Problems

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Solution:

1. \begin{align*} \sum_{n=1}^{\infty} \frac{n+1}{n!} &= \sum_{n=1}^\infty \left ( \frac{1}{(n-1)!} + \frac{1}{n!} \right) \\ &= \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=1}^\infty \frac{1}{n!} \\ &= \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=0}^\infty \frac{1}{n!} - 1 \\ &= e + e - 1 \\ &= 2e-1 \end{align*} \begin{align*} \sum_{n=1}^{\infty} \frac{(n+1)^2}{n!} &= \sum_{n=1}^{\infty} \frac{n(n-1) + 3n + 1}{n!} \\ &= \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + 3 \sum_{n=1}^\infty \frac1{(n-1)!} + \sum_{n=1}^\infty \frac{1}{n!} \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} + 3 \sum_{n=0}^\infty \frac1{n!} + \sum_{n=0}^\infty \frac{1}{n!} -1 \\ &= 5e-1 \end{align*} \begin{align*} \sum_{n=1}^\infty \frac{(2n-1)^3}{n!} &= \sum_{n=1}^\infty \frac{8n^3-12n^2+6n-1}{n!} \\ &= \sum_{n=1}^\infty \frac{8n(n-1)(n-2)+12n^2-10n-1}{n!} \\ &= \sum_{n=1}^\infty \frac{8n(n-1)(n-2)+12n(n-1)+2n-1}{n!} \\ &= 8 e+12e+2e-(e-1) \\ &=21e+1 \end{align*}
2. \begin{align*} \frac{n^2+1}{(n+1)(n+2)} &= \frac{n^2+3n+2-3n-1}{(n+1)(n+2)}\\ &= 1 - \frac{3n+1}{(n+1)(n+2)} \\ &= 1 + \frac{2}{n+1} - \frac{5}{n+2} \\ -\log(1-x) &= \sum_{n=1}^\infty \frac1{n}x^{n} \\ \log(2) &= \sum_{n=1}^\infty \frac{2^{-n}}{n} \\ \sum_{n=0}^\infty \frac{(n^2+1)2^{-n}}{(n+1)(n+2)} &= \sum_{n=0}^{\infty} 2^{-n} + 2 \sum_{n=0}^\infty \frac{2^{-n}}{n+1}-5 \sum_{n=0}^\infty \frac{2^{-n}}{n+2} \\ &= 2 + 2\log2-5 \sum_{n=2}^\infty \frac{2^{-n+2}}{n} \\ &= 2 + 2 \log 2 - 5 \left (2\log 2 - 2 \right) \\ &= 12-8\log2 \end{align*}

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Solution: Notice that the coefficient of \(x^r\) is \((-1)^r\frac{(-3) \cdot (-3-1) \cdots (-3-r+1)}{r!} = (-1)^r \frac{(-1)(-2)(-3)(-4) \cdots (-(r+2))}{(-1)(-2)r!} = (-1)^r(-1)^{r+2}\frac{(r+2)!}{2r!} = \frac{(r+2)(r+1)}2\).

1. The coefficient of \(x^r\) is \begin{align*} && c_r &=\frac{(r+1)(r+2)}{2} - \frac{(r-1+1)(r-1+2)}{2} + 2 \frac{((r-2+1)(r-2+2)}{2} \\ &&&= \frac{r^2+3r+2}{r} - \frac{r^2+r}{2} + \frac{2r^2-2r}{2}\\ &&&= \frac{2r^2+2}{2} = r^2+1 \end{align*} \begin{align*} && S & = 1+\frac22+\frac54+\frac{10}8+\frac{17}{16}+\frac{26}{32}+\frac{37}{64} +\frac{50}{128}+ \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{r^2+1}{2^r} \\ &&&= \frac{1-\tfrac12+2 \cdot \tfrac14}{(1-\tfrac12)^3} \\ &&&= 8 \end{align*}
2. \(\,\) \begin{align*} && S &= 1+2+\frac94+2+\frac{25}{16}+\frac{9}{8}+\frac{49}{64} + \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{(r+1)^2}{2^r} \\ &&&= 2 \sum_{r=0}^{\infty} \frac{(r+1)^2}{2^{r+1}} \\ &&&= 2 \sum_{r=1}^{\infty} \frac{r^2}{2^{r}} \\ &&&= 2 \left (\sum_{r=0}^{\infty} \frac{r^2+1}{2^{r}} - \sum_{r=0}^{\infty} \frac{1}{2^{r}} \right) \\ &&&= 2 (8 - 1) = 14 \end{align*}

