt

2 466 CHPTER 7 TECHNIQUES O INTEGRTION 7. cos tan cos sin sin.. sec tan. 3. tan sec 6 t dt tan 5 sec tan 3 sec tan tan3 34. cos 4 d cot 5 sin 4 d cos sin 4 3 sec 4 t dt tan tan 4 sec 4 tan 4 d tan 3 sec 5 tan 5 sec 6 tan 6 a d tan sec 53. sin 3 sin sec ind the average value of the function f sin cos 3 on the interval,. 56. Evaluate sin cos b four methods: (a) the substitution u cos the substitution u sin (c) the identit sin sin cos (d) integration b parts Eplain the different appearances of the answers ind the area of the region bounded b the given curves. 57. sin, cos, 58. sin 3, cos 3, ; 59 6 Use a graph of the integrand to guess the value of the integral. Then use the methods of this section to prove that our guess is correct. 35. sec tan 36. sin cos 3 d 59. cos 3 6. sin cos cot cot 3 csc 3 d csc sin 8 cos cot 3 csc 4 cot csc3 cos cos ind the volume obtained b rotating the region bounded b the given curves about the specified ais. 6. sin,, ; about the -ais 6. sin,, ; about the -ais 63. sin, cos, 4; about 64. sec, cos, 3; about 45. sin 5 sin d tan 48. sec 49. t sec t tan 4 t dt 5. If 4 tan 6 sec I, epress the value of 4 tan 8 sec in terms of I. ; 5 54 Evaluate the indefinite integral. Illustrate, and check that our answer is reasonable, b graphing both the integrand and its antiderivative (taking C. 5. sin 5. sin 3 cos 4 cos sin sin cos 65. particle moves on a straight line with velocit function vt sin t cos t. ind its position function s f t if f. 66. Household electricit is supplied in the form of alternating current that varies from 55 V to 55 V with a frequenc of 6 ccles per second (Hz). The voltage is thus given b the equation Et 55 sint where t is the time in seconds. Voltmeters read the RS (root-mean-square) voltage, which is the square root of the average value of Et over one ccle. (a) Calculate the RS voltage of household current. an electric stoves require an RS voltage of V. ind the corresponding amplitude needed for the voltage Et sint.

3 47 CHPTER 7 TECHNIQUES O INTEGRTION N igure 5 shows the graphs of the integrand in Eample 7 and its indefinite integral (with C ). Which is which? 3 We now substitute u sin, giving du cos d and s4 u cos, so sin cos d s3 cos _4 sin d cos C s4 u sin u C IGURE 5 _5 s3 sin C 7.3 EXERCISES 3 Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle.. ; s 9. 3 s9 ; 3. 3 ; s Evaluate the integral. 3 sec 3 sin 3 tan..6 s s s 7. s s 4 3. s dt st 6t cos t s sin t dt 4. s3 3 s s t 3 st dt s 3. (a) Use trigonometric substitution to show that s a ln( s a ) C 7. s5 9.. s s t 5 st dt Use the hperbolic substitution a sinh t to show that s a sinh a C. s sa s 9 3 a s 7 8. s 9.. s 4 3 s3 du us5 u 5 s9 a b 3 t s5 t dt 3. Evaluate These formulas are connected b ormula a 3 (a) b trigonometric substitution. b the hperbolic substitution a sinh t. 33. ind the average value of f s, ind the area of the region bounded b the hperbola and the line 3.

4 SECTION 7.4 INTEGRTION O RTIONL UNCTIONS BY PRTIL RCTIONS 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 and so 3 3 ln 3 3 C RTIONLIZING 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 n t, then the substitution u s n t ma be effective. Other instances appear in the eercises. s 4 EXPLE 9 Evaluate. SOLUTION Let u s 4. Then u 4, so u 4 and u du. Therefore s 4 u u 4 u du 4 u 4 du u u 4 du We can evaluate this integral either b factoring u 4 as u u and using partial fractions or b using ormula 6 with a : s 4 du 8 u 8 du u 4 s 4 ln ln u u C s 4 s 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) 33. (a) 4 3. (a) (a) (a) (a) Evaluate the integral t 4 t t t r r 4 dr t 4t dt

5 48 CHPTER 7 TECHNIQUES O INTEGRTION.. a b ake a substitution to epress the integrand as a rational function and then evaluate the integral s 4. 6 s [Hint: Substitute u s 6.] s s d s 3 s s s a b ds s s s 3 s 5 6 opcional Use integration b parts, together with the techniques of this section, to evaluate the integral. 5. ln 5. tan ; 53. Use a graph of f 3 to decide whether f is positive or negative. Use the graph to give a rough estimate of the value of the integral and then use partial fractions to find the eact value. ; 54. Graph both 3 and an antiderivative on the same screen Evaluate the integral b completing the square and using ormula 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 and sin t s t s t Show that (c) Show that opcional 58 6 Use the substitution in Eercise 57 to transform the integrand into a rational function of t and then evaluate the integral. 58. e e 3e cos sin sin sec t tan t 3 tan t dt e e e 3 5 sin 56. cos t t and sin t t t dt sin 4 cos sin cos

