INTEGRATION TECHNIQUES. Math 101. Contents
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1 INTEGRATION TECHNIQUES Mth 101 Contents 1. Substitution (.3,.8 1. Integrtion by prts ( x n sin x, x n cos x, x n e x.. e x sin x, e x cos x.3. x n (ln x m 3. Rtionl functions ( 7.3 [dvnced] 3.1. Step 1. Reducing n improper frction 3.. Step. Decomposing into prtil frctions 3.3. Step 3. Integrting prtil frctions. Trigonometric functions ( 7., including exercises 3.1. The universl trigonometric substitution [dvnced] 3.. Integrls of the form R(sin x, cos x sin x nd R(sin x, cos x cos x 3.3. Integrls of the form R(sin x, cos x 3.. Integrls tn n x 3.5. Integrls sin mx sin nx, sin mx cos nx, cos mx cos nx 3 5. Exponentil nd hyperbolic functions 6. Trigonometric nd hyperbolic substitutions ( Exmple 7. Other irrtionlities 8. Tble integrls 5 References: Thoms & Finney, Clculus nd Anlytic Geometry, 9th Edition:.3,.8, 7., 7.3 [dvnced], 7.. There is no generl rule for evluting n integrl. Ech time you hve to look for formul/rule tht would simplify the integrl nd eventully reduce it to tble ones. Use these formuls (direct nd bckwrd: 1. Substitution (.3,.8 [ ] u = g(x f(g(xg (x = du = g = f(u du = F (u + C = F (g(x + C, (x [ ] x = g(u f(x = = g = f(g(ug (u du (u du = F (u + C = F (g 1 (x + C, f(kx + b = 1 k F (kx + b + C, where f(x = F (x + C (importnt specil cse. For definite integrls the former formul tkes the form b f(g(xg (x = [ ] u = g(x du = g = (x g(b g( f(u du = F (u Do not forget tht lso needs substitution! (In n expression like f(g(x you cnnot substitute u = g(x! Do not forget to return to the originl vrible (or reclculte limits in cse of definite integrls! Some stndrd substitutions will be discussed lter (, 5, 6, g(b g(
2 MATH 101. Integrtion by prts ( 7. u dv = uv v du, b u dv = uv b Success of the method depends on the decomposition of the integrnd in the form u dv. Choose for dv something tht you cn esily integrte (s you need u = du for the formul! Proceed only if the resulting integrl v du is simpler tht the originl one nd you know wht to do with it! (Certin integrls, e.g., e x sin x below, re n exception: integrtion by prts my give the sme integrl, but the resulting expression cn be treted s n eqution. Below re some typicl exmples..1. x n sin x, x n cos x, x n e x. Tke u = x n. Integrtion by prts is to be pplied n times (with u = x n, ech time reducing the power of x by 1... e x sin x, e x cos x. Integrte by prts twice, ech time tking u = e x. Solve the resulting eqution in the originl integrl. b v du..3. x n (ln x m. Tke u = (ln x m. Integrtion by prts reduces m by 1. Do it m times. 3. Rtionl functions ( 7.3 [dvnced] A rtionl function P (x/q(x, where P nd Q re polynomils in x, cn be integrted in three steps: (1 Reduce the frction if it is improper (i.e., deg P deg Q; ( Decompose P (x/q(x into sum of prtil frctions; (3 Integrte ech prtil frction Step 1. Reducing n improper frction. Do not overlook this step! Otherwise, Step will produce n erroneous result! If the frction is improper, i.e., deg P deg Q, then divide P by Q nd write P/Q = F + (R/Q, where F is the rtio nd R is the reminder. The polynomil F (x cn esily be integrted. With the proper frction R(x/Q(x proceed to Step. 3.. Step. Decomposing into prtil frctions. From now on ssume tht the frction P (x/q(x is proper (i.e., deg P < deg Q, see bove. We need to decompose the denomintor Q(x into the product of liner (i.e., of the form (x r nd irreducible qudrtic (i.e., of the form (x + px + q with p q < 0 fctors. Now the decomposition into prtil frctions cn be found by the method of undetermined coefficients. Write the identity P (x/q(x =..., where the right hnd side is composed s follows: ech liner fctor (x r m contributes m terms A 1 x r + A (x r A m (x r m, nd ech qudrtic fctor (x + px + q n contributes n terms B 1 x + C 1 x + px + q + B x + C (x + px + q B nx + C n (x + px + q n. (Of course, ech coefficient in the resulting expression should be given it s own nme. The totl number of undetermined coefficients A, B, C,... must be equl to the degree of Q. In the resulting identity get rid of the denomintors (by multiplying both the sides by Q(x nd write down equtions for the undetermined coefficients. There re two wys to obtin equtions: (1 by equting the coefficients of equl powers of x (typiclly, simpler equtions re obtined from higher powers, nd ( by substituting prticulr vlues of x (usully some simple vlues like x = 0, ±1, etc. or the vlues x = r, where (x r is fctor in the decomposition of Q(x. Importnt Remrk: The resulting system of liner equtions in A, B, C, must hve unique solution! If there is no solutions, something is wrong in your clcultion. If there re mny solutions, then either something is wrong or there is not enough equtions Step 3. Integrting prtil frctions. Here re the formuls (with C omitted: = ln x r, x r (x r n = 1 (1 n(x r n 1 (substitution x r = t. For the other prtil frctions, first complete the squre: Bx + C (x + px + q n = Bt + C (t + n dt (where t = x + p, q p = > 0, nd C = C Bp.