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Solution:

1. \begin{align*} \frac{\d h}{\d t} &= \underbrace{c}_{\text{flow in}} - \underbrace{kh}_{\text{flow out}} \end{align*}. We also know that when \(h = \alpha^2 H\), \(\frac{\d h}{\d t} = 0\), ie \(c - k \alpha^2 H = 0\) therefore: \[ \frac{\d h}{\d t} = k(\alpha^2 H - h) \] \begin{align*} && \frac{\d h}{\d t} &= k(\alpha^2 H - h) \\ && \int \frac{1}{\alpha^2 H - h} \d h &= \int k \d t \\ && - \ln |\alpha^2H -h| &= kt + C \\ t = 0, h = H: && -\ln |(1-\alpha^2 )H| &= C \\ \Rightarrow && kt &= \ln \left | \frac{(\alpha^2-1)H}{h-\alpha^2 H }\right | \\ && kT &= \ln \frac{(\alpha^2-1)H}{\alpha H - \alpha^2 H} \\ &&&= \ln \frac{1+\alpha}{\alpha} \\ &&&= \ln \left (1 + \frac1{\alpha} \right) \\ &&&\approx \frac1{\alpha} - \frac12 \frac1{\alpha^2}\\ &&&\approx \alpha^{-1} \end{align*}
2. \begin{align*} && \frac{\d h}{\d t} &=c(\alpha \sqrt{H} - \sqrt{h}) \\ \Leftrightarrow && c \int_0^{T'} \d t&= \int_{H}^{\alpha H} \frac{1}{\alpha \sqrt{H}-\sqrt{h}} \d h \\ u = \sqrt{h/H}: && cT' &= \int_1^{\sqrt{\alpha}} \frac{1}{\alpha \sqrt{H} - \sqrt{H}u} 2\sqrt{H}u \d u \\ &&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u}{\alpha - u} \d u \\ &&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u - \alpha + \alpha}{\alpha - u} \d u \\ &&&= 2\sqrt{H}\left [-u - \alpha \ln |\alpha - u| \right]_1^{\sqrt{\alpha}} \\ &&&= 2\sqrt{H}\left ( -\sqrt{\alpha} + 1- \alpha \ln (\alpha - \sqrt{\alpha}) + \alpha \ln |\alpha - 1| \right) \\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\alpha-1}{\alpha - \sqrt{\alpha}} \right)\right)\\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}^2-1}{\sqrt{\alpha}(\sqrt{\alpha}-1)} \right)\right)\\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}+1}{\sqrt{\alpha}} \right)\right)\\ &&&= \boxed{2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( 1+\frac{1}{\sqrt{\alpha}} \right)\right)}\\ &&&\approx2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \left ( \frac{1}{\sqrt{\alpha}}-\frac12 \frac{1}{\alpha} \right)\right) \\ &&&=2\sqrt{H} \left ( 1 - \frac12 \right) \\ &&&= \sqrt{H} \end{align*} as required.

