13.6. Integration of Trigonometric Functions. Introduction. Prerequisites. Learning Outcomes

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1 Itegratio of Trigoometric Fuctios 13.6 Itroductio Itegrals ivolvig trigoometric fuctios are commoplace i egieerig mathematics. This is especially true whe modellig waves ad alteratig curret circuits. Whe the root-mea-square (rms) value of a waveform, or sigal is to be calculated, you will ofte fid this results i a itegral of the form si t dt I this Sectio you will lear how such itegrals ca be evaluated. Prerequisites Before startig this Sectio you should... Learig Outcomes O completio you should be able to... be able to fid a umber of simple defiite ad idefiite itegrals be able to use a table of itegrals be familiar with stadard trigoometric idetities use trigoometric idetities to write itegrads i alterative forms to eable them to be itegrated 48 HELM (8):

2 1. Itegratio of trigoometric fuctios Simple itegrals ivolvig trigoometric fuctios have already bee dealt with i Sectio See what you ca remember: Write dow the followig itegrals: (a) si x dx, (b) cos x dx, (c) si x dx, (d) cos x dx (a) (b) (c) (d) (a) cos x + c, (b) si x + c, (c) 1 cos x + c, (d) 1 si x + c. The basic rules from which these results ca be derived are summarised here: cos kx si kx dx = + c k Key Poit 8 cos kx dx = si kx k + c I egieerig applicatios it is ofte ecessary to itegrate fuctios ivolvig powers of the trigoometric fuctios such as si x dx or cos ωt dt Note that these itegrals caot be obtaied directly from the formulas i Key Poit 8 above. However, by makig use of trigoometric idetities, the itegrads ca be re-writte i a alterative form. It is ofte ot clear which idetities are useful ad each case eeds to be cosidered idividually. Experiece ad practice are essetial. Work through the followig. HELM (8): Sectio 13.6: Itegratio of Trigoometric Fuctios 49

3 Use the trigoometric idetity si θ 1 (1 cos θ) to express the itegral si x dx i a alterative form ad hece evaluate it. (a) First use the idetity: Your solutio si x dx = The itegral ca be writte 1 (1 cos x)dx. Note that the trigoometric idetity is used to covert a power of si x ito a fuctio ivolvig cos x which ca be itegrated directly usig Key Poit 8. (b) Now evaluate the itegral: 1 ( x 1 si x + c) = 1x 1 si x + K where K = c/. 4 Use the trigoometric idetity si x si x cos x to fid si x cos x dx (a) First use the idetity: Your solutio si x cos x dx = The itegrad ca be writte as 1 si x (b) Now evaluate the itegral: π si x cos x dx = π 1 si x dx = [ 14 ] π cos x + c = 1 4 cos 4π cos = = This result is oe example of what are called orthogoality relatios. 5 HELM (8):

4 Egieerig Example 3 Magetic flux Itroductio The magitude of the magetic flux desity o the axis of a soleoid, as i Figure 13, ca be foud by the itegral: B = β β 1 µ I si β dβ where µ is the permeability of free space ( 4π 1 7 H m 1 ), is the umber of turs ad I is the curret. β β 1 Problem i words Figure 13: A soleoid ad agles defiig its extet Predict the magetic flux i the middle of a log soleoid. Mathematical statemet of the problem We assume that the soleoid is so log that β 1 ad β π so that B = β β 1 µ I Mathematical aalysis si β dβ π µ I si β dβ The factor µ I ca be take outside the itegral i.e. B = µ π I si β dβ = µ [ ] π I cos β = µ I ( cos π + cos ) = µ I ( ( 1) + 1) = µ I Iterpretatio The magitude of the magetic flux desity at the midpoit of the axis of a log soleoid is predicted to be approximately µ I i.e. proportioal to the umber of turs ad proportioal to the curret flowig i the soleoid. HELM (8): Sectio 13.6: Itegratio of Trigoometric Fuctios 51

