Problems

---

Solution:

1. \(\,\) \begin{align*} && I &= \int_{-\frac12 \pi}^{\frac12\pi} \frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} \, \d\theta \\ \phi = -\theta, \d \phi = - \d \theta: &&&= \int_{\phi=\frac12\pi}^{\phi=-\frac12\pi} \frac{\cos^2(-\phi)}{1-\sin(-\phi)\sin 2 \alpha} (-1) \d \phi \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \frac {\cos^2\phi}{1+\sin\phi\sin2\alpha} \d\phi \\ \\ && 2I &= \int_{-\frac12 \pi}^{\frac12\pi} \frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} \, \d\theta +\int_{-\frac12 \pi}^{\frac12\pi} \frac {\cos^2\theta}{1+\sin\theta\sin2\alpha} \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} +\frac {\cos^2\theta}{1+\sin\theta\sin2\alpha} \right) \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {2\cos^2\theta}{1-\sin^2\theta\sin^22\alpha} \right) \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {2\cos^2\theta}{1-(1-\cos^2\theta)(1-\cos^22\alpha)} \right) \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {2\cos^2\theta}{\cos^2\theta+\cos^22\alpha-\cos^2 \theta \cos^2 2 \alpha)} \right) \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {2}{1+\cos^22\alpha(\sec^2 \theta - 1))} \right) \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {2}{1+\tan^2 \theta \cos^22\alpha} \right) \, \d\theta \\ \end{align*}
2. \(\,\) \begin{align*} && J &= \int_{-\frac12 \pi}^{\frac12\pi} \frac {\sec^2\theta}{1+\tan^2\theta\cos^22\alpha} \, \d\theta \\ &&&= \left [\sec 2 \alpha \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) \right]_{-\frac12 \pi}^{\frac12\pi} \\ &&&= \sec(2\alpha)\pi = \frac{\pi}{\cos 2 \alpha} \end{align*}
3. \(\,\) \begin{align*} && I\sin^2 2\alpha +J\cos^2 2\alpha &= \int_{-\frac12\pi}^{\frac12 \pi} \frac{\sin^2 2 \alpha+\cos^2 2 \alpha \sec^2 \theta}{1+\tan^2 \theta \cos^2 2\alpha} \d \theta \\ &&&= \int_{-\frac12\pi}^{\frac12 \pi} \frac{\sin^2 2 \alpha+\cos^2 2 \alpha (1 + \tan^2 \theta)}{1+\tan^2 \theta \cos^2 2\alpha} \d \theta \\ &&&= \pi \\ \\ \Rightarrow && I &= \frac{\pi - \pi \cos 2 \alpha}{\sin^2 2 \alpha} \\ &&&= \pi \frac{2\sin^2 \alpha}{4 \sin^2 \alpha \cos^2 \alpha} \\ &&&= \frac12 \pi \sec^2 \alpha \end{align*}
4. If \(\frac14 \pi < \alpha < \frac12 \pi\) then our calculation for \(J\) is not correct. \begin{align*} && J &= \int_{-\frac12 \pi}^{\frac12\pi} \frac {\sec^2\theta}{1+\tan^2\theta\cos^22\alpha} \, \d\theta \\ &&&= \left [\sec 2 \alpha \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) \right]_{-\frac12 \pi}^{\frac12\pi} \\ &&&= \sec(2\alpha) \left ( \lim_{\theta \to \frac{\pi}{2}} \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) - \lim_{\theta \to -\frac{\pi}{2}} \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) \right) \\ &&&= \sec(2\alpha) \left ( \tan^{-1} \left ( \lim_{x\to -\infty} x \right) - \tan^{-1} \left ( \lim_{x\to \infty} x \right) \right) \\ &&&= -\pi \sec 2 \alpha \end{align*} Still using the same logic, we can say \begin{align*} && I &= \frac{\pi+\pi\cos 2 \alpha}{\sin^2 2 \alpha} \\ &&&= \pi \frac{2 \cos^2 \alpha}{4 \sin^2 \alpha \cos^2 \alpha}\\ &&&= \frac12 \pi \cosec^2 \alpha \end{align*}

---

Solution:

1. Since \(\tan (-x) = -\tan x\), \(\tan\) is an odd function, and in particular all it's even coefficients are zero. \begin{align*} && 2 \cot 2x &\equiv \frac{2 cos 2x}{\sin 2 x} \\ &&&\equiv \frac{2(\cos^2 x- \sin^2 x)}{2 \sin x \cos x} \\ &&&\equiv \frac{\cos x}{\sin x} - \frac{\sin x}{ \cos x} \\ &&&\equiv \cot x - \tan x \end{align*} Therefore \begin{align*} && \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} - \underbrace{\sum_{n=0}^\infty a_n x^n}_{\tan x} &= 2 \left (\underbrace{\frac{1}{2x} + \sum_{n=0}^\infty b_n(2x)^n}_{\cot 2x} \right) \\ \Rightarrow && \sum_{n=0}^\infty a_n x^n &= \sum_{n=0}^\infty b_nx^n - 2\sum_{n=0}^\infty b_n(2x)^n \\ &&&= \sum_{n=0}^{\infty}b_n(1-2^{n+1})x^n \\ [x^n]: && a_n &= (1-2^{n+1})b_n \end{align*}
2. \(\,\) \begin{align*} && \cot x + \tan x &= \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \\ &&&= \frac{1}{\sin x \cos x} \\ &&&=2\cosec 2x \\ \\ \Rightarrow && \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} + \underbrace{\sum_{n=0}^\infty a_n x^n}_{\tan x} &= 2\left (\underbrace{ \frac1{2x} +\sum\limits _{n=0}^\infty c_n (2x)^n}_{\cosec 2x} \right) \\ \Rightarrow && \sum_{n=0}^\infty 2^{n+1}c_n x^n &= \sum_{n=0}^{\infty}(a_n+b_n)x^n \\ &&&= \sum_{n=0}^{\infty}\left((1-2^{n+1})b_n+ b_n\right)x^n \\ &&&= \sum_{n=0}^{\infty}\left(2-2^{n+1}\right)b_nx^n \\ [x^n]: && c_n &= (2^{-n}-1)b_n \end{align*}
3. \(\,\) \begin{align*} && \cot^2 x + 1 &= \cosec^2 x \\ \Rightarrow && x^2 \cot^2 x + x^2 &= x^2 \cosec^2 x \\ \Rightarrow && x^2 \left ( \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} \right)^2 + x^2 &= x^2 \left (\underbrace{ \frac1{x} +\sum\limits _{n=0}^\infty c_n x^n}_{\cosec x} \right)^2 \\ \Rightarrow && \left ( 1 + x\sum_{n=0}^\infty b_nx^{n} \right)^2 + x^2 &= \left ( 1 +x\sum\limits _{n=0}^\infty c_n x^{n} \right)^2 \\ \\ \Rightarrow && \left ( 1 + x(b_1x + b_3 x^3 + \cdots) \right)^2 + x^2 &= \left ( 1 + x(c_1x + c_3 x^3 + \cdots) \right)^2 \\ \Rightarrow && 1 + (1+2b_1)x^2+(2b_3+b_1^2)x^4 + \cdots &= 1 + 2c_1x^2 + (2c_3+c_1^2)x^4 + \cdots \\ \Rightarrow && 1 + 2b_1 &= 2(2^{-1}-1)b_1 \\ \Rightarrow && b_1 &= -\frac13 \\ \Rightarrow && a_1 &= (1-2^{2})(-\tfrac13) = 1 \\ && c_1 &= \frac16\\ \Rightarrow && 2b_3+\frac19&= 2c_3+\frac1{36} \\ \Rightarrow && 2b_3 -2(2^{-3}-1)b_3 &= -\frac{1}{12} \\ \Rightarrow && \frac{15}{4}b_3 &= -\frac{1}{12} \\ \Rightarrow && b_3 &= -\frac{1}{45} \\ \Rightarrow && a_3 &= -(1-2^4)\frac{1}{45} = \frac13 \end{align*}

---

Solution:

1. It is at the origin when both \(\sin 2t\) and \(\cos t = 0\), but this \(\sin 2t = 2 \sin t \cos t\) so this happens precisely when \(\cos t = 0\), ie when \(t = \frac{\pi}{2}, \frac{3\pi}{2}\)
2. \(\,\) \begin{align*} && \dot{\mathbf{r}} &= 2 \cos 2t \mathbf{i} - 2 \sin t \mathbf{j} \\ && \mathbf{r} \cdot \dot{\mathbf{r}} &= 2\cos 2t \sin 2t - 2 \sin t 2 \cos t \\ &&&= \sin 2t \left (2\cos 2t - 2 \right) \end{align*} Therefore they are perpendicular when \(\sin 2t = 0 \Rightarrow t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\) and when \(\cos 2t = 1 \Rightarrow 2t = 0, 2\pi, 4\pi \Rightarrow t = 0, \pi, 2\pi\), therefore all solutions are \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\)
3. For \(\mathbf{r}\) and \(\dot{\mathbf{r}}\) to be parallel, we would need \begin{align*} && \frac{2 \cos 2t}{\sin 2t} &= \frac{-2 \sin t}{2 \cos t}\\ && 2 \cos 2t \cos t &= - \sin t \sin 2t \\ && 0 &= 2\cos t (\cos 2t + \sin ^2 t) \\ &&&= 2 \cos t (\cos^2 t) \\ &&&= 2 \cos^3 t \end{align*} Therefore the only time we can be parallel is when \(\cos t = 0\), which is when we are at the origin.
4. \(\frac{\d }{\d t} (\mathbf{r} \cdot \mathbf{r}) = 2 \mathbf{r} \cdot \mathbf{\dot{r}}\) so we should check the values when velocity and displacement are perpendicular, ie \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\) which have values \(\mathbf{r} = \binom{0}{2}, \binom{0}{0}, \binom{0}{-2}, \binom{0}{0}, \binom{0}{2}\). Therefore the maximum distance is \(2\).
   

   + - ↺

---

Solution:

Since \(y' = \cos x - \cos x + x \sin x = x \sin x > 0\) which is positive on \((0, \frac{\pi}{2})\), so \(y\) is increasing, and therefore will achieve it's highest value at \(\frac{\pi}{2}\) which is \(y(\frac{\pi}{2}) = 1\) and it's smallest value at \(y(0) = 0\). Therefore \(0 \leq y \leq 1\)

1. \(\,\) \begin{align*} \int_0^{\frac{1}{2}\pi}\,y\;\d x &= \int_0^{\frac{1}{2}\pi} (\sin x - x \cos x) \d x \\ &= \left [-\cos x \right]_0^{\frac{1}{2}\pi} +\left [ -x \sin x \right]_0^{\frac{1}{2}\pi} + \int_0^{\frac{1}{2}\pi} \sin x \d x \\ &= 1-\frac{\pi}{2} + 1 = 2 - \frac{\pi}{2} \end{align*}
2. \(\,\) \begin{align*} \int_0^{\frac{1}{2}\pi}y^2\d x &= \int_0^{\frac{1}{2}\pi} (\sin x - x \cos x)^2 \d x \\ &= \int_0^{\frac{1}{2}\pi} (\sin^2x - 2x\sin x \cos x+x^2\cos^2 x) \d x\\ &= \int_0^{\frac{1}{2}\pi} (\sin^2x -x \sin 2x+\tfrac12x^2(\cos 2 x + 1)) \d x\\ &= \frac{\pi}{4} + \frac{\pi^3}{48} + \int_0^{\frac{1}{2}\pi} (-x \sin 2x+\tfrac12x^2\cos 2 x) \d x \\ &= \frac{\pi}{4} + \frac{\pi^3}{48} + \left [\frac12 x \cos 2x +\frac14 x^2 \sin2x\right]_0^{\frac{1}{2}\pi}-\int_0^{\frac{1}{2}\pi}(\tfrac12 \cos 2x +\tfrac12 x \sin 2x) \d x\\ &= \frac{\pi}{4} + \frac{\pi^3}{48} - \frac{\pi}{4} - \left [ \frac14 \sin 2x \right]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} \tfrac12 x \sin 2x \d x\\ &= \frac{\pi^3}{48} - \left( \left[ -\frac14 x \cos 2x \right]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} -\frac14 \cos 2x \d x \right)\\ &= \frac{\pi^3}{48} - \left( \frac{\pi}{8} + \left[ \frac18 \sin 2x \right]_0^{\frac{1}{2}\pi} \right)\\ &= \frac{\pi^3}{48} - \frac{\pi}{8} \end{align*}

Since \(y^2 < y\) on this interval, we must have \( \frac{\pi^3}{48} - \frac{\pi}{8} < 2 - \frac{\pi}{2} \Rightarrow \pi^3 +18\pi < 96\) as required.