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Solution: \begin{align*} && x &= \frac{1}{t^2-1} \\ && t &= \sqrt{\frac{x+1}{x}}\\ \Rightarrow && \frac{\d x}{\d t} &= \frac{-2t}{(t^2-1)^2} \\ \Rightarrow && I &= \int \frac{1}{\sqrt{x(x+1)}} \d x \\ &&&= \int \frac{1}{\sqrt{\frac1{t^2-1} \frac{t^2}{t^2-1}}} \cdot \frac{-2 t}{(t^2-1)^2} \d t \\ &&&= \int \frac{t^2-1}{t} \frac{-2t}{(t^2-1)^2} \d t \\ &&&= -\int \frac{2}{t^2-1} \d t \\ &&&= - \frac{2}{2 \cdot 1} \ln \left | \frac{t-1}{t+1} \right| +C \\ &&&= \ln \left | \frac{t+1}{t-1} \right| + C \\ &&&= \ln \left | \frac{\sqrt{\frac{x+1}{x}}+1}{\sqrt{\frac{x+1}{x}}-1} \right| + C \\ &&&= \ln \left | \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}-\sqrt{x}} \right| + C \\ &&&= 2\ln \left | \sqrt{x+1}+\sqrt{x}\right| + C \\ &&&= 2\ln \left ( \sqrt{x+1}+\sqrt{x}\right) + C \\ \end{align*} \begin{align*} && V&= \pi \int_{1/8}^{9/16} y^2 \d x \\ &&&= \pi \int_{1/8}^{9/16} \left ( \frac1x + \frac{1}{x+1} - \frac{2}{\sqrt{x(x+1)}}\right) \d x \\ &&&= \pi \left [ \ln x + \ln (x+1) - 4 \ln(\sqrt{x+1} + \sqrt{x}) \right]_{1/8}^{9/16} \\ &&&= \pi \left ( \ln \frac{9}{16} + \ln \frac{25}{16} - 4 \ln \left ( \frac54 + \frac34\right) \right) +\\ &&&\quad -\pi \left ( \ln \frac{1}{8} + \ln \frac{9}{8} - 4 \ln \left ( \frac1{2\sqrt{2}} + \frac3{2\sqrt{2}}\right) \right) \\ &&&= \pi \left ( 2 \ln 3 - 8 \ln 2 + 2 \ln 5 - 4\ln2 \right) - \pi \left ( -6 \ln 2 + 2\ln 3 - 2\ln 2\right) \\ &&&= \pi (2 \ln 5 - 4 \ln 2 ) \\ &&&= 2 \pi \ln \tfrac54 \end{align*}

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Solution:

1. \(\cosh a = \frac12 (e^a + e^{-a})\) \begin{align*} \int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x &= \int_0^1 \frac{1}{x^2+(e^a+e^{-a})x+e^ae^{-a}} \d x \\ &= \int_0^1 \frac{1}{e^a-e^{-a}}\left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a} \int_0^1 \left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a}\left [\ln(x+e^{-a})-\ln(x+e^a) \right]_0^1 \\ &= \frac{1}{2 \sinh a} \left (\ln(1+e^a)-\ln(1+e^{-a}) - (\ln e^{-a}-\ln e^a) \right) \\ &= \frac{1}{2\sinh a}\left (2a + \ln \frac{1+e^a}{1+e^{-a}}\right) \\ &= \frac1{2\sinh a} \left ( 2a -a \right) \\ &= \frac{a}{2 \sinh a} \end{align*}
2. \begin{align*} \int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x &= \int_1^{\infty} \frac{1}{(x+e^a)(x-e^{-a})} \d x \\ &= \int_1^{\infty} \frac{1}{e^a+e^{-a}} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \int_1^{\infty} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \left [\ln(x-e^{-a}) - \ln (x + e^{a} ) \right]_1^{\infty} \\ &= \frac1{2\cosh a} \left [ \ln \frac{x-e^{-a}}{x+e^{a}} \right]_1^{\infty} \\ &= \frac{1}{2\cosh a} \left ( 0 - \ln \frac{1-e^{-a}}{1+e^a}{}\right) \\ &= \frac{1}{2\cosh a} \ln \frac{1+e^a}{1-e^{-a}}\\ &= \frac{1}{2\cosh a} \left ( a + \ln \coth \frac{a}{2} \right) \end{align*} and \begin{align*} \int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x &= \int_0^\infty\frac{1}{(x^2+e^a)(x^2+e^{-a})} \d x \\ &= \int_0^\infty \frac{1}{e^a-e^{-a}} \left ( \frac{1}{x^2+e^{-a}} - \frac{1}{x^2+e^{a}} \right) \d x \\ &= \frac{1}{2\sinh a} \left [ \frac{1}{e^{-a/2}} \tan^{-1} \frac{x}{e^{-a/2}} - \frac{1}{e^{a/2}}\tan^{-1} \frac{x}{e^{a/2}} \right]_0^{\infty} \\ &= \frac{1}{2\sinh a} \left (e^{a/2}\frac{\pi}{2}-e^{-a/2}\frac{\pi}{2} - 0 \right) \\ &= \frac{1}{2\sinh a} \pi \sinh \frac{a}{2} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{2\sinh a} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{4\sinh \tfrac{a}{2} \cosh \tfrac{a}{2}} \\ &= \frac{\pi}{4\cosh \tfrac{a}{2}} \end{align*}

