Lecture 3. Techniques of integration (cont d) Integration by parts (cont d) Relevant section from Stewart, Eighth Edition: 7.1

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1 Lecture 3 Techniques of integrtion (cont d) Integrtion by prts (cont d) Relevnt section from Stewrt, Eighth Edition: 7. In the previous lecture, we derived the following formul: Integrtion by Prts: Formul u(x)v (x)dx = u(x)v(x) u (x)v(x)dx. () This formul is often written in n lterntive fshion, using the differentils of the functions involved in the integrnd, s opposed to functions multiplied by the differentil dx. Here: du = u (x)dx nd dv = v (x)dx, () so tht Formul my be rewritten s Integrtion by Prts: Formul udv = uv vdu. (3) Exmple : Let us revisit Exmple from the previous lecture, i.e., xe x dx, (4) nd show how it is expressed in terms of Formul. First of ll, s before, we ll let u(x) = x or simply u = x, (5) which implies tht du = u (x)dx = dx or simply du = dx. (6) Secondly, we ll let dv = e x dx. (7) But dv = v (x)dx, which implies tht v (x) = e x, (8)

2 which, in turn, implies tht v(x) = e x. (9) Note tht we did not write tht v(x) = e x +C, () where C is n rbitrry constnt. You don t need C we ll show why lter. Formul then becomes udv = uv vdu xe x dx = xe x e x dx = xe x e x +C. () Let s now see wht hppens if we included the rbitrry constnt C in our expression for v(x) bove: udv = uv vdu xe x dx = x[e x +C] [e x +C]dx = xe x +Cx e x Cx+D = xe x e x +D, () where D is n rbitrry constnt. This is the sme result s ws obtined without the use of C. As such, we see tht using the rbitrry constnt C inside the series of steps involved in integrtion by prts is redundnt. (This, of course, is the cse when you use ntiderivtives to evlute definite integrl using FTC II. You cn dd ny constnt you wish to the ntiderivtive to produce nother ntiderivtive. But when you tke the difference of the ntiderivtives evluted t x = nd x = b, this constnt disppers. Exmple 3: Find x e x dx. (3) From the experience gined in Exmples nd, it seems tht letting u = x would be good ide. Integrtion by prts will then produce n integrl with lower power of x. As such, using the differentil nottion of Formul, u = x = du = xdx, (4) 3

3 nd Using Formul, dv = e x dx = v dx = v = e x. (5) udv = uv vdu x e x dx = x e x e x xdx = x e x xe x dx = x e x (xe x e x )+C = x e x xe x +e x +C, (6) where we hve used the results of Exmples nd. Of course, it is lwys good prctice to check the result: which verifies our result. d [ x e x xe x +e x +C ] = xe x +x e x e x xe x +e x dx = x e x, (7) Exmple 4: Find xsinxdx. (8) Once gin, it seems tht it would be good ide if we set u = x so tht du = dx. Then dv = sinxdx = v (x)dx = v(x) = cosx. (9) Then, using Formul, udv = uv vdu xsinxdx = xcosx = xcosx+ ( cosx)dx cosxdx = xcosx+sinx+c. () Once gin, we verify the result, d [ xcosx+sinx+c] = cosx+xsinx+cosx dx = xsinx. () 4

4 Exmple 5: Find lnxdx. () At first sight, this might pper confusing, since there is only one function in the integrnd. How cn we express it s product. The trick is to express the integrnd s lnx = ()(lnx) = (lnx)(). (3) There re two choices for our integrtion by prts procedure. If we let u =, then du =. Furthermore, by letting dv = lnxdx, we would hve to ntidifferentite lnx, which is wht we re trying to do in the first plce! So, in wht might pper to be quite contrdictory to wht we were doing erlier, let s let dv = dx = v (x)dx = v(x) = x, (4) nd u = lnx = du = dx. (5) x Now use Formul, udv = uv vdu lnx = (lnx)(x) = xlnx dx x x dx = xlnx x+c. (6) Let s verify the result: d [xlnx x+c] = lnx+ dx = lnx. (7) Integrtion by prts pplied to definite integrls Let s return to our originl derivtion of the integrtion by prts formul: Given two functions u nd v, we note tht d dx [u(x)v(x)] = u (x)v(x)+u(x)v (x), (8) 5

