# y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx y 1 u 2 du u 1 3u 3 C

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1 Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE Evaluate cos 3 x dx. SOLUTION Simpl substituting u cos x isn t helpful, since then du sin x dx. In order to integrate powers of cosine, we would need an extra sin x factor. Similarl, a power of sine would require an extra cos x factor. Thus, here we can separate one cosine factor and convert the remaining cos 2 x factor to an expression involving sine using the identit sin 2 x cos 2 x : We can then evaluate the integral b substituting u sin x, so du cos x dx and cos 3 x dx cos 2 x cos x dx sin 2 x cos x dx In general, we tr to write an integrand involving powers of sine and cosine in a form where we have onl one sine factor (and the remainder of the expression in terms of cosine) or onl one cosine factor (and the remainder of the expression in terms of sine). The identit sin 2 x cos 2 x enables us to convert back and forth between even powers of sine and cosine. EXAMPLE 2 Find sin 5 x cos 2 x dx cos 3 x cos 2 x cos x sin 2 x cos x u 2 du u 3u 3 C sin x 3 sin 3 x C SOLUTION We could convert cos 2 x to sin 2 x, but we would be left with an expression in terms of sin x with no extra cos x factor. Instead, we separate a single sine factor and rewrite the remaining sin 4 x factor in terms of cos x: sin 5 x cos 2 x sin 2 x 2 cos 2 x sin x cos 2 x 2 cos 2 x sin x Figure shows the graphs of the integrand sin 5 x cos 2 x in Example 2 and its indefinite integral (with C ). Which is which?.2 Substituting u cos x, we have du sin x dx and so sin 5 x cos 2 x dx sin 2 x 2 cos 2 x sin x dx cos 2 x 2 cos 2 x sin x dx _π π u 2 2 u 2 du u 2 2u 4 u 6 du FIGUE _.2 u u 5 5 u 7 C 7 3 cos 3 x 2 5 cos 5 x 7 cos 7 x C

2 2 TIGONOMETIC INTEGALS In the preceding examples, an odd power of sine or cosine enabled us to separate a single factor and convert the remaining even power. If the integrand contains even powers of both sine and cosine, this strateg fails. In this case, we can take advantage of the following half-angle identities (see Equations 7b and 7a in Appendix C): sin 2 x 2 cos 2x and cos 2 x 2 cos 2x EXAMPLE 3 Evaluate. sin2 x dx SOLUTION If we write sin 2 x cos 2 x, the integral is no simpler to evaluate. Using the half-angle formula for sin 2 x, however, we have sin2 x dx 2 cos 2x dx [ 2(x 2 sin 2x)] 2( 2 sin 2) 2( 2 sin ) 2 Notice that we mentall made the substitution u 2x when integrating cos 2x. Another method for evaluating this integral was given in Exercise 33 in Section x Example 3 shows that the area of the region shown in Figure 2 is 2. π FIGUE 2 _.5 EXAMPLE 4 Find sin 4 x dx. SOLUTION We could evaluate this integral using the reduction formula for x sinnx dx (Equation 5.6.7) together with Example 3 (as in Exercise 33 in Section 5.6), but a better method is to write sin 4 x sin 2 x 2 and use a half-angle formula: sin 4 x dx sin 2 x 2 dx cos 2x cos 2x cos 2 2x dx Since cos 2 2x occurs, we must use another half-angle formula 2 dx This gives cos 2 2x 2 cos 4x sin 4 x dx 4 2 cos 2x 2 cos 4x dx 4 ( cos 2x 2 cos 4x) dx 4( 3 2 x sin 2x 8 sin 4x) C To summarize, we list guidelines to follow when evaluating integrals of the form x sin m x cos n x dx,where m and n are integers.

