Problems

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Solution:

The fluid can be divided into a cuboid parallel to the slope and a right-angled triangle. If the height of the water on the longer side is \(\ell a\), then we have \(ka^3 = (\ell a - a\tan \theta)a^2 + \frac12 a^3\tan \theta \Rightarrow \ell = k + \frac12 \tan \theta\) This is acceptable when \(k > \frac12 \tan \theta\). The centre of mass of the cuboid will be \((\frac{a}{2}, \frac12 (k - \frac12 \tan \theta))\) and of the triangle will be \((\frac13 a, \frac13 \tan \theta + (k - \frac12 \tan \theta) )\) Therefore we have: \begin{align*} && \text{COM} && \text{mass} \\ \text{cuboid} && (\frac{a}{2}, \frac{a}2 (k - \frac12 \tan \theta)) && a^3(k - \frac12 \tan \theta) \\ \text{triangle} && (\frac13 a, \frac{a}3 \tan \theta + a(k - \frac12 \tan \theta) ) && a^3\frac12 \tan \theta \\ \text{whole system} && (x, y) && a^3k \end{align*} Therefore \begin{align*} && a^3k x &= \frac{a}{2} \cdot a^3(k - \frac12 \tan \theta) + \frac13 a \cdot a^3\frac12 \tan \theta \\ &&&= a^4 \frac{k}{2} - \frac{1}{12}a^4 \tan \theta \\ \Rightarrow && \frac{x}{a} &= \frac12 - \frac{\tan \theta}{12 k} \\ \\ && a^3k y &= \frac{a}2 (k - \frac12 \tan \theta) \cdot a^3(k - \frac12 \tan \theta) + \\ &&& \qquad\qquad \cdots + \l \frac{a}3 \tan \theta + a(k - \frac12 \tan \theta) \r \cdot a^3\frac12 \tan \theta \\ &&&= \frac{a^4k^2}{2} -\frac{a^4k \tan \theta}{2} + \frac{a^4 \tan^2 \theta}{8} - \frac{a^4 \tan^2 \theta}{12} + \frac{a^4k \tan \theta}{2} \\ \Rightarrow && \frac{y}{a} &= \frac{k}2 + \frac{\tan^2 \theta}{24k} \end{align*} If the water only fills up a prism, it's sides must be \(b\) and \(b\tan \theta \), therefore the volume is \(\frac12 ab^2 \tan \theta = ka^3 \Rightarrow b = a\sqrt{\frac{2k}{\tan \theta}}\) The centre of mass will be at \(\l \frac13 a\sqrt{\frac{2k}{\tan \theta}}, a\sqrt{2k \tan \theta}\r\) The container will topple if the centre of mass is outside the base, ie if the centre of mass \((x,y)\) lies above the line \(y = \tan (90^\circ- x) = \frac{1}{\tan \theta} x\). If \(\theta > 45^\circ\) then \(\tan \theta > 1\) and so we are in the \(\frac12 \tan \theta > \frac12 > k\) and so we are in the second case. \begin{align*} \frac{y}{x} &= \frac{\frac 13 a\sqrt{2k \tan \theta}}{\frac13 a\sqrt{\frac{2k}{\tan \theta}}} \\ &= \tan \theta \end{align*} \(\tan \theta > \frac{1}{\tan \theta} \Leftrightarrow \tan \theta > 1 \Leftrightarrow \theta > 45^\circ\).

