Trigonometric Integrals - Stanford University

Trigonometric IntegralsIn this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine.EXAMPLE 1 Evaluate3y cos x dx.SOLUTION Simply substituting u cos x isn’t helpful, since then du sin x dx. In orderto integrate powers of cosine, we would need an extra sin x factor. Similarly, a power ofsine would require an extra cos x factor. Thus, here we can separate one cosine factorand convert the remaining cos2x factor to an expression involving sine using the identitysin 2x cos 2x 1:cos 3x cos 2x cos x 1 sin 2x cos xWe can then evaluate the integral by substituting u sin x, so du cos x dx andy cos x dx y cos x cos x dx y 1 sin x cos x dx322 y 1 u 2 du u 13 u 3 C sin x 13 sin 3x CIn general, we try to write an integrand involving powers of sine and cosine in a formwhere we have only one sine factor (and the remainder of the expression in terms ofcosine) or only one cosine factor (and the remainder of the expression in terms of sine).The identity sin 2x cos 2x 1 enables us to convert back and forth between even powersof sine and cosine.EXAMPLE 2 Find52y sin x cos x dxSOLUTION We could convert cos 2x to 1 sin 2x, but we would be left with an expression interms of sin x with no extra cos x factor. Instead, we separate a single sine factor andrewrite the remaining sin 4x factor in terms of cos x :sin 5x cos 2x sin2x 2 cos 2x sin x 1 cos 2x 2 cos 2x sin xFigure 1 shows the graphs of the integrandsin 5x cos 2x in Example 2 and its indefinite integral (with C 0). Which is which? Substituting u cos x, we have du sin x dx and soy sin x cos x dx y sin x 5222cos 2x sin x dx0.2 y 1 cos 2x 2 cos 2x sin x dxππ0.2FIGURE 1 y 1 u 2 2 u 2 du y u 2 2u 4 u 6 du u3u5u7 2 357 C 13 cos 3x 25 cos 5x 17 cos 7x C1

2 TRIGONOMETRIC INTEGRALSIn the preceding examples, an odd power of sine or cosine enabled us to separate asingle factor and convert the remaining even power. If the integrand contains even powersof both sine and cosine, this strategy fails. In this case, we can take advantage of the following half-angle identities (see Equations 17b and 17a in Appendix C):sin 2x 12 1 cos 2x EXAMPLE 3 Evaluatey 0cos 2x 12 1 cos 2x andsin 2x dx.SOLUTION If we write sin 2x 1 cos 2x, the integral is no simpler to evaluate. Using thehalf-angle formula for sin 2x, however, we havey 0 [ (x sin 2x dx 12 y 1 cos 2x dx 12012 0]sin 2x) 12 ( 12 sin 2 ) 12 (0 12 sin 0) 12 Notice that we mentally made the substitution u 2x when integrating cos 2x. Anothermethod for evaluating this integral was given in Exercise 33 in Section 5.6.1.5y sin@ xExample 3 shows that the area of theregion shown in Figure 2 is 2. π0FIGURE 20.5EXAMPLE 4 Find4y sin x dx.SOLUTION We could evaluate this integral using the reduction formula forx sin n x dx(Equation 5.6.7) together with Example 3 (as in Exercise 33 in Section 5.6), but a bettermethod is to write sin 4x sin 2x 2 and use a half-angle formula:y sin x dx y sin x dx42 y 21 cos 2x2 2dx 14 y 1 2 cos 2x cos 2 2x dxSince cos 2 2x occurs, we must use another half-angle formulacos 2 2x 12 1 cos 4x This givesy sin x dx y 1 2 cos 2x 414 14 y12 1 cos 4x dx( 32 2 cos 2x 12 cos 4x) dx 4 ( 2 x sin 2x 8 sin 4x) C1 31To summarize, we list guidelines to follow when evaluating integrals of the formx sin mx cos nx dx, where m 0 and n 0 are integers.