3 TIGONOMETIC INTEGALS 3 Strateg for Evaluating sin m x cos n x dx (a) If the power of cosine is odd n 2k, save one cosine factor and use cos 2 x sin 2 x to express the remaining factors in terms of sine: sin m x cos 2k x dx sin m x cos 2 x k cos x dx sin m x sin 2 x k cos x dx Then substitute u sin x. (b) If the power of sine is odd m 2k, save one sine factor and use sin 2 x cos 2 x to express the remaining factors in terms of cosine: sin 2k x cos n x dx sin 2 x k cos n x sin x dx cos 2 x k cos n x sin x dx Then substitute u cos x. [Note that if the powers of both sine and cosine are odd, either (a) or (b) can be used.] (c) If the powers of both sine and cosine are even, use the half-angle identities sin 2 x 2 cos 2x cos 2 x 2 cos 2x It is sometimes helpful to use the identit sin x cos x 2 sin 2x We can use a similar strateg to evaluate integrals of the form x tan m x sec n x dx. Since ddx tan x sec 2 x, we can separate a sec 2 x factor and convert the remaining (even) power of secant to an expression involving tangent using the identit sec 2 x tan 2 x. Or, since ddx sec x sec x tan x, we can separate a sec x tan x factor and convert the remaining (even) power of tangent to secant. EXAMPLE 5 Evaluate tan 6 x sec 4 x dx. SOLUTION If we separate one sec 2 x factor, we can express the remaining sec 2 x factor in terms of tangent using the identit sec 2 x tan 2 x. We can then evaluate the integral b substituting u tan x with du sec 2 x dx: tan 6 x sec 4 x dx tan 6 x sec 2 x sec 2 x dx tan 6 x tan 2 x sec 2 x dx u 6 u 2 du u 6 u 8 du u 7 7 u 9 9 C 7 tan 7 x 9 tan 9 x C

4 4 TIGONOMETIC INTEGALS EXAMPLE 6 Find tan 5 sec 7 d. SOLUTION If we separate a factor, as in the preceding example, we are left with a sec 5 sec 2 factor, which isn t easil converted to tangent. However, if we separate a sec tan factor, we can convert the remaining power of tangent to an expression involving onl secant using the identit tan 2 sec 2. We can then evaluate the integral b substituting u sec, so du sec tan d: tan 5 sec 7 d tan 4 sec 6 sec tan d sec 2 2 sec 6 sec tan d u 2 2 u 6 du u 2u 8 u 6 du u 2 u 9 9 u 7 7 C sec 2 9 sec 9 7 sec 7 C The preceding examples demonstrate strategies for evaluating integrals of the form x tan m x sec n x dx for two cases, which we summarize here. Strateg for Evaluating tan m x sec n x dx (a) If the power of secant is even n 2k, k 2, save a factor of sec 2 x and use sec 2 x tan 2 x to express the remaining factors in terms of tan x: tan m x sec 2k x dx tan m x sec 2 x k sec 2 x dx tan m x tan 2 x k sec 2 x dx Then substitute u tan x. (b) If the power of tangent is odd m 2k, save a factor of sec x tan x and use tan 2 x sec 2 x to express the remaining factors in terms of sec x: tan 2k x sec n x dx tan 2 x k sec n x sec x tan x dx sec 2 x k sec n x sec x tan x dx Then substitute u sec x. For other cases, the guidelines are not as clear-cut. We ma need to use identities, integration b parts, and occasionall a little ingenuit. We will sometimes need to be able to integrate tan x b using the formula established in Example 5 in Section 5.5: tan x dx ln sec x C

5 TIGONOMETIC INTEGALS 5 We will also need the indefinite integral of secant: sec x dx ln sec x tan x C We could verif Formula b differentiating the right side, or as follows. First we multipl numerator and denominator b sec x tan x: If we substitute u sec x tan x, then du sec x tan x sec 2 x dx, so the integral becomes. Thus, we have x u du ln u C sec x dx sec x sec x tan x sec x tan x dx sec2 x sec x tan x sec x tan x sec x dx ln sec x tan x C dx EXAMPLE 7 Find tan 3 x dx. SOLUTION Here onl tan x occurs, so we use tan 2 x sec 2 x to rewrite a tan 2 x factor in terms of sec 2 x: In the first integral we mentall substituted u tan x so that du sec 2 x dx. If an even power of tangent appears with an odd power of secant, it is helpful to express the integrand completel in terms of sec x. Powers of sec x ma require integration b parts, as shown in the following example. EXAMPLE 8 Find sec 3 x dx. tan 3 x dx tan x tan 2 x dx tan x sec 2 x dx tan x sec 2 x dx tan x dx tan2 x 2 ln sec x C SOLUTION Then Here we integrate b parts with u sec x dv sec 2 x dx du sec x tan x dx v tan x sec 3 x dx sec x tan x sec x tan 2 x dx sec x tan x sec x sec 2 x dx sec x tan x sec 3 x dx sec x dx