5 which implies (rther trivilly, of course, but tht s not the point) tht u(x)v(x) is the ntiderivtive of the expression on the RHS. This implies tht, by the use of the Fundmentl Theorem of Clculus II, b [u (x)v(x)+u(x)v (x)]dx = u(x)v(x) We ll rewrite the first eqution s follows, b = u(b)v(b) u()v(). (9) Definite integrtion by prts: Formul b u(x)v (x)dx = u(x)v(x) b b u (x)v(x)dx. (3) This is integrtion by prts s pplied to definite integrls. You ll see tht this is simply Formul evluted t both limits x = nd x = b. It shouldn t be too difficult to see tht Formul for definite integrls becomes Definite integrtion by prts: Formul b udv = uv b b vdu. (3) Exmple 6: Find e lnxdx. (3) We ll use two methods which, of course, re relted to ech other: Method No. : Let s simply use the ntiderivtive of the integrnd, s it ws found in Exmple 5 bove. From FTC II, e lnxdx = xlnx x e = (elne e) (ln ) = e e+ =. (33) Method No. : We ll use the definite integrtion by prts method Formul : u = lnx = du = dx, (34) x 6

6 nd dv = dx = v (x)dx = v = x, (35) so tht e lnxdx = xlnx e e dx = elne ln x e = e (e ) =, (36) in greement with Method No.. Exmple 7: (not done in clss) Find Here, we ll let π xsinxdx. (37) u = x = du = dx, nd dv = sinxdx = v (x)dx = v = cosx, (38) so tht πx sinxdx = xcosx π π ( cosx)dx = πcos(π) ( cos)+ = π +sinx π π cosxdx = π. (39) An interesting fmily of definite integrls which re connected by recursion Let s consider the following fmily of ntiderivtives, sin n xdx, n. (4) Of course, the first couple of ntiderivtives re esy, sin xdx = dx = x+c sin x dx = cosx+c. (4) 7

7 But t n =, the following ntiderivtive, sin xdx, (4) is not obvious. Actully, we ll be deling with trigonometric integrls in the next lecture. Here we wish to show tht integrtion by prts cn be used to estblish reltion between ntiderivtives with different n vlues. Let s express sin n x s product tht cn be treted with integrtion by prts, i.e., sin n xdx = sin n x sinx,dx. (43) We ll now let u = sin n x = du = (n )sin n xcosxdx, (44) so tht integrtion by prts will decrese the exponent of sinx. This implies tht nd dv = sinxdx = v = cosx. (45) Now use IP (integrtion by prts) Formul : udv = uv vdu sin n xdx = sin n xcosx ( cosx)(n )sin n x cosxdx (46) = sin n xcosx+(n ) sin n cos xdx. We ll now rewrite the cos x in the integrnd s cos x = sin x, to obtin the eqution, sin n xdx = sin n xcosx+(n ) sin n xdx (n ) sin n xdx. (47) Now notice tht the integrl in the lst term on the right is the sme s the integrl on the left side, so we ll bring it over to give n sin n xdx = sin n xcosx+(n ) sin n xdx. (48) Finlly, we ll divide by n to obtin, sin n xdx = n sinn xcosx+ (n ) n sin n xdx, n. (49) 8

8 This is kind of recursion eqution tht reltes the ntiderivtives of powers of sinx. (It is lso given in Stewrt, Eighth Edition, s Exmple 6 on Pge 475 (nd derived on Pge 476). We now show how this reltion is useful to compute vlues of the following generl fmily of definite integrls, I n = π/ First of ll, let s compute the first two integrls, nd I = I = sin n xdx, n. (5) π/ π/ dx = π, (5) sinxdx = ( cosx) π/ = ( ) ( ) =. (5) We ll now use the definite integrl form of the recursion reltion in (49), For n, π/ sin n xdx = n sinn xcosx π/ + (n ) n π/ sin n xdx, n. (53) sin n = nd cos π =, (54) so tht the first term on the right side of the eqution vnishes. As result, we hve or simply, π/ sin n xdx = (n ) n π/ sin n xdx, n, (55) I n = n n I n, n. (56) This is recursion reltion tht reltes I n to I n. From I, we cn determine I, I 4, etc.. From I, we cn determine I 3, I 5, etc.. Let s look t the first couple of terms of the even-indexed series, I = I = π I 4 = 3 4 I = 3 4 π. (57) 9

9 In generl, we hve I n = 3 5 (n ) 4 6 (n) Now exmine the first couple of terms of the odd-indexed series, π. (58) I 3 = 3 I = 3 I 5 = 4 5 I 3 = (59) In generl, we hve I n+ = 4 6 (n) (n+). (6) 3