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Solution:

1. For the first one, consider \begin{align*} && \int_0^1 \frac{1}{(1+tx)^2} \d x &= \left [ -\frac{1}{t}(1+tx)^{-1} \right]_0^1 \\ &&&= \frac{1}{t} - \frac{1}{t(1+t)} \\ &&&= \frac{t+1-1}{t(t+1)} = \frac{1}{t+1} \end{align*}
2. Consider \begin{align*} && \int_0^1 \frac{-2x}{(1+tx)^3} \d x &= \int_0^1 \frac{\frac{2}{t}(1+tx) -\frac{2}{t}}{(1+tx)^3} \d x \\ &&&= -\frac{2}{t} \int_0^1 \left (\frac{1}{(1+tx)^2}- \frac{1}{(1+tx)^3} \right) \d x \\ &&&= -\frac{2}{t} \frac{1}{t+1} + \frac{2}{t} \left [ \frac{1}{-2t}(1+tx)^{-2}\right]_0^1 \\ &&&= -\frac{2}{t(t+1)} + \frac2t\left (\frac{1}{2t} - \frac{1}{2t(1+t)^2} \right) \\ &&&= -\frac{2}{t} \left ( \frac{1}{t+1} + \frac{1}{2t(1+t)^2} - \frac{1}{2t}\right) \\ &&&= -\frac{2}{t} \frac{2t(1+t)+1-(1+t)^2}{2t(1+t)^2} \\ &&&= -\frac{2}{t} \frac{2t^2+2t+1-1-2t-t^2}{2t(1+t)^2} \\ &&&= -\frac{1}{(1+t)^2} \end{align*}

I would expect it to be \(\frac{2}{(1+t)^3}\). This is actually an application of differentiating under the integral sign and is completely valid where functions are well behaved.

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Solution:

1. \(\,\)
   

   + - ↺
   Notice the tangent must hit the \(y\)-axis above the origin, ie \begin{align*} && 0 &< R'(t)(0-t) + R(t) \\ \Rightarrow && R'(t) t &< R(t) \\ \Rightarrow && t &< \frac{R(t)}{R'(t)} = \frac{1}{H(t)} \end{align*}
2. Suppose \(H(t) = a/t\) then \begin{align*} && \frac{R'}{R} &= \frac{a}{t} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{a}{t} \d t \\ \Rightarrow && \ln R &= a \ln t + C \tag{t, R > 0} \\ \Rightarrow && R &= Kt^a \end{align*} Since we need \(R(t) > 0\), \(K > 0\), since \(R'(t) > 0\) we need \(a > 0\), since \(R''(t) < 0\) we need \(a(a-1) < 0\) ie \(0 < a < 1\)
3. Suppose instead \(H(t) = bt^{-2}\) then \begin{align*} && \frac{R'}{R} &= \frac{b}{t^2} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{b}{t^2} \d t \\ \Rightarrow && \ln R &= -bt^{-1} + C \tag{R > 0} \\ \Rightarrow && R &= Ke^{-b/t} \end{align*} Since \(R > 0\) we must have \(K > 0\). \begin{align*} R' > 0: && R' &= K(b/t^2)e^{-b/t} > 0 \\ \Rightarrow && b &> 0 \\ R'' < 0: && R'' &= K(b^2/t^4)e^{-b/t} -K2b/t^3 e^{-b/t} \\ &&&= Kb/t^4 (b-2t)e^{-b/t} < 0 \\ \Rightarrow && b &< 2t\\ \Rightarrow && b &< 2t \end{align*} which cannot be true for all \(t\), ie there is no \(b\) which satisfies this.

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Solution: \begin{align*} && \int_0^1 \frac{x^4}{1+x^2} \d x &= \int_0^1 \frac{(x^2-1)(1+x^2)+1}{x^2+1} \d x\\ &&&= \int_0^1 \frac{1}{1+x^2} \d x -\int_0^1 (1-x^2) \d x \\ &&&= \left [\tan^{-1}x \right]_0^1 - \left [x - \tfrac13x^3 \right]_0^1 \\ &&&= \frac{\pi}{4} - \frac23 \end{align*}

