Problems

Solution: Let \(f(x) = x^4 + ax^3 + bx^2 + ax + 1\), and suppose \(f(k) = 0\). Since \(f(0) = 1\), \(k \neq 0\), therefore we can talk about \(k^{-1}\). \begin{align*} && f(k^{-1}) &= k^{-4} + ak^{-3} + bk^{-2} + ak^{-1} + 1 \\ &&&= k^{-4}(1 + ak + bk^2 + ak^3 + k^4) \\ &&&= k^{-4}(k^4+ak^3+bk^2+ak+1) \\ &&&= k^{-4}f(k) = 0 \end{align*} Therefore \(k^{-1}\) is also a root of \(f\)

1. If \(f\) has only on distinct root, we must have \(f(x) = (x+k)^4\) therefore \(k = k^{-1} \Rightarrow k^2 = 1 \Rightarrow k = \pm1\), or \(a = 4, b = 6\) or \(a = -4, b = 6\)
2. If \(f\) has exactly three distinct roots then one of the roots must be a repeated \(1\) or \(-1\), ie \(0 = f(1) = 1 + a + b + a + 1 = 2 + b +2a \Rightarrow b = -2a-2\) or \(0 = f(-1) = 1 -a + b -a + 1 \Rightarrow b = 2a - 2\)
3. If \(b = 2a-2\), we have \begin{align*} && f(x) &= 1 + ax + (2a-2)x^2 + ax^3 + x^4 \\ &&&= (x^2+2x+1)(1+(a-2)x+x^2) \\ \Rightarrow && x &= \frac{2-a \pm \sqrt{(a-2)^2 - 4}}{2} \\ &&&= \frac{2-a \pm \sqrt{a^2-4a}}{2} \end{align*}

Solution:

1. \begin{align*} && \frac1{1+x} &= 1 - x+ x^2 - x^3+ \cdots \\ \Rightarrow && \int_0^x \frac{1}{1+t} \d t &= \int_0^x \sum_{n=0}^{\infty} (-t)^n \d t \\ &&&= \left [\sum_{n=0}^{\infty} -\frac{(-t)^{n+1}}{n+1} \right]_0^x \\ \Rightarrow &&\ln(1+x)&=- \sum_{n=1}^\infty \frac{(-x)^n}{n} \end{align*}
2. \begin{align*} && e^{-ax} &= \sum_{n=0}^\infty \frac{(-a)^n}{n!} x^n \\ \Rightarrow && \int_0^{\infty} \frac{1}{x} \left (1-e^{-ax} \right)e^{-x} \d x &= \int_0^{\infty} \frac{1}{x} \left (-\sum_{n=1}^\infty \frac{(-a)^n}{n!}x^n \right)e^{-x} \d x \\ &&&= -\int_0^{\infty} \sum_{n=1}^\infty \frac{(-a)^n}{n!} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} \int_0^{\infty} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} (n-1)! \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n} \\ &&&= \ln (1+a) \end{align*}
3. \begin{align*} && \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x &= \int_0^1 \frac{x^p(1 - x^{q-p})}{\ln x} \, \d x \\ e^{-u} = x, \d x = -e^{-u} \d u: &&&=\int_{u=\infty}^{0} \frac{e^{-pu}-e^{-qu}}{-u} (-e^{-u})\d u \\ &&&= \int_0^\infty \frac{e^{-u}(e^{-qu}-e^{-pu})}{u} \d u \\ &&&= \int_0^\infty \frac{e^{-(1+q)u}(1-e^{-(p-q)u})}{u} \d u \\ v = (1+q)u, \d v = (1+q) \d u: &&&=\int_0^{\infty} \frac{e^{-v}(1-e^{-\left(\frac{p-q}{1+q}\right)v}}{v}\d v \\ &&&= \ln \left(1 + \frac{p-q}{1+q} \right) \\ &&&= \ln \left ( \frac{1+p}{1+q} \right) \end{align*}

Solution:

1. Suppose \(v = \sqrt{y} \Rightarrow v^2 = y \Rightarrow 2v v' = y'\) so \begin{align*} && 2vv' &= \alpha v-\beta v^2 \\ \Rightarrow && 2v' &= \alpha - \beta v \\ \Rightarrow && v' + \frac{\beta}{2} v &= \frac{\alpha}{2} \\ \Rightarrow && \frac{\d}{\d t} \left (e^{\beta t/2} v \right) &= \frac{\alpha}{2} e^{\beta t/2} \\ \Rightarrow && e^{\beta t/2} v &=C+ \frac{\alpha}{\beta}e^{\beta t /2} \\ \Rightarrow && v &= Ce^{-\beta t/2} + \frac{\alpha}{\beta} \\ \Rightarrow && \sqrt{y} &= Ce^{-\beta t/2} + \frac{\alpha}{\beta} \\ y(0) = 0: && 0 &= C+\frac{\alpha}{\beta} \\ \Rightarrow && \sqrt{y} &= \frac{\alpha}{\beta} \left (1-e^{-\beta t/2} \right) \\ \Rightarrow && y &= \frac{ \alpha^2}{\beta^2} \left (1-e^{-\beta t/2} \right)^2 \end{align*}
2. Try \(v = y^{1/3} \Rightarrow v^3 = y \Rightarrow 3v^2 v' = y'\) so \begin{align*} && y' &= \alpha v^2 - \beta y \\ \Rightarrow && 3v^2v' &= \alpha v^2 - \beta v^3 \\ \Rightarrow && v' +\frac{\beta}{3} v &= \frac{\alpha}{3} \\ \Rightarrow && (v e^{\beta t/3})' &= \frac{\alpha}{3}e^{\beta t/3} \\ \Rightarrow && v &= Ce^{-\beta t/3} + \frac{\alpha}{\beta} \\ v(0) = 0: && v &= \frac{\alpha}{\beta} \left (1 - e^{-\beta t/3} \right) \\ \Rightarrow && y &= \frac{\alpha^3}{\beta^3} \left (1 - e^{-\beta t/3} \right) ^3 \end{align*}
3. \(y_1 = (1-e^{-\beta t/2})^2, y_2 = (1-e^{-\beta t/3})^3\)
   