6 PPENDIX I NSWERS TO ODD-NUBERED EXERCISES 93 EXERCISES 6.5 N PGE e (a), 4 (c) (a) (c).4, cos sin cos 3 C 9. ln C. t arctan 4t 8 ln 6t C 3. t tan t 4 lnsec t C 5. ln ln C 7. 3 e sin 3 3 cos 3 C e ln 4 e 7. 6 ( 6 3s3) 9. sin ln sin C ln 64 5 ln s sin s cos s C ln C 39. e C 7 ƒ kgm L CHPTER 6 REVIEW N PGE 446 Eercises ( )(cos 3 ah h 3 3 4) 5. (a) 5 6 (c) (a) Solid obtained b rotating the region cos, about the -ais. Solid obtained b rotating the region, sin about the -ais s3 m J 9. (a) ft-lb. ft 3. f PROBLES PLUS N PGE s 6.5 min. (a) f t 3t f s (c).6736 m (d) (i) 5.3 ins (ii) (a) V h f d (c) f skc 4 dvantage: the markings on the container are equall spaced. 3. b a 5. B 6 CHPTER 7 EXERCISES 7. N PGE ln 9 3 C r e r C sin 5 5 cos 5 C cos sin sin C 45. 3, ln 3 3ln 6 ln 6 C e ,.873; e 6. ln e t t t m 65. EXERCISES 7. N PGE C. 5 cos 5 3 cos 3 C sin 3 sin 5 sin 7 C ssin 45 8 sin 5 sin 4 C 7. cos ln cos C 9. ln sin sin C. tan C 3. tan C tan 5 t 3 tan 3 t tan t C sec 3 sec C 3. 4 sec 4 tan ln sec C tan 6 4 tan 4 C 35. sec ln sec tan C 37. s ln csc cot C 3 csc 3 5 csc 5 C sin 4 6 cos 3 6 cos 3 C sin 6 C 47. sin C 49. tan 5 t C 3 _ 5 f 4 _ sin 4 sin C

7 94 PPENDIX I NSWERS TO ODD-NUBERED EXERCISES sin cos C s cos 3t3 EXERCISES 7.3 N PGE 47. s 99 C s 9 C 5. 4 s s5 5 C 9. ln(s 6 ) C. 4 sin s 4 C 3. 6 sec 3 s 9 C 5. 6a ln. 9 4 s 7 C (s ) s C 3. sin 3 s5 4 C 5. s ln(s ) C 7. s ln s C 9. 4 sin 4 s 4 C (s48 sec 7) 37..8, ;. 4. rsr r r R arcsinrr 43. EXERCISES 7.4 N PGE 48. (a) 3 B 3 3. (a) B C D E B 3 5. (a) B C D t B Ct D t t 4 Et t ln 6 C 9. ln 5 ln C ln ln 9 5 ln 3 (or 9 5 ln 8 3) _π a ln b C 36 ln 5 6 ln 4 tan C ln 3 ln C ln ln 9 3 tan 3 C ln 5 3 tan 3 ln 6 ln tan s3 4 ln f π _π C 3 B D ln C C ln (s ) tan (s ) C π 4 6 ln 3 ln 4 6 sin 3 8 sin 9 C _ ln 3 ƒ (s 5 ) 5 Rr C C s3 8 4 C s tan s ln s s C 3 4. ln C 45. s 3s 47. ln 6s e 6 ln s 6 C C e 49. ln tan t ln tan t C 5. ( ) ln s7 tan C s7 53. ln 3.55 tan 55. ln ln tan C C 6. 4 ln ln 65. t ln P 9 ln.9p 9 C, where C.3 4, (a) ,55 3 7,98 48,935 6, ln ln,49 6,5 ln 5 s9 The CS omits the absolute value signs and the constant of integration. EXERCISES 7.5 N PGE 488 sin 3 sin 3 C sin ln csc cot C ln 9 7. e4 e ln ln 4 5 tan C 3. 8 cos 8 6 cos 6 C (or 4 sin 4 3 sin 6 8 sin 8 C) 5. s C 7. 4 sin cos 4 sin C (or 4 4 sin 8 cos C) 9. e e C. arctan s s C ln ln C ln e C ln 7 3. sin s 33. sin C s3 C ln sec ln sec C 4. C e 3 C e 3 C 47. ln 3 s ln s4 ln C s C C 53. m coshm m sinhm 3 coshm C m tan ln sec C 75,77 tan 6,5s ,55 ln 3 7 C

y sin n x dx cos x sin n 1 x n 1 y sin n 2 x cos 2 x dx y sin n xdx cos x sin n 1 x n 1 y sin n 2 x dx n 1 y sin n x dx

y sin n x dx cos x sin n 1 x n 1 y sin n 2 x cos 2 x dx y sin n xdx cos x sin n 1 x n 1 y sin n 2 x dx n 1 y sin n x dx SECTION 7. INTEGRATION BY PARTS 57 EXAPLE 6 Prove the reduction formula N Equation 7 is called a reduction formula because the eponent n has been reduced to n and n. 7 sin n n cos sinn n n sin n where