Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.ICN Textbook: Calculus Early Vectors Preliminary Edition By Stewart. Monday, February 17 8.2 Trig Integrals -------------------------------------------------------------------- Mon, Feb 17 8.2 Trig Integrals p470: 1,4,(12),13,18,(24),28,(36),(48),49,53,54 Main Idea: Save out something for the dx. m n Key Words: INT Sin[x] Cos[x] dx m n INT Sec[x] Tan[x] dx INT Sec[x] dx INT Tan[x] dx 2 INT Sin[x] dx 2 INT Cos[x] dx Goal: We learn how to integrate the above types of trigonometric integrals. Previous Assignment Page 463 Problem 6 2 INT x Sin[2x] dx 2 x -Cos[2x]/2 2x dx Sin[2x] dx 2 = -x Cos[2x]/2 + INT x Cos[2x] dx x Sin[2x]/2 dx Cos[2x] dx 2 = -x Cos[2x]/2 + x Sin[2x]/2 - 1/2 INT Sin[2x] dx 2 = -x Cos[2x]/2 + x Sin[2x]/2 + 1/4 Cos[2x] + C <===== ANSWER check 2 -x Cos[2x] + x Sin[2x] +x Cos[2x] Sin[2x]/2 -Sin[2x]/2 -------------------------------------------------- 2 x Sin[2x] It checks ======================================================== Page 463 Problem 18 t=4 INT Sqrt[t] ln(t) dt t=1 3/2 ln(t) t /(3/2) dt/t Sqrt[t] dt 3/2 1/2 = t ln(t)/(3/2) - (2/3) INT t dt 3/2 3/2 = t ln(t)/(3/2) - (2/3) t /(3/2) + C | 3/2 3/2 | t=4 =|(2/3) t ln(t) -(4/9) t | | | t=1 (2/3) 8 ln(4) -(4/9) 8 - (-4/9) 16/3 ln(4) -28/9 <===answer. check: (2/3)(3/2) t^(1/2) ln(t) + (2/3) t^(3/2) 1/t -(4/9)(3/2) t^(1/2) t^(1/2) ln(t) +(2/3) t^(1/2) -(2/3) t^(1/2) t^(1/2) ln(t) It checks. ===================================================================== Page 463 Problem 44 Use integration by parts to prove the reduction formula. n INT Sec[x] dx Sec[x]^(n-2) Tan[x] (n-2) Sec[x]^(n-3) Sec[x] Tan[x] dx Sec[x]^2 dx INT Sec[x]^n dx = Sec[x]^(n-2) Tan[x] - INT (n-2) Sec[x]^(n-2) Tan[x]^2 dx INT Sec[x]^n dx = Sec[x]^(n-2) Tan[x] - INT (n-2) Sec[x]^(n-2) (Sec[x]^2-1) dx INT Sec[x]^n dx = Sec[x]^(n-2) Tan[x] - (n-2)INT Sec[x]^n dx +(n-2) INT Sec[x]^(n-2) dx (n-1) INT Sec[x]^n dx = Sec[x]^(n-2) Tan[x] + (n-2) INT Sec[x]^(n-2) dx n-2 n Sec[x] Tan[x] n-2 n-2 INT Sec[x] dx = ------------------- + ----- INT Sec[x] dx n-1 n-1 ===================================================================== New Material: Integrating the trig functions. (* Integral of Sin[x]^2 *) INT Sin[x]^2 dx We use the trigonometric identity 2 1-Cos[2x] Sin [x] = ---------- 2 2 1+Cos[2x] Cos [x] = ---------- 2 2 2 Notice that these are compatible with the identity that Sin[x] + Cos[x] = 1. Also notice that the Cos uses the plus while the Sin uses the minus. ----------------------------------------------------------------------- 2 (* Integral of Sin[x] *) ========================== 1-Cos[2x] x Sin[2x] INT Sin[x]^2 = INT ---------- dx = --- - --------- + C 2 2 4 2 Cos[2x] 1 - 2 Sin[x]^2 Check: 1/2 - ----------- = 1/2 - ------------ = Sin[x]^2 4 2 ------------------------------------------------------------------ 2 (* Integral of Cos[x] *) ========================== 1 + Cos[2x] x Sin[2x] INT Cos[x]^2 dx = INT ------------- = --- + ---------- + C 2 2 4 2 Cos[2x] 2 Cos[x]^2 -1 Check: 1/2 + ------------ = 1/2 + ------------- = Cos[x]^2 4 2 ----------------------------------------------------------------- (* INT Sin[x]^m Cos[x]^n dx *) ================================ (a) If one of m and n are odd, you convert all but one to the other and it easily integrates. (b) If both m and n are even, you can use the half angle formulas to replace the powers by m/2 and n/2. ------------------------------------------------------------- Case(a): where one of them is odd =================================== INT Sin[x]^5 