1. \(\,\) \begin{align*} && I &= \int_0^1 x^3 \; \tan ^{-1} \left(\frac {1-x} {1+x} \right) \,\d x \\ &&&= \left [ \frac{x^4}{4}\tan ^{-1} \left(\frac {1-x} {1+x} \right) \right]_0^1 -\int_0^1 \frac{x^4}{4} \frac{1}{1 +\left(\frac {1-x} {1+x} \right) ^2 } \cdot \frac{-2}{(1+x)^2} \d x \\ &&&= \frac{1}{2} \int_0^1 \frac{x^4}{(1+x)^2+(1-x)^2} \d x \\ &&&= \frac{1}{4} \int_0^1 \frac{x^4}{1+x^2} \d x \\ &&&= \frac{\pi}{16} - \frac{1}{6} \end{align*}
2. \(\,\) \begin{align*} && J &= \int_0^1 \frac {(1-y)^3} {(1+y)^5} \; {{\tan}^{-1} y}\, \d y \\ &&&= \left [ \frac {(y(1+y^2)} {(1+y)^4} \tan^{-1}y \right]_0^1 - \int_0^1 \frac {(y(1+y^2)} {(1+y)^4} \frac{1}{1+y^2} \d y \\ &&&= \frac{\pi}{32} - \int_0^1 \frac{y}{(1+y)^4} \d y \\ &&&= \frac{\pi}{32} - \left[ - \frac{3y+1}{6(1+y)^3} \right]_0^1 \\ &&&= \frac{\pi}{32} +\frac{4}{6 \cdot 8} - \frac{1}{6} \\ &&&= \frac{\pi}{32} - \frac{1}{12} \end{align*}

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Solution: \begin{align*} && x &= \cosh y \\ \Rightarrow && x &= \tfrac12 (e^y + e^{-y} ) \\ \Rightarrow && 0 &= e^{2y} - 2xe^y + 1 \\ \Rightarrow && e^y &= \frac{2x \pm \sqrt{4x^2-4}}{2} \\ &&&= x \pm \sqrt{x^2-1} \\ \Rightarrow &&e^y &= x + \sqrt{x^2-1} \tag{by convention \(\cosh^{-1} > 0\)} \\ \Rightarrow && y &= \ln (x + \sqrt{x^2-1}) \end{align*}

\begin{align*} && A &= 4 \left ( \int_0^3 \frac15\sqrt{16+x^2} \d x - \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x \right) \\ \\ x = 4\sinh u: && \int_0^3 \sqrt{4^2+x^2} \d x &= \int_{u=0}^{u=\sinh^{-1}(3/4)} \sqrt{4^2 (1+\sinh^2 u)} 4 \cosh u \d u \\ &&&= \int_0^{\sinh^{-1}(3/4)} 16 \cosh^{2}u \d u \\ &&&= 8\int_0^{\sinh^{-1}(3/4)} (1+\cosh 2u) \d u \\ &&&= 8 \left[u + \frac12 \sinh 2u\right]_0^{\sinh^{-1}(3/4)} \\ &&&= 8 \left (\sinh^{-1}(3/4) + \frac12 \sinh \left ( 2 \sinh^{-1}(3/4) \right) \right) \\ \\ && \sinh^{-1}(3/4) &= \ln\left ( \frac34 + \sqrt{\left ( \frac{3}{4} \right)^2 + 1} \right) \\ &&&= \ln \left ( \frac34 +\frac{5}{4} \right) \\ &&&= \ln 2 \\ \\ \Rightarrow && \int_0^3 \sqrt{4^2+x^2} \d x &= 8 \ln 2 + 4 \left ( \frac{e^{2 \ln 2} - e^{-2\ln2}}{2} \right) \\ &&&= 8 \ln 2 + 2 \cdot 4 - 2\cdot \frac{1}{4} \\ &&&= 8 \ln 2 + \frac{15}{2} \end{align*} \begin{align*} x = 2\sqrt{2} \cosh u: && \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x &= \int_{u=0}^{u = \cosh^{-1} \frac{3}{2\sqrt{2}}} \sqrt{8(\cosh^2 u - 1)} 2 \sqrt{2} \sinh u \d u \\ &&&= \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} 8\sinh^2 u \d u \\ &&&= 4 \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} 2\sinh^2 u \d u \\ &&&= 4 \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} \cosh 2 u -1 \d u \\ &&&= 4 \left [\frac12 \sinh 2u - u \right]_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} \\ \\ && \cosh^{-1} \frac{3}{2\sqrt{2}} &= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\left ( \frac{3}{2\sqrt{2}} \right)^2-1} \right) \\ &&&= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\frac{9}{8} - 1} \right) \\ &&&= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\frac{1}{8} } \right) \\ &&&= \ln \frac{4}{2\sqrt{2}} \\ &&&= \frac12 \ln 2 \\ \\ && \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x &= 4 \left ( \frac12 \frac{e^{\ln2} - e^{-\ln2}}{2} - \frac12 \ln 2\right) \\ &&&= 2 - \frac12 -2 \ln 2 \\ &&&= \frac32 - 2 \ln 2 \end{align*} \begin{align*} A &= 4 \left (\frac15\left(8\ln 2 + \frac{15}2 \right)- \left ( \frac32 - 2 \ln 2\right)\right) \\ &=4\cdot \left( \frac{8}{5} + 2 \right) \ln 2 \\ &= \frac{72}{5} \ln 2 \end{align*}