---

Solution:

1. \(P\) is travelling in a unit circle about the origin in the \(\mathbf{i}-\mathbf{j}\) plane. \(Q\) is travelling in a circle (also about the origin, but in a different plane with radius \(3\)).
2. \(\,\) \begin{align*} && \mathbf{p}\cdot \mathbf{q} &= |\mathbf{p}||\mathbf{q}| \cos \theta \\ \Rightarrow && \cos \theta &= \frac{\tfrac32\cos t \cos(t + \tfrac{\pi}4)+3\sin t \sin (t + \tfrac{\pi}{4})}{3} \\ &&&= \tfrac12\cos t \cos(t + \tfrac{\pi}4)+\sin t \sin (t + \tfrac{\pi}{4}) \\ &&&= \tfrac14 (\cos (2t + \tfrac{\pi}{4}) + \cos(\tfrac{\pi}{4} ))+\tfrac12(\cos(\tfrac{\pi}{4})-\cos(2t + \tfrac{\pi}{4})) \\ &&&= \tfrac{3\sqrt{2}}8 - \tfrac14 \cos ( 2t +\tfrac{\pi}{4}) \end{align*}
3. If \(\theta \geq \frac14\pi\), then \(\cos \theta \leq \frac{\sqrt{2}}2\) \begin{align*} && \frac{\sqrt{2}}2 & \geq \frac{3\sqrt{2}}8 - \frac14 \cos ( 2t +\tfrac{\pi}{4}) \\ \Rightarrow && \frac{\sqrt{2}}2 &\geq -\cos(2t + \tfrac{\pi}{4}) \\ \Rightarrow && \cos(2t + \tfrac{\pi}{4}) &\geq -\frac{1}{\sqrt{2}} \\ \Rightarrow && 2t + \tfrac{\pi}{4} &\not\in (\tfrac{3\pi}{4},\tfrac{5\pi}{4}) \cup (\tfrac{11\pi}{4},\tfrac{13\pi}{4}) \\ \Rightarrow && t &\not\in (\tfrac{\pi}{4}, \tfrac{\pi}{2})\cup (\tfrac{5\pi}{4}, \tfrac{3\pi}{2}) \end{align*} which is is a time of \(\frac{\pi}{2}\), therefore the left over time is \(\frac32\pi\)

---

Solution: \begin{align*} && I &= \int_0^\pi x \sin x \d x \\ &&&= \left [ -x \cos x \right]_0^\pi + \int_0^{\pi} \cos x \d x \\ &&&= \pi \\ \\ && J &= \int_0^\pi x \cos x \d x \\ &&&= \left [ x \sin x \right]_0^\pi - \int_0^\pi \sin x \d x \\ &&&= -2 \end{align*} \begin{align*} && f(t) &= t + \int_0^\pi f(x) \sin (x+t) \d x \\ &&&= t + \int_0^\pi f(x) \left ( \sin t \cos x + \cos t \sin x \right) \d x \\ &&&= t + \sin t \int_0^{\pi} f(x) \cos x \d x + \cos t \int_0^{\pi} f(x) \sin x \d x \\ \\ && A &= \int_0^\pi (x + A \sin x + B \cos x) \cos x \d x \\ &&&= -2+ \frac{\pi}{2} B \\ && B &= \int_0^{\pi} (x + A \sin x + B \cos x ) \sin x \d x \\ &&&= \pi + \frac{\pi}{2} A \\ \Rightarrow && (A,B) &= (-2,0) \end{align*}