5 . Orthogoality relatios I geeral two fuctios f(x), g(x) are said to be orthogoal to each other over a iterval a x b if b a f(x)g(x) dx = It follows from the previous that si x ad cos x are orthogoal to each other over the iterval x π. This is also true over ay iterval α x α + π (e.g. π/ x 5π, or π x π). More geerally there is a whole set of orthogoality relatios ivolvig these trigoometric fuctios o itervals of legth π (i.e. over oe period of both si x ad cos x). These relatios are useful i coectio with a widely used techique i egieerig, kow as Fourier aalysis where we represet periodic fuctios i terms of a ifiite series of sies ad cosies called a Fourier series. (This subject is covered i 3.) We shall demostrate the orthogoality property I m = π si mx si x dx = where m ad are itegers such that m. The secret is to use a trigoometric idetity to covert the itegrad ito a form that ca be readily itegrated. You may recall the idetity si A si B 1 (cos(a B) cos(a + B)) It follows, puttig A = mx ad B = x that provided m π I m = 1 [cos(m )x cos(m + )x] dx = 1 [ ] π si(m )x si(m + )x (m ) (m + ) = because (m ) ad (m + ) will be itegers ad si(iteger π) =. Of course si =. Why does the case m = have to be excluded from the aalysis? (left to the reader to figure out!) The correspodig orthogoality relatio for cosies J m = π cos mx cos x dx = follows by use of a similar idetity to that just used. Here agai m ad are itegers such that m. 5 HELM (8):

6 Example 3 Use the idetity si A cos B 1 (si(a + B) + si(a B)) to show that K m = π si mx cos x dx = m ad itegers, m. Solutio π K m = 1 [si(m + )x + si(m )x] dx = 1 [ ] π cos(m + )x cos(m )x (m + ) (m ) = 1 [ ] cos(m + )π 1 cos(m )π 1 + = (m + ) (m ) (recallig that cos(iteger π) = 1) Derive the orthogoality relatio K m = π si mx cos x dx = m ad itegers, m = Hit: You will eed to use a differet trigoometric idetity to that used i Example 3. HELM (8): Sectio 13.6: Itegratio of Trigoometric Fuctios 53

7 K m = π si mx cos mx dx Puttig m =, ad the usig the idetity si A si A cos A we get K mm = = 1 = 1 π π Puttig m = = gives K = 1 si mx cos mx dx si mx dx [ ] π cos mx = 1 1 (cos 4mπ cos ) = (1 1) = m 4m 4m π si cos dx =. Note that the particular case m = = 1 was cosidered earlier i this Sectio. 3. Reductio formulae You have see earlier i this Workbook how to itegrate si x ad si x (which is (si x) ). Applicatios sometimes arise which ivolve itegratig higher powers of si x or cos x. It is possible, as we ow show, to obtai a reductio formula to aid i this. Give I = si (x) dx write dow the itegrals represeted by I, I 3, I 1 I = I 3 = I 1 = I = si x dx I 3 = si 3 x dx I 1 = si 1 x dx To obtai a reductio formula for I we write si x = si 1 (x) si x ad use itegratio by parts. 54 HELM (8):

8 I the otatio used earlier i this Workbook for itegratio by parts (Key Poit 5, page 31) put f = si 1 x ad g = si x ad evaluate df dx ad g dx. df dx = ( 1) si x cos x (usig the chai rule of differetiatio), g dx = si x dx = cos x Now use the itegratio by parts formula o secod itegral that you obtai.] si 1 x si x dx. [Do ot attempt to evaluate the df si 1 x si x dx = si 1 (x) g dx g dx dx = si 1 (x)( cos x) + ( 1) si x cos x dx We ow eed to evaluate si x cos xdx. Puttig cos x = 1 si x this itegral becomes: si (x) dx si (x) dx But this is expressible as I I so fially, usig this ad the result from the last we have I = si 1 (x) si x dx = si 1 (x)( cos x) + ( 1)(I I ) from which we get Key Poit 9: HELM (8): Sectio 13.6: Itegratio of Trigoometric Fuctios 55

9 Give I = si xdx Key Poit 9 Reductio Formula I = 1 si 1 (x) cos x + 1 I This is our reductio formula for I. It eables us, for example, to evaluate I 6 i terms of I 4, the I 4 i terms of I ad I i terms of I where I = si x dx = 1 dx = x. Use the reductio formula i Key Poit 9 with = to fid I. i.e. I = 1 [si x cos x] + 1 I = 1 [1 si x] + x + c si x dx = 1 4 si x + x + c as obtaied earlier by a differet techique. 56 HELM (8):

10 Use the reductio formula i Key Poit 9 to obtai I 6 = si 6 x dx. Firstly obtai I 6 i terms of I 4, the I 4 i terms of I : Usig Key Poit 9 with = 6 gives I 6 = 1 6 si5 x cos x I 4. The, usig Key Poit 9 agai with = 4, gives I 4 = 1 4 si3 x cos x I Now substitute for I from the previous to obtai I 4 ad hece I 6. I 4 = 1 4 si3 x cos x 3 16 si x + 3 x+ costat 8 I 6 = 1 6 si5 x cos x 5 4 si3 x cos x 5 3 si x x + costat Defiite itegrals ca also be readily evaluated usig the reductio formula i Key Poit 9. example, I = π/ si x dx so I = We obtai, immediately I = 1 [ si 1 (x) cos x ] π/ π/ + 1 I si x dx or, sice cos π = si =, I ( 1) = I This simple easy-to-use formula is well kow ad is called Wallis formula. For HELM (8): Sectio 13.6: Itegratio of Trigoometric Fuctios 57

