Problems

Solution:

1. \begin{align*} \int_{1}^{\mathrm{e}}\frac{\ln x}{x^{2}}\,\mathrm{d}x &= \left [-\frac{\ln x}{x} \right]_1^e + \int_1^e \frac{1}{x^2} \, \d x \\ &= -\frac{1}{e} + \left [ -\frac{1}{x} \right]_1^e \\ &= 1 - \frac{2}{e} \end{align*}
2. \begin{align*} \int\frac{\cos x}{\sin x\sqrt{1+\sin x}}\,\mathrm{d}x &= \int \frac{2u}{(u^2-1)u} \d u \tag{\(u^2 = 1+\sin x\)} \\ &= \int \frac{1}{u-1} - \frac{1}{u+1} \d u \\ &= \ln(u-1) - \ln (u+1) + C \\ &= \ln \l \frac{u-1}{u+1} \r + C \\ &= \ln \l \frac{\sqrt{\sin x + 1} + 1}{\sqrt{\sin x + 1} -1} \r + C \end{align*}

Solution: \begin{align*} && f(x) &= \sin 2x \cos x \\ &&&= \frac12 \l \sin 3x + \sin x \r \\ \Rightarrow && f^{(1988)}(x) &= \frac12 \l 3^{1988} (-1)^{994} \sin 3x+ (-1)^{994} \sin x \r \\ &&&= \boxed{\frac12 \left (3^{1998} \sin 3x + \sin x \right)} \\ \\ f^{(1988)}(x) = 0: && 0 &= 3^{1988} \sin 3x + \sin x \\ \Rightarrow && 0 &= 3^{1988} ( 3\sin x-4\sin^3 x) + \sin x \\ \Rightarrow && 0 &= \sin x \left (1+3^{1989}-4\cdot 3^{1988}\sin^{2} x \right) \end{align*} Since \(\sin x\) will first contribute a zero when \(x = \frac{\pi}{2}\) we focus on the second bracket, in particular, we need: \begin{align*} && \sin^2 x &= \frac{3}{4} \left ( 1 + \frac{1}{3^{1988}} \right) \\ \Rightarrow && \sin x &= \frac{\sqrt{3}}2 \left (1 + \frac{1}{2 \cdot 3^{1988}} + \cdots \right ) \end{align*} Since near \(\frac{\pi}{3}\), \begin{align*} \sin (\frac{\pi}{3} + \epsilon) &= \sin \frac{\pi}{3} \cos \epsilon + \cos \frac{\pi}{3} \sin \epsilon \\ &\approx \frac{\sqrt{3}}{2} (1-\epsilon^2 + \cdots ) + \frac{1}{2}(\epsilon + \cdots) \\ &= \frac{\sqrt{3}}2 + \frac12 \epsilon + \cdots \end{align*} Therefore by comparison we can see that \(x = \frac{\pi}{3} + \frac{\sqrt{3}}{2} 3^{-1988}\) will be a very good approximation for the root.

Solution: Let \(f(x) = e^{ax} \cos bx\) then, \(f'(x) = ae^{ax} \cos bx - be^{ax} \sin bx = e^{ax} \l a\cos bx - b \sin bx \r\). Therefore the stationary points are where \(f'(x) = 0 \Leftrightarrow \tan bx = \frac{b}a\), ie \(x = \tan^{-1} \frac{a}{b} + \frac{n}{b} \pi, n \in \mathbb{Z}\). \begin{align*} f(\tan^{-1} \frac{a}{b} + \frac{n}{b} \pi) &= e^{a \tan^{-1} \frac{a}{b} + \frac{an}{b} \pi} \cos \l b \tan^{-1} \frac{a}{b} +n \pi\r \\ &= e^{a \tan^{-1} \frac{a}{b}} \cos \l b \tan^{-1} \frac{a}{b}\r \cdot e^{\frac{an}{b} \pi}(-1)^n \\ &= e^{a \tan^{-1} \frac{a}{b}} \cos \l b \tan^{-1} \frac{a}{b}\r \cdot (-e^{\frac{a}{b} \pi})^n \\ \end{align*} showing the form the desired geometric progression.

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Solution: Consider the angle from the origin, then \(P = (\cos \theta, \sin \theta)\) where \(\theta \sim U(0, 2\pi)\), and \(X = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta}\) \begin{align*} \mathbb{E}[X] &= \int_0^{2\pi} \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} \frac1{2\pi} \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} \sqrt{2 - 2\cos \theta} \d \theta \\ &= \frac{1}{2\pi}\int_0^{2\pi} \sqrt{4\sin^2 \frac{\theta}{2}} \d \theta \\ &= \frac{1}{\pi}\int_0^{2\pi} \left |\sin \frac{\theta}{2} \right| \d \theta \\ &= \frac{1}{\pi} \left [ -2\cos \frac{\theta}{2} \right]_0^{2\pi} \\ &= \frac1{\pi} \l 2 + 2\r \\ &= \frac{4}{\pi} \end{align*} \begin{align*} \mathbb{E}(X^2) &= \frac1{2\pi}\int_0^{2\pi} (\cos \theta - 1)^2 + \sin^2 \theta \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} 2 - 2 \cos \theta \d \theta \\ &= \frac{4\pi}{2\pi} \\ &= 2 \\ \end{align*} \(\Rightarrow\) \(\mathrm{Var}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2 = 2 - \frac{16}{\pi^2} = \frac{2\pi^2 - 16}{\pi^2}\).