TRIGONOMETRIC INTEGRALS 3Strategy for Evaluatingy sinmx cos nx dx(a) If the power of cosine is odd n 2k 1 , save one cosine factor and usecos 2x 1 sin 2x to express the remaining factors in terms of sine:y sinmx cos 2k 1x dx y sin m x cos 2x k cos x dx y sin m x 1 sin 2x k cos x dxThen substitute u sin x.(b) If the power of sine is odd m 2k 1 , save one sine factor and usesin 2x 1 cos 2x to express the remaining factors in terms of cosine:y sinx cos n x dx y sin 2x k cos n x sin x dx2k 1 y 1 cos 2x k cos n x sin x dxThen substitute u cos x. [Note that if the powers of both sine and cosine areodd, either (a) or (b) can be used.](c) If the powers of both sine and cosine are even, use the half-angle identitiessin 2x 12 1 cos 2x cos 2x 12 1 cos 2x It is sometimes helpful to use the identitysin x cos x 12 sin 2xWe can use a similar strategy to evaluate integrals of the form x tan mx sec nx dx. Since d dx tan x sec 2x, we can separate a sec 2x factor and convert the remaining (even)power of secant to an expression involving tangent using the identity sec 2x 1 tan 2x.Or, since d dx sec x sec x tan x, we can separate a sec x tan x factor and convert theremaining (even) power of tangent to secant.EXAMPLE 5 Evaluate64y tan x sec x dx.SOLUTION If we separate one sec 2x factor, we can express the remaining sec 2x factor interms of tangent using the identity sec 2x 1 tan 2x. We can then evaluate the integralby substituting u tan x with du sec 2x dx :y tan x sec x dx y tan x sec x sec x dx64622 y tan 6x 1 tan 2x sec 2x dx y u 6 1 u 2 du y u 6 u 8 du u7u9 C79 17 tan 7x 19 tan 9x C

4 TRIGONOMETRIC INTEGRALSEXAMPLE 6 Findy tan sec d .57SOLUTION If we separate a sec 2 factor, as in the preceding example, we are left witha sec 5 factor, which isn’t easily converted to tangent. However, if we separate asec tan factor, we can convert the remaining power of tangent to an expressioninvolving only secant using the identity tan 2 sec 2 1. We can then evaluate theintegral by substituting u sec , so du sec tan d :y tan 5sec 7 d y tan 4 sec 6 sec tan d y sec 2 1 2 sec 6 sec tan d y u 2 1 2 u 6 du y u 10 2u 8 u 6 du u 11u9u7 2 C1197 111 sec 11 29 sec 9 17 sec 7 CThe preceding examples demonstrate strategies for evaluating integrals of the formx tan mx sec nx dx for two cases, which we summarize here.Strategy for Evaluatingy tanmx sec nx dx(a) If the power of secant is even n 2k, k 2 , save a factor of sec 2x and usesec 2x 1 tan 2x to express the remaining factors in terms of tan x :y tanmx sec 2kx dx y tan m x sec 2x k 1 sec 2x dx y tan m x 1 tan 2x k 1 sec 2x dxThen substitute u tan x.(b) If the power of tangent is odd m 2k 1 , save a factor of sec x tan x anduse tan 2x sec 2x 1 to express the remaining factors in terms of sec x :y tanx sec n x dx y tan 2x k sec n 1x sec x tan x dx2k 1 y sec 2x 1 k sec n 1x sec x tan x dxThen substitute u sec x.For other cases, the guidelines are not as clear-cut. We may need to use identities, integration by parts, and occasionally a little ingenuity. We will sometimes need to be able tointegrate tan x by using the formula established in Example 5 in Section 5.5:y tan x dx ln sec x C

TRIGONOMETRIC INTEGRALS 5We will also need the indefinite integral of secant:y sec x dx ln sec x tan x C1We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x tan x :sec x tan xy sec x dx y sec x sec x tan x dx ysec 2x sec x tan xdxsec x tan xIf we substitute u sec x tan x, then du sec x tan x sec 2x dx, so the integralbecomes x 1 u du ln u C. Thus, we have y sec x dx ln sec x tan x CEXAMPLE 7 Find3y tan x dx.SOLUTION Here only tan x occurs, so we use tan 2x sec 2x 1 to rewrite a tan 2x factor interms of sec 2x :y tan x dx y tan x tan x dx32 y tan x sec 2x 1 dx y tan x sec 2x dx y tan x dx tan 2x ln sec x C2 In the first integral we mentally substituted u tan x so that du sec 2x dx.If an even power of tangent appears with an odd power of secant, it is helpful to expressthe integrand completely in terms of sec x. Powers of sec x may require integration byparts, as shown in the following example.EXAMPLE 8 Find3y sec x dx.SOLUTION Here we integrate by parts withu sec xdu sec x tan x dxThendv sec 2x dxv tan xy sec x dx sec x tan x y sec x tan x dx32 sec x tan x y sec x sec 2x 1 dx sec x tan x y sec 3x dx y sec x dx