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Solution:

\begin{align*} \int_0^5 [x]\;\d x &= 0 \cdot 1 + 1 \cdot 1 + 2 \cdot 1 + 3 \cdot 1 + 4 \cdot 1 \\ &= 10 \end{align*} \begin{align*} \int_{0}^{\ln n}[\e^{x}]\, \d x &= \sum_{k=1}^{n-1} \int_{\ln k}^{\ln (k+1)}[\e^{x}]\, \d x \\ &= \sum_{k=1}^{n-1} k \l \ln (k+1) - \ln (k) \r\\ &= \sum_{k=1}^{n-1} \l( (k+1) \l \ln (k+1) - \ln (k) \r - \ln(k+1) \r \\ &= \sum_{k=1}^{n-1} (k+1) \ln (k+1) - \sum_{k=1}^{n-1} k \ln (k) - \sum_{k=1}^{n-1} \ln (k+1) \\ &= n \ln n - 1 \ln 1 - \sum_{k=1}^{n-1} \ln (k+1) \\ &= n \ln n - \ln \l \prod_{k=1}^{n-1} (k+1)\r \\ &= n \ln n - \ln (n!) \end{align*}

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Solution:

1. \(x^2 + 1 \geq 2x \Rightarrow \frac{1}{x^2+1} \leq \frac1{2x} \Rightarrow \frac{x^6}{(x^2+1)^4}\leq \frac{x^6}{16x^2} = \frac1{16}x^2 \leq \frac1{16}\) with equality when \(x = 1\)
2. \(\,\) \begin{align*} && RHS &= \frac{\d \ }{\d x} \left( \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3}\right) +\frac{Dx^6}{(x^2+1)^4} \\ &&&= \frac{(5Ax^4+3Bx^2+C)(x^2+1)^3-6x(x^2+1)^2(Ax^5+Bx^3+Cx)}{(x^2+1)^6}+\frac{Dx^6}{(x^2+1)^4} \\ &&&= \frac{(5Ax^4+3Bx^2+C)(x^2+1)-6x(Ax^5+Bx^3+Cx)+Dx^6}{(x^2+1)^4} \\ &&&= \frac{(D-A) x^6 + (5A-3B) x^4 + (3B-5C)x^2 + C}{(x^2+1)^4} \\ \Rightarrow && C &= 1 \\ && 3B &= 5 \quad\quad\quad\Rightarrow B = \frac53 \\ && 5A &= 3B = 5\Rightarrow A = 1 \\ && D &= A \quad\quad \Rightarrow D = 1 \end{align*}
3. So \begin{align*} && I &= \int_{0}^{1}\frac{1}{(x^{2}+1)^{4}}\, \d x \\ &&&= \int_{0}^{1}\frac{\d \ }{\d x} \left( \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3}\right) +\frac{Dx^6}{(x^2+1)^4} \, \d x \\ &&&= \left [ \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3} \right]_0^1 + \int_{0}^{1}\frac{Dx^6}{(x^2+1)^4} \, \d x \\ &&& \leq \frac{A+B+C}{8} + \frac1{16} \\ &&&= \frac{2+\frac53}{8} + \frac1{16} = \frac{11}{24} + \frac{1}{16} \\ && I &\geq \left [ \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3} \right]_0^1 = \frac{11}{24} \end{align*}

