\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

2014.3.2 Question 2

1. Since \(u = \cosh x\), \(\cosh 2x = 2 \cosh ^2 x - 1 = 2 u^2 - 1\), and \(\sinh x \Diff x = \Diff \cosh x = \Diff u\). Hence, \begin {align*} \int \frac {\sinh x}{\cosh 2x} \Diff x & = \int \frac {\Diff u}{2 u^2 - 1} \\ & = \int \frac {1}{2} \left (\frac {1}{\sqrt {2}u - 1} - \frac {1}{\sqrt {2}u + 1}\right ) \Diff u \\ & = \frac {1}{2\sqrt {2}} \left (\ln \abs *{\sqrt {2}u - 1} - \ln \abs *{\sqrt {2}u + 1}\right ) + C \\ & = \frac {1}{2\sqrt {2}} \ln \abs *{\frac {\sqrt {2}\cosh x - 1}{\sqrt {2}\cosh x + 1}} + C, \end {align*}

   as desired.

2. Let \(u = \sinh x\), \(\cosh 2x = 1 + 2\sinh ^2 x = 1 + 2u^2\), and \(\cosh x \Diff x = \Diff \sinh x = \Diff u\). Hence, \begin {align*} \int \frac {\cosh x}{\cosh 2x} \Diff x & = \int \frac {\Diff u}{1 + 2u^2} \\ & = \frac {1}{\sqrt {2}}\arctan ()\sqrt {2}u) + C \\ & = \frac {1}{\sqrt {2}} \arctan ()\sqrt {2} \sinh x) + C. \end {align*}

3. Notice that \[ \frac {\cosh x}{\cosh 2x} - \frac {\sinh x}{\cosh 2x} = \frac {2e^{-x}}{e^{2x} + e^{-2x}} = \frac {2e^{x}}{1 + e^{4x}}. \]

   Let \(u = e^x\), \(\Diff u = \Diff e^x = e^x \Diff x\), and therefore \begin {align*} \int _{0}^{1} \frac {\Diff u}{1 + u^4} & = \int _{-\infty }^{0} \frac {e^x \Diff x}{1 + x^4} \\ & = \frac {1}{2} \int _{-\infty }^{0} \frac {\cosh x}{\cosh 2x} - \frac {\sinh x}{\cosh 2x} \Diff x \\ & = \frac {1}{2} \left [\frac {1}{\sqrt {2}} \arctan (\sqrt {2} \sinh x) - \frac {1}{2\sqrt {2}} \ln \abs *{\frac {\sqrt {2}\cosh x - 1}{\sqrt {2}\cosh x + 1}}\right ]_{-\infty }^{0} \\ & = \frac {1}{4\sqrt {2}} \left [2\arctan (\sqrt {2} \sinh x) - \ln \abs *{\frac {\sqrt {2}\cosh x - 1}{\sqrt {2}\cosh x + 1}}\right ]_{-\infty }^{0} \\ & = \frac {1}{4\sqrt {2}} \left [\left (0 - \ln \abs *{\frac {\sqrt {2} - 1}{\sqrt {2} + 1}}\right ) - \left (2 \cdot (-\frac {\pi }{2}) - \ln \abs *{1}\right )\right ] \\ & = \frac {1}{4\sqrt {2}} \left [\pi - 2\ln (\sqrt {2} - 1)\right ] \\ & = \frac {\pi + 2\ln (\sqrt {2} + 1)}{4\sqrt {2}}, \end {align*}

   as desired.