10 Appendix: Some mteril covered in the Mondy Section 5 Tutoril The definite integrl nd its pplictions (cont d) Both of these topics re prticulr exmples of the Spirit of Clculus t work. They re tken from ERV s lecture notes for MATH 37P Fll (Week ). The work done by nonconstnt force We strt with result tht is well-known to you from high school physics. Suppose tht constnt force F = Fi cts on mss m, cusing it to move long the x-xis from position x = to x = b. Then the totl work done W by the force is given by the product of the mgnitude of the force nd the displcement of the mss, i.e., W = F(b ) (6) This is specil cse of the more generl result in which constnt force F moves the mss in stright line tht is not necessrily prllel to F. If the displcement vector of the mss is d, then the totl work W done by F is W = F d. (6) In the discussion tht follows, it will be sufficient to consider Eq. (6). Now suppose tht the force F is no longer constnt, i.e., F = f(x)i, where the function f(x) is not necessrily constnt. If the mss m is moved from position x = to x = b, wht is the totl work W done by the force? You hve most probbly seen the nswer in your first-yer Physics course. It is given by the definite integrl, W = b f(x)dx. (63) We now derive this result mthemticlly in terms of our Riemnn sum definition of the definite integrl. And our derivtion will be done by employing wht we hve previously clled the Spirit of Clculus. Very briefly, we ll subdivide the intervl [,b] into tiny subintervls I k of length x, nd then pproximte the force function f(x) s constnt over ech subintervl. We then use the constnt-force result from Eq. (6) over ech subintervl I k, to pproximte the work W k done in moving the mss over the subintervl I k. Finlly, we sum over the contributions from ll subintervls. 3

11 As before, we first consider n n > (with the ide of letting n ) nd define Then define the prtition points, x = b n. (64) x k = +k x, k =,,,. (65) Note tht x = nd x n = b. These prtition points define set of n subintervls I k = [x k,x k ], k =,, n, of equl length x. Now select smple point x k [x k,x k ] from ech subintervl I k. Then evlute the force function f t ech smple point x k. We now consider ech vlue f(x k ) s the pproximtion of f(x) over the subintervl I k. In other words, the function f(x) is pproximted by constnt function f(x k ). In this wy, we my use Eq. (6) to pproximte the work W k done by the function f(x) in moving the mss over the subintervl I k, i.e., from x k to x k, s follows, W k = f(x k ) x (constnt force strength displcement). (66) The totl work done by the force in moving the mss from x = to x n = b will then be pproximted s follows, W = n n W k = f(x k ) x. (67) k= But by construction, the RHS of this eqution is Riemnn sum for the definite integrl of f(x) from x = to x = b. Assuming tht the definite integrl of f exists (which is ensured if f is continuous or piecewise continuous), we hve W = lim = k= n k= b n f(x k ) x f(x)dx. (68) This concludes our mthemticl justifiction of the definite integrl formul for work. An importnt note regrding the dimensions of work nd the integrl formul ThedimensionlityofforceisMLT. (ThinkofF = mndthedimensionsofmssndccelertion. Thereforethedimensionlity of workis forcetimes distnce, or ML T. Note tht thedimensionlity 3

12 of the Riemnn sum in Eq. (67) is lso force times distnce. Since the definite integrl in Eq. (68) is the limit of Riemnn sums with this dimensionlity, it follows tht the definite integrl hs the dimension of work. Bsiclly, we cn think of the integrnd s hving the dimensionlity of force nd the infinitesiml dx s the dimensionlity of length. The extension of the work integrl to severl dimensions nd motion long curves In future course on dvnced clculus tht includes the subject of vector clculus (e.g., AMATH 3 or MATH 7), you will consider the more generl cse of nonconstnt force F(r) in R 3 cting on mss m s the mss moves long curve C from point P to point Q. The sitution is sketched in the digrm below. z Q P m r(t) F(r(t)) x y The gol is once gin to compute the totl mount of work W done by the force. Once gin, in the Spirit of Clculus, the ide is to brek up the motion into tiny pieces over which we cn use the constnt-force-stright-line formul W = F(b ) to pproximte the work over these pieces. We then sum up, i.e., integrte over ll contributions to obtin W. In this cse, since the force vectors F(r) will not, in generl, be prllel to the motion of the mss, we ll hve to tke sclr products of these vectors F with the instntneous direction of motion of the mss m in other words, the velocity vectors v = r long the curve. The net result is tht we hve n integrl of the following form F dr, (69) C which is known s the line integrl of the vector field F over the curve C. You my lredy hve seen this integrl in your Physics course. 33

13 The definite integrl nd its pplictions (cont d) The verge vlue of function Suppose tht thin, stright wire is locted on the x-xis, specificlly on the intervl [, b]. Furthermore, suppose tht the function f(x) represents the temperture of the wire t point x [,b]. The question is, Wht is the verge temperture of the wire? This is specific exmple of the more generl question: Wht is the verge vlue of function f over the intervl [,b]? We ll ddress this problem in the usul wy, i.e., by mens of the Spirit of Clculus. We ll divide up the intervl [,b] into n subintervls I k, tke smples of the function f(x) on these subintervls, nd then compute the verge of these smple vlues. The verge vlue of f over the intervl [,b] will be the limit n of these verge vlues, provided tht the limit exists. So, s before, let n > nd define x = b. Then define the prtition points, n x k = +k x, k =,,,n, (7) so tht x = nd x n = b. These points define the n subintervls I k = [x k,x k ], k =,,,n. From ech subintervl I k, choose smple point x k I k. Then evlute the function t this smple point. The result is set of n function vlues f(x k ). These my be viewed s smples of the function f(x) over the intervl [,b]. It seems resonble to tke the verge of these n function vlues we ll denote this verge s f n = n n k= f(x k ). (7) Now the sum on the RHS looks lmost like Riemnn sum to the definite integrl of f. However, x is missing. So let s multiply nd divide by x s follows, f n = n x n f(x k ) x Here, S n is Riemnn sum corresponding to the definite integrl question bout wht to do bout the fctor = x = b n k= n x S n. (7) b. Reclling the definition of x: n x f(x)dx. There remins the n x = b. (73) 34