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Solution: \begin{align*} && \int_{-1}^1 |x e^x |\d x &= \int_{-1}^0 |xe^x| \d x + \int_0^1 |xe^x| \d x \\ &&&= \int_{-1}^0 -xe^x \d x + \int_0^1 x \e^x \d x \\ &&&= -\int_{-1}^0 xe^x \d x + \int_0^1 x \e^x \d x \\ \\ && \int xe^x \d x &= xe^x - \int e^x \d x \\ &&&= xe^x - e^x \\ \\ \Rightarrow && \int_{-1}^1 |x e^x |\d x &= \left [ xe^x - e^x \right]_0^{-1}+ \left [ xe^x - e^x \right]_0^{1} \\ &&&= -e^{-1}-e^{-1} +e^{0} + e^1 - e^1 +e^0 \\ &&&= 2-2e^{-1} \end{align*}

1. \(\,\) \begin{align*} && I &= \int_0^4 | x^3-2x^2-x+2| \d x \\ &&&= \int_0^4 |(x-2)(x-1)(x+1)| \d x\\ &&&= \int_0^1( x^3-2x^2-x+2) \d x- \int_1^2 ( x^3-2x^2-x+2) \d x + \int_2^4 ( x^3-2x^2-x+2) \d x \\ &&&= \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_0^1 - \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_1^2 + \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_2^4 \\ &&&= 2 \left ( \frac14 - \frac23 -\frac12 + 2\right) - 2 \left ( \frac14 2^4 - \frac23 2^3 -\frac12 2^2 + 2 \cdot 2\right)+ \left ( \frac14 4^4 - \frac23 4^3 -\frac12 4^2 + 2 \cdot 4\right) \\ &&&= \frac{133}{6} \end{align*}
2. \(\,\) \begin{align*} && J &= \int_{-\pi}^\pi | \sin x + \cos x | \d x \\ &&&= \int_{-\pi}^{\pi} | \sqrt{2} \sin(x + \tfrac{\pi}{4})| \d x \\ &&&= 2\sqrt{2}\int_0^\pi \sin x \d x \\ &&&= 4\sqrt{2} \end{align*}