6 6 TIGONOMETIC INTEGALS Using Formula and solving for the required integral, we get sec 3 x dx 2 (sec x tan x ln sec x tan x ) C Integrals such as the one in the preceding example ma seem ver special but the occur frequentl in applications of integration, as we will see in Chapter 6. Integrals of the form x cot m x csc n x dx can be found b similar methods because of the identit cot 2 x csc 2 x. Finall, we can make use of another set of trigonometric identities: These product identities are discussed in Appendix C. 2 To evaluate the integrals (a) x sin mx cos nx dx, (b) x sin mx sin nx dx, or (c) x cos mx cos nx dx, use the corresponding identit: (a) sin A cos B 2sinA B sina B (b) sin A sin B 2cosA B cosa B (c) cos A cos B 2cosA B cosa B EXAMPLE 9 Evaluate sin 4x cos 5x dx. SOLUTION This integral could be evaluated using integration b parts, but it s easier to use the identit in Equation 2(a) as follows: sin 4x cos 5x dx 2 sinx sin 9x dx 2 sin x sin 9x dx 2(cos x 9 cos 9x C Exercises A Click here for answers. S Click here for solutions. 5. sin 3 x scos x dx 6. cos cos 5 sin d 47 Evaluate the integral. 7. cos 2 x tan 3 x dx 8. cot 5 sin 4 d. sin 3 x cos 2 x dx sin 5 x cos 3 x dx 4. sin 6 x cos 3 x dx 2 cos 5 x dx 9. sin x dx cos x sec 2 x tan x dx 22. cos 2 x sin 2x dx 2 sec 4 t2 dt 5. cos 5 x sin 4 x dx 6. sin 3 mx dx 23. tan 2 x dx 24. tan 4 x dx 2 7. cos 2 d 8. 2 sin 2 2 d 25. sec 6 t dt sec 4 tan 4 d 9.. sin4 3t dt cos6 d tan 5 x sec 4 x dx 28. tan 3 2x sec 5 2x dx. cos 2 d 2. x cos 2 x dx 29. tan 3 x sec x dx 3. 3 tan 5 x sec 6 x dx 4 3. sin 4 x cos 2 x dx 4. 2 sin 2 x cos 2 x dx 3. tan 5 x dx 32. tan 6 a d

7 TIGONOMETIC INTEGALS tan3 34. cos 4 2 d cot2 x dx tan 2 x sec x dx 2 4 cot3 x dx ; 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 cos 3 x dx sin 2x cos 5x dx 37. cot 3 csc 3 d 38. csc 4 x cot 6 x dx 39. csc x dx sin 5x sin 2x dx 42. sin 3x cos x dx 43. cos 7 cos 5 d tan2 x dx 46. sec 2 x If x4 tan 6 x sec x dx I, express the value of ; Evaluate the indefinite integral. Illustrate, and check that our answer is reasonable, b graphing both the integrand and its antiderivative (taking C. 49. sin 5 x dx 5. sin 4 x cos 4 x dx 5. sin 3x sin 6x dx 52. sec x 4 2 dx 53. Find the average value of the function f x sin 2 x cos 3 x on the interval,. 54. Evaluate x sin x cos x dx b four methods: (a) the substitution u cos x, (b) the substitution u sin x, (c) the identit sin 2x 2 sin x cos x, and (d) integration b parts. Explain the different appearances of the answers Find the area of the region bounded b the given curves. 55. sin x, sin 3 x, x, x t sec 2 t 2 tan 4 t 2 dt x4 tan 8 x sec x dx in terms of I. sin x, 2 sin 2 x, x, x csc 3 x dx cos x sin x sin 2x dx cos x dx Find the volume obtained b rotating the region bounded b the given curves about the specified axis. 59. sin x, x 2, x, ; about the x-axis 6. tan 2 x,, x, x 4; about the x-axis 6. cos x,, x, x 2; about 62. cos x,, x, x 2; about 63. A particle moves on a straight line with velocit function vt sin t cos 2t. Find its position function s f t if f. 64. 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 sin2t where t is the time in seconds. Voltmeters read the MS (rootmean-square) voltage, which is the square root of the average value of Et 2 over one ccle. (a) Calculate the MS voltage of household current. (b) Man electric stoves require an MS voltage of 22 V. Find the corresponding amplitude A needed for the voltage Et A sin2t Prove the formula, where m and n are positive integers sin mx cos nx dx sin mx sin nx dx cos mx cos nx dx 68. A finite Fourier series is given b the sum f x N n a n sin nx a sin x a 2 sin 2x a N sin Nx Show that the mth coefficient a m if m n if m n if m n if m n a m is given b the formula f x sin mx dx