14 Therefore, the verge vlue in (7) becomes f n = b S n. (74) Assuming tht f is continuous (or t lest piecewise continuous), the limit of the Riemnn sums S n exists, nd we hve lim f n = n b lim S n = n b b f(x)dx. (75) This is the verge vlue of f over the intervl [,b], which we shll denote s follows, f [,b] = b b f(x)dx. (76) In other words, we compute the definite integrl of f over the intervl [,b] nd then divide by the length of the intervl, b. Let s now rewrite Eq. (76) s follows, b f(x)dx = f [,b] (b ). (77) If we ssume, for the moment - for the ske of simplicity - tht f(x) > on [,b], then Eq. (77) is stting tht the re enclosed by the grph of f(x), the lines x = nd x = b nd the x xis is given by the verge vlue of f on [,b] multiplied by the length of the intervl (b ). In other words, s sketched in the figure below, we hve replced the re enclosed by the grph, etc., by rectngle of height f [,b]. The rectngulr region is shded. y = f(x) f [,b] b x Tht being sid, we my now relx the restriction tht f(x) be strictly positive on [,b]. In this cse, the verge vlue of f on [,b] times the length (b ) will be signed re. 35

15 Some simple, yet illuminting, exmples: (Note: These my not hve been covered in the tutoril, but re dded for your informtion.). The function f(x) = over the intervl [,b] = [,]. Since f(x) ssumes only one vlue over the entire intervl, nmely, the vlue, we expect tht its verge vlue is. Let s check this. Since =, b = nd f(x) =, we hve f [,] = s expected. dx = [x] =, (78). The function f(x) = x over the intervl [,b] = [,]. From look t the grph of f over [,], we might guess tht the verge vlue is its verge vlue is /. Since =, b = nd f(x) =, we hve Our intuition ws correct. f [,] = [ x xdx = ] =. (79) 3. The function f(x) = x over the intervl [,b] = [,]. A look t the grph of f(x) = x shows tht there re mny more x-vlues for which f(x) < / thn in the previous cse, f(x) = x. Therefore, we would expect the verge vlue to be less thn /. Since =, b = nd f(x) =, we hve f [,] = [ x x 3 dx = 3 ] = 3. (8) 4. In generl, the function f(x) = x n over the intervl [,b] = [,], where n >. Since =, b = nd f(x) =, we hve f [,] = [ x x n n+ dx = n+ ] = n+. (8) Note tht s n, the verge vlue of the function x n behves s follows,. Does n+ this mke sense? For ny x such tht x <, rising it to higher powers mkes it smller, i.e., x n s n. (Think of x = /.) Tht mens tht the grph of f(x) = x n gets fltter nd fltter s n increses, except t x =, since n = lwys. This is illustrted below. Since ll vlues of x n for x [,) note tht we exclude the cse x = pproch zero s n, we expect the verge vlue of x n to pproch zero in the limit n. 36

16 An importnt piece of dvice: When you encounter new concept in mthemtics, it is often most helpful to pply tht concept to set of cses, perhps one-prmeter fmily of functions, nd to observe the behviour of the results s you vry the prmeter. We hve done this with the concept of the verge vlue of function, pplying it to the one-prmeter fmily of functions x n on [,]. 37

17 Lecture 4 Trigonometric integrls Relevnt section from Stewrt, Eighth Edition: 7. In this section we consider the problem of finding ntiderivtives of intgrnds tht re products of trigonometric functions. These integrls re ctully encountered when the method of trigonometric substitution, to be discussed the next lecture, is used. Integrnds involving powers of sin nd cos Here we consider integrls of the form sin m xcos n xdx. (8) The simplest cses re well known to us, sinxdx = cosx+c cosxdx = sinx+c. (83) Exmple : (This cse ws not done in clss.) The next simplest cse is sinxcosxdx = udu (u = sinx, du = cosxdx) = u +C = sin x+c. (84) Note tht we could lso hve performed the substitution u = cosx, du = sinxdx, sinxcosxdx = udu = u +C = cos x+c. (85) At first glnce, this result is different from the first result, but it is equivlent since cos x+c = ( sin x)+c = sin x+d, (86) 38