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Solution: \begin{align*} && g(\lambda) &= \int_0^1 \big(\f(x)-\lambda x \big)^{\!2} \,\d x \\ &&&= \int_0^1 \left ( f(x)^2 -2\lambda xf(x) + \lambda^2 x^2\right) \d x \\ &&&= \frac13\lambda^2 - 2\lambda \int_0^1 x f(x) \d x + \int_0^1 f(x)^2 \d x \\ \end{align*} Differentiating (or completing the square) it is clear the minimum occurs when \(\displaystyle \lambda = 3 \int_0^1 xf(x) \d x\) \begin{align*} && R^2 &= \int_0^1 (f(x) - \lambda x )^2 \d x \\ &&&= \frac13\lambda^2 - 2\lambda \int_0^1 x f(x) \d x + \int_0^1 f(x)^2 \d x \\ &&&= \frac13 \left (\lambda -3\int_0^1 xf(x) \d x \right)^2 -\frac13 \left ( 3\int_0^1 xf(x) \d x \right)^2+\int_0^1 f(x)^2 \d x \\ \end{align*} When \(\lambda = 3\int_0^1 xf(x) \d x \) clearly this is the desired result. \begin{align*} && \lambda &= 3\int_0^1 xf(x) \d x \\ &&&= 3\int_0^1 x \sin(\pi x /n) \d x \\ &&&= 3 \left [-x \frac{n}{\pi} \cos (\pi x /n) \right]_0^1 + \frac{3n}{\pi} \int_0^1 \cos(\pi x /n) \d x \\ &&&= -\frac{3n}{\pi}\cos(\pi/n) + \frac{3n}{\pi} \left [ \frac{n}{\pi} \sin(\pi x /n)\right]_0^1 \\ &&&= -\frac{3n}{\pi} \cos(\pi/n) + \frac{3n^2}{\pi^2} \sin(\pi /n) \\ \text{for large }n: &&&\approx -\frac{3n}{\pi}\left ( 1 - \frac12\frac{\pi^2}{n^2} + o(1/n^4)\right) + \frac{3n^2}{\pi^2} \left (\frac{\pi}{n} - \frac16 \frac{\pi^3}{n^3} +o(1/n^5) \right) \\ &&&= \left (\frac32 -\frac12\right)\frac{\pi}{n} + o(1/n^3) \\ &&&= \frac{\pi}{n} + o(1/n^2) \end{align*} Therefore for large \(n\), \(\lambda \approx \frac{\pi}n\) \begin{align*} && \int_0^1 \sin^2(\pi x/n) \d x &= \frac12\int_0^1(1- \cos(2\pi x/n)) \d x\\ &&&= \frac12\left ( 1 - \frac{n}{2\pi}\left[\sin(2\pi x/n) \right]_0^1 \right) \\ &&&= \frac12 -\frac{n}{4\pi}\sin(2\pi /n) \\ \\ && R^2 &= \frac12 -\frac{n}{4\pi}\sin(2\pi /n) - \frac13 \left ( \frac{\pi}{n}+o(1/n^2)\right)^2 \\ &&&= \frac12 - \left ( \frac{1}{2} -\frac16\frac{\pi}{n}+o(1/n^3) \right) - o(1/n^2) \\ &&& = \frac16 \frac{\pi}{n} + o(1/n^2) \\ &&&\to 0 \text{ as } n \to \infty \end{align*} We should expect these results as for \(n\) very large \(\sin(\pi x/n) \approx \frac{\pi }{n}x\) so the best linear approximation is likely to be \(\lambda = \frac{\pi}{n}\) and we should expect it to improve to the point that we cannot tell the difference, ie \(R^2 \to 0\)