6 TRIGONOMETRIC INTEGRALSUsing Formula 1 and solving for the required integral, we gety sec x dx (sec x tan x ln sec x tan x ) C123Integrals such as the one in the preceding example may seem very special but theyoccur frequently in applications of integration, as we will see in Chapter 6. Integrals ofthe form x cot m x csc n x dx can be found by similar methods because of the identity1 cot 2x csc 2x.Finally, we can make use of another set of trigonometric identities: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 identity:(a) sin A cos B 12 sin A B sin A B These product identities are discussed inAppendix C. (b) sin A sin B 12 cos A B cos A B (c) cos A cos B 12 cos A B cos A B EXAMPLE 9 Evaluatey sin 4x cos 5x dx.SOLUTION This integral could be evaluated using integration by parts, but it’s easier to usethe identity in Equation 2(a) as follows:y sin 4x cos 5x dx y12 sin x sin 9x dx 12 y sin x sin 9x dx 12 (cos x 19 cos 9x CExercisesA Click here for answers.1–47Evaluate the integral.321.y sin x cos x dx3.y 3 4 2sin 5x cos 3x dx545.y cos x sin x dx7.y9.11.13.Click here for solutions.S 20y 0cos2 d sin 4 3t dt2y 404.yd sin 4x cos 2x dx 20cos 5x dx6.y sin mx dx8.y12.14. 2y 0sin 2 2 d cos6 d 2y x cos x dxy 20sin 2x cos 2x dx18.y cot sin d 20.y cos x sin 2x dx22.y224.y tan x dx626.ytan 5 x sec 4 x dx28.y tan 2x sec 2x dx330.y32.y tan ay dy17.y cos x tan x dx19.y21.y sec x tan x dx23.y tan x dx25.y sec t dt27.y29.y tan x sec x dx31.y tan x dx231 sin xdxcos x230y cos cos sin d y sin x scos x dx3y sin x cos x dx10.y 1 cos 62.16.315. 3055542 20sec 4 t 2 dt4 40sec 4 tan 4 d 3 305tan 5x sec6x dx6

TRIGONOMETRIC INTEGRALS 7tan 3 33.y cos d 35.y 4 2 6cot 2x dx; 57–58Use a graph of the integrand to guess the value of theintegral. Then use the methods of this section to prove that yourguess is correct.234.y tan x sec x dx36.y 2 4cot 3x dx57.y2 0 37.y cot 3 csc 3 d 38.y csc 4 x cot 6 x dx39.y csc x dx40.y 41. 3 642.y sin 5x sin 2x dx43.y cos 7 cos 5 d 45.y1 tan xdxsec 2x47.y t sec tycos x sin xdxsin 2x2 2 62. y cos x, y 0, x 0, x 2;about y 151. y sec 4 65–67 53. Find the average value of the function f x sin 2x cos 3x onthe interval , .54. Evaluate x sin x cos x dx by four methods: (a) the substitutionProve the formula, where m and n are positive integers.y sin mx cos nx dx 066.y sin mx sin nx dx 67.y cos mx cos nx dx 0 if m nif m nif m nif m n0 68. A finite Fourier series is given by the sumNf x ansin nxn 1y sin 3x,56. y sin x,y 2 sin x, a 1 sin x a 2 sin 2x a N sin NxFind the area of the region bounded by the given curves.55. y sin x, x 0,2 x 2 Show that the mth coefficient a m is given by the formulax 2x 0, am 65. u cos x, (b) the substitution u sin x, (c) the identitysin 2x 2 sin x cos x, and (d) integration by parts. Explain thedifferent appearances of the answers.55–56 where t is the time in seconds. Voltmeters read the RMS (rootmean-square) voltage, which is the square root of the averagevalue of E t 2 over one cycle.(a) Calculate the RMS voltage of household current.(b) Many electric stoves require an RMS voltage of 220 V.Find the corresponding amplitude A needed for the voltageE t A sin 120 t .xdx2 current that varies from 155 V to 155 V with a frequencyof 60 cycles per second (Hz). The voltage is thus given bythe equationE t 155 sin 120 t 4 64. Household electricity is supplied in the form of alternatingy sin x cos x dx52.y sin 3x sin 6x dx 4 f 0 0.Evaluate the indefinite integral. Illustrate, and check thatyour answer is reasonable, by graphing both the integrand and itsantiderivative (taking C 0 .50. about y 1; 49–525 61. y cos x, y 0, x 0, x 2;x0 4 tan 8 x sec x dx in terms of I.y sin x dx about the x-axis48. If x0 4 tan 6 x sec x dx I , express the value of49. 63. A particle moves on a straight line with velocity functionv t sin t cos 2 t. Find its position function s f t ify cos x 1 60. y tan 2x, y 0, x 0, x 4;2 sin 2 x cos 5 x dxabout the x-axis tan t dt4 2059. y sin x, x 2, x , y 0; dx46. yFind the volume obtained by rotating the region boundedby the given curves about the specified axis.csc 3x dx2 58.59–62y sin 3x cos x dx44. cos 3x dx 1 y f x sin mx dx