18 where D is n rbitrry constnt. Note tht from the double ngle formul, sinx = sinxcosx, (87) the bove integrl cn lso be evluted s follows, sinxcosxdx = = sinxcosxdx sinxdx = cosx +C. (88) 4 This result ppers to be quite different from the erlier two results. But with the double ngle formul for the cosine function (below), it cn be shown tht this result is equivlent to the other two. We ll leve this s n exercise for the reder. Exmple : The next simplest cses re sin xdx cos xdx. (89) In these cses, the following double ngle formul for the cosine is useful: cosx = cos x sin x. (9) If we rewrite the bove formul s follows, cos x = ( sin x) sin x = sin x, (9) we then obtin the result, sin x = cosx. (9) We my then use this result to obtin, sin xdx = [ ] cosx dx = x sinx+c. (93) 4 39

19 We cn use the following double ngle formul for the sine function, sinx = sinxcosx, (94) to rewrite the bove result s sin xdx = x sinxcosx+c. (95) If we rewrite the formul in (9) s follows, cos x = cos x ( cos x) = cos x, (96) we obtin cos x = + cosx. (97) We then use this result to obtin cos xdx = [ + ] sinx dx which cn lso be written s = x+ sinx+c, 4 (98) cos xdx = x+ sinxcosx+c. (99) Exmple 3(): Let us now consider the following integrl, cos 3 xdx. () We ll express the integrnd s the following product, cos 3 x = cos xcosx () to perform the following steps, cos 3 xdx = = = cos xcosxdx ( sin x)cosxdx cosxdx sin cosxdx = sinx 3 sin3 x+c. () 4

20 The finl integrl ws obtined by the method of substitution: Let u = sinx so tht du = cosxdx. Then sin cosxdx = u du = 3 u3 +C = 3 sin3 x+c. (3) Note tht the integrl in the second line of the originl derivtion could lso be treted in one line using substitution s follows: Letting u = sinx so tht du = cosxdx, ( sin x)cosxdx = ( u )du = u 3 u3 +C = sinx 3 sin3 +C, (4) in greement with the previous result. This is often the wy tht solutions re presented in Stewrt s text. Exmple 3(b): We cn evlute the following ntiderivtive, sin 3 xdx, (5) in similr mnner: sin 3 xdx = = = sin sinxdx ( cos x)sinxdx sinxdx cos xsinxdx = cosx+ 3 cos3 x+c. (6) The bove two exmples re specil cses of more generl method of treting n integrl of the form sin m xcos n xdx. (7) 4

21 In the cse tht one of m or n is odd, then one tries, using the reltion sin x+cos x =, (8) to express the integrl in one of the following forms, [ polynomil in cosx] sinxdx (9) which cn be treted by the substitution u = cosx, du = sinxdx, or or [ polynomil in sinx] cosxdx, () which cn be treted by the substitution u = sinx, du = cosxdx. A summry of the entire procedure is given in the box with heding, Strtegy for Evluting sin m xcos n xdx, on Pge 48 of Stewrt s textbook (Eighth Edition). Exmple 3() revisited: Let us now reconsider the following integrl, cos 3 dx = cos cosxdx. () This time, however, insted of expressing cos x ( sin x) s before, we ll try the method of integrtion by prts: u = cos x = du = cosxsinxdx, () dv = cosxex = v = sinx. (3) Now use Formul of Integrtion by Prts, udv = uv vdu, (4) to obtin cos 3 xdx = cos xcosxdx = cos xsinx+ sin xcosxdx = cos xsinx+ 3 sin3 x+c. (5) At first glnce, this does not look like the nswer obtined in our first tretment of this integrl, Eq. (). With little rewriting, however, we obtin the sme result, i.e., cos xsinx+ 3 sin3 x+c = [ sin x]sinx+ 3 sin3 +C = sinx 3 sin3 x+c. (6) 4

22 The morl of the story: Sometimes, you my obtin result tht does not gree with the result presented in book or tble of integrls. This does not necessrily men tht your result is incorrect - it my simply be the correct result, but cst in slightly different form. Exmple 4: Find sin 5 xcos xdx. (7) If we substitute cos x = sin x, we then obtin the following sum, sin 5 xdx sin 7 xdx. (8) In other words, we re in deeper mess! Insted, we ll do the following, sin 5 xcos x = sin 4 xcos xsinx = ( cos x) cos xsinx = ( cos x+cos 4 x)cos xsinx = (cos x cos 4 x+cos 6 x)sinx. (9) This is in keeping with the Strtegy from Stewrt s textbook, where the power of the sine function, 5, is n odd number. As result, our integrnd is now in the form, [ polynomil in sinx]cosx. () We my now use the substitution method, sin 5 xcos xdx = (cos x cos 4 x+cos 6 x)sinxdx = (u u 4 +u 6 )du (u = cosx, du = sinxdx) = 3 u3 + 5 u5 7 u7 +C = 3 cos3 x+ 5 cos5 x 7 cos7 x+c (resubstituing u = cosx). () Integrls involving powers of tn nd sec Here we consider integrls of the form tn m xsec n xdx. () 43

