Problems

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Solution: \begin{align*} \sum_{r=0}^{n-1} \e^{2i(\alpha + r\pi/n)} &= e^{2i\alpha} \sum_{r=0}^{n-1} \left (\e^{2i\pi/n} \right)^r \\ &= e^{2i\alpha} \frac{1-\left (\e^{2i\pi/n} \right)^n}{1-\e^{2i\pi/n} } \\ &= 0 \end{align*} \(d = s + r \cos \theta\) ie \(s = d - r \cos \theta\) Therefore \(d = \frac{r}{k} + r \cos \theta \Rightarrow r = \frac{kd}{1+k \cos \theta}\). The \(l_j\) will come from \(r(\alpha + \frac{j \pi}{n} )+r(\alpha + \pi + \frac{j \pi}{n} )\) \begin{align*} && l_j &= r(\alpha + \frac{(j-1) \pi}{n} )+r(\alpha + \pi + \frac{(j-1) \pi}{n} ) \\ &&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1+k \cos \left ( \alpha+\pi+ \frac{(j-1) \pi}{n}\right)}\\ &&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1-k \cos \left ( \alpha+ \frac{(j-1) \pi}{n}\right)}\\ &&&= \frac{2kd}{1-k^2 \cos^2 \left ( \alpha + \frac{(j-1) \pi}{n}\right)}\\ \Rightarrow && \sum_{j=1}^n \frac 1 {l_j} &= \sum_{j=0}^{n-1} \frac{1-k^2 \cos^2 \left ( \alpha + \frac{j \pi}{n}\right)}{2kd} \\ &&&= \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \cos^2 \left ( \alpha + \frac{j \pi}{n}\right) \\ &&&= \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \frac{1+ \cos \left ( 2\alpha + \frac{2j \pi}{n}\right)}{2} \\ &&&= \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \sum_{j=0}^{n-1}\cos \left ( 2\alpha + \frac{2j \pi}{n}\right) \\ &&&= \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \underbrace{\textrm{Re} \left ( \sum_{j=0}^{n-1}e^{ 2i(\alpha + \frac{j \pi}{n})} \right)}_{=0} \\ &&&= \frac{n}{2kd} - \frac{nk^2}{4kd} \\ &&&= \frac{n(2-k^2)}{4kd} \end{align*}

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Solution:

1. \(\,\)
   

   + - ↺
   The area under the blue curve is \(1-\cos b\). The area under the green line is \(\frac12 b \sin b\) The area under the red line is \(\frac12 b^2\) Therefore \(\frac12 b \sin b < 1- \cos b < \frac12 b^2 \Rightarrow 1- \frac12 b^2 < \cos b < 1 - \frac12 b \sin b\)
2. \(\,\)
   

   + - ↺
   \begin{align*} &&\text{Area under blue curve}: &= \int_0^1 a^x \d x\\ &&&= \left [ \frac{1}{\ln a}e^{x \ln a} \right]_0^1 \\ &&&= \frac{a-1}{\ln a} \\ \\ &&\text{Area under green line}: &=\frac12 \cdot 1 \cdot (a + 1)\\ &&&= \frac{a+1}{2} \\ \\ &&\text{Area under tangent}: &=\frac12 \cdot 1 \cdot (1+\ln a + 1)\\ &&&= \frac{\ln a+2}{2} \\ \\ \Rightarrow && \frac{a+1}{2} & > \frac{a-1}{\ln a} \\ \Rightarrow && \ln a& > \frac{2(a-1)}{a+1} \\ \\ \Rightarrow && \frac{a-1}{\ln a} &> \frac{\ln a +2}{2} \\ \Rightarrow && 2(a-1) -2\ln a - (\ln a)^2 &> 0 \\ \Rightarrow && \ln a & < -1 + \sqrt{2a-1} \end{align*}

