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Further Techniques and App Icat Ions Chapter 10 Further Techniques and App icat ions Some simple geometric problems require advanced methods of integration. Besides the basic methods of integration associated with reversing the differen- tiation rules, there are special methods for integrands of particular forms. Using these methods, we can solve some interesting length and area problems. 10.1 Trigonometric Integrals The key to evaluating many integrals is a trigonometric identity or substitution. The integrals treated in this section fall into two groups. First, there are some purely trigonometric integrals that can be evaluated using trigonometric identities. Second, there are integrals involving quadratic functions and their square roots which can be evaluated using trigonometric substitutions. We begin by considering integrals of the form J'sinmx cosnxdx, where m and n are integers. The case n = 1 is easy, for if we let u = sinx, we find I sinm+ ' (x) J'sinmxcosxdx= umdu=- +C= +C J' m+l m+l (or lnlsinxl + C, if m = - 1). The case m = 1 is similar: c0sn+ l(x) Jsin x cosnxdx = - +C n+l (or - lnlcosxl + C, if n = - 1). If either m or n is odd, we can use the identity sin2x + cos2x = 1 to reduce the integral to one of the types just treated. Example 1 Evaluate Jsin2x cos3x dx. Solution Jsin2x cos3x dx = Jsin2x cos2xcosx dx = J(sin2x)(l - sin2x)cosx dx, which can be integrated by the substitution u = sinx. We get Copyright 1985 Springer-Verlag. All rights reserved. 458 Chapter 10 Further Techniques and Applications of Integration If m = 2k and n = 21 are both even, we can use the half-angle formulas sin2x = (1 - cos2x)/2 and cos2x = (1 + cos2x)/2 to write where y = 2x. Multiplying this out, we are faced with a sum of integrals of the form Jcosmydy, with m ranging from zero to k + I. The integrals for odd m can be handled by the previous method; to those with even m we apply the half-angle formula once again. The whole process is repeated as often as necessary until everything is integrated. Example 2 Evaluate Jsin2x cos2x dx. Solution J sin2xcos2x dx = 1 +Y~x)~, To evaluate Jsinmxcosnx dx: 1. If m is odd, write m = 2k + 1, and k = J ( 1 - cos2x) cosnxsin x dx. Now integrate by substituting u = cosx. 2. If n is odd, write n = 21 + 1, and Now integrate by substituting u = sinx. 3. (a) If m and n are even, write m = 2k and n = 21 and Substitute y = 2x. Expand and apply step 2 to the odd powers of (b) Apply step 3(a) to the even powers of cosy and continue until the integration is completed. Copyright 1985 Springer-Verlag. All rights reserved. 10.1 Trigonometric Integrals 459 Example 3 Evaluate: (a) sin4x cos2x dx (b) I(sin2x + sin3x cos2x)dx. (c) Stan3@sec38 dB. Solution (a) Substitute sin2x = (1 - cos 2x)/2 and cos2x = (1 + cos 2x)/2 to get (1 - cos 2~)~(1 + cos 2x) Jsin4x cos2xdx = I4 2 dx = -1 l(l- 2 cos 2x cos22x)(1 cos 2x) dx 8 + + 1 = - l(l - cos y - cos? cos?) dy, 16 + where y = 2x. Integrating the last two terms gives and sin? Icos$dy= (I-sin cosydy=siny- -+ C. I 4) 3 Thus sin 2y sin? ~sin4xcos2xdx= - siny - 2 - -+ sin y - - 16 2 4 3 sin2y sin? sin 4x sin32x ) + c, 16 4 3 and so L2'sin4x cos2x dx = . =8 (b) I(sin2x + sin3xcos2x) dx = Isin2x dx + Isin3x cos2x dx =I(1-~2x)dx + I(1 - cos2x)cos2xsin x dx - x 2 4 I(1- u2)u2du (u = cosx) x sin2x cos3x + cos5x + C. 2 4 3 5 (c) Method 1. Rewrite in terms of sec 8 and its derivative tan 8 sec 8: Itan38sec30d0 = I(tan 8sec 0 )(tan2@sec28 ) dB Copyright 1985 Springer-Verlag. All rights reserved. 460 Chapter 10 Further Techniques and Applications of Integration Method 2. Convert to sines and cosines: sin 0(1 - cos20) J tan3e sec3edo = do = do c0s6e Certain other integration problems yield to the use of the addition formulas: sin(x + y) = sin x cos y + cos x sin y, (la) cos(x + y) = cos x cos y - sinx sin y (9 and the product formulas: sinxcos = 4 [sin(x - y) + sin(x + y)], (2a) sinx sin y = t [cos(x - y) - cos(x + y)], (2b) COSX COS y = + [COS(X- y) + COS(X+ y)]. (2~) Exarnple 4 Evaluate (a) Jsin x cos 2x dx and (b) Jcos 3x cos 5x dx. Solution (a) Jsinx cos2xdx= (sin3x - sinx) dx= - -cos3x + cosx + c. 6 2 (see product formula (2a)). (b) ~cos3xcos5xdx= (cos8x + cos2x)dx= + +C 16 4 (see product formula (2c)). la Example 5 Evaluate ssinax sin bx dx, where a and b