23 Once gin, these integrls re often encountered when we employ the method of trigonometric substitution. First of ll, let s review the bsic differentition results of tn nd sec: d dx tnx = sec x d secx = secxtnx. (3) dx If you hve these results memorized, then fine. If not, they cn lwys be derived quickly from first principles: d dx tnx = d dx [ ] sinx cosx = cos x+sin x cos x = cos x (quotient rule) = sec x. (4) d dx secx = d [ ] dx cosx = cos x ( sinx) = cosx sinx cosx = secxtnx. (5) Also useful will be the following reltionship between tnx nd secx, tn x = sec x or sec x tn x = or tn x+ = sec x. (6) Once gin, if you hve one or more of thse results memorized, fine. But if not, they cn esily be derived from the Pythgoren reltion, sin x+cos x =. (7) Divide both sides by cos x: sin x cos x + = cos x, (8) 44

24 which, of course, is tn x+ = sec x. (9) Exmple No. : The simplest integrls in this clss, which you my hve seen in MATH 37, re tnxdx = ln secx +C (3) nd secxdx = ln secx+tnx +C. (3) Let s derive these results. The first one is quite strightforwrd: sinx tnxdx = cosx dx = du (u = cosx du = sinxdx) u = ln cosx +C = ln cosx +C = ln (cosx) +C = ln secx +C. (3) The second result is obtined by mens of clever trick: secxdx = = = ( ) secx+tnx secx dx (multipliction by ) secx+tnx sec +secxtnx dx secx+tnx du (u = [secx+tnx] ) u = ln u +C = ln secx+tnx +C. (33) (As mentioned in the lecture, some people would refer to this trick s cheesy. But it does chieve the correct result.) Exmple No. : The next simplest integrl involving tn nd sec is quite esy, secxtnxdx = secx+c. (34) 45

25 Exmple No. : Let us now consider the integrls, tn xdx nd sec xdx. (35) These re rther strightforwrd: tn xdx = (sec x)dx = tnx x+c (36) nd sec xdx = tnx+c. (37) For higher powers of tn nd sec, there is strtegy similr to tht for powers of tn nd sec: See the box entitled, Strtegy for evluting tn m xsec m x. This strtegy does not cover ll cses nd, s the text indictes, one my need to use identities, integrtion by prts, nd occsionlly little ingenuity. We consider few exmples below. Exmple No. 3: The ntiderivtive, tn 3 xdx. (38) Let s write tn 3 x = tn xtnx nd then convert the tn x term: tn 3 xdx = tn xtnxdx = (sec )tnxdx = tnxsec xdx tnxdx = tn x ln secx +C. (39) The first integrl ws obtined from the method of substitution: u = tnx, du = sec xdx. Exmple No. 4: The ntiderivtive, sec 3 dx. (4) Method No. : (This method ws strted in clss but not finished.) We ll write sec 3 x = sec xsecx 46

26 nd then convert the sec x term: sec 3 xdx = = = sec xsecxdx (4) (tn x+)secxdx (4) tn xsecxdx+ secxdx. (43) The second integrl is esy we computed it erlier, secxdx = ln secx+tnx +C. (44) As for the first integrl, we ll try integrtion by prts, tn xsecxdx = tnx(secxtnx)dx (45) with u = tnx nd du = secxtnxdx, implying tht v = secx nd du = sec x. Reclling tht udv = uv vdu, (46) we hve tnx(secxtnx)dx = tnxsecx sec 3 xdx. (47) If we now substitute the results of (44 nd (47) into (4), we obtin sec 3 xdx = secxtnx sec 3 xdx+ln secx+tnx +C. (48) It might pper tht we hve goofed, since the sme integrl ppers on both sides of the eqution! But the sving grce is tht the integrl on the right side is multiplied by (-). As such, we cn bring it over to the left side to give sec 3 xdx = secxtnx+ln secx+tnx +C. (49) nd then divide by to rrive t the finl result, sec 3 xdx = secxtnx+ ln secx+tnx +C. (5) Method No. : (This ws the method shown in clss.) We ll once gin express the integrnd s sec 3 x = sec xsec x but this time, insted of using trigonometric identity to re-express the sec x function, we ll use integrtion by prts: sec 3 xdx = sec xsecxdx. (5) 47