8 8 TIGONOMETIC INTEGALS Answers S Click here for solutions.. 5 cos 5 x 3 cos 3 x C sin 5 x 2 7 sin 7 x 9 sin 9 x C ( 2 7 cos 3 x 2 3 cos x)scos x C 7. 2 cos 2 x ln cos x C 9. ln sin x C 2. 2 tan 2 x C 23. tan x x C tan 5 t 2 3 tan 3 t tan t C sec 3 x sec x C 3. 4 sec 4 x tan 2 x ln sec x C tan s csc 3 4 tan 4 C 5 csc 5 C 39. ln csc x cot x C sin 2 6 sin 3x 4 sin 7x C 24 sin 2 C sin 2x C 47. tan 5 t 2 C 2 2 sin 4 sin 2 C cos 5 x 2 3 cos 3 x cos x C _2π 6 sin 3x 8 sin 9x C F _ s cos 3t3.. ƒ ƒ F 2π 2 2 4

9 TIGONOMETIC INTEGALS 9 Solutions: Trigonometric Integrals The smbols s = and c = indicate the use of the substitutions {u =sinx, du =cosxdx} and {u =cosx, du = sin xdx}, respectivel.. sin 3 x cos 2 xdx = sin 2 x cos 2 x sin xdx= cos 2 x cos 2 x sin xdx c = u 2 u 2 ( du) = u 2 u 2 du = u 4 u 2 du = 5 u5 3 u3 + C = 5 cos5 x 3 cos3 x + C 3. 3π/4 sin 5 x cos 3 xdx = 3π/4 sin 5 x cos 2 x cos xdx= 3π/4 sin 5 x sin 2 x cos xdx π/2 π/2 π/2 s = 2/2 u 5 u 2 du = 2/2 ³ = /8 / = 384 u 5 u 7 du = 6 u6 8 u8 2/2 5. cos 5 x sin 4 xdx = cos 4 x sin 4 x cos xdx= sin 2 x 2 sin 4 x cos xdx s = u 2 2 u 4 du = 2u 2 + u 4 u 4 du = u 4 2u 6 + u 8 du = 5 u5 2 7 u7 + 9 u9 + C = 5 sin5 x 2 7 sin7 x + 9 sin9 x + C 7. π/2 cos 2 θ dθ = π/2 ( + cos 2θ) dθ [half-angle identit] 2 = 2 θ + 2 sin 2θ π/2 = π ( + ) = π 4 9. π sin4 (3t) dt = π sin 2 (3t) 2 dt = π ( cos 2 6t) 2 dt = π ( 2cos6t 4 +cos2 6t) dt = π 4 2cos6t + ( + cos 2t) dt = π 3 2cos6t + cos 2t dt = 3 t sin 6t + sin 2t π = 3π + ( +) = 3π ( + cos θ) 2 dθ = ( + 2 cos θ +cos 2 θ) dθ = θ +2sinθ + 2 ( + cos 2θ) dθ = θ +2sinθ + 2 θ + 4 sin 2θ + C = 3 2 θ +2sinθ + 4 sin 2θ + C 3. π/4 sin 4 x cos 2 xdx = π/4 sin 2 x (sin x cos x) 2 dx = π/4 = π/4 8 = π/4 6 π/4 2 ( cos 2x) 2 sin 2x 2 dx ( cos 2x) sin 2 2xdx= sin 2 2xdx 8 8 ( cos 4x) dx 6 3 sin3 2x π/4 = 6 = 6 π 4 3 = 92 (3π 4) π/4 sin 2 2x cos 2xdx x sin 4x 4 3 sin3 2x π/4 5. sin 3 x cos xdx = cos 2 x cos x sin xdx c = u 2 u /2 ( du) = ³ u 5/2 u /2 du 7. = 2 7 u7/2 2 3 u3/2 + C = 2 7 (cos x)7/2 2 3 (cos x)3/2 + C = 2 7 cos3 x 2 cos x cos x + C 3 sin cos 2 x tan 3 3 x u xdx = cos x dx = c 2 ( du) = u u + u du = ln u + 2 u2 + C = 2 cos2 x ln cos x + C