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Solution: \begin{align*} &&\int_0^{\frac14\pi} \sin (2x) \ln(\cos x)\, \d x &= \int_0^{\frac14 \pi} 2 \sin x \cos x \ln (\cos x) \d x \\ u = \cos \theta :&&&= \int_{u=1}^{u=\frac1{\sqrt2}} -2u \ln u \d u \\ &&&= \int_{\frac1{\sqrt{2}}}^1 2u \ln u \d u \\ &&&= \left [u^2 \ln u \right]_{\frac1{\sqrt{2}}}^1-\int_{\frac1{\sqrt{2}}}^1 u \d u \\ &&&= -\frac12 \ln \frac{1}{\sqrt{2}} - \l\frac12 - \frac14 \r \\ &&&= \frac14 (\ln 2 - 1) \end{align*} \begin{align*} && \int_0^{\frac14\pi} \cos (2x) \ln(\cos x)\, \d x &= \left [ \frac12 \sin 2x \ln (\cos x) \right]_0^{\frac14\pi}- \int_0^{\frac14\pi} \frac12 \sin 2x \frac{-\sin x}{\cos x} \d x \\ &&&=\frac12 \ln \frac{1}{\sqrt{2}}+\int_0^{\frac14\pi} \sin^2 x \d x \\ &&&= -\frac14 \ln 2 + \int_0^{\frac14\pi} \frac{1-\cos 2x }{2} \d x \\ &&&= -\frac14 \ln 2 +\frac{\pi}{8} -\frac{1}{4} \\ &&&= \frac18 (\pi - 2\ln 2 - 2) \\ &&&= \frac18 (\pi - \ln 4 - 2) \\ \end{align*} Notice that \(\cos x + \sin x = \sqrt{2} \cos (x -\frac{\pi}{4})\), so: \begin{align*} &&\int_{\frac14\pi}^{\frac12\pi} \big ( \cos(2x) + \sin (2x)\big) \ln \big( \cos x + \sin x\big)\d x &= \int_{\frac14\pi}^{\frac12\pi} \big ( \cos(2x) + \sin (2x)\big) \ln (\sqrt{2} \cos ( x - \frac{\pi}{4}) ) \d x \\ &&&= \int_{u=0}^{u=\frac{\pi}{4}} \l \cos(2u+\frac{\pi}{2})+\sin(2u+\frac{\pi}{2}) \r \cdot \l \frac12 \ln 2 +\ln (\cos u) \r \d x \\ &&&= \int_{0}^{\frac{\pi}{4}} \l -\sin 2u+\cos 2u \r \cdot \l \frac12 \ln 2 +\ln (\cos u) \r \d x \\ &&&= \frac14 \ln 2\left [ \cos 2u + \sin 2u \ \right]_{0}^{\frac{\pi}{4}} - \frac14(\ln2 - 1) + \frac18\pi - \frac14(\ln 2 +1) \\ &&&= \frac{\pi}{8}-\frac12 \ln 2 \end{align*}

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Solution: \begin{align*} && s &= ut + \frac12 gt^2 \\ \Rightarrow && -2.5 &= 10 \sin \theta \, T - 5 T^2 \\ \Rightarrow && T &= \frac{10\sin \theta \pm \sqrt{100\sin^2 \theta - 4 \cdot 5 \cdot (-2.5)}}{10} \\ &&&= \sin \theta +\sqrt{\sin^2 \theta + \frac12} \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{2} \sin \theta +\sqrt{2 \sin^2 \theta +1} \right) \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{2 (1-\cos^2 \theta)} + \sqrt{2-\cos 2\theta} \right) \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{1-\cos2 \theta} + \sqrt{2-\cos 2\theta} \right) \\ &&&= \frac{1}{\sqrt{2}}\left ( \sqrt{1-c}+\sqrt{2-c} \right)\\ \\ && s &= 10 \cos \theta T \\ &&&= 10 \sqrt{\frac{\cos 2 \theta +1}{2}}\frac{1}{\sqrt{2}}\left ( \sqrt{1-c}+\sqrt{2-c} \right) \\ &&&= 5 \sqrt{c+1}\left ( \sqrt{1-c}+\sqrt{2-c} \right) \\ \\ && \frac15\frac{\d s}{\d c} &= \frac12(c+1)^{-1/2}((1-c)^{1/2} + (2-c)^{1/2}) - \frac12(c+1)^{1/2}\left ((1-c)^{-1/2}+(2-c)^{-1/2} \right) \\ &&&= \frac{((1-c)(2-c)^{1/2}+(2-c)(1-c)^{1/2})-((c+1)(2-c)^{1/2}+(c+1)(1-c)^{1/2})}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ &&&= \frac{\sqrt{2-c}\left (1-c-c-1 \right)+\sqrt{1-c}\left(2-c-c-1) \right)}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ &&&= \frac{\sqrt{1-c}\left(1-2c\right)-2c\sqrt{2-c}}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ \\ \frac{\d s}{\d c} =0: && \sqrt{1-c}\left(1-2c\right)&=2c\sqrt{2-c} \\ \Rightarrow && (1-c)(1-2c)^2&=4c^2(2-c) \\ \Rightarrow && 1-5c+8c^2-4c^3 &= 8c^2-4c^3 \\ \Rightarrow && 0 &= -5c+1 \\ \Rightarrow && c &= \frac15 \end{align*} When \(\theta = 45^{\circ}, c = 0\), so \(s_{45^{\circ}} = 5(1+\sqrt{2})\) When \(c = \frac15\), \begin{align*} s &= 5 \sqrt{\frac15+1}\left ( \sqrt{1-\frac15}+\sqrt{2-\frac15} \right) \\ &= 5 \sqrt{\frac65} \left ( \sqrt{\frac45} + \sqrt{\frac95} \right) \\ &= 2\sqrt{6}+3\sqrt{6} = 5\sqrt{6} \end{align*} Therefore the additional distance is \(5(\sqrt{6}-\sqrt{2}-1)\)