are constants. Solution If we use identity (2b), we get 1 Isin ax sin bx dx = - J [cos(a - b)x - cos(a + b)x] dx 2 1 sin(a - b)x sin(a + b)x - - - +C ifafkb, 2 a-b 2 a+b - 1 - - - sin 2ax C, if a = b, 2 4a + 1sin 2ax - 4a 2 + C, ifa= -b. [The difference between the case a .f + b and the other two should be noted. The first case is "pure oscillation" in that it consists of two sine terms. The others contain the nonoscillating linear term x/2, called a secular term. This example is related to the phenomena of resonance: when an oscillating system is subjected to a sinusoidally varying force, the oscillation will build up indefinitely if the force has the same frequency as the oscillator. See Review Exercise 56, Chapter 8 and the discussion in the last part of Section 12.7, following equation (14).] A Copyright 1985 Springer-Verlag. All rights reserved. 10.1 Trigonometric Integrals 461 Many integrals containing factors of the form dm,d=, or a2+ x2 can be evaluated or simplified by means of trigonometric substitutions. In order to remember what to substitute, it is useful to draw the appropriate right-angle triangle, as in the following box. Trigonometric Substitutions 1. If occurs, try x = a sin 6; then dx = a cos 6 dB and = a cos6; (a > 0 and 6 is an acute angle). 4asin.=. acose ZJP-2 2. If d- occurs, try x = a sec 6; then dx = a tan 6 sec6 d6 and dm= a tan@. A =utane a 3. If or a2+ x2 occurs, try x = a tan6; then dx = a sec26d6 and = a sec 6 (one can also use x = a sinh 6; then {a- = a cosh 6). @T7 =asece Aatan0 = x U Example 6 Evaluate: (a) dx Solution (a) Let x = 3 sin 6, so d9 - x2 = 3 cos 6. Thus dx = 3 cos 6 d6 and In the last line, we used the first figure in the preceding box to get the identity cot 6 = Jm/xwith a = 3. Copyright 1985 Springer-Verlag. All rights reserved. 462 Chapter 10 Further Techniques and Applications of Integration (b) Let x = $ sec 8, so dx = 4tan 8 sec 8 dB, and JG= tan 8. Thus = I J tan'sec' do = 2 tan 8 Here is a trick' for evaluating sec 8: Jsec 8 d8 = sec 8 sec 8 + tan 8 de = J sec28 + sec 8 tan 8 sec8 + tan8 sec8 + tan8 = lnlsec8 + tan81 + C (substituting u = sec8 + tan8). A Thus Figure 10.1.1. Geometry of (see Fig. lo. the substitution x = + sec 6. If you have studied the hyperbolic functions you should note that this integral can also be evaluated by means of the formula ~[du/Jm] = cosh- 'u + C. A These examples show that trigonometric substitutions work quite well in the presence of algebraic integrands involving square roots. You should also keep in mind the possibility of a simple algebraic substitution or using the direct integration formulas involving inverse trigonometric and hyperbolic functions. Example 7 Evaluate: X dx; Solution (a) Let u = 4 - x2, so du = - 2x dx. Thus (no trigonometric function appears). = 1 + ) + C (see p. 396). You may use the method of Example 6(b) if you are not familiar with hyperbolic functions. (c) To evaluate ](x2/J=) dx, let x = 2 sin 8; then dx = 2 cos 8 dB, and = 2 cos 8. Thus x 2 cos 0 d8 = 4 sin28d8 dx=12coss. J =4J -y2'd8=28- sin28+ C = 28 - 2 sin 8cos 8 + C. ' The same trick shows that Jcsch @dB= -1nlcsch B + coth 81 + C. Copyright 1985 Springer-Verlag. All rights reserved. 10.1 Trigonometric Integrals 463 " A From Fig. 10.1.2 we get Figure 10.1.2. Geometry of the substitution x = 2 sin 0. Completing the square can be useful in simplifying integrals involving the expression ax2 + bx + c. The following two examples illustrate the method. Example 8 Evaluate J dx 10+4x-x Solution To complete the square, write 10 + 4x - x2 = - (x + a)2 + b; solving for a and b, we find a = -2 and b = 14, so 10 + 4x - x2 = -(x - 2)2 + 14. Hence where u = x - 2. This integral is sin-'(u/m) + C, so our final answer is If an integral involves ax2 + bx + c, complete the square and then use a trigonometric substitution or some other method to evaluate the integral. Example 9 Evaluate (a) dx ~olution (a) J dx = dx x2+ x + 1 J (x + 1/2)~+ 3/4 Copyright 1985 Springer-Verlag. All rights reserved. 464 Chapter 10 Further Techniques and Applications of Integration Applications of the kind encountered in earlier chapters may involve integrals of the type in this section. Here is an example. Example 10 Find the average value of sin2xcos2x on the interval [O, 2771.
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