   + - ↺
   By considering the differential equation, notice that \(0 < y_i < 1\) so \(y^{1/2} > y^{2/3}\) and therefore \(y_1' > y_2'\) and so \(y_1\) increases faster.

Solution:

1. \begin{align*} && I &= \int_0^1 \frac{f(x^{-1})}{1+x} \d x \\ u = x^{-1}, \d u = -x^{-2} \d x: &&&= \int_{\infty}^1 \frac{f(u)}{1+\frac{1}{u}} \frac{-1}{u^2} \d u \\ &&&= \int_1^{\infty} \frac{f(u)}{u(1+u)} \d u \\ &&&= \sum_{n=1}^{\infty} \int_n^{n+1} \frac{f(u)}{u(u+1)} \d u \\ \\ \text{if} f(x) = f(x+1)\, \forall x && &=\sum_{n=1}^{\infty} \int_{0}^1 \frac{f(x+n)}{(x+n)(x+n+1)} \d x \\ &&&= \sum_{n=1}^\infty \int_0^1 \frac{f(x)}{(x+n)(x+n+1)} \d x \\ &&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \frac{1}{(x+n)(x+n+1)}\r \d x \\ &&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \l \frac{1}{x+n} - \frac{1}{x+n+1} \r\r \d x \\ &&&= \int_0^1 f(x) \l \frac{1}{x+1} \r \d x \\ &&&= \int_0^1\frac{f(x)}{x+1} \d x \\ \end{align*}
2. Since the fractional part is periodic with period \(1\), we can say \begin{align*} && \int_0^1 \frac{\{ x^{-1} \} }{1+x} \d x &= \int_0^1 \frac{\{ x\}}{x+1} \d x \\ &&&= \int_0^1 \frac{x}{x+1} \d x \\ &&&= \int_0^1 1-\frac{1}{x+1} \d x \\ &&&= [x - \ln (1+x) ]_0^1 \\ &&&= 1 - \ln 2 \end{align*} \begin{align*} && \int_0^1 \frac{\{ 2x^{-1}\}}{1+x} \d x &= \int_0^1 \frac{\{2x\}}{x+1} \d x \\ &&&= \int_0^{1/2} \frac{2x}{x+1} \d x +\int_{1/2}^{1} \frac{2x-1}{x+1} \d x \\ &&&= 2\int_0^1 \frac{x}{x+1} \d x + \int_{1/2}^1 \frac{-1}{x+1} \d x \\ &&&= 2 - 2\ln 2 - \l \ln 2 - \ln \tfrac32 \r \\ &&&= 2 - 4 \ln 2 + \ln 3 \\ &&&= 2 + \ln \tfrac {3}{16} \end{align*}
3. \begin{align*} && \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x\r \end{align*} Consider for \(f\) periodic with period \(1\) \begin{align*} \int_0^1 \frac{f(x^{-1})}{1-x} \d x &= \int_1^{\infty} \frac{f(u)}{u(u-1)} \d u \\ &= \sum_{n=1}^{\infty}\int_{n}^{n+1} \frac{f(u)}{u(u-1)} \d u \\ &= \sum_{n=1}^{\infty}\int_{0}^{1} \frac{f(u)}{(u+n)(u+n-1)} \d u \\ &= \int_{0}^{1} \sum_{n=1}^{\infty}\frac{f(u)}{(u+n)(u+n-1)} \d u \\ &= \int_{0}^{1} f(u) \sum_{n=1}^{\infty}\l\frac{1}{u+n-1} - \frac{1}{u+n} \r\d u \\ &= \int_0^1 \frac{f(u)}{u} \d u \end{align*} So we have \begin{align*} && \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x \r \\ &&&= \frac12 \int_0^1 \frac{\{ x \}}{x} \d x - \frac12 (1 - \ln 2) \\ &&&= \frac12 - \frac12 + \frac12 \ln 2 \\ &&&= \frac12 \ln 2 \end{align*}

Solution:

1. \(\,\) \begin{align*} && I &= \int \frac{x}{x \tan x + 1} \d x \\ &&&= \int \frac{x \cos x}{x \sin x + \cos x} \d x \\ u = x \sin x + \cos x , \d u = x \cos x \d x: &&&= \int \frac{\d u}{u} \\ &&&= \ln u + C \\ &&&= \ln (x \sin x + \cos x) + C \\ \\ && J &= \int \frac{x}{x \cot x - 1} \d x \\ &&&= \int \frac{x \sin x }{x \cos x - \sin x} \d x \\ u = x \cos x - \sin x, \d u = x \sin x \d x: &&&= \int \frac{1}{u} \d u \\ &&&= \ln u + K \\ &&&= \ln (x \cos x -\sin x) + K \end{align*}
2. \(\,\) \begin{align*} && I &= \int \frac{x \sec^2 x \tan x}{x \sec^2 x - \tan x} \d x \\ u = x\sec^2 x-\tan x, \d u = 2x \sec^2 x \tan x&&&= \frac12 \int \frac{1}{u} \d u \\ &&&= \frac12 \ln (x \sec^2 x - \tan x) + C \\ \\ && J &= \int \frac{x \sin x \cos x}{(x - \sin x \cos x)^2} \d x \\ u = x \sec^2 x -\tan x, \d u=2x \frac{\sin x}{\cos^3 x} &&&= \int \frac{x \sin x \cos x}{\cos^4x(x\sec^2 x -\tan x)^2} \d x \\ &&&= \frac12 \int \frac{1}{u^2} \d u \\ &&&= -\frac12u^{-1} + K \\ &&&= \frac{1}{2(\tan x - x \sec^2 x)} + K \end{align*}

Solution:

1. \(\,\) \begin{align*} && T(x) &= \int_0^x \! \frac 1 {1+u^2} \, \d u \\ &&&= \int_0^{\infty} \frac{1}{1+u^2} \d u - \int_x^\infty \frac{1}{1+u^2} \d u \\ &&&= T_\infty - \int_x^\infty \frac{1}{1+u^2} \d u \\ u = 1/v, \d u = -1/v^2 \d v: &&&= T_\infty - \int_{v=x^{-1}}^{v=0} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\ &&&= T_\infty - \int_{0}^{x^{-1}} \frac{1}{1+v^2} \d v \\ &&&= T_\infty - T(x^{-1}) \end{align*}
2. Let \(v = \frac{u+a}{1-au}\) then \begin{align*} && \frac{\d v}{\d u} &= \frac{(1-au) \cdot 1 - (u+a)\cdot(-a)}{(1-au)^2} \\ &&&= \frac{1-au+au+a^2}{(1-au)^2} \\ &&&= \frac{1+a^2}{(1-au)^2} \\ \\ && \frac{1+v^2}{1+u^2} &= \frac{1 + \left ( \frac{u+a}{1-ua} \right)^2}{1+u^2} \\ &&&= \frac{(1-ua)^2+(u+a)^2}{(1-ua)^2(1+u^2)} \\ &&&= \frac{1+u^2a^2+u^2+a^2}{(1-ua)^2(1+u^2)} \\ &&&= \frac{(1+u^2)(1+a^2)}{(1-ua)^2(1+u^2)} \\ &&&= \frac{1+a^2}{(1-ua)^2} \end{align*} if \(a > 0, x < \frac1a\) then \begin{align*} && T(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ &&&= \int_{v=a}^{v=\frac{a+x}{1-ax}} \frac{1}{1+u^2} \frac{1+u^2}{1+v^2} \d v \\ &&&= T\left ( \frac{x+a}{1-ax} \right) - T(a) \\ \\ \Rightarrow && T(x^{-1}) &= T_\infty - T(x) \\ &&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T(a) \\ &&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T_\infty-T(a^{-1}) \\ &&&= 2T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) -T(a^{-1}) \end{align*} \(b > 0, y > \frac1b\) then \(y> 0, b > \frac1y\) (same as letting \(x = \frac1y, a = \frac1b\) \begin{align*} && T(y) &= 2T_\infty - 2T \left ( \frac{\frac1y+\frac1b}{1-\frac1{by}} \right) + T(b) \\ \Rightarrow && T(y) &= 2T_\infty - 2T \left ( \frac{b+y}{by-1} \right) + T(b) \\ \end{align*}
3. Letting \(y = b = \sqrt{3}\) in the final equation \begin{align*} && T(\sqrt{3}) &= 2T_{\infty} - T \left ( \frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}\sqrt{3}-1} \right) -T (\sqrt{3}) \\ &&&= 2T_\infty - 2T(\sqrt{3}) \\ \Rightarrow && T(\sqrt{3}) &= \tfrac23 T_\infty \end{align*} Let \(x = \sqrt2 - 1, a = 1\) so, \begin{align*} && T(\sqrt2 -1) &= T \left ( \frac{\sqrt2-1+1}{1-\sqrt2+1} \right)-T(1) \\ &&&= T \left ( \frac{\sqrt{2}}{2-\sqrt{2}} \right) - T(1) \\ &&&= T(\frac{\sqrt{2}(2+\sqrt{2})}{2}) - T(1) \\ &&&= T(\sqrt{2}+1) - T(1) \\ &&&= T_\infty - T(\sqrt2-1)-T(1) \\ \Rightarrow && T(\sqrt{2}-1) &= \frac12T_\infty-\frac12T(1) \\ && T(1) &= T_\infty - T(1) \\ \Rightarrow && T(1) &= \frac12 T_\infty \\ \Rightarrow && T(\sqrt2-1) &= \frac12T_\infty - \frac14T_\infty \\ &&&= \frac14 T_\infty \end{align*}