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Solution: \begin{align*} && s &= ut \\ \Rightarrow && a &= u \cos \alpha t\\ \Rightarrow && t &= \frac{a}{u \cos \alpha}\\ && s &= ut+ \frac12at^2 \\ \Rightarrow && -h &< -u\sin \alpha \frac{a}{u \cos \alpha}-\frac12 g \left (\frac{a}{u \cos \alpha} \right)^2 \\ &&&= -a \tan \alpha-\frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha \\ \Rightarrow && \frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha &< h -a\tan \alpha \\ \Rightarrow &&\frac{1}{u^2} &< \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \end{align*} \begin{align*} && s &= ut + \frac12a t^2 \\ \Rightarrow && 2h &= u\sin \alpha t + \frac12 gt^2 \\ \Rightarrow && t &= \frac{-u\sin \alpha \pm \sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && t &= \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && s &= ut \\ \Rightarrow && b &> u \cos \alpha t \\ \Rightarrow && \frac{b}{u \cos \alpha} &> \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g} \\ \Rightarrow && \sqrt{u^2 \sin^2 \alpha+4hg} &< \frac{bg}{u \cos \alpha} + u \sin \alpha \\ \end{align*} \begin{align*} \Rightarrow && u^2 \sin^2 \alpha+4hg &< \frac{b^2g^2}{u^2 \cos^2 \alpha} +u^2 \sin^2 \alpha + 2bg \tan \alpha \\ \Rightarrow && 4hg - 2bg \tan \alpha &< \frac{b^2g^2}{u^2 \cos^2 \alpha} \\ &&&< \frac{b^2g^2}{\cos^2 \alpha} \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \\ &&&= \frac{2b^2g(h-a\tan \alpha)}{a^2} \\ \Rightarrow && \tan \alpha \left (\frac{2b^2g}{a} - 2bg \right) &< \frac{2b^2gh}{a^2} - 4hg \\ \Leftrightarrow && \tan \alpha \left (\frac{2b^2g- 2abg}{a} \right) &< \frac{2b^2gh- 4hga^2}{a^2} \\ \Leftrightarrow && \tan \alpha \left (\frac{2bg(b- a)}{a} \right) &< \frac{2hg(b^2- 2a^2)}{a^2} \\ \Rightarrow && \tan \alpha &< \frac{h(b^2-2a^2)}{ab(b-a)} \end{align*}

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Solution:

Let \(M\) be the midpoint of \(AB\), then \begin{align*} \overset{\curvearrowright}{M}: && R_1 \sin \phi-\mu R_1 \cos \phi &= R_2 \sin \phi+\mu R_2 \cos \phi \\ \Rightarrow && (R_1-R_2) \tan \phi &= \mu(R_1+R_2) \end{align*} As required. \begin{align*} && \cos \phi = \frac{a}{r} &,\,\, \sin \phi = \frac{\sqrt{r^2-a^2}}{r} \\ \text{N2}(\rightarrow): && R_1\cos(\phi + \theta)+\mu R_1 \sin(\phi + \theta) &= R_2 \cos(\theta - \phi) + \mu R_2 \sin(\theta - \phi) \\ \Rightarrow && R_1(\cos \theta \cos \phi - \sin \theta \sin \phi)+ \mu R_1 (\sin \theta \cos \phi + \cos \theta \sin \phi) &= R_2 (\cos\theta \cos \phi + \sin \theta \sin \phi)+ \mu R_2 (\sin \theta \cos \phi - \cos \theta \sin \phi) \\ && R_1 (1 - \tan \theta \tan \phi)+\mu R_1 (\tan \theta + \tan \phi) &= R_2(1 + \tan \theta \tan \phi) +\mu R_2 (\tan \theta - \tan \phi) \\ && 0 &= (R_1-R_2)(1+\mu \tan \theta)+(R_1+R_2)(-\tan \theta \tan\phi+\mu \tan \phi) \\ \Rightarrow && \frac{R_1+R_2}{R_1-R_2} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\ \Rightarrow && \frac{\tan \phi}{\mu} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\ \Rightarrow && \tan^2 \phi &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\ \Rightarrow && \frac{r^2-a^2}{a^2} &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\ \Rightarrow && \tan \theta (r^2-a^2-a^2\mu^2) &= \mu a^2+\mu(r^2-a^2) \\ \Rightarrow && \tan \theta &= \frac{\mu r^2}{r^2-(1+\mu^2)a^2} \end{align*} Since \(\mu r^2 > 0\) we must also have \(r^2 > a^2(1+\mu^2)\) ie \(\\sec^2 \phi > 1 + \mu^2 = \sec^2 \lambda\) and the result follows.

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Solution:

Note that the line \(y = x\) is the tangent \((0,0)\) and \(y = \sin x \) is always below it. For any other line through the origin with gradient \(0 < k < 1\) it must start below \(y = \sin x\), but finish above it at \(x = \pi\). It also can only cross once due the the convexity of \(\sin\) in this interval. \begin{align*} && I &= \int_0^\pi | \sin x -kx | \d x \\ &&&= \int_0^\alpha (\sin x -k x) \d x + \int_{\alpha}^\pi (kx - \sin x) \d x \\ &&&= \left [ -\cos x - \frac{kx^2}{2} \right]_0^{\alpha} + \left [ \cos x +\frac{kx^2}{2} \right]_{\alpha}^\pi \\ &&&= -\cos \alpha - \frac{k \alpha^2}{2} +1+(-1)+\frac{k\pi^2}{2} - \cos \alpha - \frac{k\alpha^2}{2} \\ &&&= -2\cos \alpha - k\left (\alpha^2 - \frac{\pi^2}{2} \right) \\ &&&= -2\cos \alpha - \frac{\sin \alpha}{\alpha}\left (\alpha^2 - \frac{\pi^2}{2} \right) \\ &&&= \frac{\pi^2 \sin \alpha}{\alpha} - \alpha \sin \alpha - 2\cos \alpha \end{align*} \begin{align*} && \frac{\d I}{\d \alpha} &= \frac{\pi^2(\alpha \cos \alpha - \sin \alpha)}{2\alpha^2} + 2 \sin \alpha - \sin \alpha - \alpha \cos \alpha \\ &&&= \frac{-2\alpha^3 \cos \alpha + 2\alpha \sin\alpha + \pi^2 \alpha \cos \alpha - \pi^2 \sin \alpha}{2\alpha^2} \\ \\ &&&= \left ( \alpha \cos \alpha - \sin \alpha\right) \left ( \frac{\pi^2}{2\alpha^2}-1 \right)\ \end{align*} Therefore \(I' = 0\) if \(\tan \alpha = \alpha\) or \(\alpha = \frac{\pi}{\sqrt{2}}\). Since \(\tan \alpha = \alpha\) only at \(\alpha = 0\) (between \(0 \leq \alpha < \pi\) (by considering the tangent), we must have a unique turning point when \(\alpha = \frac{\pi}{\sqrt{2}}\). Note that \(I(\frac{\pi}{\sqrt{2}}) = \frac{\pi^2 \sqrt{2} \sin \alpha}{2\pi} - \frac{\pi}{\sqrt{2}} \sin \alpha - 2\cos \frac{\pi}{\sqrt{2}}=- 2\cos \frac{\pi}{\sqrt{2}}\). Notice that \(I(0) = \frac{\pi^2}2 - 2 > 2\) and \(I(\pi) = 2\), but \(-2\cos \frac{\pi}{\sqrt{2}} < 2\) so we must be a at a minimum