11 Give I = π/ si x dx or I = Key Poit 1 Reductio Formula - Wallis Formula π/ cos x dx I = ( 1) I If I = π/ si x dx calculate I 1 ad the use Wallis formula, without further itegratio, to obtai I 3 ad I 5. I 1 = π/ [ si x dx = ] π/ cos x = 1 The usig Wallis formula with = 3 ad = 5 respectively I 3 = I 5 = π/ π/ si 3 x dx = 3 I 1 = 3 1 = 3 si 5 x dx = 4 5 I 3 = = HELM (8):

12 The total power P of a atea is give by P = π ηl I π 4λ si 3 θ dθ where η, λ, I are costats as is the legth L of atea. Usig the reductio formula for si x dx i Key Poit 9, obtai P. Igorig the costats for the momet, cosider I 3 = I 1 = π π si 3 θ dθ which we will reduce to I 1 ad evaluate. [ si θ dθ = ] π cos θ = so by the reductio formula with = 3 I 3 = 1 [ ] π si x cos x I 1 = + 3 = 4 3 We ow cosider the actual itegral with all the costats. Hece P = ηl I π 4λ π si 3 θ dθ = ηl I π 4λ 4 3, so P = η L I π 3λ. A similar reductio formula to that i Key Poit 9 ca be obtaied for at the ed of this Workbook). I particular if J = π/ cos x dx the J = ( 1) J i.e. Wallis formula is the same for cos x as for si x. cos x dx (see Exercise 5 HELM (8): Sectio 13.6: Itegratio of Trigoometric Fuctios 59

13 4. Harder trigoometric itegrals The followig seemigly iocet itegrals are examples, importat i egieerig, of trigoometric itegrals that caot be evaluated as idefiite itegrals: (a) si(x ) dx ad cos(x ) dx These are called Fresel itegrals. (b) si x x dx This is called the Sie itegral. Defiite itegrals of this type, which are what ormally arise i applicatios, have to be evaluated by approximate umerical methods. Fresel itegrals with limits arise i wave ad atea theory ad the Sie itegral with limits i filter theory. It is useful sometimes to be able to visualize the defiite itegral. For example cosider F (t) = t Clearly, F () = si x x dx t > si x x si x dx =. Recall the graph of x si x x agaist x, x > : t π π x Figure 14 For ay positive value of t, F (t) is the shaded area show (the area iterpretatio of a defiite itegral was covered earlier i this Workbook). As t icreases from to π, it follows that F (t) icreases from to a maximum value F (π) = π si x x dx whose value could be determied umerically (it is actually about 1.85). As t further icreases from π to π the value of F (t) will decrease to a local miimum at π because the si x curve is below x the x-axis betwee π ad π. Note that the area below the curve is cosidered to be egative i this applicatio. Cotiuig to argue i this way we ca obtai the shape of the F (t) graph i Figure 15: (ca you 6 HELM (8):

14 see why the oscillatios decrease i amplitude?) 1.85 F (t) π π π t Figure 15 si x The result dx = π is clearly illustrated i the graph (you are ot expected to kow x how this result is obtaied). Methods for solvig such problems are dealt with i 31. HELM (8): Sectio 13.6: Itegratio of Trigoometric Fuctios 61

15 Exercises You will eed to refer to a Table of Trigoometric Idetities to aswer these questios. π/ 1. Fid (a) cos xdx (b) cos tdt (c) (cos θ + si θ)dθ. Use the idetity si(a + B) + si(a B) si A cos B to fid 3. Fid (1 + ta x)dx. si 3x cos xdx 4. The mea square value of a fuctio f(t) over the iterval t = a to t = b is defied to be 1 b a b a (f(t)) dt Fid the mea square value of f(t) = si t over the iterval t = to t = π. 5. (a) Show that the reductio formula for J = cos x dx is (b) J = 1 cos 1 (x) si x + ( 1) J Usig the reductio formula i (a) show that cos 5 x dx = 1 5 cos4 x si x cos x si x si x (c) Show that if J = π/ ( ) 1 cos x dx, the J = J (Wallis formula). (d) s Usig Wallis formula show that π/ cos 6 x dx = 5 3 π. 1. (a) 1x + 1 si x + c (b) π/4 (c) θ + c cos 5x 1 cos x + c ta x + c HELM (8):