Solution: Suppose \(c = a+b\) then \begin{align*} (a+b+c)^3 &-6(a+b+c)(a^2+b^2+c^2) +8(a^3+b^3+c^3) \\ &= (2(a+b))^3-6(2(a+b))(a^2+b^2+(a+b)^2) + 8(a^3+b^3+(a+b)^3) \\ &=16(a+b)^3 - 24(a+b)(a^2+b^2+ab)+8(a^3+b^3) \\ &= 8(a+b)(2(a+b)^2-3(a^2+b^2+ab)+(a^2-ab+b^2)) \\ &= 0 \end{align*} Therefore \(a+b-c\) is a factor. By symmetry \(a-b+c\) and \(-a+b+c\) are also factors. Since our polynomial is degree \(3\) it must be \(K(a+b-c)(b+c-a)(c+a-b)\) for some \(K\). Since the coefficient of \(a^3\) is \(3\), \(K = 3\). so we have: \(3(a+b-c)(b+c-a)(c+a-b)\)

1. We want \(x + a + b = x+1\), \(x^2 + a^2 + b^2 = x^3+\frac52, x^3 + a^3 + b^3 = x^3+ \frac{13}{4}\). \(a+b = 1, a^2 + b^2 = 5/2\) so \(a = \frac32, b = -\frac12\) \begin{align*} 0 &= (x+1)^3 - 3(x+1)(2x^2+5)+2(4x^3+13) \\ &= 3(x +\frac{3}{2}+\frac{1}{2})(x - \frac{3}{2} - \frac{1}{2})(-x + \frac{3}{2} - \frac{1}{2}) \\ &= 3(x+2)(x-2)(1-x) \end{align*} and so the roots are \(x = 1, 2, -2\)
2. Letting \(c = d+e\) we have \begin{align*} (a+b+d+e)^3 &-6(a+b+d+e)(a^2+b^2+d^2+e^2) +8(a^3+b^3+d^3+e^3) \\ &= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2-2de) +8(a^3+b^3+c^3 - 3cde) \\ &= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)+12(a+b+c)de - 24cde \\ &= \underbrace{(a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)}_{\text{has a factor of }a+b-c} + 12(a+b-c)de \end{align*} Therefore there is a factor of \(a+b-c\) or \(a+b-d-e\). By symmetry we must have the factors: \((a+b-d-e)(a-b-d+e)(a-b+d-e)\) and so the final expression must be: \(K(a+b-d-e)(a-b-d+e)(a-b+d-e)\) The coefficient of \(a^3\) is \(3\), therefore \(K = 3\) We want \(x+a+b+c = x + 6\), \(x^2+a^2+b^2+c^2 = 14\) and \(x^3 + a^3+b^3+c^3 = 36\), ie \(a = 1,b=2,c=3\) would work, so \begin{align*} 0 &= (x+6)^3 - 6(x+6)(x^2+14) +8(x^3+36) \\ &= 3(x+1-2-3)(x-1+2-3)(x-1-2+3) \\ &= 3x(x-4)(x-2) \end{align*} ie the roots are \(x = 0, 2, 4\)

Solution:

1. When \(k =1\), we have \((1-x^2)(y')^2 + y^2 = 1\). Notice that if \(y_1 = x\) we have \(y_1' = 1\) and \((1-x^2) \cdot 1 + x^2 = 1\) so \(y_1\) is a solution, and we are allowed to assume this is the only solution. And notice that \(y_1(1) = 1\).
2. When \(k = 2\) we have \((1-x^2)(y')^2 + 4y^2 = 4\). Trying \(y_2 = 2x^2-1\) we see that \(y_2' = 4x\) and \((1-x^2)(4x)^2 + 4(2x^2-1)^2 = 16x^2-16x^4+16x^4-16x^2+4 = 4\). We can also check that \(y(1) = 2 \cdot 1^2 - 1 = 1\)
3. Let \(z(x) = 2(y_n(x))^2-1\), then \begin{align*} && \frac{\d z}{\d x} &= 4y'_n(x)y_n(x) \\ \Rightarrow && LHS &= (1-x^2)\left ( \frac{\d z}{\d x} \right)^2 + 4n^2 z^2 \\ &&&= (1-x^2)16(y'_n(x))^2(y_n(x))^2 + 4n^2(2(y_n(x))^2-1)^2 \\ &&&= 16y_n^2(1-x^2) \left [\frac{n^2-n^2y_n^2}{(1-x^2)} \right] + 16n^2y_n^4-16n^2y_n^2+4n^2 \\ &&&= 4n^2 = RHS \end{align*} Therefore \(y_{2n}(x) = 2(y_n(x))^2-1 = y_2(y_n(x))\) (notice also that \(z(1) = 2(y_n(1))^2-1 = 2-1 = 1\)).
4. Let \(v(x) = y_n(y_m(x))\) so \begin{align*} && (y_m')^2 &= \frac{m^2(1-y_m^2)}{1-x^2} \\ && (y_n')^2 &= \frac{n^2(1-y_n^2)}{1-x^2} \\ \\ && \frac{\d v}{\d x} &= y_n'(y_m(x)) \cdot y_m'(x) \\ && (1-x^2)(v')^2 &= (1-x^2) \cdot (y_n'(y_m(x)))^2 (y_m')^2 \\ &&&= (1-x^2) \cdot (y_n'(y_m(x)))^2 \left ( \frac{m^2(1-y_m^2)}{1-x^2} \right) \\ &&&= (y_n'(y_m(x)))^2 m^2(1-y_m^2) \\ &&&= \frac{n^2(1-(y_n(y_m(x)))^2)}{1-y_m^2}m^2(1-y_m^2) \\ &&&= n^2m^2(1-v^2) \end{align*} Therefore \(v\) satisfies our differential equation and \(v(1) = y_n(y_m(1)) = y_n(1) = 1\) so it must be our desired solution.

