Problems

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Solution: \begin{align*} && \int_0^x \mathrm{sech}\, t \d t &= \int_0^x \frac{2}{e^t+e^{-t}} \d t \\ &&&= \int_0^x \frac{2e^t}{e^{2t}+1} \d t \\ &&&= \left [2 \arctan e^t \right ]_0^x \\ &&&= 2\tan^{-1}e^x- \frac{\pi}{2} \end{align*} \begin{align*} && I_n &= \int_0^x \mathrm{sech}^n\, t \d t \\ &&&= \int_0^x \mathrm{sech}^{n-2}\, t \mathrm{sech}^2\, t \d t \\ &&&= \left [ \mathrm{sech}^{n-2}\, t \cdot \tanh t\right ]_0^x - \int_0^x (n-2) \mathrm{sech}^{n-3} \, t \cdot (- \tanh t \mathrm{sech}\, t) \tanh t \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t \tanh^2 t \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t (1-\mathrm{sech}^2 \, t) \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2}-(n-2)I_n \\ \Rightarrow && (n-1)I_n &= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \\ \Rightarrow && I_n &= \frac{1}{n-1} \left ( \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \right) \\ \end{align*} \begin{align*} I_1 &= 2\tan^{-1}e^x- \frac{\pi}{2} \\ I_3 &= \frac12 \left ( \mathrm{sech}\, x \cdot \tanh x+ 2\tan^{-1}e^x- \frac{\pi}{2}\right) \\ &= \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \\ I_5 &= \frac14 \left (\mathrm{sech}^3\, x \cdot \tanh x + 3 \left ( \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \right) \right) \\ &= \frac14 \mathrm{sech}^3\, x \cdot \tanh x +\frac38 \mathrm{sech}\, x \cdot \tanh x+\frac34 \tan^{-1}e^x- \frac{3\pi}{16} \\ \\ I_2 &= \tanh x \\ I_4 &= \frac13 \left ( \mathrm{sech}^2 x\tanh x +2\tanh x\right) \\ &= \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \\ I_6 &= \frac15 \left ( \mathrm{sech}^4 x \tanh x+4\left ( \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \right) \right) \\ &= \frac15 \mathrm{sech}^4 x \tanh x + \frac4{15}\mathrm{sech}^2 x\tanh x + \frac{8}{15} \tanh x \end{align*}

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Solution: \begin{align*} && I &= \int_0^{\pi} x f(\sin x) \d x \\ t = \pi - x, \d t = -\d t : &&&= \int_{t = \pi}^{t = 0} (\pi - t) f(\sin (\pi - t)) -\d t \\ &&&= \int_0^{\pi} (\pi - t) f(\sin t) \d t \\ \Rightarrow && 2 I &= \pi \int_0^\pi f(\sin t) \d t \\ \Rightarrow && I &= \frac{\pi}{2} \int_0^{\pi} f(\sin x) \d x \end{align*} \begin{align*} && I &= \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ &&&= \frac{\pi}{2}\int_0^\pi \frac{\sin x}{1 + \cos^2 x} \d x \\ &&&= \frac{\pi}{2}\left [ -\tan^{-1} \cos x\right]_0^{\pi} \\ &&&= \tan 1 - \tan (-1) = \frac{\pi^2}{4} \\ \\ && I &= \int_{0}^{2\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ &&&= \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x + \int_{\pi}^{2\pi} \frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ u = x - \pi, \d u = \d x: &&&= \frac{\pi^2}{4} + \int_{0}^{\pi} \frac{(u+\pi)(-\sin u)}{1 + \cos^2 u}\d u \\ &&&= \frac{\pi^2}{4} -\frac{3\pi}{2} \int_0^{\pi} \frac{\sin u}{1+\cos^2 u} \d u \\ &&&= - \frac{\pi^2}2 \end{align*}

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Solution: While the force is constant, the train is accelerating at \(\frac{F}{M}\), and since \(u = \frac{F}{M} t_1 \Rightarrow t_1 = \frac{Mu^2}{Fu} = \frac{Mu^2}{P}\). Once the train is being driven at a constant rate, we can observe that change in energy will be power times time, ie \(Pt_2 = \frac{1}{2}M(v^2 - u^2) \Rightarrow t_2 = \frac{M}{2P} ( v^2 - u^2)\). Therefore the total time will be \(t_1 + t_2 = \frac{M}{2P} ( u^2 + v^2)\). During the first period, the distance will be: \(s_1 = \frac12 \frac{F}{M} t_1^2 = \frac12 \frac{F}{M} \frac{M^2u^2}{F^2} = \frac{Mu^3}{2P}\) In the second period, \(P = Fu\) and so \(\text{Force} = \frac{P}{v} \Rightarrow M v \frac{\d v}{\d x} = \frac{P}{v} \Rightarrow M \l \frac{v^3}{3} - \frac{u^3}{3}\r = Ps_2\) and therefore total distance will be: \(\frac{M}{6P}(2v^{3}+u^{3})\)

