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.166.08 Office Hours 9:15-10:15 MTWThF Textbook: Calculus by Varberg, Purcell, Rigdon, ninth edition. Monday, February 11 7.3 Trig. Int. p399: 6,12,24,32 Main Idea: Save out a Cosine and change everything else to Sine. 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: p391: (32),(42),(46),(50) Page 391 Problem 32 Use integration by parts to evaluate the integral. ln(x) INT --------- dx Sqrt[x] 1/2 x ln(x) ------ 1/2 1/x dx 1/Sqrt[x] dx ln(x) 1/2 -1/2 INT --------- dx = 2 x ln(x) - INT 2 x dx Sqrt[x] ln(x) 1/2 1/2 INT --------- dx = 2 x ln(x) - 4 x + C <==== ANSWER Sqrt[x] / -1/2 1/2 -1/2 -1/2 \/ x ln(x) + 2 x (1/x) -2 x = x ln(x) ------------------------------------------------------------- Integrate[ Log[x]/Sqrt[x],x] = -4 Sqrt[x] + 2 Sqrt[x] Log[x] ------------------------------------------------------------- Get["font.math"]; f[x_] := Log[x]/Sqrt[x]; F[x_] := -4 Sqrt[x] + 2 Sqrt[x] Log[x]-5; fp[x_] = D[f[x],x]; P1 = Plot[f[x],{x,0.1,20},PlotStyle->{Thickness[0.005]}]; P2 = Plot[F[x],{x,0.1,20},PlotStyle->{RGBColor[1,0,0],Thickness[0.005]}]; P3 = ListPlot[ { {E^2,f[E^2]},{E^2,F[E^2]} }, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P4 = Plot[f[E^2],{x,0,20},PlotStyle->{RGBColor[0,0,1]}]; P5 = Show[P1,P2,P3,P4, PlotLabel->"P391 p32: f[x] = Log[x]/Sqrt[x]; red=INT f(x)", PlotRange->All,AspectRatio->Automatic]; Display["p32.ps",P5]; ---------------------------------------------- Page 391 Problem 42 Apply integration by parts twice to evaluate each integral. at INT e Sin[t] dt at e -Cos[t] at a e dt Sin[t] dt at at at INT e Sin[t] dt = - e Cos[t] + a INT e Cos[t] dt at e Sin[t] at a e dt Cos[t] dt _ _ at at | at at | INT e Sin[t] dt = - e Cos[t] + a | e Sin[t] - a INT e Sin[t] dt | |_ _| 2 at at (a + 1) INT e Sin[t] dt = e ( - Cos[t] + a Sin[t] ) at at e ( - Cos[t] + a Sin[t] ) INT e Sin[t] dt = ------------------------------- <=== ANSWER 2 a + 1 at at a e ( - Cos[t] + a Sin[t] ) + e ( Sin[t] + a Cos[t] ) / ----------------------------------------------------------- \/ 2 a + 1 at 2 e ( a + 1) Sin[t] at ------------------------- = e Sin[t] 2 a + 1 ------------------------------------------------------------------ a t E (-Cos[t] + a Sin[t]) Integrate[ E^(a t) Sin[t], t] = ---------------------------- 2 1 + a Get["font.math"]; a = 1/3; b = ArcTan[-3]; b = 0; f[t_] := E^(a t) Sin[t]; F[t_] := E^(a t) (-Cos[t] + a Sin[t] )/(a^2+1); P1 = Plot[f[t], {t,-2 Pi,2 Pi}]; P2 = Plot[F[t], {t,-2 Pi,2 Pi},PlotStyle->{RGBColor[1,0,0]}]; P3 = ListPlot[{{b-2 Pi,f[b-2 Pi]},{b-Pi,f[b-Pi]},{b,f[b]},{b+Pi,f[b+Pi]}, {b+2 Pi,f[b+2 Pi]}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P4 = ListPlot[{{b-2 Pi,F[b-2 Pi]},{b-Pi,F[b-Pi]},{b,F[b]},{b+Pi,F[b+Pi]}, {b+2 Pi,F[b+2 Pi]}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P5 = Show[P1,P2,P3,P4,PlotLabel->"P391 