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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*}

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Solution:

1. \(\,\) \begin{align*} && \int_{1/m}^m \frac{x^2}{x+1} \d x &= \int_{1/m}^m \left ( x- 1 + \frac{1}{x+1} \right) \d x \\ &&&= \left [ \frac{x^2}{2} - x + \ln (x+1) \right]_{1/m}^m \\ &&&= \left ( m^2/2 - m + \ln(m+1) \right)- \left ( \frac{1}{2m^2} - \frac{1}{m} + \ln\left(\frac1m+1\right) \right) \\ &&&= \frac{m^4-2m^3-1+2m}{2m^2} + \ln (m+1) - \ln(m+1) + \ln m \\ &&&= \frac{(m-1)^3(m+1)}{2m^2} + \ln m \end{align*}
2. \(\,\) \begin{align*} u = \frac{1}x, \d x = -\frac{1}{u^2} \d u:&& \int_{1/m}^m \frac1 {x^n(x+1)}\,\d x &= \int_{u=m}^{u=1/m} \frac{1}{u^{-n}(u^{-1}+1)} \frac{-1}{u^2} \d u \\ &&&= \int_{1/m}^m \frac{u^{n-1}}{u+1} \d u \end{align*}
3. • \(\bf (a)\) \(\,\) \begin{align*} && I &= \int_{1/2}^2 \frac {x^5+3}{x^3(x+1)}\,\d x \\ &&&= \int_{1/2}^2 \left ( \frac{x^2}{x+1} + \frac{3}{x^3(x+1)} \right) \d x \\ &&&= \int_{1/2}^2 \frac{x^2}{x+1} \d x + 3 \int_{1/2}^2 \frac{x^2}{x+1} \d x \\ &&&= 4 \left ( \frac{(2-1)^3(2+1)}{2 \cdot 2^2} + \ln 2 \right) \\ &&&= \frac32+4 \ln 2 \end{align*}
   • \(\bf (b)\) \(\,\) \begin{align*} && J &= \int_1^2 \frac{x^5+x^3 +1}{x^3(x+1)}\, \d x \\ && K &= \int_1^2 \frac{x^5 +1}{x^3(x+1)}\, \d x\\ u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=1}^{u=1/2} \frac{u^{-5}+1}{u^{-3}(u^{-1}+1)} \frac{-1}{u^2} \d u \\ &&&= \int_{1/2}^1 \frac{1 + u^5}{u^3(u+1)} \d u \\ \Rightarrow && K &= \frac12 \int_{1/2}^2 \frac{x^5+1}{x^3(x+1)} \d x \\ &&&= \frac{(2-1)^3(2+1)}{2 \cdot 2^2} + \ln 2 \\ &&&= \frac38 + \ln 2 \\ && L &= \int_1^2 \frac{x^3}{x^3(x+1)} \d x \\ &&&= \ln (3) - \ln 2 \\ \Rightarrow && J &= \frac38 + \ln 3 \end{align*}