10 TIGONOMETIC INTEGALS 9. Or: sin x cos x dx = sin x cos x dx = (sec x tan x) dx =ln sec x +tanx ln sec x + C =ln (sec x +tanx)cosx + C =ln +sinx + C =ln(+sinx)+c since +sinx sin x +sinx cos x +sinx dx = = dw w sin 2 x dx cos x ( + sin x) = [where w =+sinx, dw =cosxdx] =ln w + C =ln +sinx + C =ln(+sinx)+c b () and the boxed formula above it cos xdx +sinx 2. Let u =tanx, du =sec 2 xdx.then sec 2 x tan xdx= udu= 2 u2 + C = 2 tan2 x + C. Or: Let v =secx, dv =secx tan xdx.then sec 2 x tan xdx= vdv= 2 v2 + C = 2 sec2 x + C. 23. tan 2 xdx= sec 2 x dx =tanx x + C 25. sec 6 tdt = sec 4 t sec 2 tdt= (tan 2 t +) 2 sec 2 tdt= (u 2 +) 2 du [u =tant, du =sec 2 tdt] = (u 4 +2u 2 +)du = 5 u u3 + u + C = 5 tan5 t tan3 t +tant + C 27. π/3 tan 5 x sec 4 xdx = π/3 tan 5 x (tan 2 x +)sec 2 xdx = 3 u 5 (u 2 +)du [u =tanx, du =sec 2 xdx] = 3 (u 7 + u 5 ) du = 8 u8 + 6 u6 3 = = = = 7 8 Alternate solution: π/3 tan 5 x sec 4 xdx = π/3 tan 4 x sec 3 x sec x tan xdx= π/3 (sec 2 x ) 2 sec 3 x sec x tan xdx = 2 (u2 ) 2 u 3 du [u =secx, du =secx tan xdx] = 2 (u4 2u 2 +)u 3 du = 2 (u7 2u 5 + u 3 ) du = 8 u8 3 u6 + 4 u4 2 = = tan 3 x sec xdx = tan 2 x sec x tan xdx= sec 2 x sec x tan xdx = (u 2 ) du [u =secx, du =secx tan xdx] = 3 u3 u + C = 3 sec3 x sec x + C 3. tan 5 xdx = sec 2 x 2 tan xdx= sec 4 x tan xdx 2 sec 2 x tan xdx+ tan xdx = sec 3 x sec x tan xdx 2 tan x sec 2 xdx+ tan xdx = 4 sec4 x tan 2 x +ln sec x + C [or 4 sec4 x sec 2 x +ln sec x + C ] 33. tan 3 θ cos 4 θ dθ = tan 3 θ sec 4 θ dθ = tan 3 θ (tan 2 θ +) sec 2 θ dθ = u 3 (u 2 +)du [u =tanθ, du =sec 2 θ dθ] = (u 5 + u 3 ) du = 6 u6 + 4 u4 + C = 6 tan6 θ + 4 tan4 θ + C 35. π/2 π/6 cot2 xdx= π/2 π/6 csc 2 x dx =[ cot x x] π/2 π/6 = π 2 3 π 6 = 3 π cot 3 α csc 3 α dα = cot 2 α csc 2 α csc α cot α dα = (csc 2 α ) csc 2 α csc α cot α dα = (u 2 )u 2 ( du) [u =cscα, du = csc α cot α dα] = (u 2 u 4 ) du = 3 u3 5 u5 + C = 3 csc3 α 5 csc5 α + C