27 Let u = secx nd dv = sec xdx so tht du = secxtnxdx nd v = tnx. Then from the integrtion by prts formul, udv = uv vdu, (5) we obtin the following, sec 3 xdx = secxtnx tnxsecxtnxdx. (53) This result is ctully equivlent to Eq. (47). We ll work on the second integrl by combining the two tnx functions nd using the trig identity: tn xsecxdx = (sec x )secxdx = sec 3 xdx secxdx = sec 3 xdx ln secx+tnx +C. (54) Inserting this result into (53) yields sec 3 xdx = secxtnx sec 3 dx+ln secx+tnx +C, (55) which is identicl to Eq. (48) from Method No.. As such, we shll be ble to rrive t the sme result, i.e., sec 3 xdx = secxtnx+ ln secx+tnx +C. (56) 48

28 Lecture 5 Trigonometric substitution Relevnt section from Stewrt, Eighth Edition: 7.3 In science nd engineering, we often encounter integrls with integrnds composed of squre roots, or integer powers of these squre roots, of qudrtic functions, e.g., A x, A+x, x A. (57) Such integrls cn often be hndled by mens of pproprite trigonometric substitutions, the subject of this section. As we ll see below, the gol is to remove the squre root terms from the integrnds nd this cn be ccomplished using pproprite trigonometric substitutions. These substitutions will be bsed upon the following two importnt trigonometric reltions, sin x+cos x =, tn x+ = sec x. (58) As we discussed erlier, the second reltion cn be obtined from the first one by dividing both sides of the first one by cos x. We ll see tht it is convenient to replce the constnts A in the bove expresssions with, i.e., x, +x, x. (59) Integrls which contin x Exmple : The ntiderivtive, x dx. (6) Let s consider the chnge of vrible, x = sinθ = dx = cosθdθ. (6) We would like this to be reltionship between x nd θ, which implies tht the ngle θ should be restricted to the intervl [ π/, π/]. More on this lter. 49

29 In ddition, we ll ssume, for simplicity, tht >. With this chnge of vrible, the squre root term becomes, x = sin θ = sin θ = cosθ. (6) Techniclly, the lst line should be x = cosθ, (63) since the function is, by definition, non-negtive. If we restrict the rnge of the θ vrible to [ π/,π/], then cosθ nd the bsolute vlue sign cn be omitted. Inserting the expressions for the squre root function nd the dx term into the originl integrl, we obtin x = [cosθ][cosθ]dθ = cos θdθ. (64) As you see, we ve produced trigonometric integrl, the subject of the previous lecture. You my recll tht we cn esily evlute this integrl using the double-ngle formul for the cosine function, [ cos θdθ = + ] cosθ dθ = θ + 4 sinθ = θ + sinθcosθ +C. (65) But we must now express this result in terms of the originl vrible x. How do we do this? Let us recll tht which, in turn, implies tht This tkes cre of θ nd sinθ. But wht bout cosθ? One wy to obtin cosθ is to simply use the trigonometric reltion, x = sinθ = sinθ = x, (66) θ = Sin ( x ). (67) cos θ = sin θ = cos θ = x. (68) 5

30 This, in turn, implies tht cosθ = x = Since θ [ π/,π/], we cn tke the positive squre root. x. (69) The bove procedure of trnslting expressions involving θ to those involving x is simplified if we drw digrm tht is ssocited with the chnge of vrible, x = sinθ = sinθ = x. (7) θ x x Note: This prticulr instructor thinks tht it is lwys good ide to drw the digrm. We now combine ll of our results to obtin the desired ntiderivtive, x dx = cos θ = θ+ sinθcosθ+c = Sin ( x )+ = Sin ( x ( x ) x +C ) + x x +C. (7) It s good ide to check this result by differentition (omitting the rbitrry constnt C): [ d dx Sin ( x ) + x ] ( ) x = + x x + x x ( x) = [ x + x x ] x = [ x ] x + x = x. (7) Exmple revisited - An ppliction: Let us now pply this result to the following definite integrl, x dx, (73) 5

31 which is the re of the semi-circulr region enclosed by the curve y = x nd the x-xis between x = nd x = b. This is one-hlf the re of circle of rdius, nmely π. From the FTC II nd our erlier result, x dx = Sin ( x [ = = Sin ()+ ( π ) ( π ) + x x x ] ) [ Sin ( )+ x ] = π. (74) Thebove resultws mde possibleby the fct tht we hd previously done ll of the work to compute the ntiderivtive of x. If, on the other hnd, you were given the definite integrl to evlute, nd then strted to employ the trigonometric substitution method, some time nd work could be sved by stopping fter obtining the ntiderivtive in terms of θ nd evluting it t pproprite limits, s opposed to expressing your results in terms of x nd then evluting them. We illustrte this with the problem ddressed bove. First of ll, note tht the lower nd upper limits of integrtion re nd, respectively. With the chnge of vrible, x = sinθ, (75) this implies tht the lower nd upper limits of integrtion re π/ nd π/. Then, π/ x dx = cos θdθ π/ = [ θ + 4 sinθ ] π/ π/ = [ π + ] [ π + ] = π, (76) in greement with our erlier result. Exmple : The ntiderivtive, dx. (77) x 5