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Solution: \begin{align*} && I &= \int_0^1 \cos^{-1} t \d t \\ \cos y = t: -\sin y \d y = \d t: &&&= \int_{\frac{\pi}{2}}^0 -y \sin y \d y \\ &&&= \int_0^{\pi/2} y \sin y \d y \\ &&&= \left [-y \cos y \right]_0^{\pi/2} + \int_0^{\pi/2} \cos y \d y \\ &&&= \left [ \sin y \right]_0^{\pi/2} = 1 \end{align*}

If \(X = x\) then the rod will cross the line if \(\frac{x}{\sin \theta} < a\) or \(\frac{2b-x}{\sin \theta} < a\), ie \(a\sin \theta > \max (x, 2b-x)\). Therefore the probability is \(\frac{2\sin^{-1} \left (\max(\frac{x}{a}, \frac{2b-x}{a}) \right)}{\pi}\). Therefore the probability the pin crosses a line is: \begin{align*} \mathbb{P} &= \frac{1}{2b}\int_{x=0}^{x=2b} \frac{2\sin^{-1} \left (\max(\frac{x}{a}, \frac{2b-x}{a}) \right)}{\pi} \d x \\ &= \frac{2a}{b\pi} \end{align*}

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Solution: While descending, the body experiences the force \(-\frac15mg - 2mv^2\). \begin{align*} \text{N2:} && m \dot{v} &= -\frac15 mg - 2mv^2 \\ \Rightarrow && \frac{\dot{v}}{\frac15g + 2v^2} &= -1 \\ \Rightarrow && \frac{1}{2}\tan^{-1} v_1 - \frac{1}{2}\tan^{-1} {v_0} &= -T \end{align*} We care about when \(v_1 = 0\), ie \(\displaystyle T = \frac{1}{2}\tan^{-1} {v_0} < \frac12 \frac{\pi}2 = \frac{\pi}4\) seconds. If the body enters at speed \(1\ \mathrm{ms}^{-1}.\) then for the first part of it's journey it will experience forces \(-\frac15mg - 2mv^2\) and so: \begin{align*} \text{N2:} && m v \frac{\d v}{\d x} &= -\frac15 mg - 2mv^2 \\ \Rightarrow && \int \frac{v}{2(1 + v^2)} \d v &= \int -1 \d x \\ \Rightarrow && \frac14 \ln (1 + v^2) &= -x \end{align*} Therefore the depth is \(\frac14 \ln 2\) metres. When the body is rising, it experiences forces of: \(\frac15mg - 2mv^2\) and so: \begin{align*} \text{N2:} && m v \frac{\d v}{\d x} &= \frac15mg - 2mv^2 \\ \Rightarrow && \int \frac{v}{2(1 - v^2)} \d v &= \int -1 \d x \\ \Rightarrow && -\frac14 \ln (1 - v^2) &= \frac14 \ln 2 \\ \Rightarrow && 1-v^2 &= \frac12 \\ \Rightarrow && v &= \frac{1}{\sqrt{2}} \ \mathrm{ms}^{-1} \end{align*} This will take \begin{align*} \text{N2:} && m \dot{v} &= \frac15mg - 2mv^2 \\ \Rightarrow && \frac{\dot{v}}{2(1-v^2)} &= -1 \\ \Rightarrow && \dot{v} \frac{1}{4}\l \frac{1}{1 - v} + \frac{1}{1+v} \r &= -1 \\ \Rightarrow && \frac{1}{4} \l -\ln(1 - v) + \ln(1 + v)\r &= -T \end{align*} Since \(v = \frac{1}{\sqrt{2}}\) \begin{align*} T &= \frac{1}{4} \ln \l \frac{1+ \frac1{\sqrt{2}}}{1 - \frac1{\sqrt{2}}}\r \\ &= \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \end{align*} and therefore the total time will be: \begin{align*} & \frac12 \tan^{-1} 1 + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \\ =& \frac{\pi}{8} + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \end{align*}

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Solution: First notice that \(y = \cos 2 \theta = 1 - 2\sin^2 \theta = 1- 2x^2\), therefore \(P\) is lies on that parabola.

