Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://orion.math.iastate.edu/hentzel/class.166.05 Textbook: Calculus by Varberg, Purcell, Rigdon, eight edition. February 2 8.2 Trig Integrals p380: 1-3,(4),(10),15, 17, 18, 23, (26),(34) Main Idea: When its odd, save out one, and convert the others. When everything is even, shrink it. m n Key Words: INT Sin x Cos x dx m INT Sin x dx n INT Cos x dx Goal: Learn to integrate powers of sines and cosines. -------------------------------------------------------- Previous assignment: January 31 8.1 Substitution p375: 1-10, (28),30,(32),49,(50), 55-58,(68) Page 375 Problem 28 Sin[4t-1] INT --------------- dt 2 1-Sin [4t-1] Sin[4t-1] INT --------------- dt 2 Cos [4t-1] Let u = Cos[4t-1] du = -Sin[4t-1] 4 dt -(1/4) du u^(-1) 1 INT ----------------- = -(1/4) ------- + C = ---- +C u^2 -1 4u 1 = --------------- + C 4 Cos[4t-1] / -Sin[4t-1] 4 Sin[4t-1] \/ ---------------------- = -------------- -4 Cos[4t-1]^2 1-Sin[4t-1]^2 ------------------------------- Page 375 Problem 32 (6t-1) Sin[ Sqrt[ 3t^2 - t - 1] ] INT -------------------------------------- dt Sqrt[3t^2 - t - 1] u = Sqrt[3t^2 - t - 1] du = (1/2)/Sqrt[3t^2 -t -1] (6t -1) dt INT 2 Sin[u] du = -2 Cos[u] + C = -2 Cos[ Sqrt[3t^2 -t -1] ] + C / \/ 2 Sin[ Sqrt[3t^2 -t -1]] (1/2) (3t^2 -t -1)^(-1/2) (6t -1 ) Sin[ Sqrt[3t^2 -t - 1]] (6t -1 ) --------------------------------------------- Sqrt[ 3t^2 -t - 1] ------------------------------------- Page 375 Problem 50 dx INT ------------------ Sqrt[16 + 6x - x^2] dx INT ---------------------------- Sqrt[16 - ( -6x + x^2 ) ] dx INT ---------------------------- Sqrt[25 - ( 9 -6x + x^2 ) ] dx INT ------------------- Sqrt[25 - (x-3)^2 ] 1 dx ---- INT ------------------- 5 2 |x-3| Sqrt[1 - |---| ] | 5 | dx/5 INT ------------------- 2 |x-3| Sqrt[1 - |---| ] | 5 | ArcSin[ Sqrt[ (x-3)/5 ] + C / dx/5 \/ ---------------------------- Sqrt[ 1 - ((x-3)/5)^2 ] dx ---------------------------- Sqrt[25 - (x-3)^2 ] dx ---------------------------- Sqrt[16 +6x - x^2 ] -------------------------------- Page 375 Problem 68 Sin[x] Cos[x] Show Sec x = --------- + ----------- Cos[x] 1+Sin[x] Sin[x] + Sin[x]^2 + Cos[x]^2 = --------------------- Cos[x] (1+Sin[x] ) 1 + Sin[x] = ---------------------- Cos[x]( 1 + Sin[x] ) = Sec[x] Show INT Sec[x] dx = ln| Sec[x] + Tan[x] | + C Sin[x] Cos[x] INT Sec x dx = INT --------- dx + INT ----------- dx Cos[x] 1+Sin[x] = - ln|Cos[x]| + ln(1+Sin[x]) + C = ln | Sec[x] | + ln |1 + Sin[x] | + C 1 + Sin[x] = ln | ---------------- | + C Cos[x] = ln | Sec[x] + Tan[x] | + C ------------------------------------------------------- ------------------------------------------------------- New Material 2 2 Sin x + Cos x = 1. Sin[A+B] = Sin[A] Cos[B] + Cos[A] Sin[B] Cos[A+B] = Cos[A] Cos[B] - Sin[A] Sin[B] (Notice that these multiply like Complex numbers) (Cos[A]+i Sin[A])(Cos[B]+i Sin[B]) = Cos[A+B] + i Sin[A+B] 2 1-Cos[2x] Sin x = ------------ 2 2 1+Cos[2x] Cos x = ----------- 2 --------------------------------- 5 INT Sin x dx Ans - Cos[x] + 2/3 Cos[x]^3 - 1/5 Cos[x]^5 + C ------------------------------------------------------ 4 INT Cos x dx Ans 3/8 x + 1/4 Sin[2x] + 1/32 Sin[4x] + C -------------------------------------------------- 3 -4 INT Sin x Cos x dx Ans 1/3 Sec[x]^3 - Sec[x] + C ------------------------------------- 2 4 INT Sin x Cos x dx