32 (In the cse tht =, you my remember tht the ntiderivtive is Sin (x).) Once gin, we consider the chnge of vrible, x = sinθ = dx = cosθdθ. (78) Then, s before, x = cosθ. (79) Inserting these expressions into the integrl, cosθ x dx = cosθ dθ = dθ = θ+c. (8) But from the chnge of vrible, x = sinθ = sinθ = x = θ = Sin ( x ). (8) Therefore, Exmple revisited: Suppose tht you remembered tht x dx = Sin ( x ) +C. (8) x dx = Sin (x). (83) You could use this fct to find the ntiderivtive, s follows. First, fctor out the from the squre root: Now mke the chnge of vrible, dx, (84) x x dx = ( ) dx. (85) x u = x = du = dx = dx = du. (86) 53

33 The integrl t the right then becomes in greement with our erlier result. ( ) dx = x = u du u du = Sin u+c = Sin ( x ) +C, (87) Exmple : It s possible tht we lso hve powers of x present in the integrl, e.g., x (88) xdx. This integrl is ctully quite esy since the top is derivtive of the inside of the bottom. Let u = x = du = xdx = xdx = du, (89) so tht the bove integrl becomes u / du = ()u/ +C = x +C. (9) Exmple 4: The ntiderivtive, As before, let x (9) xdx. x = sinθ = dx = cosθ. (9) 54

34 The bove integrl becomes x x dx = sin θ cosθ cosθdθ = sin θdθ [ ] = cosθ dθ (93) = θ 4 sin(θ) = θ = Sin ( x = Sin ( x sinθcosθ ) ( x ) x +C ) x x +C. (94) Integrls which contin +x Exmple 5: The ntiderivtive, +x dx. (95) With n eye to the trigonometric reltion, tn x+ = sec x, (96) let us consider the chnge of vrible, x = tnθ = dx = sec θdθ. (97) With this chnge of vrible, the squre root term becomes +x = + tn θ = +tn θ = secθ. (98) As erlier, the secθ term should techniclly be secθ, but we ll restrict the rnge of θ once gin to [ π/, π/]. And while we re t it, let s drw the tringle tht is ssocited with this chnge of vrible, strting with the fct tht tnθ = x. (99) 55

35 + x x θ We now perform the bove chnge of vrible in the integrl, +x dx = (secθ)sec θdθ = sec 3 θ. () We evluted this ntiderivtive in the previous lecture nd simply employ the result here: +x dx = sec 3 θ = [secθtnθ+ln secθ+tnθ ]+C. () We must now trnslte the functions in θ into functions involving x. From the digrm, tnθ = x secθ = x +. () Inserting these results into our expression, we obtin +x dx = [secθtnθ +ln secθ+tnθ ]+C = x +x + ln +x + x +C. (3) Note tht we cn write the logrithmic term s follows, ln +x + x = ln +x +x = ln +x +x ln. (4) The constnt term ln cn be bsorbed into the rbitrry constnt C in the previous expression. As such, our finl result is +x dx = x +x + ln +x +x +C. (5) The result cn be checked by differentition. (But we won t do it here!) 56

36 Exmple 6: The ntiderivtive dx. (6) +x Once gin, we consider the following chnge of vrible, x = tnθ = dx = sec θdθ. (7) The bove integrl becomes +x dx = = secθ (sec θ)dθ secθdθ = ln secθ +tnθ +C = ln +x + x +C = ln +x +x +C, (8) where we hve once gin bsorbed the ln term into the rbitrry constnt. Integrls which contin x Exmple 7: The ntiderivtive, x dx. (9) Once gin with n eye to the trigonometric reltion, tn x+ = sec x = sec x = tn x, () we consider the following chnge of vrible, x = secθ = dx = secθtnθdθ. () This implies tht Let s drw the tringle ssocited with this chnge of vrible: x = secθ = θ = Tn ( x ). () 57

37 x x θ Performing the chnge of vrible in the bove integrl yields, x dx = (tnθ)(secθtnθ),dθ = tn θsecθdθ = (sec θ )secθdθ = sec 3 θdθ secθdθ = [ secθtnθ + ln secθ+tnθ ln secθ+tnθ ]+C = [secθtnθ ln secθ+tnθ ]+C [ (x = ) ( ) x ] x x ln + +C = x x ln x+ x +C, (3) where, once gin, we hve bsorbed ln into the rbitrry constnt. 58