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Solution:

1. Given \(\frac{\d}{\d x} (\arctan x) \leq 1\) we must have \(\frac{\d}{ \d x} (x-\arctan x) \geq 0\) for \(x \geq 0\), but since \( 0 - \arctan 0 = 0\) this means that \(x - \arctan x \geq 0\), ie \( \arctan x \geq x\) for \(x \geq 0\) \(g(x) = x^{-1} \tan x \Rightarrow g'(x) = -x^{-2}\tan x +x^{-1} \sec^2 x\). If we can show \(f(x) = x \sec ^2 x - \tan x\) is positive that would be great. However \(f'(x) = x 2 \tan x \sec^2 x \geq 0\) and \(f(0) = 0\) so \(f(x)\) is positive and \(g'(x)\) is positive and hence increasing, therefore \(g(x) \geq g(\arctan x) \Rightarrow \frac{\tan x}{x} \geq \frac{x}{\arctan x}\) from which the result follows.

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Solution: If \(f''(x) = kf(x)f'(x)\) then we can see \begin{align*} && I &= \int [\f'(x)]^2 \,[\f(x)]^n \d x \\ &&&= \int f'(x) \cdot f'(x) [f(x)]^n \d x \\ &&&= \left[ f'(x) \cdot \frac{[f(x)]^{n+1}}{n+1} \right] - \int f''(x) \frac{[f(x)]^{n+1}}{n+1} \d x \\ &&&= \frac{1}{n+1} \left (f'(x) [f(x)]^{n+1} - \int kf'(x) [f(x)]^{n+2} \d x \right) \\ &&&= \frac{1}{n+1} \left (f'(x) [f(x)]^{n+1} - k \frac{[f(x)]^{n+3}}{n+3} \right) +C\\ &&&= \frac{[f(x)]^{n+1}}{n+1} \left ( f'(x) - \frac{k[f(x)]^2}{n+3} \right) + C \end{align*}

1. If \(f(x) = \tan x, f'(x) = \sec^2 x, f''(x) = 2 \sec^2 x \tan x = 2 \cdot f(x) \cdot f'(x)\), so \(\tan\) satisfies the conditions for the theorem. \begin{align*} && I &= \int \sec^4 x \tan^n x \d x \\ &&&= \left [\sec^2x \cdot \frac{\tan^{n+1} x}{n+1} \right] - \int 2 \sec^2 x \tan x \cdot \frac{\tan^{n+1} x}{n+1} \d x \\ &&&= \left [\sec^2x \cdot \frac{\tan^{n+1} x}{n+1} \right] - 2 \cdot \frac{\tan^{n+3} x}{(n+1)(n+3)} \\ \end{align*} So \begin{align*} && I &= \int \frac{\sin^4 x}{\cos^8 x} \d x \\ &&&= \int \tan^4 x \sec^4 x \d x \\ &&&= \int [\sec^2 x]^2 [\tan x]^4 \d x \\ &&&= \frac{\tan^{5}x}{5} \left ( \sec^2 x - \frac{2 \tan^2 x}{7} \right) + C \end{align*}
2. \begin{align*} && I &= \int \sec^2x\, (\sec x + \tan x)^6\,\d x \\ &&&= \int (\sec x (\sec x + \tan x))^2 \cdot (\sec x + \tan x )^4 \d x \\ &&&= \frac{(\sec x + \tan x)^5}{5} \left ( \sec x (\sec x + \tan x) - \frac{(\sec x + \tan x)^2}{7} \right) + C \end{align*}

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Solution: The condition is that we match the first and second derivative (as well as passing through the point in question, which is \((\frac{\pi}{4}, 2 - \frac{\pi}{4})\) The gradient is \(y' = -1 + \sec^2 x\), so the value is \(1\). The second derivative is \(y'' = 2 \sec^2 x \tan x\), which is \(4\) If we have a circle, radius \(r\), so \((x-a)^2 + (y-b)^2 = r^2\) then \(2(x-a) + 2(y-b) \frac{\d y}{\d x} = 0\) and \(2 + 2 \left ( \frac{\d y}{\d x} \right)^2 + 2(y-b) \frac{\d^2y}{\d x^2} = 0\). Therefore we must have \(1+1+(2-\frac{\pi}{4}-b)4 = 0 \Rightarrow b =\frac52-\frac{\pi}{4}\) We know that the centre lies on the line \(y = 2-x\), so we must have \(a = \frac{\pi}{4}-\frac12\) and so the centre is \(( \frac{\pi}{4} - \frac12,\frac52 - \frac{\pi}{4})\) and the radius is \(\sqrt{\frac14 + \frac14} = \frac{\sqrt{2}}{2}\)

