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. ********************************************************* * * * TEST FRIDAY * * * ********************************************************* Monday, February 24 8.5 Rational Substitution p488: 1,4,(6),(10),15,(19),(25) Main Idea: You can integrate any rational function of trig identities. Key Words: t = Tan[x/2] Goal: Learn to remove roots by substitution of a power. Learn how to integrate things like Csc[x] Tan[x]+Sin[x] --------------------------------- 4 2 Sec[x] + 9 Tan[x] + 6 Cos[x] ================================================================= Previous Assignment Page 485 Problem 12 Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients. 1+16x ------------------------------------------ 2 2 (2x-3)(x+5) (x +x+1) A B C Dx+E ----------- + ------------ + ----------- + -------------- Answer. 2 2 x-3/2 (x+5) x+5 x + x + 1 =================================================================== Page 485 Problem 22 Evaluate the integral. 1 1 ------ ----- 1 -a+b -b+a INT ---------- dx = INT -------- + ------ dx (x+a)(x+b) x+a x+b -1 1 -------- ln(x+a) + ------ ln(x+b) + C <== Answer. a-b a-b -(x+b)+(x+a) ------------ (x+b)/(-a+b)+(x+a)/(-b+a) a-b 1 check ------------------------- = ---------- = ------- (x+a)(x+b) (x+a)(x+b) (x+a)(x+b) =================================================================== Page 485 Problem 34 Evaluate the integral dx A B C D INT -------------- = INT --- + --- + ------- + --- dt 2 2 2 x 2 x-1 x (x-1) x (x-1) ---------------------------------------------------------------------- 1 = A(x-1)^2 + B x(x-1)^2 + C x^2 + D x^2 (x-1) 1 = A(x^2 - 2x + 1) + B(x^3 -2x^2 + x) + C x^2 + D(x^3 - x^2) x^3 B D x^2 A -2 B C -D x -2 A B 1 A B +D = 0 A -2B +C -D = 0 -2A +B = 0 A = 1 A = 1 B = 2 D = -2 C = -A+2B+D = -1+4-2 = 1 -2A-2D = 0 A = 1 B = 2 C = 1 D = -2 dx 1 2 1 -2 INT -------------- = INT --- + --- + ------- + --- dt x^2 (x-1)^2 x^2 x (x-1)^2 x-1 x^(-1)/(-1) +2ln(x) +(x-1)^(-1)/(-1) -2ln(x-1) + C -1/x + 2 ln(x) -1/(x-1) - 2 ln(x-1) + C =================================================================== New Material Page 486 Example 1. Sqrt[x+4] INT ----------- dx x ------------------------------------------------------------------ Let u = Sqrt[x+4] then u^2 = x+4 x = u^2 -4 2u du = dx u INT ------------ 2 u du u^2 - 4 2 u^2 INT ------------- du u^2 -4 2(u^2 - 4) + 8 INT ----------------- du u^2 - 4 8 du INT 2 du + INT ------------ (u+2)(u-2) -2 +2 2 u + INT -------- + -------- du u+2 u-2 2 u -2 ln(u+2) +2 ln(u-2) u-2 2 u +2 ln(--------) u+2 Sqrt[x+4]-2 2 Sqrt[x+4] + 2 ln------------ + C Sqrt[x+4]+2 ==================================================================== Page 487 Example 2 dx INT ------------------ x^(1/2) - x^(1/3) -------------------------------------------------- x = u^6 dx = 6u^5 du 6 u^5 du INT ----------------- u^3 - u^2 6 u^3 INT-------- du u - 1 u^3 -1 + 1 6 INT ---------- du u-1 1 6 INT u^2 + u + 1 + --- du u-1 6[ u^3/3 + u^2/2 + u + ln(u-1) ] + C 6[ x^(1/2)/3 + x^(1/3)/2 + x^(1/6) + ln(x^(1/6)-1) + C. ================================================================= Finally, the substitutions that change any rational expression in Sines and Cosines into rational expressions in t. 2t 1-t^2 2 Sin[x] = -------- Cos[x] = ------- dx = ---------- 1+t^2 1+t^2 1+t^2 t = Tan[x/2] dx INT --------------------------- my way. 3 Sin[x] - 4 Cos[x] 1 1 --- ----- dt 5 -5 INT --------------- === --------- + ------------- 2t^2 + 3t -2 t-1/2 t+2 Book's way. || (2/5) -(1/5) || INT ---------- + -------- dt 2t-1 t+2 | 2t-1 | | t-1/2 | 1/5 ln |------| + C 1/5 ln |-------| + C | t+2 | | t+2 | | 2 Tan[x/2] -1 | | Tan[x/2]-1/2 | 1/5 ln |---------------| 1/5 ln |------------- | + C | Tan[x/2] + 2 | | Tan[x/2]+2 | How the substitutions are derived. /| / | / | Sqrt[1+t^2] / | / |t / | /x/2 | /_______| 1 t Sin[x] = 2 Sin[x/2] Cox[x/2] = 2 --------- 1+t^2 1 1-t^2 Cos[x] = 2 Cos[x/2]^2 -1 = 2 --------- - 1 = -------- 1+t^2 1+t^2 x = ArcTan[t] 1 dx = ----------- 1+t^2