p42: f(x) = E^(a t) Sin[t]; red = INT f", PlotRange->{{-2 Pi,2 Pi},{-6,6}}]; Display["p42.ps",P5]; -------------------------------------------------- Page 391 Problem 46 INT Cos( ln(x) ) dx Cos( ln(x) ) x -Sin( ln(x) )/x dx dx INT Cos( ln(x) ) dx = x Cos( ln(x) ) + INT Sin( ln(x) ) dx Sin( ln(x) ) x Cos( ln(x) )/x dx INT Cos( ln(x) ) dx = x Cos( ln(x) ) + x Sin( ln(x) ) - INT Cos( ln(x) ) 2 INT Cos( ln(x) ) dx = x Cos( ln(x) ) + x Sin( ln(x) ) INT Cos( ln(x) ) dx = (1/2) x ( Cos( ln(x) ) + Sin( ln(x) ) )+ C <====ANSWER / \/ (1/2) ( Cos( ln(x) ) + Sin( ln(x) ) ) (1/2) x( -Sin( ln(x) )/x + Cos( ln(x) )/x ) = Cos( ln(x) ) ------------------------------------------ Get["font.math"]; f[x_] := Cos[ Log[x] ]; F[x_] := (1/2) x ( Cos[ Log[x] ] + Sin[ Log[x] ]); Q1 = Plot[f[x],{x,0.1,E^( 5 Pi/2)}]; P1 = Plot[f[x],{x,0.1,E^( 5 Pi/2)},PlotPoints->2000]; A = Table[ {E^(n*Pi/2.0),f[E^(n*Pi/2.0)]},{n,1, 5}]; P2 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; Q3 = Show[Q1,P2,PlotLabel->"P391 P46: f(x) = Cos[ Log[x] ],25-points" ]; P3 = Show[P1,P2,PlotLabel->"P391 P46: f(x) = Cos[ Log[x] ],2000-points" ]; Display["p46x.ps",Q3]; Display["p46y.ps",P3]; ----------------------------------------------------------------- Page 391 Problem 50 Use integration by parts to derive the given formula. INT Cos[5x] Sin[7x] dx = -7/24 Cos[5x] Cos[7x] - 5/24 Sin[5x] Sin[7x] + C INT Cos[5x] Sin[7x] dx Cos[5 x] -(1/7) Cos[7 x] -5 Sin[5 x] dx Sin[7x] dx INT Cos[5x] Sin[7x] dx = -(1/7) Cos[5x] Cos[7x] - 5/7 INT Sin[5x] Cos[7x] Sin[5x] (1/7) Sin[7x] 5 Cos[5x] dx Cos[7x] dx INT Cos[5x] Sin[7x] dx = -(1/7) Cos[5x] Cos[7x] _ _ | | - 5/7 | (1/7) Sin[5x] Sin[7x] - (5/7) INT Cos[5 x] Sin[7 x] dx | |_ _| (1-25/49) Cos[5x] Sin[7x] = -(1/7) Cos[5x] Cos[7x] -(5/49) Sin[5x] Sin[7x] (24/49) Cos[5x] Sin[7x] = -(1/7) Cos[5x] Cos[7x] -(5/49) Sin[5x] Sin[7x] INT Cos[5x] Sin[7x] dx = -(7/24) Cos[5x] Cos[7x] -(5/24) Sin[5x] Sin[7x] ANSWER ------------------------------------------------------------ Get["font.math"]; f[x_] := Cos[5x] Sin[7x]; F[x_] := -7/24 Cos[5x] Cos[7x] - 5/24 Sin[5x] Sin[7x]; P1 = Plot[f[x],{x,-Pi/2, Pi/2}]; P2 = Plot[F[x],{x,-Pi/2, Pi/2},PlotStyle->{RGBColor[1,0,0]}]; A = Table[{n Pi/10, F[n Pi/10]},{n,-5,5,2}]; B = Table[{ n Pi/7, F[n Pi/7]},{n,-3,3}]; P3 = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; P4 = ListPlot[B,PlotStyle->{RGBColor[0,1,0]}]; P5 = Show[P1,P2,P3,P4,PlotLabel->"P391 P50: f[x] = Cos[5x] Sin[7x], red = INT f", PlotRange->All, AspectRatio->Automatic]; Display["p50x.ps",P5]; P6 = Plot[1/2 Sin[12x],{x,-Pi/2,Pi/2}, PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; P7 = Plot[1/2 Sin[ 2x],{x,-Pi/2,Pi/2}, PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; P8 = Show[P5,P6,P7,PlotLabel->"1/2 Sin[12 x];1/2 Sin[2x]"]; Display["p50y.ps",P8]; Sin[7x] Cos[5x] = 1/2( Sin[12 x]) + Sin[2x] ) ----------------------------------------------------------- New Material 2 2 Sin x + Cos x = 1. 