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Solution: \begin{align*} \cos 3x &\equiv \cos (2x + x) \\ &\equiv \cos 2x \cos x - \sin 2x \sin x \\ &\equiv (2\cos^2 x - 1) \cos x - 2 \sin x \cos x \sin x \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (\sin^2 x) \\ &\equiv 2 \cos^3 x - \cos x - 2\cos x (1- \cos^2 x) \\ &\equiv 4\cos^3 x - 3\cos x \end{align*} Similarly, \begin{align*} \sin 3x &\equiv \sin (2x + x) \\ &\equiv \sin 2x \cos x + \cos 2x \sin x \\ &\equiv 2 \sin x \cos x \cos x + (1-2\sin^2 x) \sin x \\ &\equiv 2 \sin x (1-\sin^2 x) + \sin x - 2 \sin^3 x \\ &\equiv 3 \sin x -4 \sin ^3 x \end{align*}

1. \begin{align*} I(\alpha) &= \int_0^{\alpha} (7 \sin x - 8 \sin^3 x) \d x \\ &= \int_0^{\alpha} (7 \sin x - (6\sin x-2 \sin 3x) ) \d x \\ &= \int_0^{\alpha} (\sin x +2 \sin 3x ) \d x \\ &= -\cos \alpha - \frac23 \cos 3\alpha +1+\frac23 \\ &= -c - \frac23 (4c^3-3c) + \frac53 \\ &= -\frac83 c^3 +c + \frac53 \end{align*} as required. When \(c = -1\) this value is \(0\). Eustace will obtain the value \(\frac{7}{2} \sin^2 \beta - 2 \sin^4 \beta = \frac72 (1-\cos^2 \beta) - 2(1-\cos^2 \beta)^2 = \frac32 + \frac12\cos^2 \beta -2\cos^4 \beta\) So if \(\cos \beta = -\frac16\) he will obtain \(\frac32 + \frac{1}{2\cdot36} - \frac{2}{6^4}\) and he should obtain \(\frac{8}{3} \frac{1}{6^3} - \frac{1}{6} + \frac{5}{3}\) which are equal. We want to find all roots of: \begin{align*} && \frac32 + \frac12 c^2 - 2c^4 &= -\frac83 c^3+ c + \frac53 \\ \Rightarrow && 0 &=2c^4-\frac83c^3-\frac12 c^2+c +\frac{1}{6} \\ &&&= 12c^4-16c^3-3c^2+6c+1\\ &&&= (6c+1)(2c^3-3c^2+1) \\ &&&= (6c+1)(2c+1)(c-1)^2 \end{align*} Therefore \(\cos \alpha = - \frac16, -\frac12, 1\) will give the correct answers.

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Solution:

1. \(\,\) \begin{align*} && I &= \int_0^a \frac{f(x)}{f(x)+f(a-x)} \d x \\ u =a-x, \d u = - \d x: &&& \int_{u=a}^{u=0} \frac{f(a-u)}{f(a-u)+f(u)} (-1) \d u \\ &&&= \int_0^a \frac{f(a-u)}{f(u)+f(a-u)} \d u \\ &&&= \int_0^a \frac{f(a-x)}{f(x)+f(a-x)} \d x \\ \Rightarrow && 2 I &= \int_0^a \left ( \frac{f(x)}{f(x)+f(a-x)} + \frac{f(a-x)}{f(x)+f(a-x)} \right) \d x \\ &&&= \int_0^a 1 \d x \\ &&&= a \\ \Rightarrow && I &= \frac{a}{2} \end{align*} \begin{align*} && J &= \int_0^1 \frac{\ln (x+1)}{\ln (2+x-x^2)}\, \d x \\ &&&= \int_0^1 \frac{\ln (x+1)}{\ln((x+1)(2-x))} \d x \\ &&&= \int_0^1 \frac{\ln (x+1)}{\ln(x+1) + \ln ((1-x)+1)} \d x \\ &&&= \frac{1}{2} \tag{\(f(x) = \ln (x+1)\)} \\ \\ && K &= \int_0^{\frac\pi 2} \frac{\sin x } {\sin(x+\frac \pi 4 )} \, \d x \\ &&&= \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}} \\ &&&= \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x + \sin (\frac{\pi}{2}-x)} \d x\\ &&&= \frac{\pi}{2\sqrt{2}} \end{align*}
2. \(\,\) \begin{align*} &&I &= \int_{\frac12}^2 \frac{\sin x }{x(\sin x + \sin \frac1x)} \d x \\ u = 1/x, \d u = -1/x^2 \d x : &&&= \int_{u = 2}^{u=\frac12} \frac{\sin \frac1u}{\frac{1}{u}(\sin \frac1u + \sin u)} (-\frac{1}{u^2} ) \d u \\ &&&= \int_{\frac12}^2 \frac{\sin \frac1u}{u (\sin u + \sin \frac1u)} \d u \\ \Rightarrow && 2I &= \int_{\frac12}^2 \left ( \frac{\sin x }{x(\sin x + \sin \frac1x)} + \frac{\sin \frac1x }{x(\sin x + \sin \frac1x)}\right) \d x \\ &&&= \int_{\frac12}^2 \frac{1}{x} \d x\\ &&&= 2\ln2 \\ \Rightarrow && I &= \ln 2 \end{align*}

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Solution:

The curves touch when \(\sin x = 1\) ie \(x = \frac{4n+1}{2} \pi\). therefore \begin{align*} && I &= \int e^{-x} \sin x \d x \\ &&&= -e^{-x} \sin x + \int e^{-x} \cos x \d x \\ &&&= -e^{-x} \sin x -e^{-x} \cos x - I \\ && I &= -\frac{e^{-x}}2 ( \sin x + \cos x)\\ \\ && A_n &= \int_{\frac{4n+3}{2}\pi}^{\frac{4n-1}{2}\pi} e^{-x} (1-\sin x) \d x \\ &&&= \left [ -e^{-x} - \frac{e^{-x}}2(\sin x + \cos x) \right]_{\frac{4n+1}{2}\pi}^{\frac{4n+5}{2}\pi} \\ &&&= -\frac12\exp \left ( -\frac{4n+1}{2}\pi\right)+ \frac12\exp \left ( -\frac{4n-3}{2}\pi\right) \\ &&&= \frac12 (e^{2 \pi}-1) \left ( -\frac{4n+1}{2}\pi\right) \\ \\ && \sum_{n=1}^\infty A_n &= \frac12(e^{2\pi}-1)e^{-\pi/2} \sum_{n=1}^\infty e^{-2n\pi} \\ &&&= \frac12(e^{2\pi}-1)e^{-\pi/2} \frac{e^{-2\pi}}{1-e^{-2\pi}} \\ &&&= \frac12e^{-\pi/2} \end{align*}

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Solution: \begin{align*} && y &= \cos(m \arcsin x) \\ && y' &= -m \sin (m \arcsin x) \cdot (1-x^2)^{-\frac12} \\ && y'' &= -m^2 \cos(m \arcsin x) \cdot (1-x^2)^{-1} -m \sin(m \arcsin x) \cdot (1-x^2)^{-\frac32} \cdot (-x) \\ &&&= -m^2 y (1-x^2)^{-1} + x(1-x^2)^{-1} y' \\ \Rightarrow && 0 &= (1-x^2)y'' - x y' + m^2y \\ \\ && 0 &= (1-x^2)y^{(3)} -2xy'' - xy''-y' + m^2y' \\ &&&= (1-x^2)y^{(3)} - 3xy'' + (m^2-1)y' \\ \\ && 0 &= (1-x^2)y^{(4)} - 2xy^{(3)} - 3xy^{(3)} - 3y^{(2)} + (m^2-1)y^{(2)} \\ &&&= (1-x^2)y^{(4)}- 5xy^{(3)} - (m^2-4)y^{(2)} \end{align*} Claim: \(0 = (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n)}\) Proof: (By induction) Clearly the first few base cases are true. Suppose it is true for some \(n\), then \begin{align*} && 0 &= (1-x^2)y^{(n+2)} - (2n+1)xy^{(n+1)} + (m^2-n^2)y^{(n)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+3)} - 2xy^{(n+2)} - (2n+1)xy^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+3)} - (2n+3)xy^{(n+2)} + (m^2-n^2-2n-1)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+1+2)} - (2(n+1)+1)xy^{(n+1+1)} +(m^2-(n+1)^2)y^{(n)} \end{align*} And so we can conclude the result by induction. Notice that \begin{align*} && y(0) &= \cos(m 0) = 1 \\ && y'(0) &= -m\sin(m 0) = 0 \\ && y''(0) &= -m^2 y(0) = -m^2\\ \end{align*} Notice that \(y^{(n+2)}(0) + (m^2-n^2)y^{(n)} = 0\) so in particular all the odd terms will be \(0\) and the even terms will be \(1, -m^2, m^2(m^2-2^2), \cdots\), therefore \begin{align*} && \cos (m \arcsin x) &= 1 -\frac{m^2}{2!} x^2 + \frac{m^2(m^2-2^2)}{4!}x^4 - \cdots \\ \Rightarrow && \cos(m \theta) &= 1 - \frac{m^2}{2!} \sin^2 \theta + \frac{m^2(m^2-2^2)}{4!} \sin^4 \theta \end{align*} Notice that if \(m\) is even, then at some point we will have \(m^2-m^2\) appearing in our expansion and all remaining terms will be zero. Therefore we will end up with a polynomial series, of degree \(m\) in \(\sin \theta\)