The arc length is \begin{align*} && s &= \int_{-\pi/2}^{\pi/2} \sqrt{\left ( \frac{\d x}{\d \theta} \right)^2+\left ( \frac{\d y}{\d \theta} \right)^2} \d \theta\\ && &= \int_{-\pi/2}^{\pi/2} \sqrt{\cos^2 \theta+16 \sin^2 \theta \cos^2 \theta } \d \theta\\ && &= \int_{-\pi/2}^{\pi/2} \cos \theta\sqrt{1+16 \sin^2 \theta} \d \theta\\ u = \sin \theta, \d u = \cos \theta \d \theta && &= \int_{u=-1}^{u=1} \sqrt{1+16 u^2} \d u\\ 4u = \sinh v, 4\d u = \cosh v: && &= \int_{v=-\sinh^{-1} 4}^{v=\sinh^{-1} 4} \sqrt{1+\sinh^2 v} \tfrac14\cosh v \d v\\ && &= \frac14 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4} \cosh^2 v \d v\\ && &= \frac18 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4} (1 + \cosh 2v) \d v\\ && &= \frac14 \sinh^{-1} 4 + \frac18\left [ \frac12\sinh 2v \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\ && &= \frac14 \sinh^{-1} 4 + \frac18\left [ \sinh v \sqrt{1 + \sinh^2 v} \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\ && &= \frac14 \sinh^{-1} 4 + \left (\frac18 \cdot 4 \sqrt{17} \right) - \left (\frac18 \cdot (-4) \sqrt{17} \right)\\ && &= \frac14 \sinh^{-1} 4 + \sqrt{17}\\ \end{align*} The volume of revolution is \begin{align*} && V &=\pi \int_{-1}^1 (2-2x^2)^2 \d x \\ &&&= \pi \left [4x-\frac83x^3+\frac45x^5 \right]_{-1}^1 \\ &&&= \pi \left ( 8-\frac{16}3+\frac85 \right) \\ &&&= \frac{64}{15}\pi \end{align*}

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Solution:

1. \begin{align*} \int_{-\pi}^\pi |\sin x | \d x &= \int_{-\pi}^{0} - \sin x \d x + \int_0^\pi \sin x \d x \\ &= \left [\cos x \right]_{-\pi}^{0} +[-\cos x]_0^{\pi} \\ &= 1-(-1)+(1)-(-1) \\ &= 4 \end{align*}
2. \begin{align*} \int_{-\pi}^\pi \sin | x | \d x &= \int_{-\pi}^0 - \sin x \d x + \int_0^\pi \sin x \d x \\ &= 4 \end{align*}
3. \begin{align*} \int_{-\pi}^\pi x \sin x \d x &= \left [ -x \cos x \right]_{-\pi}^\pi + \int_{-\pi}^{\pi} \cos x \d x \\ &= \pi -(-\pi) + \left [\sin x \right]_{-\pi}^\pi \\ &= 2\pi \end{align*}
4. \begin{align*} \int_{-\pi}^{\pi} x^{10} \sin x \d x &\underbrace{=}_{x^{10}\sin x \text{ is odd}} 0 \end{align*}

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Solution: Consider \(E = \frac12 m \dot{x}^2 - \frac{Gm}{x}\), notice that \begin{align*} && \dot{E} &= m \dot{x} \ddot{x} + \frac{Gm}{x^2} \dot{x} \\ &&&= \dot{x} \underbrace{\left (m\ddot{x} + \frac{Gm}{x^2} \right)}_{=0 \text{ by N2}} \end{align*} Therefore \(E\) is conserved. Therefore if \(x \to \infty\) \(\frac12 m V^2 - \frac{Gm}{h} = \frac12 m u^2 - 0 \geq 0\) so \(V^2 \geqslant 2G/h\) Since \(E \approx 0\) we want to solve \begin{align*} && \dot{x} &= -\sqrt{\frac{2G}{x}} \\ \Rightarrow && -\int_h^0 \sqrt{x} \d x &= \int_0^T \sqrt{2G} \d t \\ \Rightarrow && \frac{2h^{3/2}}{3} &= \sqrt{2G}T \\ \Rightarrow && T &= \frac{\sqrt{2}h^{3/2}}{3\sqrt{G}} = \frac13 \sqrt{\frac{2h^3}{G}} \end{align*}

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Solution:

1. \begin{align*} \int\frac{x}{(x-1)(x^{2}-1)}\,\mathrm{d}x &= \int \frac{x}{(x-1)^2 (x+1)} \d x \\ &= \int \frac{1}{2(x-1)^2} + \frac{1}{4(x-1)} - \frac{1}{4(x+1)} \d x \\ &= -\frac12 (x-1)^{-1} + \frac14 \ln(x-1) - \frac14 \ln (x+1) + C \end{align*}
2. \begin{align*} \int \frac{1}{3 \cos x + 4 \sin x } \d x &= \int \frac{1}{5 \cos (x - \cos^{-1}(3/5))} \d x \\ &= \frac15 \int \sec (x - \cos^{-1}(3/5)) \d x\\ &= \frac15 \left (\ln | \sec (x - \cos^{-1}(3/5)) + \tan (x - \cos^{-1}(3/5)) | \right) + C \end{align*}
3. \begin{align*} \int \frac{1}{\sinh x} \d x &= \int \frac{2}{e^x - e^{-x}} \\ &= \int \frac{2e^x}{e^{2x}-1} \d x \\ &=\int \frac{e^x}{e^x-1} - \frac{e^x}{e^x+1} \d x \\ &= \ln (e^x - 1) + \ln (e^x+1) + C \end{align*}