2 1-Cos[2x] Sin x = ------------ 2 2 1+Cos[2x] Cos x = ----------- 2 --------------------------------- Page 304 Example 1: 5 INT Sin x dx 4 INT (Sin x) Sin[x] dx 2 2 INT (1-Cos [x]) Sin[x] dx 2 4 INT (1 - 2 Cos [x] + Cos [x] ) Sin[x] dx 3 5 Cos [x] Cos [x] - Cos[x] +2 ------- - ----------- + C <=== Answer 3 5 3 5 Ans - Cos[x] + 2/3 Cos [x] - 1/5 Cos [x] + C ------------------------------------------------------- Page 394 Example 2; 4 INT Cos x dx _ _ 2 | 1 + Cos[2x] | INT | ------------ | dx |_ 2 _| 2 (1/4) INT 1 + 2 Cos[2x] + Cos [2x] dx 1 + Cos[4x] (1/4) INT 1 + 2 Cos[2x] + ------------ dx 2 1/4( x + Sin[2x] + x/2 + Sin[4x]/8 ) + C 3/8 x + 1/4 Sin[2x] + 1/32 Sin[4x] + C Ans 3/8 x + 1/4 Sin[2x] + 1/32 Sin[4x] + C -------------------------------------------------- Page 394 Example 3 3 -4 INT Sin x Cos x dx 2 -4 INT Sin [x] Cos [x] Sin[x] dx 2 -4 INT (1-Cos [x]) Cos [x] Sin[x] dx -4 -2 INT (Cos [x] - Cos [x] ) Sin[x] dx -3 -1 Cos [x] Cos [x] ---------- - --------- + C (-3)(-1) (-1)(-1) Ans 1/3 Sec[x]^3 - Sec[x] + C ------------------------------------- Page 395 Example 4 2 4 INT Sin x Cos x dx _ _ _ _ 2 | 1-Cos[2x] || 1 + Cos[2x] | INT | ---------------||-----------------| dx |_ 2 _||_ 2 _| 2 (1/8) INT (1-Cos[2x]) ( 1+ 2 Cos[2x] + Cos 2x] ) dx 2 (1/8) INT 1 + 2 Cos[2x] + Cos [2x] 2 3 - Cos[2x] -2 Cos [2x] - Cos [2x] dx 2 3 (1/8) INT 1 + Cos[2x] - Cos [2x] - Cos [2x] dx 1 + Cos[4x] 2 (1/8) INT 1 + Cos[2x] - ---------------- - (1-Sin [2x]) Cos[2x] dx 2 _ 3 _ | Sin [2x] | (1/8)| x +(1/2) Sin[2x] - x/2 -(1/8) Sin[4x] -1/2 Sin[2x] + ----------- | |_ 6 _| 3 1/8( 1/2 x - (1/8) Sin[4x] + (1/6) Sin [2x] ) + C 3 Ans 1/8 ( 1/2 x - 1/8 Sin[4x] + 1/6 Sin [2x] ) + C 2 For Secants and Tangents, either save out a Sec [x] and change everything else to Tangents, OR Save out a Sec[x] Tan[x] and change everything else to Secants. The only thing that cannot be solved this way is equivalent to n Sec [x] which has the reduction formula suggested at the end of the last lecture. Page 398 Example 9 5 INT Tan [x] dx 3 2 INT Tan [x] (Sec [x] -1 ) 3 2 3 INT Tan [x] Sec [x] dx - INT Tan [x] dx 4 Tan [x] 2 -------- - INT Tan[x] (Sec [x] -1) dx 4 4 Tan [x] 2 -------- - INT Tan[x] Sec [x] dx + INT Tan[x] dx 4 4 2 Tan [x] Tan [x] -------- - -------- - ln(Cos[x]) + C 4 2 Page 398 Example 10. -3/2 4 INT Tan [x] Sec [x] dx -3/2 2 2 INT Tan [x] (1+Tan [x] )Sec [x] dx -3/2 2 1/2 2 INT Tan [x] Sec [x] dx + INT Tan [x] Sec [x] dx -1/2 3/2 Tan [x] Tan [x} ----------- + ----------- + C -1/2 3/2 Page 398 Example 11 3 -1/2 INT Tan [x] Sec [x] dx 2 -3/2 INT Tan [x] Sec [x] Sec[x] Tan[x]dx 2 -3/2 INT (Sec [x]-1) Sec [x] Sec[x] Tan[x]dx 1/2 -3/2 INT (Sec [x] Sec[x] Tan[x]dx - INT Sec [x] Sec[x]Tan[x] dx 3/2 -1/2 Sec [x] Sec [x] ---------- - ----------- + C 3/2 -1/2