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Solution: \begin{align*} && \frac{2t}{1+t^2} &= \frac{2 \sin \tfrac12 \theta \cos\tfrac12 \theta }{\cos^2 \tfrac12 \theta + \sin^2 \tfrac12 \theta} \\ &&&= \frac{\sin \theta}{1} = \sin \theta \\ \\ && \frac{1-t^2}{1+t^2} &= \frac{\cos^2 \tfrac12 \theta - \sin^2 \tfrac12 \theta}{\cos^2 \tfrac12 \theta + \sin^2 \tfrac12 \theta} \\ &&&= \frac{\cos \theta }{1} = \cos \theta \\ \\ && \tan(\tfrac12 \pi - \tfrac12 \theta) &= \frac{1}{t} \\ && \frac{1+\cos \theta}{\sin \theta} &= \frac{1 + \frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}} \\ &&&= \frac{2}{2t} = \frac1t = \tan(\tfrac12\pi - \tfrac12 \theta) \end{align*} Notice also that \(\frac{\d t}{\d \theta} = \tfrac12 \sec^2 \tfrac12 \theta = \tfrac12(1 + t^2)\) so \begin{align*} && I &= \int_0^{\frac12 \pi} \frac{1}{1 + \cos \alpha \sin \theta} \d \theta \\ t = \tan \tfrac12 \theta, \d \theta = \frac{2}{1+t^2} \d t: &&&= \int_{0}^{1} \frac{1}{1 + \cos \alpha \frac{2t}{1+t^2}}\frac{2}{1+t^2} \d t \\ &&&= \int_0^1 \frac{2}{1+t^2 + 2\cos \alpha t} \d t \\ &&&= \int_0^1 \frac{2}{(t + \cos \alpha)^2+\sin^2 \alpha} \d t \\ &&&= \left [ \frac{2}{\sin \alpha} \tan^{-1} \left ( \frac{t+ \cos \alpha}{\sin \alpha} \right) \right]_0^1 \\ &&&= \frac{2}{\sin \alpha} \left ( \tan^{-1} \left ( \frac{1+ \cos \alpha}{\sin \alpha} \right) - \tan^{-1} \left ( \frac{ \cos \alpha}{\sin \alpha} \right) \right) \\ &&&= \frac{2}{\sin \alpha} \left ( \tan^{-1} \left (\tan (\tfrac12 \pi - \tfrac12 \alpha \right) - \tan^{-1} \left (\tan(\tfrac12\pi - \alpha )\right) \right) \\ &&&= \frac{2}{\sin \alpha} \left ( \tfrac12 \pi - \tfrac12 \alpha - \tfrac12 \pi + \alpha \right) \\ &&&= \frac{\alpha}{\sin \alpha} \end{align*} \begin{align*} && J &= \int_0^{\tfrac12 \pi} \frac{1}{1 + \sin \alpha \cos \theta} \d \theta \\ &&&= \int_0^{\tfrac12 \pi} \frac{1}{1 + \cos (\tfrac12 \pi - \alpha) \sin \theta} \d \theta \\ &&&= \frac{\tfrac12 \pi - \alpha}{\cos \alpha} \end{align*}

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Solution: Looking at the forces on Jane, \(kv < g \Rightarrow v < \frac{g}{k}\). For Karen we have \begin{align*} kv + (2k^2/g)v^2 &< g\\ -g^2 + gkv + (2k^2)v^2 &< 0 \\ (2kv-g)(kv+g) &< 0\\ \Rightarrow v &< \frac{g}{2k} \end{align*} \begin{align*} && \dot{v} &= g - kv \\ \Rightarrow && \frac{\dot{v}}{g - kv} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv\\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \left [-\frac{1}{k} \ln \l g - kv \r \right ]_0^{g/(3k)} \\ && &= \frac{1}{k} \ln \l g \r - \frac{1}{k} \ln \l \frac{2}{3}g \r\\ &&&= \frac{1}{k} \ln \l \frac{3}{2} \r \end{align*} \begin{align*} && \dot{v} &= g - kv - (2k^2/g)v^2 \\ \Rightarrow && \frac{\dot{v}}{g - kv - (2k^2/g)v^2} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv - (2k^2/g)v^2} dv \\ && &= \int_0^{g/(3k)} \frac{g}{(g-2kv)(kv+g)} dv\\ && &= \int_0^{g/(3k)} \l \frac{2}{3(g-2kv)} + \frac{1}{3(kv+g)} \r dv\\ && &= \left [ \l -\frac{1}{3k} \ln (g-2kv) + \frac{1}{3k}\ln(kv+g) \r \right ]_0^{g/(3k)} \\ && &= \left [ \l -\frac{1}{3k}\ln \l \frac{g}{3} \r + \frac{1}{3k}\ln \l \frac{4g}{3} \r \r \right ] - \left [- \frac1{3k} \ln(g) + \frac{1}{3k} \ln (g) \right ] \\ && &= \frac{1}{3k} \ln \l 4 \r \end{align*} NB: \(\sqrt[3]{4} \approx 1.58 > \frac{3}{2}\) so Karen has been in free-fall for longer, but not \emph{much} longer than Jane.