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Solution: \begin{align*} && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx+D}{1+x^2} \\ \Rightarrow && 1 + x &= A(1-x)(1+x^2) + B(1+x^2) + Cx(1-x)^2 + D(1-x)^2 \\ \Rightarrow && 2 &= 2B \tag{\(x = 1\)} \\ \Rightarrow && 1 &= B \\ \Rightarrow && 1 &= A+B+D \tag{\(x = 0\)}\\ \Rightarrow && A &= -D \\ \Rightarrow && 0 &= 4A+2B-4C+4D \tag{\(x = -1\)}\\ \Rightarrow && C &= \frac12\\ \Rightarrow && 3 &= -5A+5B+2C+D \tag{\(x=2\)} \\ \Rightarrow && 3 &= -6A+6 \\ \Rightarrow && A,D &=-\frac12,\frac12 \\ \Rightarrow && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{1}{(1-x)^2} +\frac{1}{2(1-x)}+ \frac{x-1}{2(1+x^2)} \\ &&&=\sum_{k=0}^{\infty}(k+1)x^k + \sum_{k=0}^{\infty}\frac12 x^k + \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k+1} - \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k} \end{align*} Therefore the coefficient of \(x^n\) is \(n+1\) or \(n+2\) depending on whether the coefficients from the final series add constructively \(n \equiv 1, 2 \pmod{4}\) or destructively. \begin{align*} \frac{11\, 000}{8181} &= \frac{(10+1) \cdot 1000}{(10-1)^2 \cdot (100+1)} \\ &= \frac{(1+\frac{1}{10})}{(1-\frac{1}{10})^2 \cdot (1+\frac1{10})} \\ &= 1 + \frac3{10} + \frac4{10^2} + \frac{4}{10^3}+\frac{5}{10^4} + \frac{7}{10^5} + \frac{8}{10^6} + \cdots \\ & \quad \quad \cdots + \frac{8}{10^7} + \frac{9}{10^8} + \frac{11}{10^9} + \frac{12}{10^{10}} + \cdots \\ &= 1.34457890 + \frac{12}{10^{10}} + \cdots \end{align*} \begin{align*} && \sum_{k=m}^{\infty} (k+2)x^k &= x^m \sum_{k=0}^{\infty} (k+m+2)x^{k} \\ && &= \frac{x^k}{(1-x)^2} + \frac{(m+2)x^k}{1-x} \\ \Rightarrow && |\sum_{k=m}^{\infty} a_k \left ( \frac1{10} \right )^k | &\leq \frac{1}{10^m}\left ( \frac{1}{(1-\frac1{10})^2} + \frac{m+2}{1-\frac1{10}} \right) \\ &&&= \frac{1}{10^{m-1}} \left ( \frac{9m+28}{81}\right ) \end{align*} Therefore for this will be less than \(10^{-9}\), when \(m = 11\), so our approximation is valid to 9sf

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Solution: \begin{align*} && I &= \int \frac{1}{a^2+x^2} \d x\\ x = a \tan \theta, \d x =a \sec^2 \theta \d \theta &&&= \int \frac{1}{a^2+a^2\tan^2 x} a \sec^2 \theta \d \theta \\ &&&=\int \frac{\sec^2 \theta}{a \sec^2 \theta} \d \theta \\ &&&= \frac1a \theta + C \\ &&&= \frac1a \arctan \frac{x}{a} + C \end{align*}

