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 4 8.3 Rat'n Subst's p385: (1),(8),9-13,(20),21,27,29,(30) Main Idea: Draw a triangle. Sines and Cosines are your friends. Much better than radicals. Key Words: Triangular substitution. Goal: Learn to integrate semi-complicated things with Square roots in them. -------------------------------------------------------- Previous assignment: February 2 8.2 Trig Integrals p380: 1-3,(4),(10),15, 17, 18, 23, (26),(34) Page 380 Problem 4 3 INT Cos [x] dx 2 INT (1-Sin [x]) Cos[x] dx 3 Sin[x] - 1/3 Sin [x] + C ------------------------------------------------ Page 380 Problem 10 1/2 3 INT Sin 2z Cos 2z dz 1/2 2 INT Sin 2z (1-Sin 2z) Cos[2z] dz 1/2 5/2 INT Sin 2z Cos[2z] - Sin 2z Cos[2z] dz 3/2 7/2 Sin 2z Sin 2z ---------- - ------------ 3/2 2 7/2 2 1/3 Sin[2z]^(3/2) -1/7 Sin[2z]^(7/2) + C ----------------------------------------------------- Page 380 Problem 26 3 -1/2 INT Tan [x] Sec [x] dx 2 -1/2 INT Tan[x] (Sec [x] -1)Sec [x] dx 1/2 -3/2 INT Sec [x] (Sec[x] Tan[x] ) - Sec [x] Sec[x] Tan[x] dx 3/2 -1/2 Sec [x] Sec [x] --------- - ----------- + C 3/2 -1/2 2/3 Sec[x]^(3/2) + 2 Sec[x]^(-1/2) ---------------------------------------------------------------------- Page 380 Problem 34 The shaded region between one arch of y = Sin[x] 0 <= x <= Pi and the line y = k 0 <= x <= k is revolved about the line y = k generating a solid S. Determine k so that S has (a) Minimum volume. (b) Maximum volume. x=Pi V = INT (Sin[x]-k)^2 dx x=0 x=Pi V = INT Sin[x]^2 -2 k Sin[x] + k^2 dx x=0 x=Pi 1-Cos[2x] V = INT ------------ -2 k Sin[x] + k^2 dx x=0 2 | | x=Pi V = | x/2 - 1/4 Sin[2x] + 2 k Cos[x] + k^2 x | | | x=0 V = Pi/2 - 2 k -2 k + k^2 Pi V = Pi/2 - 4 k + k^2 Pi dV/k = -4 + 2 k Pi Max/Min occurs when k = 4/(2 Pi) = 2/Pi \_/ must be a minimum. The Maximum must occur at the endpoint which is when k = 0 or k = 1. At k = 0 the area is Pi/2. At k = 1 the area is Pi/2 -4 + Pi which is less than Pi/2 So the maximum comes at k = 0. ---------------------------------------------------------------- f[k_] := Integrate[ (k-Sin[x])^2,{x,0,Pi}]; g[k_] = f[k]; a = Plot[g[k],{k,0,1}]; b = Show[a,PlotLabel->"Page 380 Problem 34",PlotRange->All]; Display["34.ps",b]; p[k_] := ParametricPlot3D[{ x,(k-Sin[x])*Cos[t], (k-Sin[x])*Sin[t]},{t,0,2 Pi},{x,0,Pi}]; ========================================================== New Material: Triangular substitution. Example 4 Page 382 INT Sqrt[a^2 - x^2] dx ans: a^2/2 ArcSin[x/a] + x/2 Sqrt[a^2 - x^2] + C --------------------------------------------------------------- Example 5 Page 383 dx INT ----------- Sqrt[9+x^2] ans: ln | Sec[t] + Tan[t] | ln| Sqrt[9+x^2] + x | + C ---------------------------------------------------- Example 6 Page 383 x=4 Sqrt[x^2-4] INT ------------ dx x=2 x ans: 2 Sqrt[3] - 2 Pi/3 --------------------------------------------------------- Example 7 Page 384 dx INT --------------------- Sqrt[x^2 + 2x + 26] ans: 2 Sqrt[x^2 + 2x + 26] - 2 ln | Sqrt[x^2 + 2x + 26] + x + 1] + C --------------------------------------------------------- Rationalizing substitution: dx INT -------- x-Sqrt[x] 1/3 INT x (x-4) dx ----------------------------------------------------