\(% 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}}\)

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2023.2.1 Question 1

  1. If \(x = \frac {1}{t}\), we have \[ \DiffFrac {x}{t} = - \frac {1}{t^2} \] and hence \[ \Diff x = - \frac {\Diff t}{t^2}. \]

    Hence, \begin {align*} \int _{a}^{b} \frac {\Diff x}{\left (1 + x^2\right )^{\frac {3}{2}}} & = \int _{a^{-1}}^{b^{-1}} \frac {- \Diff t}{t^2 \left (1 + \frac {1}{t^2}\right )^{\frac {3}{2}}} \\ & = \int _{a^{-1}}^{b^{-1}} \frac {-t \Diff t}{t^3 \left (1 + \frac {1}{t^2}\right )^{\frac {3}{2}}} \\ & = \int _{a^{-1}}^{b^{-1}} \frac {-t \Diff t}{\left (1 + t^2\right )^{\frac {3}{2}}} \end {align*}

    as desired.

  2. We have \[ \int _{a^{-1}}^{b^{-1}} \frac {-t \Diff t}{\left (1 + t^2\right )^{\frac {3}{2}}} = \left [\left (1 + t^2\right )^{-\frac {1}{2}}\right ]_{a^{-1}}^{b^{-1}}. \]

    1. \begin {align*} \int _{\frac {1}{2}}^{2} \frac {\Diff x}{\left (1 + x^2\right )^{\frac {3}{2}}} & = \int _{2}^{\frac {1}{2}} \frac {-t \Diff t}{\left (1 + t^2\right )^{\frac {3}{2}}} \\ & = \left [\left (1 + t^2\right )^{- \frac {1}{2}}\right ]_{2}^{\frac {1}{2}} \\ & = \left (1 + \left (\frac {1}{2}\right )^2\right )^{-\frac {1}{2}} - \left (1 + \left (2\right )^2\right )^{-\frac {1}{2}} \\ & = \frac {1}{\sqrt {\frac {5}{4}}} - \frac {1}{\sqrt {5}} \\ & = \frac {2}{\sqrt {5}} - \frac {1}{\sqrt {5}} \\ & = \frac {1}{\sqrt {5}}. \end {align*}
    2. Notice that the integrand is even, we have \begin {align*} \int _{-2}^{2} \frac {\Diff x}{\left (1 + x^2\right )^{\frac {3}{2}}} & = 2 \int _{0}^{2} \frac {\Diff x}{\left (1 + x^2\right )^{\frac {3}{2}}} \\ & = 2 \lim _{u \to 0^{+}} \int _{u}^{2} \frac {\Diff x}{\left ((1 + x^2)\right )^{\frac {3}{2}}} \\ & = 2 \lim _{u \to 0^{+}} \int _{\frac {1}{u}}^{\frac {1}{2}} \frac {-t \Diff t}{\left (1 + t^2\right )^{\frac {3}{2}}} \\ & = 2 \lim _{u \to \infty } \int _{u}^{\frac {1}{2}} \frac {-t \Diff t}{\left (1 + t^2\right )^{\frac {3}{2}}} \\ & = 2 \lim _{u \to \infty } \left [\left (1 + t^2\right )^{-\frac {1}{2}}\right ]_{u}^{2} \\ & = 2 \cdot \left (\frac {2}{\sqrt {5}} - \lim _{u \to \infty } \frac {1}{\sqrt {1 + u^2}}\right ) \\ & = 2 \cdot \left (\frac {2}{\sqrt {5}} - 0\right ) \\ & = \frac {4}{\sqrt {5}}. \end {align*}

    1. Starting from the left, we have \begin {align*} \int _{\frac {1}{2}}^{2} \frac {\Diff x}{\left (1 + x^2\right )^2} & = \int _{2}^{\frac {1}{2}} \frac {- \frac {1}{t^2} \Diff t}{\left (1 + \frac {1}{t^2}\right )^2} \\ & = \int _{\frac {1}{2}}^{2} \frac {\frac {1}{t^2} \cdot t^4 \Diff t}{t^4 \left (1 + \frac {1}{t^2}\right )^2} \\ & = \int _{\frac {1}{2}}^{2} \frac {t^2 \Diff t}{\left (1 + t^2\right )^2}, \end {align*}

      and therefore the first equal sign is true.

      As for the second equal sign, we notice that \begin {align*} \int _{\frac {1}{2}}^{2} \frac {\Diff x}{\left (1 + x^2\right )^2} + \int _{\frac {1}{2}}^{2} \frac {x^2\Diff x}{\left (1 + x^2\right )^2} & = \int _{\frac {1}{2}}^{2} \frac {\left (1 + x^2\right ) \Diff x}{\left (1 + x^2\right )^2} \\ & = \int _{\frac {1}{2}}^{2} \frac {\Diff x}{1 + x^2}, \end {align*}

      which means that \[ \int _{\frac {1}{2}}^{2} \frac {\Diff x}{\left (1 + x^2\right )^2} = \int _{\frac {1}{2}}^{2} \frac {x^2 \Diff x}{\left (1 + x^2\right )^2} = \frac {1}{2} \int _{\frac {1}{2}}^{2} \frac {\Diff x}{1 + x^2}. \]

      Hence, \begin {align*} \int _{\frac {1}{2}}^{2} \frac {\Diff x}{\left (1 + x^2\right )^2} & = \frac {1}{2} \int _{\frac {1}{2}}^{2} \frac {\Diff x}{1 + x^2} \\ & = \frac {1}{2} \left [\arctan x\right ]_{\frac {1}{2}}^{2} \\ & = \frac {1}{2} \arctan 2 - \frac {1}{2} \arctan \frac {1}{2} \\ & = \frac {1}{2} \arctan 2 - \frac {1}{2} \left (\frac {\pi }{2} - \arctan 2\right ) \\ & = \arctan 2 - \frac {\pi }{4}. \end {align*}

    2. Let \(x = \frac {1}{u}\), we have \(\Diff x = - \frac {1}{u^2} \Diff u\).

      Let the integral be \(I\), and we have \begin {align*} I & = \int _{\frac {1}{2}}^{2} \frac {1 - x}{x \left (1 + x^2\right )^{\frac {1}{2}}} \Diff x \\ & = \int _{\frac {1}{2}}^{2} \frac {1 - \frac {1}{u}}{\frac {1}{u} \left (1 + \frac {1}{u^2}\right )^{\frac {1}{2}}} \cdot \frac {1}{u^2} \Diff u \\ & = \int _{\frac {1}{2}}^{2} \frac {u - 1}{u^2 \left (1 + \frac {1}{u^2}\right )^{\frac {1}{2}}} \Diff u \\ & = \int _{\frac {1}{2}}^{2} \frac {u - 1}{u \left (1 + u^2\right )^{\frac {1}{2}}} \Diff u \\ & = -I. \end {align*}

      This therefore means \[ I = \int _{\frac {1}{2}}^{2} \frac {1 - x}{x \left (1 + x^2\right )^{\frac {1}{2}}} \Diff x = 0. \]

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