Cos[x]^2 dx INT Sin[x]^4 Cos[x]^2 (Sin[x] dx) INT (1-Cos[x]^2)^2 Cos[x]^2 (Sin[x] dx) INT (1-2 Cos[x]^2 + Cos[x]^4) Cos[x]^2 (Sin[x] dx) INT (Cos[x]^2 -2 Cos[x]^4 + Cos[x]^6) Sin[x] dx Cos[x]^3 2 Cos[x]^5 Cos[x]^7 - --------- + --------- - ---------- + C 3 5 7 ------------------------------------------------------------------- Case(b) where both of them are even ====================================== INT Sin[x]^4 Cos[x]^2 dx ======================== 1-Cos[2x] 2 1+Cos[2x] INT (---------) (----------) dx 2 2 1/8 INT (1-Cos[2x])^2 (1+Cos[2x]) dx 2 3 1/8 INT 1 - Cos[2 x] - Cos[2 x] + Cos[2 x] dx 1+Cos[4x] 1/8 ( x - Sin[2 x]/2 - INT --------- + INT (1-Sin[2x]^2) Cos[2x]dx ) 2 x Sin[4x] Sin[2x] Sin[2x]^3 1/8 ( x - Sin[2 x]/2 - --- - ------- + ------- - --------- ) 2 8 2 6 x Sin[2x] Sin[4x] Sin[2x] Sin[2x]^3 ---- - ------ - ------- + ------------ - ----------- + C 16 16 64 16 48 same same x Sin[4x] Sin[2x]^3 ---- - ------- - ----------- + C 16 64 48 ------------------------------------------------------------------------- (* Check and memorize the integral of Sec[x] *) ============================================== | | INT Sec[x] dx = ln | Sec[x]+Tan[x] | + C | | Sec[x]Tan[x]+Sec[x]^2 Sec[x](Tan[x]+Sec[x]) check: ------------------- = --------------------- = Sec[x] Sec[x]+Tan[x] Sec[x]+Tan[x] ----------------------------------------------------------------- (* The integral of the tangent is obvious once you see it *) =========================================================== | | | | INT Tan[x] = -ln( |Cos[x]| ) + C = ln( |Sec[x]| ) + C | | | | -Sin[x] check: - --------- = Tan[x] Cos[x] ---------------------------------------------------------------- m n INT Tan [x] Sec[x] dx ================================= The idea is to convert it to one of three types. (a) INT Sec[x]^n (Sec[x] Tan[x] dx) ***************** (b) INT Tan[x]^n (Sec[x]^2 dx) *********** (c) INT Sec[x]^n dx If the tangents are even, change them all to Secants and use (b). If the tangents are odd, change all but one to Secants and use (a). If the Secants are odd, change all but one to Tangents and use (b). The idea is to factor out a Sec[x]^2 and change everything to Tangents or, factor out a Sec[x] Tan[x] and change everything to Secants. INT Tan[x]^6 Sec[x]^4 dx Save out a Sec[x]^2 Save out a Sec[x] Tan[x] INT Tan[x]^6 Sec[x]^2 (Sec[x]^2 dx) INT Tan[x]^5 Sec[x]^3 (Sec[x] Tan[x])dx INT Tan[x]^6 (1+Tan[x]^2) Sec[x]^2 dx This will not work since we have to INT (Tan[x]^6 + Tan[x]^8 ) Sec[x]^2 dx change everything to Secants and that Tan[x]^7 Tan[x]^9 requires the tangents to be even power. -------- + --------- + C 7 9 So the other way is the right way. -------------------------------------------------------------------------- INT Tan[x]^5 Sec[x]^7 dx INT Tan[x]^5 Sec[x]^5 (Sec[x]^2 dx) INT Tan[x]^4 Sec[x]^6 (Sec[x] Tan[x]) dx Now we would have to convert the We convert all the Tangents to Secants Secants to Tangents. This we INT (Sec[x]^2-1)^2 Sec[x]^6 (Sec[x]Tan[x] dx cannot do since the exponent is Odd. INT (Sec[x]^10 -2 Sec[x]^8 + Sec[x]^6) (Sec[x] Tan[x] dx) This is not the way to go. Sec[x]^11 Sec[x]^9 Sec[x]^7 --------- -2 --------- + -------- + C 11 9 7 ------------------------------------------------------------------------------ odd even The remaining case is Sec[x] Tan[x] We simply express everything in terms of Sec[x] and use a reduction formula. (1) INT Sin[3x]^2 dx (2) INT Cos[x]^4 dx (3) INT Tan[x]^4 Sec[x]^2 (4) Find the reduction formula for INT Sec[x]^n dx (5) Find the reduction formula for INT Tan[x]^n dx