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. Wednesday, February 13 7.4 Rational Substitution p403: 4,18,32,33 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: p399: (6),(12),(24),(32) Page 399 Problem 6 x=Pi/2 6 INT Sin[x] dx x=0 _ _ 3 x=Pi/2 | 1-Cos[2x] | INT | ----------- | dx x=0 |_ 2 _| x=Pi/2 3 1/8 INT (1-Cos[2x]) dx x=0 x=Pi/2 2 3 1/8 INT 1 - 3 Cos[2x] + 3 Cos [2x] - Cos [2x] dx x=0 x=Pi/2 1+Cos[4x] 2 1/8 INT 1 - 3 Cos[2x] + 3 ---------- - (1-Sin [2x]) Cos[2x] dx x=0 2 _ _ | 3 | x=Pi/2 1/8 | x - 3 Sin[2x]/2 + 3/2 x + 3/2 Sin[4x]/4 -Sin[2x]/2 + Sin [2x]/6 | |_ _| x=0 1/8 ( Pi/2 - 0 + 3Pi/4 + 0 0 0 ) = 5 Pi/32 = 0.490874 --------------------------------------------------------------- Integrate[Sin[x]^6,{x,0,Pi/2}] -------------------------------------------------------------------- Get["font.math"]; P1 = Plot[Sin[x]^6,{x,0,Pi/2},PlotStyle->{Thickness[0.01]}]; P2 = ParametricPlot[{Pi/2,y},{y,0,1},PlotStyle->{Thickness[0.01]}]; P3 = ListPlot[ {{ (Pi/2-5 Pi/16),0},{Pi/2,1}}, PlotStyle->{RGBColor[1,0,0],Thickness[0.003]},PlotJoined->True]; P4 = Show[P1,P2,P3,PlotLabel->P399 P6: "y=Sin[x]^6", PlotRange->All,AspectRatio->Automatic]; Display["p6.ps",P4]; ----------------------------------------------------------------- Page 399 Problem 12 6 2 INT Cos [x] Sin [x] dx _ _ 3 _ _ | 1+Cos[2x] | | 1-Cos[2x] | INT |----------- | | ------------- | dx |_ 2 _| |_ 2 _| 2 2 1/16 INT (1+Cos[2x]) ( 1- Cos [2x] ) dx 2 2 1/16 INT (1 + Cos [2x] + 2 Cos[2x])(1-Cos [2x]) dx 4 3 1/16 INT 1 - Cos [2x] + 2 Cos[2x] - 2 Cos [2x] dx _ _ 2 | 1 + Cos[4x] | 2 1/16 INT 1 - | ------------- | + 2 Cos[2x] -2 (1- Sin [2x]) Cos[2x] dx |_ 2 _| 2 2 1/16 INT 1-1/4-2 Cos[4x]/4-Cos [4x]/4+2 Cos[2x]-2 Cos[2x]+2 Sin [2x] Cos[2x] dx 2 2 1/16 INT 3/4 - 2/4 Cos[4x] - Cos [4x]/4 + 2 Sin [2x] Cos[2x] dx 1 + Cos[8x] 2 1/16 INT 3/4 - 2/4 Cos[4x] - ----------- + 2 Sin [2x] Cos[2x] dx 8 2 1/16 INT 5/8 -2/4 Cos[4x] - Cos[8x]/8 + 2 Sin [2x] Cos[2x] dx _ _ | 3 | 1/16 | 5/8 x -2/16 Sin[4x] -1/64 Sin[8x] + Sin [2x]/3 + C | |_ _| ----------------------------------------------------------------------- Get["font.math"]; F[x_] := 1/16( 5/8 x - 2/16 Sin[4x] -1/64 Sin[8x] + (1/3) Sin[2x]^3); f[x_] := Cos[x]^6 Sin[x]^2; P1 = Plot[f[x],{x,-Pi,Pi}]; P2 = Plot[F[x],{x,-Pi,Pi},PlotStyle->{RGBColor[1,0,0]}]; P3 = Show[P1,P2,PlotLabel->"P399 P12, y = Cos[x]^6 Sin[x]^2,red=INT f"]; Display[ "p12x.ps",P3]; delta := 0.1; g[x_] := (F[x+delta]-F[x])/delta; P4 = Plot[g[x],{x,-Pi,Pi},PlotStyle->{RGBColor[0,1,0]}]; P5 = Show[P1,P4, PlotLabel->"P399 P12 y = Cos[x]^6 Sin[x]^2, Approximate F' step=0.1", PlotRange->All]; Display["p12y.ps",P5] --------------------------------------------------------------------------- Page 399 Problem 24 5 INT Ctn [2t] dt _ _ 2 | 2 | INT | Ctn [2t] | Ctn[2t] dt |_ _| _ _ 2 | 2 | INT | Csc [2t] -1 | Ctn[2t dt |_ _| _ _ | 4 2 | INT | Csc [2t] - 2 Csc [2t] + 1 | Ctn[2t] dt |_ _| _ _ | 3 | INT | Csc [2t] - 2 Csc [2t] | Csc[2t] Ctn[2t] dt + INT Ctn[2t] dt |_ _| 4 2 Csc [2t] Csc [2t] ln(Sin[2t]) - --------- + 2 --------- + ----------- + C 4*2 2*2 2 ---------------------------------------------------------------------------- Integrate[ Cot[2t]^5,t ] = -(1/8) Csc[2t]^4 +(1/2) Csc[2t]^2 + 1/2 Log[Sin[2t]] ------------------------------------------------------------------------------ Page 399 Problem 32 2 2 The region bounded by y = Sin [x ] , y = 0, and x = Sqrt[ Pi/2] is revolved about the y-axis. Find the volume of the resulting solid. --------------------------------- Get["font.math"]; f[x_] := Sin[x^2]^2; P1 = Plot[f[x],{x,0,Sqrt[Pi/2]}]; P2 = ParametricPlot[ {Sqrt[Pi/2],y}, {y,0, f[ Sqrt[Pi/2] ] }]; P3 = Show[P1,P2,PlotLabel->"P399 P32 y = Sin[x^2]^2"]; Display["p32x.ps",P3]; P4 = ParametricPlot3D[{x Cos[t], x Sin[t],f[x]},{x,0,Sqrt[Pi/2]},{t,0,3 Pi/2}]; P5 = ParametricPlot3D[{x,0,f[x],{RGBColor[1,0,0],Thickness[0.01]}}, {x,0,Sqrt[Pi/2]}]; P6 = ParametricPlot3D[{x,0,1,{RGBColor[1,0,0],Thickness[0.01]}}, {x,0,Sqrt[Pi/2]}]; P7 = ParametricPlot3D[{0,0,z,{RGBColor[1,0,0],Thickness[0.01]}}, {z,0,1}]; P8 = Show[P4,P5,P6,P7,PlotLabel->"P399 P32 y = Sin[x^2]^2"]; Display["p32y.ps",P8]; ------------------------------------------------- x=Sqrt[Pi/2] 2 2 INT 2 Pi x Sin[x ] dx x=0 x=Sqrt[Pi/2] 2 2 2 Pi INT Sin[x ] x dx x=0 2 x=Sqrt[Pi/2] 1 - Cos[2 x ] 2 Pi INT -------------- x dx x=0 2 x=Sqrt[Pi/2] 2 Pi INT (1 - Cos[2 x ] ) x dx x=0 _ _ | 2 2 | x = Sqrt[Pi/2] | x Sin[2 x ] | Pi | ---- - -------------- | |_ 2 4 _| x=0 _ _ 2 | Pi/2 | Pi Pi | ----- | = ------ |_ 2 _| 4 -------------------------------------------------------- Integrate[ 2 Pi x Sin[x^2]^2,{x,0,Sqrt[Pi/2]}]; ========================================================== New Material: Triangular substitution. Example 4 Page 401 2 2 INT Sqrt[a - x ] dx /| / | / | 2 2 a / | Sqrt[a -x ] / | / | / | /t | /________| x x = a Cos[t] dx = -a Sin[t] 2 2 Sqrt[a - x ] = a Sin[t] INT a Sin[t] (-a Sin[t]) dt 2 2 -a INT Sin [t] dt 2 1-Cos[2t] -a INT --------- dt 2 2 -a /2 (t - Sin[2t]/2) 2 -a /2 t + (1/2) a Sin[t] a Cos[t] + C 2 2 2 -a /2( ArcCos[x/a] + (1/2) x Sqrt[a - x ]) + C ------------------------------------------------- ans: a^2/2 ArcSin[x/a] + x/2 Sqrt[a^2 - x^2] + C -------------------------------------------------- Example 5 Page 401 dx INT ----------- 2 Sqrt[9+x ] /| / | 2 / | Sqrt[9+x ] / | 3 / | / | / | /t | /________| x x = 3 Ctn[t] 2 Sqrt[9+x ] = 3 Csc[t] 2 dx = -3 Csc [t] dt 2 -3 Csc [t] dt INT -------------- 3 Csc[t] - INT Csc[t] dt ln( Csc[t] + Ctn[t] ) + C _ 2 _ | Sqrt[9+x ] x | ln|------------ + ---- | + C |_ 3 3 _| ----------------------------------- | | ans: ln | Sqrt[9+x^2] + x | + C | | ----------------------------------- Example 6 Page 402 2 x=4 Sqrt[x - 4] INT ------------ dx x=2 x /| / | / | 2 x / | Sqrt[x - 4] / | / | / | /t | /________| 2 x = 2 Sec[t] dx = 2 Sec[t] Tan[t] dt 2 Sqrt[x - 4] = 2 Tan[t] x=4 2 Tan[t] INT --------------- 2 Sec[t] Tan[t] dt x=2 2 Sec[t] x=4 2 INT 2 Tan [t] dt x=2 x=4 2 2 INT Sec [t] - 1 dt x=2 t=Pi/3 _ _ x=4 | | 2 | Tan[t] - t | |_ _| x=2 t=0 2 (Tan[Pi/3]-Pi/3) 2 Sqrt[3] - 2 Pi/3 ------------------------------------- ans: 2 Sqrt[3] - 2 Pi/3 = 1.36971 ------------------------------------ -------------------------------------------------------------------- Get["font.math"]; f[x_] := Sqrt[x^2-4]/x; a = Plot[f[x],{x,2,5}]; b = ParametricPlot[{4,y},{y,0,f[4]},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->"P383 ex6: y=Sqrt[x^2 - 4]/x", AspectRatio->Automatic]; Display["ex6x.ps",c]; d = Table[ ParametricPlot[{i/4,y},{y,0,f[i/4]}, PlotStyle->{RGBColor[1,0,0]}],{i,9,16}]; e = Table[ Plot[f[i/4],{x,i/4,(i+1)/4},PlotStyle->{RGBColor[1,0,0]}],{i,9,15}]; h = Show[a,b,d,e,PlotLabel->"P383 ex6: y=Sqrt[x^2 - 4]/x", AspectRatio->Automatic]; Display["ex6y.ps",h]; uppersum = Sum[ f[i/4],{i,9,16}] 1/4; lowersum = Sum[ f[i/4],{i,8,15}] 1/4; ------------------------------------------------------------------------- Lower Sum = 1.23615 < 1.36971 < 1.45266 = Upper Sum ------------------------------------------------------------------------- Example 7 Page 402 dx INT --------------------- 2 Sqrt[x + 2x + 26] dx INT --------------------- 2 Sqrt[ (x+1) + 25] /| / | 2 / | Sqrt[ (x+1) +25] / | / | 5 / | / | /t | /________| x+1 x+1 = 5 Ctn[t] 2 dx = -5 Csc [t] dt Sqrt[(x+1)^2+25] = 5 Csc[t] 2 -5 Csc [t] dt INT ----------------- 5 Csc[t] -INT Csc[t] dt _ _ | | ln| Csc[t] + Ctn[t] | + C |_ _| _ 2 _ | Sqrt[ (x+1) + 25] x+1 | ln| ------------------- + ------ | + C |_ 5 5 _| ----------------------------------------------- ans: n | Sqrt[x^2 + 2x + 26] + x + 1] + C ---------------------------------------------- ###################################################################### Rationalizing substitution: Example 1 Page 400 dx INT -------- x-Sqrt[x] u = Sqrt[x] -1/2 du = 1/2 x dx 2 u du INT ------------- 2 u - u 2 du INT --------- u - 1 2 ln(u-1) + C 2 ln( Sqrt[x] - 1) + C Example 2 Page 400 1/3 INT x (x-4) dx 1/3 u = (x-4) 3 u = x-4 2 3 u du = dx 3 2 INT (u + 4)u 3 u du 6 3 3 INT u + 4 u du 7 4 3 ( u /7 + u ) _ _ | 7/3 | | (x-4) 4/3 | 3 | ---------- + (x-4) | + C | 7 | |_ _| Example 3 Page 400 2/5 INT x (x+1) dx 1/5 u = (x+1) 5 u = x+1 4 5 u du = dx 5 2 4 INT (u - 1) u 5 u du 11 6 5 INT u - u du _ _ | 12 7 | | u u | 5 | ------ - ---- | + C |_ 12 7 _| 12/5 7/5 5/12 (x+1) -5/7 (x+1) + C