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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 12729, 529]*) (*NotebookOutlinePosition[ 13560, 558]*) (* CellTagsIndexPosition[ 13516, 554]*) (*WindowFrame->Normal*) Notebook[{ Cell["p. 591", "Subsubsection"], Cell[CellGroupData[{ Cell["2.", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\[DifferentialD]x\/\(x \@\( x + 4\)\)\)], "Input"], Cell[BoxData[ \(\(-ArcTanh[\@\(4 + x\)\/2]\)\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["8.", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\[DifferentialD]x\/\(\(x\^2\) \@\(4 x - 9\)\)\)], "Input"], Cell[BoxData[ \(\@\(\(-9\) + 4\ x\)\/\(9\ x\) + 4\/27\ ArcTan[1\/3\ \@\(\(-9\) + 4\ x\)]\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["16.", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\(q\^2\) \(\@\(25 - 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n)\)\ \((1 + n)\)\) + \(x\ Log[x]\)\/\(1 + n\)) \)\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["110.", "Subsubsection"], Cell["a)", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\(\(\ Log[x]\)\/x\^2\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(\(-\(1\/x\)\) - Log[x]\/x\)], "Output"] }, Open ]], Cell["b)", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\(\(\ Log[x]\)\/x\^3\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(\(-\(1\/\(4\ x\^2\)\)\) - Log[x]\/\(2\ x\^2\)\)], "Output"] }, Open ]], Cell["c)", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\(\(\ Log[x]\)\/x\^4\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(\(-\(1\/\(9\ x\^3\)\)\) - Log[x]\/\(3\ x\^3\)\)], "Output"] }, Open ]], Cell[TextData[{ "d) The pattern looks like ", Cell[BoxData[ \(TraditionalForm \`\[Integral]\(\(ln(x)\)\/x\^n\) \[DifferentialD]x = \ \(-\(1\/\(\(\((n - 1)\)\^2\) x\^\(n - 1\)\)\)\) - \(1\/\(n - 1\)\) \(ln(x)\)\)]], ", so for ", Cell[BoxData[ \(TraditionalForm\`n = 5\)]], "the answer should be ", Cell[BoxData[ \(TraditionalForm\`\(-\ \(1\/\(16 x\^4\)\)\) - \(1\/4\) \(ln(x)\)\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\(\(\ Log[x]\)\/x\^5\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(\(-\(1\/\(16\ x\^4\)\)\) - Log[x]\/\(4\ x\^4\)\)], "Output"] }, Open ]], Cell[TextData[{ "e) ", StyleBox["Mathematica", FontSlant->"Italic"], " knows the general formula:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\ \(Log[x]\/x\^n\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(x\^\(-n\)\ \((x\/\(\((1 - n)\)\ \((\(-1\) + n)\)\) + \(x\ Log[x]\)\/\(1 - n\)) \)\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["111.", "Subsubsection"], Cell[TextData[{ "a) ", StyleBox["Mathematica", FontSlant->"Italic"], " cannot compute:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(Sin[x]\^n\/\(Cos[x]\^n + Sin[x]\^n\)\) \[DifferentialD]x\)], "Output"] }, Open ]], Cell["b)", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) \[DifferentialD]x\ /. \ n \[Rule] 1\)], "Input"], Cell[BoxData[ \(\[Pi]\/4\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) \[DifferentialD]x\ /. \ n \[Rule] 2\)], "Input"], Cell[BoxData[ \(\[Pi]\/4\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) \[DifferentialD]x\ /. \ n \[Rule] 3\)], "Input"], Cell[BoxData[ \(1\/3\ \((\(-I\)\ \[Pi] - Log[2])\) + 1\/12\ \((\((3 + 4\ I)\)\ \[Pi] + Log[16])\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[%]\)], "Input"], Cell[BoxData[ \(\[Pi]\/4\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) \[DifferentialD]x\ /. \ n \[Rule] 4\)], "Input"], Cell[BoxData[ \(\[Pi]\/4\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) \[DifferentialD]x\ /. \ n \[Rule] 5\)], "Input"], Cell[BoxData[ \(1\/20\ \((5\ \[Pi] + 2\ \((\(-1\) + \@5)\)\ Log[\(-1\) + \@5] - 2\ \((1 + \@5)\)\ Log[1 + \@5])\) + 1\/10\ \(( \(-\((\(-1\) + \@5)\)\)\ Log[\(-1\) + \@5] + \((1 + \@5)\)\ Log[1 + \@5])\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[%]\)], "Input"], Cell[BoxData[ \(\[Pi]\/4\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) \[DifferentialD]x\ /. \ n \[Rule] 6\)], "Input"], Cell[BoxData[ \(\[Pi]\/4\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) \[DifferentialD]x\ /. \ n \[Rule] 7\)], "Input"], Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(Sin[x]\^7\/\(Cos[x]\^7 + Sin[x]\^7\)\) \[DifferentialD]x\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " gives up at ", Cell[BoxData[ \(TraditionalForm\`n = 7\)]], "." }], "Text"], Cell[TextData[{ "c) Substituting ", Cell[BoxData[ \(TraditionalForm\`x\ = \ \[Pi]/2\ - \ u\)]], ", ", Cell[BoxData[ \(TraditionalForm \`\[DifferentialD]x\ = \ \(-\ \[DifferentialD]u\)\)]], ", and using the identities ", Cell[BoxData[ \(TraditionalForm\`sin(\[Pi]/2 - u)\ = \ cos(u)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`cos(\[Pi]/2 - u)\ = \ sin(u)\)]], "we get ", Cell[BoxData[ \(I\ = \ \ \(\[Integral]\_0 \%\(\[Pi]/2 \)\(\(\(sin\^n\) \((x)\)\)\/\(\(sin\^n\) \((x)\)\ + \ \(cos\^n\) \((x)\)\)\) \[DifferentialD]x\ = \ \(\(-\(\[Integral]\_\(\[Pi]/2\)\%0 \(\(\( cos\^n\) \((u)\)\)\/\(\(cos\^n\) \((u)\)\ + \ \(sin\^n\) \((u)\)\ \)\) \[DifferentialD]u\)\)\ = \[Integral]\_0 \%\(\[Pi]/2 \)\(\(\(cos\^n\) \((u)\)\)\/\(\(sin\^n\) \((u)\)\ + \ \(cos\^n\) \((u)\)\ \)\) \[DifferentialD]u\)\)\)]], "\nTherefore,\n", Cell[BoxData[ \(I\ + \ I\ = \ \ \(\[Integral]\_0 \%\(\[Pi]/2 \)\(\(\(sin\^n\) \((t)\)\)\/\(\(sin\^n\) \((t)\)\ + \ \(cos\^n\) \((t)\)\)\) \[DifferentialD]t\ + \[Integral]\_0 \%\(\[Pi]/2 \)\(\(\(cos\^n\) \((t)\)\)\/\(\(sin\^n\) \((t)\)\ + \ \(cos\^n\) \((t)\)\)\) \[DifferentialD]t\ = \ \(\[Integral]\_0 \%\(\[Pi]/2 \)\(\(\(sin\^n\) \((t)\)\ + \ \(cos\^n\) \((t)\)\)\/\(\(sin\^n\) \((t)\)\ + \ \(cos\^n\) \((t)\)\)\) \[DifferentialD]t\ = \ \(\[Integral]\_0\%\(\[Pi]/2\)\[DifferentialD]t = \ \[Pi]\/2\)\)\)\)]], "\nand ", Cell[BoxData[ \(I\ = \ \[Pi]\/4\)]], "." }], "Text"] }, Open ]] }, FrontEndVersion->"X 3.0", ScreenRectangle->{{0, 1024}, {0, 768}}, ScreenStyleEnvironment->"Working", WindowSize->{520, 600}, WindowMargins->{{Automatic, 67}, {12, Automatic}}, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PaperSize"->{612, 792}, "PaperOrientation"->"Portrait", "Magnification"->1}, ShowCellLabel->False ] (*********************************************************************** Cached data follows. 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