---

Solution: \begin{array}{c|c|c|c|c|c} n & a_n & b_n & c_n & f(a_n) & f(c_n) & f(b_n) \\ \hline 0 & \tfrac12 \pi & \pi & \tfrac34\pi & 2\sin(\tfrac12\pi)-\tfrac12\pi = 2-\tfrac12\pi & 2\sin(\tfrac34\pi)-\tfrac34\pi = \frac{2}{\sqrt{2}}-\tfrac34\pi & 2\sin(\pi)-\pi =-\pi \\ 0 & \tfrac12 \pi & \pi & \tfrac34\pi & >0 & 2-\frac{9}{16}10 < 0& <0 \\ \hline 1 & \frac12 \pi & \frac34\pi & \frac58\pi & >0 &2\sin \tfrac58\pi - \tfrac58\pi & < 0\\ 1 & \frac12 \pi & \frac34\pi & \frac58\pi & >0 & \approx 1.4 \cdot \sqrt{1.7} -\frac58\sqrt{10} < 0 & <0 \\ \hline 2 & \frac12 \pi & \frac58\pi & \frac9{16}\pi & >0 & 2\sin \frac{9}{16}\pi-\frac{9}{16}\pi & <0 \\ 2 & \frac12 \pi & \frac58\pi & \frac9{16}\pi & >0 & > 0 & <0 \\ \end{array} Threfore \(I_3 = [\frac9{16}\pi,\frac58\pi]\) \(\sin \frac{5\pi}{8} = \cos \frac{\pi}{8} = \sqrt{\frac12(\cos \frac{\pi}{4}+1)} = \frac{1}{\sqrt{2}}\sqrt{1 + \frac{1}{\sqrt{2}}} \approx 0.7 \cdot \sqrt{1.7}\) \(\sin \frac{9\pi}{16} = \cos \frac{\pi}{16} = \sqrt{\frac12\left ( \cos \frac{\pi}{8}+1 \right)} \) So we are comparing \(2\cos \frac{\pi}{16}\) with \(\frac{9}{16}\pi\) or \(4 \cos^2 \frac{\pi}{16} = 2\cos \frac{\pi}{8}+2\) with \(\frac{90}{16}\)

---

Solution:

1. \(\,\) \begin{align*} && \dot{x} &= (1-x) \\ \Rightarrow &&\int \frac{1}{1-x} \d x &= \int \d t \\ \Rightarrow && -\ln |1-x| &= t + C \\ t=0, x = 0: && -\ln 1 &= C \Rightarrow C = 0\\ \Rightarrow && -\ln|1-x| &= t \\ \Rightarrow && 1-x&= e^{-t} \\ \Rightarrow && x &= 1-e^{-t} \end{align*}
   

   + - ↺
2. Notice that \((1-x^2)^{1/2} = (1-x)^{1/2}(1+x)^{1/2} > (1-x)^{1/2} > 1-x\)
   

   + - ↺
   \begin{align*} && \dot{x} &= \sqrt{1-x^2} \\ \Rightarrow && \int \frac{1}{\sqrt{1-x^2}} \d x &= t + C \\ x = \sin y, \d x = \cos y && \int \frac{\cos y}{\cos y} \d y &= t + C \\ \Rightarrow && y &= t + C \\ \Rightarrow && \sin^{-1} x &= t + C \\ t = 0, x = 0: && x &= \sin t \end{align*}
3. \(\,\)
   

   + - ↺
   We know the gradient is steeper, so the solution must always be above \(\sin t\), which means it reaches \(1\) before \(\frac{\pi}{2}\)

---

Solution:

If \(n\) is even, and \(0 < x < n\) then \(g_n(x) = \begin{cases} \{x \} & \text{if }\lfloor x \rfloor\text{ is even} \\ 1-\{x \} & \text{if }\lfloor x \rfloor\text{ is odd} \\\end{cases}\), in other words, there are \(\frac{n}{2}\) triangles, with height \(1\) and base \(2\), giving total area of \(\frac{n}{2}\). Each section of \(|\sin (\frac{n \pi}{2})|\) will have area \(\frac{2}{\pi}\) and there will be \(n\) of them, therefore \(\frac{2n}{\pi} - \frac{n}{2}\)