1. 1. \(\,\) \begin{align*} && I &= \int_0^{\frac{1}{2}\pi} \frac {\cos x}{1+\sin^2 x} \d x \\ &&&= \left [ \arctan (\sin x) \right]_0^{\pi/2} \\ &&&= \arctan(1) - \arctan(0) = \frac{\pi}{4} \end{align*}
   2. \(\,\) \begin{align*} && t &= \tan \frac{x}{2} \\ \Rightarrow && \sin x &= \frac{2t}{1+t^2} \\ && \cos x &= \frac{1-t^2}{1+t^2} \\ && \d x &= \frac{2}{1+t^2} \d t \\ \Rightarrow && I &= \int_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x } \d x \\ &&&= \int_{t=0}^{t = 1} \frac{\frac{1-t^2}{1+t^2}}{1 + \left (\frac{2t}{1+t^2} \right)^2} \frac{2}{1+t^2} \d t \\ &&&= 2 \int_0^1 \frac{1-t^2}{(1+t^2)^2+(2t)^2} \d t\\ &&&= 2 \int_0^1 \frac{1-t^2}{1+6t^2+t^4} \d t\\ \end{align*} From which the conclusion follows
2. \(\,\) \begin{align*} && J &= \int_0^1 \frac {1-t^2}{1+14t^2+t^4} \, \d t \\ &&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{1+14t^2+t^4}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t \\ &&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{(t^2+1)^2+3(2t)^2}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t \\ &&&= \frac12\int_{x=0}^{x=\pi/2} \frac {\cos x}{1+3 \sin^2 x} \d x \\ &&&= \frac{1}{6}\left[ \sqrt{3} \arctan(\sin \sqrt{3}x)\right]_0^{\pi/2} \\ &&&= \frac16 \sqrt{3} \frac{\pi}{3} \\ &&&= \frac{\sqrt{3}\pi}{18} \end{align*}

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Solution:

1. \begin{align*} && I &= \int \frac{3}{(x-4)\sqrt{2x+1}}\, \d x \\ u^2 =2x+1, 2u \frac{\d u}{\d x}=2: && &= \int \frac{3}{\left(\frac{u^2-1}{2}-4\right)u} u \d u \\ &&&= \int \frac{6}{u^2-9} \d u \\ &&&= \int \frac{6}{(u-3)(u+3)} \d u\\ &&&= \int \left ( \frac{1}{u-3} - \frac{1}{u+3} \right )\d u \\ &&&= \ln (u-3) - \ln (u+3) + C \\ &&&= \ln \frac{u-3}{u+3} + C \\ &&&= \ln \left (\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3} \right) + C \end{align*}
2. \begin{align*} && I &= \int_{\ln 3}^{\ln 8} \frac{2}{e^x\sqrt{e^x+1}} \d x \\ u = e^x, \frac{\d u}{\d x} = e^x: &&&= \int_{u=3}^{u=8} \frac{2}{u\sqrt{u+1}} \frac{1}{u} \d u \\ v^2=u+1, 2v \frac{\d v}{\d u} = 1: &&&= \int_{v=2}^{v=3} \frac{2}{v(v^2-1)^2} \d v \\ &&&= \int_2^3 \left ( \frac{2}{v} - \frac{}{v-1} - \frac{1}{2(v-1)^2} - \frac{1}{v+1} - \frac{1}{2(v+1)^2}\right) \d v \\ &&&= \left [2\ln v - \ln(v^2-1)+\frac12(v-1)^{-1}+\frac12(v+1)^{-1} \right]_2^3 \\ &&&= \left ( 2\ln3-\ln 8+\frac14+\frac18\right)-\left ( 2\ln2-\ln 3+\frac12+\frac16\right) \\ &&&= \end{align*}

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Solution:

1. \(\,\) \begin{align*} \prod^n_{r=1} \left ( \frac{r+ 1 }{ r } \right) &= \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{n-1}{n-2} \cdot \frac{n}{n-1} \cdot \frac{n+1}{n} \\ &= \frac{n+1}{1} = n+1 \end{align*}
2. \(\,\) \begin{align*} \prod^n_{r=2} \left ( \frac{r^2 -1}{r^2 } \right) &= \prod^n_{r=2} \left ( \frac{(r -1)(r+1)}{r^2 } \right) \\ &= \left ( \frac{1}{2} \cdot \frac{3}{2} \right) \cdot \left ( \frac{2}{3} \cdot \frac{4}{3} \right) \cdots \left ( \frac{r-1}{r} \cdot \frac{r+1}{r}\right) \cdots \frac{n-1}{n} \cdot \frac{n+1}{n} \\ &= \frac{1}{n} \cdot \frac{n+1}{2} \\ &= \frac{n+1}{2n} \end{align*}
3. When \(n\) is odd, the product is undefined, since we have a \(\cot \pi\) lurking in there. \begin{align*} \prod^n_{r=1} \left ( {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \cot \frac{ (2r-1 ) \pi }{ n} } \right) &= \prod^n_{r=1} \left ( {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \frac{\cos \frac{ (2r-1 ) \pi }{ n}}{\sin\frac{ (2r-1 ) \pi }{ n}} } \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \left ( {\cos \frac{2\pi }{ n} \sin\frac{ (2r-1 ) \pi }{ n} + \sin \frac{2\pi}{ n} \cos \frac{ (2r-1 ) \pi }{ n} } \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \sin \left ( \frac{2\pi}{n} + \frac{(2r-1)\pi}{n} \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \sin \left ( \frac{(2r+1)\pi}{n} \right) \\ &= \frac{\sin \frac{3\pi}{n}}{\sin \frac{\pi}{n}} \cdot \frac{\sin \frac{5\pi}{n}}{\sin \frac{3\pi}{n}} \cdots \frac{\sin \frac{(2n+1)\pi}{n}}{\sin \frac{(2n-1)\pi}{n}} \\ &= \frac{\sin \frac{(2n+1)\pi}{n}}{\sin \frac{\pi}{n}} \\ &= 1 \end{align*}

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Solution: \begin{align*} && {\frac{\d y} {\d x}} &= \frac{x^3y^2}{(1 + x^2)^{5/2}} \\ \Rightarrow &&\int \frac{1}{y^2} \d y &= \int \frac{x^3}{(1+x^2)^{5/2}} \d x \\ \Rightarrow && -\frac1y &= \int \frac{x^3+x-x}{(1+x^2)^{5/2}} \d x \\ &&&= \int \left ( \frac{x}{(1+x^2)^{3/2}}-\frac{x}{(1+x^2)^{5/2}} \right) \d x \\ &&&= \frac{-1}{(1+x^2)^{1/2}} + \frac{1}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{1-3(1+x^2)}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{-3x^2-2}{3(1+x^2)^{3/2}} + C \\ (x,y) = (0,1): &&-1 &= -\frac23 + C \\ \Rightarrow && C &= -\frac13 \\ \Rightarrow && \frac1y &= \frac{3x^2+2}{3(1+x^2)^{3/2}} + \frac13 \end{align*} \begin{align*} y &= \frac{1}{\frac13 +\frac{3x^2+2}{3(1+x^2)^{3/2}} } \\ &= \frac{3}{1+ \frac{3x^2+2}{3(1+x^2)^{3/2}}} \\ &= \frac{3}{1+ \frac{3}{x} + \cdots} \\ &\approx 3 - \frac{9}{x} \end{align*}

\begin{align*} && \frac {\d z} {\d x} &= \frac{x^3z^3}{2(1 + x^2)^{5/2}} \\ \Rightarrow && \int \frac{1}{z^3} \d z &= \int \frac{x^3}{2(1+x^2)^{5/2}} \\ && -\frac{1}{2z^2} &= -\frac{3x^2+2}{3(1+x^2)^{3/2}} - C \\ (x,z) = (0, \pm 1): && \frac{1}{2} &= \frac{2}{3} + C \\ \Rightarrow && C &= -\frac16 \\ \Rightarrow && \frac{1}{z^2} &= \frac{6x^2+4}{3(1+x^2)^{3/2}} - \frac13 \end{align*} So as \(x \to \infty\) \(z \sim \pm (3 + \frac{2}{x} + \cdots)\) and so:

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Solution: \begin{align*} u = x^m, \d u = m x^{m-1} && \int \frac{m}{x f(x^m)} \d x &= \int \frac{m x^{m-1}}{uf(u)} \d x \\ &&&= \int \frac{1}{u f(u)} \d u \\ &&&= F(u) + c \\ &&&= F(x^m) + c \end{align*}

1. \begin{align*} && \int \frac{1}{u(u-1)} \d u &= \int \left ( \frac{1}{u-1}-\frac{1}{u} \right ) \d u \\ &&&= \ln \left ( \frac{u-1}{u} \right) + c \\ &&&= \ln \left ( 1 - \frac{1}{u} \right) + c \\ && \int \frac{1}{x^n - x} \d x &= \int \frac{1}{x (x^{n-1}-1)} \d x \\ f(u) = u - 1: && &= \frac{1}{n-1} \ln \left ( 1 - \frac{1}{x^{n-1}} \right) + c \end{align*}
2. \begin{align*} v = \sqrt{u+1}, \d v = \tfrac12 (u+1)^{-1/2} \d u && \int \frac{1}{u\sqrt{u+1}} \d u &= \int \frac{1}{(v^2-1)} (u+1)^{-1/2} \d u \\ &&&= \int \frac{2}{v^2-1} \d v \\ &&&=\ln \frac{1-v}{1+v} + c \\ &&&= \ln \left (\frac{1-\sqrt{u+1}}{1+\sqrt{u+1}} \right)+ c \\ f(u) = \sqrt{x+1}:&& \int \frac{1}{\sqrt{x^n + x^2}} \d x &= \int \frac{1}{x\sqrt{x^{n-2}+1}} \d x \\ &&&= \frac{1}{n-2} \ln \left ( \frac{1-\sqrt{x^{n-2}+1}}{1+\sqrt{x^{n-2}+1}} \right)+c \end{align*}

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Solution:

1. \begin{align*} && f(x) -7xf(x)+10x^2f(x) &= \sum_{n=0}^\infty a_n x^n - 7x \sum_{n=0}^{\infty} a_n x^n + 10x^2 \sum_{n=0}^{\infty} a_nx^n \\ &&&= \sum_{n=2}^\infty (a_n-7a_{n-1}+10a_{n-2})x^n + a_0+a_1x-7a_0x \\ &&&= 0 + 2-7x \\ \\ \Rightarrow && f(x) &= \frac{2-7x}{1-7x+10x^2} \\ &&&= \frac{2-7x}{(1-5x)(1-2x)} \\ &&&= \frac{1}{1-2x} + \frac{1}{1-5x} \\ &&&= \sum_{n=0}^{\infty} (2^n + 5^n)x^n \end{align*} Therefore \(a_n = 2^n +5^n\)
2. \(\,\) \begin{align*} && 40 &= 10p + 5 q \\ && 100 &= 40p+10q \\ && 10 &= 4p + q \\ \Rightarrow && (p,q) &= (1,6) \\ \\ && g(x) -xg(x)-6x^2g(x) &= 5+5x \\ \Rightarrow && g(x) &= \frac{5+5x}{1-x-6x^2} \\ &&&= \frac{5+5x}{(1-3x)(1+2x)} \\ &&&= \frac{4}{1-3x} + \frac{1}{1+2x} \\ &&&= \sum_{n=0}^{\infty} (4 \cdot 3^n + (-2)^n)x^n \\ \Rightarrow && b_n &= 4 \cdot 3^n + (-2)^n \end{align*}