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Solution: \begin{align*} && \int \frac{{\rm P} ( x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x &= \int \frac{{\rm Q} (x){\rm R}'(x) - {\rm Q}'(x){\rm R}(x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x \\ &&&= \int \frac{\d}{\d x} \left ( \frac{R(x)}{Q(x)} \right) \d x \\ &&&= \frac{R(x)}{Q(x)} + C \end{align*}

1. Suppose \(Q(x) = 1 + 2x + 3x^2, P(x) = 5x^2-4x-3\), and \(R(x) = a +bx + cx^2\), then \begin{align*} && 5x^2-4x-3 &= (1 + 2x + 3x^2)(2cx+b) - (6x+2)(a+bx+cx^2) \\ &&&= (6c-6c)x^3 + (3b+4c-6b-2c)x^2 + \\ &&&\quad+(2c+2b-6a-2b)x + (b-2a) \\ \Rightarrow && 2c-3b &= 5 \\ && 2c-6a &= -4 \\ && b - 2a &= -3 \\ \Rightarrow && b &= 2a - 3\\ && c &= 3a-2 \end{align*} So say \(a = 0, b = -3, c = -2\) we will have \begin{align*} \int \frac{5x^2 - 4x - 3} {(1 + 2x + 3x^2 )^2 } \, \d x &= \frac{-3x-2x^2}{1+2x+3x^2} + C \end{align*} Suppose we have a different value of \(a\), then we end up with: \begin{align*} \frac{a+(2a-3)x+(3a-2)x^2}{1+2x+3x^2} = a +\frac{-3x-2x^2}{1+2x+3x^2} \end{align*} So the different antiderivatives differ by a constant.
2. \(\,\) \begin{align*} && \frac{\d }{\d x} \left ( \frac{1}{1+\cos x + 2 \sin x } y\right) &= \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \\ \end{align*} We want to find \(R(x) = A \cos x + B\sin x + C\) such that \begin{align*} &&5-3\cos x + 4 \sin x &= (1+\cos x + 2 \sin x)R'(x) - R(x)(-\sin x + 2 \cos x) \\ &&&= (1+\cos x + 2 \sin x)(-A\sin x + B \cos x) \\ &&&\quad- (A\cos x + B \sin x + C)(-\sin x + 2 \cos x) \\ &&&=(-A+C) \sin x + (B-2C)\cos x +\\ &&&\quad\quad (2B-A+A-2B)\sin x \cos x \\ &&&\quad\quad (-2A+B)\sin^2x+(B-2A)\cos^2x \\ &&&= (-A+C)\sin x + (B-2C)\cos x +(B-2A) \\ \Rightarrow && B-2A &= 5\\ && C-A &= 4 \\ && B-2C &= -3 \\ \end{align*} There are many solutions so WLOG \(C=4, A = 0, B = 5\) and so \begin{align*} && \int \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \d x &= \frac{5\sin x +4}{1+\cos x + 2 \sin x} + K \\ \Rightarrow && y &= 5\sin x + 4 + K(1 + \cos x + 2 \sin x) \end{align*}