11 TIGONOMETIC INTEGALS 39. I = csc xdx= csc x (csc x cot x) csc x cot x dx = csc x cot x +csc 2 x csc x cot x du = csc x cot x +csc 2 x dx. ThenI = du/u =ln u =ln csc x cot x + C. dx. Let u =cscx cot x 4. Use Equation 2(b): sin 5x sin 2xdx = [cos(5x 2x) cos(5x +2x)] dx = 2 2 (cos 3x cos 7x) dx = 6 sin 3x 4 sin 7x + C 43. Use Equation 2(c): cos 7θ cos 5θ dθ = 2 [cos(7θ 5θ)+cos(7θ +5θ)] dθ = 2 (cos 2θ +cos2θ) dθ 45. tan 2 x sec 2 x = 2 2 sin 2θ + 2 sin 2θ + C = 4 sin 2θ + 24 sin 2θ + C cos dx = 2 x sin 2 x dx = cos 2xdx= sin 2x + C Let u =tan(t 2 ) du =2t sec 2 (t 2 ) dt. Then t sec 2 t 2 tan 4 t 2 dt = u 4 du = 2 u5 + C = tan5 (t 2 )+C. 49. Let u =cosx du = sin xdx. Then sin 5 xdx= cos 2 x 2 sin xdx= u 2 2 ( du) = +2u 2 u 4 du = 5 u u3 u + C = 5 cos5 x cos3 x cos x + C Notice that F is increasing when f (x) >, so the graphsserve as a check on our work. 5. sin 3x sin 6xdx = [cos(3x 6x) cos(3x +6x)] dx 2 (cos 3x cos 9x) dx = 2 = sin 3x sin 9x + C 6 8 Notice that f(x) =whenever F has a horizontal tangent. 53. f ave = π 2π π sin2 x cos 3 xdx= 2π u2 u 2 du = 2π = π π sin2 x sin 2 x cos xdx [where u =sinx] 55. For <x< π,wehave < sin x<, 2 sosin3 x<sin x. Hence the area is π/2 sin x sin 3 x dx = π/2 sin x sin 2 x dx = π/2 cos 2 x sin xdx.nowletu =cosx du = sin xdx. Thenarea = u2 ( du) = u2 du = 3 u3 = It seems from the graph that 2π cos 3 xdx=, since the area below the x-axis and above the graph looks about equal to the area above the axis and below the graph. B Example, the integral is sin x 3 sin3 x 2π =. Note that due to smmetr, the integral of an odd power of sin x or cos x between limits which differ b 2nπ (n an integer) is.

12 2 TIGONOMETIC INTEGALS 59. V = π π π/2 sin2 xdx= π π π/2 2 ( cos 2x) dx = π 2 x 4 sin 2x π = π π π/2 2 π 4 + = π2 4 2cosx +cos 2 x dx 6. Volume = π π/2 ( + cos x) 2 2 dx = π π/2 = π 2sinx + 2 x + 4 sin 2x π/2 = π 2+ π 4 =2π + π s = f(t) = t sin ωu cos2 ωudu.let =cosωu d = ω sin ωudu.then s = cos ωt 2 d = 3 cos ωt = ω ω 3 3ω cos 3 ωt. 65. Just note that the integrand is odd [f( x) = f(x)]. Or: If m 6= n, calculate π sin mx cos nx dx = π [sin(m n)x +sin(m + n)x] dx π π 2 = 2 cos(m n)x m n If m = n, then the first term in each set of brackets is zero. π cos(m + n)x = m + n π 67. π cos mx cos nx dx = π π π 2 [cos(m n)x +cos(m + n)x] dx. Ifm 6= n, this is equal to π sin(m n)x sin(m + n)x + =.Ifm = n, weget 2 m n m + n π π π 2 [ + cos(m + n)x] dx = π 2 x π sin(m + n)x + = π +=π. π 2(m + n) π

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