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Solution: \begin{align*} \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\cos^{2}\theta}{1-\sin\theta\sin2\alpha}\,\mathrm{d}\theta &= \int_{u = \frac12 \pi}^{u = -\frac12 \pi} \frac{\cos^2 (-u)}{1-\sin(-u) \sin 2 \alpha} -\d u \tag{\(u = -\theta\)} \\ &= \int_{\frac12 \pi}^{-\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} -\d u \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} \d u \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta \\ \end{align*} Since \(\displaystyle \frac{1}{(1-a^2u^2)} = \frac12 \l \frac{1}{1+au} + \frac1{1-au} \r\) \begin{align*} \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{1-\sin^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{(1-\sin ^2\theta \sin^2 2 \alpha) \frac{1}{\sin^2 2\alpha} + 1 - \cosec^2 2\alpha}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta \sin^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta (1-\cos^2 2 \alpha)} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{\cos^2 \theta +\sin^2 \theta \cos^2 2 \alpha} \d \theta \\ &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta \\ \end{align*} Finally, using the substitution \(u =|\cos 2 \alpha | \tan \theta, \d u = |\cos 2 \alpha |\sec^2 \theta \d \theta\) \begin{align*} \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta &= |\sec 2\alpha|\int_{u = -\infty}^{u = \infty} \frac{1}{1 + u^2} \d u \\ &= |\sec 2 \alpha|\pi \end{align*} and so \begin{align*} I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha |\sec 2 \alpha|\pi \\ &= \frac{\pi}{\sin^2 2\alpha} \l 1-\cos 2\alpha \r \\ &= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \sin^2 \alpha \r \\ &= \frac{\pi}{2 \cos^2 \alpha} = \frac{\pi}{2} \sec^2 \alpha \end{align*} When \(\alpha\) small enough that the modulus doesn't flip the sign. When if \(\frac{1}{4}\pi<\alpha<\frac{1}{2}\pi\) we have: \begin{align*} I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha |\sec 2 \alpha|\pi \\ &= \frac{\pi}{\sin^2 2\alpha} \l 1+\cos 2\alpha \r \\ &= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \cos^2 \alpha \r \\ &= \frac{\pi}{2 \sin^2 \alpha} = \frac{\pi}{2} \cosec^2 \alpha \end{align*}

Solution: Let \({\displaystyle I_{m,n}=\int\cos^{m}x\sin nx\,\mathrm{d}x,}\) Then \begin{align*} && I_{m,n} &= \int\cos^{m}x\sin nx\,\mathrm{d}x \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] - \frac{m}{n} \int \sin^{m-1} x \cos x \cos n x \d x \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} \int \sin^{m-1} x (\cos (n-1)x -\sin x \sin nx) \d x\\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} \int \sin^{m-1} x \cos (n-1)x \d x-\frac{m}{n} I_{m,n} \\ &&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} I_{m-1,n-1} -\frac{m}{n} I_{m,n} \\ \Rightarrow && nI_{m,n} &= -\cos^m x \cos n x + mI_{m-1,n-1} -mI_{m,n}\\ \Rightarrow && (m+n)I_{m,n} &= -\cos^m x \cos n x + mI_{m-1,n-1} \end{align*}

1. Note that \(I_{2m,0} = 0\) (the integrand is 0) and \(I_{0, 2m} = 0\) (symmetry for our limits). \(\displaystyle \left [-\cos^m x \cos n x \right]_0^\pi = \l - (-1)^m (-1)^n \r - \l -1 \r = 1 - (-1)^{m+n} = 0\) since \(m+n\) is even. Therefore all reductions are \(I_{m,n} = \frac{I_{m-1,n-1}}{m+n}\) terminating at \(0\), so all values are zero
2. \begin{align*} \int_{0}^{\frac{\pi}{2}}\sin^{2}x\sin3x\,\mathrm{d}x &= \int_{0}^{\frac{\pi}{2}}(1-\cos^2x)\sin3x\,\mathrm{d}x \\ &= I_{0,3} - I_{2,3} \\ &= \frac13 - \frac15 \l \left [-\cos^2 x \cos 3 x \right]_0^{\pi/2} + 2 \cdot I_{1,2} \r \\ &= \frac13 - \frac15 \l 1 + \frac23 \l \left [-\cos x \cos 2 x \right]_0^{\pi/2} + 1\cdot I_{0,1} \r \r \\ &= \frac13 - \frac15 -\frac2{15} - \frac2{15} \\ &= \frac{5}{15} - \frac{3}{15} - \frac{4}{15} \\ &= -\frac2{15} \end{align*}

Solution: Letting \(\mathbf{x} =\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}\) and \(\mathbf{A} = \begin{pmatrix} -2 & 5 \\ -a & 2 \end{pmatrix}\) then our differential equation is \(\mathbf{x}' = \mathbf{Ax} + \begin{pmatrix} 0 \\2 \cos t \end{pmatrix}\). Looking at the eigenvalues of \(\mathbf{A}\), we find: \begin{align*} && \det \begin{pmatrix} -2-\lambda & 5 \\ -a & 2 -\lambda \end{pmatrix} &= (\lambda^2-4)+5a\\ &&&= \lambda^2 +5a-4 \end{align*} Therefore if \(a = 1\), \(\lambda = \pm i\). In which case we should expect the complementary solutions to be of the form \(\mathbf{x} = \begin{pmatrix} A \sin t + B \cos t \\ C \sin t + D \cos t \end{pmatrix}\). The first equation tells us that \((A-5D+B)\cos t + (-B+5C)\sin t=0\) so the complementary solution is:\(\mathbf{x} = \begin{pmatrix} 5(D-C) \sin t + 5C \cos t \\ C \sin t + D \cos t \end{pmatrix}\). Looking for a particular integral, we should expect to try something like \(\mathbf{x} = \begin{pmatrix} Et\cos t+Ft\sin t\\ Gt\cos t+Ht \sin t\end{pmatrix}\) and we find