Solution: \begin{align*} u = a-x, \d u = - \d x: && \int_0^a f(x) \d x &= \int_{u=a}^{u=0} f(a-u) (-1) \d u \\ &&&= \int_0^a f(a-u) \d u \\ &&&= \int_0^a f(a-x) \d x \end{align*}

1. \begin{align*} && I &= \int_0^{\frac12 \pi} \frac{\sin x}{\cos x + \sin x } \d x\\ &&&= \int_0^{\frac12 \pi} \frac{\sin (\frac12 \pi - x)}{\cos (\frac12 \pi-x) + \sin (\frac12 \pi-x) } \d x\\ &&&= \int_0^{\frac12 \pi} \frac{\cos x}{\sin x + \cos x } \d x\\ \Rightarrow && 2I &= \int_0^{\frac12 \pi} 1 \d x \\ \Rightarrow && I &= \frac{\pi}{4} \end{align*}
2. \begin{align*} && I &= \int_0^{\frac14 \pi} \frac{\sin x}{\cos x + \sin x } \d x\\ &&&= \int_0^{\frac14 \pi} \frac{\sin (\frac14 \pi - x)}{\cos (\frac14 \pi-x) + \sin (\frac14 \pi-x) } \d x\\ &&&= \int_0^{\frac14 \pi} \frac{\frac1{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}{\frac1{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x + \frac1{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x} \d x \\ &&&= \int_0^{\frac14 \pi} \frac{\cos x - \sin x}{2 \cos x} \d x \\ &&&= \left [\frac12 x + \ln(\cos x) \right]_0^{\pi/4} \\ &&&= \frac{\pi}{8} -\frac12\ln2 - 1 \end{align*}
3. \begin{align*} && I &= \int_0^{\frac14\pi} \ln (1+\tan x) \, \d x \\ &&&= \int_0^{\frac14 \pi} \ln \left (1 + \tan \left(\frac{\pi}{4} - x\right) \right) \, \d x\\ &&&= \int_0^{\frac14 \pi} \ln \left (1 +\frac{1 - \tan x}{1+ \tan x} \right) \, \d x\\ &&&= \int_0^{\frac14 \pi} \ln \left (\frac{2}{1+ \tan x} \right) \, \d x\\ &&&= \frac{\pi}{4} \ln 2 - I \\ \Rightarrow && I &= \frac{\pi}{8} \ln 2 \end{align*}
4. \begin{align*} && I &= \int_0^{\frac14 \pi} \frac x {\cos x \, (\cos x + \sin x)}\, \d x \\ &&&= \int_0^{\frac14 \pi} \frac {\frac14 \pi - x} {(\frac1{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x) \, (\frac{2}{\sqrt{2}}\cos x)}\, \d x \\ &&&= \int_0^{\frac14 \pi} \frac {\frac14 \pi - x} {\cos x \, (\cos x + \sin x)}\, \d x \\ \\ \Rightarrow && I &= \frac{\pi}{8} \int_0^{\pi/4} \frac{\sec^2 x}{1 + \tan x} \d x\\ &&&= \frac{\pi}{8} \left [\ln (1 + \tan x) \right]_0^{\pi/4} \\ &&&= \frac{\pi}{8} \ln 2 \end{align*}

Solution:

1. \(\,\) \begin{align*} && \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}} &= \frac{1+x^{r+1}-1-x^r}{(1+x^r)(1+x^{r+1})} \\ &&&= \frac{x^r(x-1)}{(1+x^r)(1+x^{r+1})} \\ \\ && \sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \sum_{r=1}^N \frac{1}{x-1} \left ( \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}}\right) \\ &&&= \frac{1}{x-1} \Bigg ( \frac{1}{1+x} + \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^2} + \frac{1}{1+x^2} + \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^3} + \frac{1}{1+x^3} + \cdots \\ &&& \qquad \qquad \quad - \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^{N+1}} \Bigg ) \\ &&&= \frac{1}{x-1} \left (\frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right) \\ \\ && \sum_{r=1}^{\infty} \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \lim_{N\to \infty} \frac{1}{x-1} \left (\frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right) \\ &&&= \frac{1}{x-1} \left ( \frac{1}{1+x} - 1\right) \\ &&&= \frac{1}{x-1} \left ( \frac{-x}{1+x} \right) \\ &&&= \frac{x}{1-x^2} \end{align*}
2. \(\,\) \begin{align*} && \sum_{r=1}^\infty \textrm{sech}(ry)\textrm{sech}((r + 1)y) &= \sum_{r=1}^\infty \frac{4}{(e^{ry}+e^{-ry})(e^{(r+1)y}+e^{-(r+1)y})} \\ &&&=\sum_{r=1}^\infty \frac{4e^{-(2r+1)y}}{(1+e^{-2ry})(1+e^{-2(r+1)y})} \\ x = e^{-2y}: &&&= \frac{4e^{-y}e^{-2y}}{1-e^{-4y}} \\ &&&= \frac{4e^{-y}e^{-2y}}{e^{-2y}(e^{2y}-e^{-2y})} \\ &&&=2e^{-y}\textrm{cosech}(2y) \end{align*} \begin{align*} && \sum_{r=-\infty}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) &= \sum_{r=1}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) + \sum_{r=-\infty}^0 \textrm{sech}(ry) \textrm{sech}((r + 1)y) \\ &&&= 2e^{-y}\textrm{cosech}(2y) + \sum_{r=0}^\infty \textrm{sech}(-ry) \textrm{sech}(-(r-1)y) \\ &&&= 2e^{-y}\textrm{cosech}(2y) + \sum_{r=0}^\infty \textrm{sech}((r-1)y) \textrm{sech}(ry) \\ &&&= 4e^{-y}\textrm{cosech}(2y) + \textrm{sech}(y) + \textrm{sech}(-y) \\ &&&= 4e^{-y}\textrm{cosech}(2y)+2\textrm{sech}(y) \\ &&&= 4e^{-y} \frac12 \textrm{sech}(y) \textrm{cosech}(y) + 2 \textrm{sech}(y) \\ &&&= 2\textrm{sech}(y) \left ( e^{-y} \textrm{cosech}(y)+1 \right) \\ &&&= 2\textrm{sech}(y) \left ( \frac{2}{e^{2y}-1} + 1 \right) \\ &&&= 2\textrm{sech}(y) \left ( \frac{e^{2y}+1}{e^{2y}-1} \right) \\ &&&= 2 \textrm{cosech}(y) \end{align*}

Solution:

1. \(\,\) Let \(P(x)\) mean the proposition that \(f(x) + (1-x)f(-x) = x^2\) so \begin{align*} P(x): && f(x) + (1-x)f(-x) &= x^2 \\ P(-x): && f(-x)+(1+x)f(x) &= (-x)^2 = x^2 \\ \Rightarrow && f(x)+(1-x)\left (x^2-(1+x)f(x) \right) &= x^2 \\ \Rightarrow && f(x) \left (1 -(1-x^2) \right) &= x^2 + (x-1)x^2 \\ \Rightarrow && f(x)x^2 &= x^3 \\ \Rightarrow && f(x) &= x \end{align*} Notice that \(x + (1-x)(-x) = x^2\) so it does satisfy the functional equation.
2. Let \(K(x) = \frac{x+1}{x-1}\) if \(x \neq 1\) so \begin{align*} && K(K(x)) &= \frac{K(x)+1}{K(x)-1} \\ &&&= \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} \\ &&&= \frac{\frac{2x}{x-1}}{\frac{2}{x-1}} \\ &&&= x \end{align*} Let \(Q(x)\) denote the proposition that \(g(x) + xg(K(x)) = x\) so \begin{align*} Q(x): && g(x) + xg(K(x)) &= x \\ Q(K(x)): && g(K(x)) + K(x)g(x) &= K(x) \\ \Rightarrow && g(x) +xK(x)[1-g(x)] &= x \\ \Rightarrow && g(x)[1-xK(x)] &= x(1-K(x)) \\ \Rightarrow && g(x) \frac{x-1-x^2-x}{x-1} &= \frac{-2x}{x-1} \\ \Rightarrow && g(x) &= \frac{2x}{x^2+1} \end{align*}. And notice that \(\frac{2x}{x^2+1} + x \frac{2\frac{x+1}{x-1}}{\left( \frac{x+1}{x-1}\right)^2+1} = \frac{2x}{x^2+1} + \frac{2x(x^2-1)}{2x^2+2} = x\)
3. Consider \(H(x) = \frac{1}{1-x}\) then notice that \(H(H(x)) = \frac{1}{1-\frac{1}{1-x}} = \frac{x-1}{x}\) and \(H^3(x) = \frac{\frac{1}{1-x}-1}{\frac{1}{1-x}} = 1-(1-x) = x\). So So letting \(S(x)\) be the statement that \(h(x) + h(H(x)) = 1 - x - \frac{1}{1-x}\) we have \begin{align*} S(x): && h(x) + h(H(x)) &= 1 - x - H(x) \\ S(H(x)): && h(H(x)) + h(H^2(x)) &= 1 - H(x) - H^2(x) \\ S(H^2(x)): && h(H^2(x)) + h(x) &= 1 - H^2(x) - x \\ S(x) - S(H(x)) + S(H^2(x)): && 2h(x) &= 1 - 2x \\ \Rightarrow && h(x)& = \frac12 - x \end{align*} and notice that \(\frac12 -x +\frac12 - \frac{1}{1-x} = 1 - x - \frac{1}{1-x}\) so it does satisfy the equation.

