Problems

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Solution:

1. \(\,\) \begin{align*} && I &= \int_0^1 \frac 1 {x^2 + (a+2)x +2a} \; \d x \\ &&&= \int_0^1 \frac{1}{(x+a)(x+2)} \d x\\ \end{align*} Case 1: \(a = 2\) \begin{align*} && I &= \int_0^1 \frac{1}{(x+2)^2} \d x \\ &&&= \left [ -(x+2)^{-1}\right]_0^1 = \frac12 - \frac13 = \frac16 \end{align*} Case 2: \(a \neq 2, a \not \in [0,1]\) \begin{align*} && I &=\frac{1}{a-2} \int_0^1 \left ( \frac{1}{x+2} - \frac{1}{x+a} \right) \d x \\ &&&= \frac{1}{a-2} \left [ \ln |x+2| - \ln |x + a|\right]_0^1 \\ &&&= \frac{1}{a-2} \left ( \ln \frac{3}{|1+a|} - \ln \frac{2}{|a|} \right) \\ &&&= \frac{1}{a-2} \ln \frac{3|a|}{2|a+1|} \end{align*} Case 3: \(a \in [0, 1]\), \(I\) does not converge
2. \(\,\) \begin{align*} && J &= \int _{1}^2\frac 1 {u^2 +au +a-1} \; \d u \\ &&&= \int_1^2 \frac{1}{(u+a-1)(u+1)} \d u \\ x = u-1:&&&= \int_0^1 \frac{1}{(x+a)(x+2)} \d x \end{align*} So it's the same as the previous integral

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Solution:

1. \begin{align*} \frac{1-\cos \alpha}{\sin \alpha} &= \frac{1-(1-2\sin^2 \frac{\alpha}{2})}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\ &= \frac{2 \sin^2 \frac \alpha2}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\ &= \frac{\sin \frac \alpha2}{ \cos \frac\alpha2} \\ &= \tan \tfrac{\alpha}{2} \end{align*}
2. \begin{align*} \int\frac{\mathrm{d}x}{1-2kx+x^{2}} &= \int \frac{\d x}{(x-k)^2+1-k^2} \\ &= \frac{1}{1-k^2}\int \frac{\d x}{\left (\frac{x-k}{\sqrt{1-k^2}} \right)^2+1} \\ &= \frac{1}{\sqrt{1-k^2}} \tan^{-1} \left (\frac{x-k}{\sqrt{1-k^2}} \right)+C \end{align*}

\begin{align*} \int_{0}^{1}\frac{\sin\alpha}{1-2x\cos\alpha+x^{2}}\,\mathrm{d}x &= \sin \alpha \left [\frac{1}{\sqrt{1-\cos ^2\alpha}} \tan^{-1} \left ( \frac{x - \cos \alpha}{\sqrt{1-\cos^2\alpha}} \right) \right]_0^1 \\ &= \tan^{-1} \left ( \frac{1 - \cos \alpha}{\sin \alpha} \right) -\tan^{-1} \left ( \frac{- \cos \alpha}{\sin \alpha} \right) \\ &= \tan^{-1} \tan \tfrac{\alpha}{2} + \tan^{-1} \cot \alpha \\ &= \frac{\alpha}{2} + \frac{\pi}{2} - \alpha \\ &= \frac{\pi-\alpha}{2} \end{align*}

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Solution:

In the coordinate system where \(A\) is \((0,0)\) and \(AD\) is the \(x\)-axis and \(AB\) the \(y\)-axis and all side lengths are \(2\), we find the centre of mass of each of the sides are: \begin{align*} AB :& (0,1) \\ BC :& (1,2) \\ CD :& (2,1) \\ \\ ABCD:& \l 1, \frac{4}{3} \r \end{align*} When hung from \(A\), the angle \(AB\) makes to the vertical is \(\alpha\) and the angle \(BC\) makes to the vertical will be \(90^{\circ} + \alpha\). When hung from \(B\) the angle \(BC\) makes to the vertical will be \(\beta\). The value we are interested in therefore is \(\beta + 90^{\circ} + \alpha\) \begin{align*} && \tan \alpha &= \frac{1}{\frac{4}{3}} \\ &&& = \frac{4}{3} \\ \\ && \tan \beta &= \frac{\frac{2}{3}}{1} \\ &&&= \frac{2}{3} \\ \\ && \tan \l \beta + (90^{\circ} + \alpha) \r &= \frac{\tan \beta + \tan\l 90^{\circ} + \alpha \r}{1 - \tan \beta \tan\l 90^{\circ} + \alpha \r} \\ &&&= \frac{\frac23 + \frac43}{1- \frac23 \frac43} \\ &&&= \frac{2}{1 - \frac89} \\ &&&= 18 \end{align*}