3 Then t dt t + = 1 ln(t +, INTEGRATION TECHNIQUES 3 t dt (t + n = 1 (1 n(t + n 1 (substitution t + = u, dt t + = 1 t tn 1 is tble integrl, nd the remining integrl I n = steps vi integrtion by prts nd reducing n: I n = 1 t + t (t + n dt = 1 I 1 n 1 (n 1 = [by prts] = 1 I n 1 ( t d. Trigonometric functions ( 7., including exercises dt (t + is evluted in n n 1 (t + n 1 1 t (n 1 (t + n (n 1 I n 1. An integrl of rtionl function of sin x nd cos x cn lwys be reduced to integrting rtionl function. The best thing to try is using trigonometric identities (see trnsc.pdf to convert products to sums nd to reduce powers. Below re few stndrd hints..1. The universl trigonometric substitution [dvnced]. One lets t = tn(x/. Then x = tn 1 t nd sin x = t 1 t dt, cos x =, = 1 + t 1 + t 1 + t re expressed in terms of t rtionlly. Importnt Remrk: This substitution is universl. However, typiclly it leds to huge mount of clcultions. Thus, use it only if you cnnot think of better pproch! Importnt Remrk: Alwys try to simplify the expression using trigonometric identities. The generl rule is the following: the bigger the rgument of the functions (sin nd cos, the lower the degree of the expression. However, keep in mind tht in generl you wnt to hve ll functions of the sme rgument! Below re few exmples... Integrls of the form R(sin x, cos x sin x nd R(sin x, cos x cos x, where R is rtionl function. Use the substitution cos x = t, sin x = 1 t, sin x = dt (in the former cse or sin x = t, cos x = 1 t, cos x = dt (in the ltter cse. Exmple. sin x cos 3 x = sin x(1 sin x cos x = (where u = sin x nd du = cos x. (u u 3 du = sin x sin x.3. Integrls of the form R(sin x, cos x, where R is rtionl function whose ll terms hve even degree In other words, the integrnd cn be expressed in terms of sin x, cos x, nd sin x cos x. Reduce the degree using the formuls Exmple. sin x = sin x cos 3 x = 1 cos x, cos x = 1 + C 1 + cos x, sin x cos x = 1 sin x (see trnsc.pdf. ( + cos x sin x sin x1 = + sin x cos x cos x = + C Integrls tn n x. One hs tn x = ln cos x, tn n x = ( tn n x 1 1 cos = x tn n x tnn 1 x n Integrls sin mx sin nx, sin mx cos nx, cos mx cos nx. Convert products of functions to sums (see trigonometric identities in trnsc.pdf.
4 MATH Exponentil nd hyperbolic functions An integrl of the form R(e x (where R is rtionl function cn be reduced to integrting rtionl function by the substitution e x = t, x = ln t, = dt/t. An integrl of the form R(sinh x, cosh x cn be treted either by expressing sinh x nd cosh x in terms of e x or, better yet, similr to, using corresponding hyperbolic identities (see trnsc.pdf. 6. Trigonometric nd hyperbolic substitutions ( 7. Assume tht we need to integrte n expression rtionl in x nd the rdicl x + bx + c. complete the squre First, ( ( x + bx + c = x + b D, where, s usul, D = b c, nd mke the substitution u = x + b, du =. Now, depending on the signs of nd D, we need to tret one of the following three irrtionlities: u + r, u r, or r u (where r = D/ Exmple x 3x = 3(x x 1 = 3((x 1 = 3 u. Here r =. Now we mke n pproprite trigonometric or hyperbolic substitution nd reduce the given integrl to trigonometric or hyperbolic one, which cn be treted s in or 5, respectively. Irrtionlity Trigonometric substitution Hyperbolic substitution u = r sin t, π/ t π/ r u du = r cos t dt, r u = r cos t t = sin 1 (u/r u = r tn t, π/ < t < π/ u = r sinh t, t ny number u + r du = r dt/ cos t, u + r = r/ cos t du = r cosh t dt, u + r = r cosh t t = tn 1 (u/r t = sinh 1 (u/r u = r/ cos t, 0 t < π/ or π/ < t π u = ±r cosh t, t 0 u r du = r sin t dt/ cos t, u r = r tn t du = ±r sinh t dt, u r = r sinh t t = cos 1 (r/u t = cosh 1 (±u/r Importnt Remrk: To my opinion, in the lst two cses the hyperbolic substitutions re usully simpler. Importnt Remrk: In the lst cse one should be very creful bout the signs. The integrnd is defined on two disjoint intervls, x 1 nd x 1, which should be treted differently. (The upper sign corresponds to the first intervl. One should keep in mind tht cosh t 1 for ny t nd tht, by definition, u r 0 for ny u. 7. Other irrtionlities x + b x + b An expression rtionl in x nd m is integrted by the substitution cx + d cx + d = tm. Then x = dtm b ct m nd the integrnd becomes rtionl (see 3.
5 INTEGRATION TECHNIQUES 5 8. Tble integrls x n = xn+1 + C (n 1, n + 1 e x = e x + C, sin x = cos x + C, sec x = tn x + C, sec x tn x = sec x + C, sinh x = cosh x + C, x = x sin 1 + C ( > 0, x x = 1 x sec 1 + C ( > 0, x + = x sinh 1 + C ( > 0, x = 1 x tnh 1 + C ( > 0. = ln x + C, x x = x ln + C, cos x = sin x + C, csc x = cot x + C, csc x cot x = csc x + C, cosh x = sinh x + C, + x = 1 x tn 1 + C ( > 0, x = x cosh 1 + C ( > 0, (see trnsc.pdf for the expressions for sinh 1, cosh 1, nd tnh 1 in terms of ln. [dvnced] This topic hs been omitted or moved to Mth 10
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