Solution:

1. \(\,\) \begin{align*} && f(x) &= \int_1^3 (t-1)^{x-1} \d t \\ &&&= \left [ \frac1x(t-1)^{x} \right]_1^3 \\ &&&= \frac{2^x}{x} \end{align*}
   

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2. \(\,\) \begin{align*} && g(x) &= \int_{-1}^1 \frac{1}{\sqrt{1-2xt+x^2}} \d t \\ &&&= \left [ -\frac{1}{x}(1 +x^2 - 2xt)^{\frac12} \right]_{-1}^1 \\ &&&= \frac1x \left ( \sqrt{1+x^2+2x}-\sqrt{1+x^2-2x}\right) \\ &&&= \frac1x \left ( |1+x|-|1-x| \right) \end{align*}
   

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Solution:

1. \(\,\) \begin{align*} && \sec^2\left(\tfrac14\pi-\tfrac12 x\right) &= \frac{1}{\cos^2 \left(\tfrac14\pi-\tfrac12 x\right)} \\ &&&= \frac{1}{\frac{1+\cos 2\left(\tfrac14\pi-\tfrac12 x\right)}{2}} \\ &&&= \frac{2}{1 + \cos \left(\tfrac12\pi- x\right)} \\ &&&= \frac{2}{1+\sin x} \\ \\ && \int \frac{1}{1+\sin x} \d x &= \int \tfrac12\sec^2\left(\tfrac14\pi-\tfrac12 x\right) \d x\\ &&&= - \tan\left(\tfrac14\pi-\tfrac12 x\right) + C \end{align*}
2. \(\,\) \begin{align*} && I &= \int_0^{\pi} x f(\sin x) \d x \\ y = \pi - x, \d y = - \d x: &&&= \int_{y=\pi}^{y = 0} (\pi - y) f(\sin(\pi - y))(-1) \d y \\ &&&= \int_0^\pi (\pi - y) f(\sin y) \d y \\ &&&= \pi \int_0^\pi f(\sin y) \d y - I \\ \Rightarrow && I &= \frac{\pi}{2} \int_0^\pi f(\sin x) \d x \\ \\ \Rightarrow && \int_0^{\pi} \frac{x}{1 + \sin x} \d x &= \frac{\pi}{2} \int_0^{\pi} \frac{1}{1 + \sin x} \d x\\ &&&=\frac{\pi}{2} \left [- \tan\left(\tfrac14\pi-\tfrac12 x\right) \right]_0^{\pi} \\ &&&= \frac{\pi}{2} \left (-\tan (-\tfrac{\pi}{4}) + \tan \tfrac{\pi}{4} \right) \\ &&&= \pi \end{align*}
3. \(\,\) \begin{align*} && J &= \int_0^{\pi} \frac{2x^3-3\pi x^2}{(1+\sin x)^2} \d x \\ y = \pi - x: &&&= \int_0^{\pi} \frac{2(\pi-y)^3-3\pi (\pi - y)^2}{(1+\sin x)^2 } \d y \\ &&&= \int_0^{\pi} \frac{-2 y^3 + 3 \pi y^2 - \pi^3}{(1+ \sin x)^2}\\ &&&= -\pi^3 \int_0^{\pi} \frac{1}{(1 + \sin x)^2} \d x -J \\ \Rightarrow && J &= -\frac{\pi^3}{2} \int_0^{\pi} \frac{1}{(1 + \sin x)^2} \d x\\ &&&= -\frac{\pi^3}{2} \int_0^\pi \tfrac14 \sec^4\left(\tfrac14\pi-\tfrac12 x\right) \d x \\ &&&= -\frac{\pi^3}{8} \int_0^\pi \sec^2\left(\tfrac14\pi-\tfrac12 x\right)\left (1 + \tan^2\left(\tfrac14\pi-\tfrac12 x\right) \right) \d x \\ &&&= -\frac{\pi^3}{8} \left [-\frac23 \tan^3\left(\tfrac14\pi-\tfrac12 x\right) - 2 \tan\left(\tfrac14\pi-\tfrac12 x\right) \right]_0^{\pi} \\ &&&= -\frac{\pi^3}{8} \left (\frac43+4 \right) \\ &&&= -\frac{2\pi^3}{3} \end{align*}

Solution: \begin{align*} I_n - I_{n+1} &= \int_0^\infty \frac 1 {(1+u^2)^n}\, \d u - \int_0^\infty \frac 1 {(1+u^2)^{n+1}}\, \d u \\ &= \int_0^\infty \l \frac 1 {(1+u^2)^n}- \frac 1 {(1+u^2)^{n+1}} \r\, \d u \\ &= \int_0^\infty \frac {u^2} {(1+u^2)^{n+1}} \, \d u \\ &= \left [ u \frac{u}{(1+u^2)^{n+1}} \right]_0^{\infty} - \frac{-1}{2n}\int_0^{\infty} \frac{1}{(1+u^2)^n} \d u \tag{\(IBP: u = u, v' = \frac{u}{(1+u^2)^{n+1}}\)}\\ &= \frac{1}{2n} I_n \end{align*} \(\displaystyle I_1 = \int_0^{\infty} \frac{1}{1+u^2} \d u = \left [ \tan^{-1} u \right]_0^\infty = \frac{\pi}{2}\) as expected. We also have, \(I_{n+1} = \frac{2n(2n-1)}{2n \cdot 2n} I_n \), by rearranging the recurrence relation. Therefore, when we multiply out the top we will have \(2n!\) and the bottom we will have two factors of \(n!\) and two factors of \(2^n\) combined with the original \(\frac{\pi}{2}\) we get \[ I_{n+1} = \frac{(2n)! \pi}{2^{2n+1} (n!)^2} \] \begin{align*} J = \int_0^\infty f\big( (x- x^{-1})^2\big ) \, \d x &= \int_{u = \infty}^{u = 0} f((u^{-1}-u)^2)(-u^{-2} )\d u \tag{\(u = x^{-1}, \d u = -x^{-2} \d x\)} \\ &= \int^{u = \infty}_{u = 0} f((u^{-1}-u)^2)u^{-2} \d u \\ &= \int^{\infty}_{0} u^{-2}f((u-u^{-1})^2) \d u \\ \end{align*} Therefore adding the two forms for \(J\) we have \begin{align*} 2 J &= \int_0^\infty f\big( (x- x^{-1})^2\big ) \, \d x + \int_0^\infty x^{-2} f\big( (x- x^{-1})^2\big ) \, \d x \\ &= \int_0^\infty (1+x^{-2}) f\big( (x- x^{-1})^2\big ) \, \d x \end{align*} And letting \(u = x - x^{-1}\), we have \(\d u = (1 + x^{-2}) \d x\), and \(u\) runs from \(-\infty\) to \(\infty\) so we have: \begin{align*} \int_0^\infty (1+x^{-2}) f\big( (x- x^{-1})^2\big ) \, \d x &= \int_{-\infty}^\infty f(u^2) \, \d u \\ &=2 \int_{0}^\infty f(u^2) \, \d u \end{align*} Since both of these are \(2J\) we have the result we are after. Finally, \begin{align*} \int_0^\infty \frac {x^{2n-2}}{(x^4-x^2+1)^n} \, \d x &= \int_0^{\infty} \frac{x^{2n-2}}{x^{2n}(x^2-1+x^{-2})^n} \d x \\ &= \int_0^{\infty} \frac{x^{-2}}{((x-x^{-1})^2+1)^n} \d x \\ &= \int_0^{\infty} \frac{1}{(x^2+1)^n} \d x \tag{Where \(f(x) = (1+x^2)^{-n}\) in \(J\) integral} \\ &= I_n = \frac{(2n-2)! \pi}{2^{2n-1} ((n-1)!)^2} \end{align*}

Solution:

1. \(\,\) \begin{align*} && \int_0^b x^2 \d x &= \left ( \int_0^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} &= \left ( \frac{b^2}{2} \right)^2 \\ \Rightarrow && b &= \frac{4}{3} \end{align*}
2. \(\,\) \begin{align*} && \int_1^b x^2 \d x &= \left ( \int_1^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{1}{3} &= \left ( \frac{b^2}{2} - \frac{1}{2} \right)^2 \\ \Rightarrow && 4(b^3 - 1) &= 3(b^2-1)^2 \\ \Rightarrow && 4(b^3-1) &= 3(b^4-2b^2+1) \\ \Rightarrow && 0 &= 3b^4-4b^3-6b^2+7 \\ &&&= (b-1)(3b^3-b^2-7b-7) \\ \Rightarrow && 0 &= 3b^3-b^2-7b-7 \end{align*}
   

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   Let \(f(x) = 3x^3-x^2-7x-7\) then \(f(2) = 3 \cdot 8 - 4 - 14 - 7 = -1 < 0\), \(f(3) = 3 \cdot 27 - 9 - 21 - 7 = 44 > 0\) therefore the root must lie between \(2\) and \(3\).
3. \(,\) \begin{align*} && \int_a^b x^2 \d x &= \left ( \int_a^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{a^3}{3} &= \left ( \frac{b^2}{2} - \frac{a^2}{2} \right)^2 \\ \Rightarrow && 4(b^3 - a^3) &= 3(b^2-a^2)^2 \\ \Rightarrow && 4(b^2+ab+a^2) &= 3(b-a)(b+a)^2 \\ \Rightarrow && 4 \left ( \left ( \frac{p+q}{2}\right)^2+\left ( \frac{p+q}{2}\right)\left ( \frac{p-q}{2}\right)+\left ( \frac{p-q}{2}\right)^2\right) &= 3qp^2 \\ \Rightarrow && 3p^2 + q^2 &= 3qp^2 \\ \Rightarrow && 3p^2(q-1) &= q^2 \\ \Rightarrow && p^2 &= \frac{q^2}{3(q-1)} \\ \Rightarrow && 1 &\leq \frac{1}{3(q-1)} \\ \Rightarrow && 3(q-1) &\leq 1 \\ \Rightarrow && q & \leq \frac{4}{3} \\ \end{align*}