Problems

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Solution: \begin{align*} && G(a) &= \frac{F(a)}{2-F(a)}\\ &&&= 0 \tag{\(F(a)= 0\)}\\ \\ && G(b) &= \frac{F(b)}{2-F(b)} \\ &&&= \frac{1}{2-1} = 1 \tag{\(F(b)=1\)}\\ \\ && G'(y) &= \frac{F'(y)(2-F(y))+F(y)F'(y)}{(2-F(y))^2} \\ &&&= \frac{2F'(y)}{(2-F(y))^2} \geq 0 \tag{\(F'(y) \geq 0\)} \end{align*} \begin{align*} && 0 \leq F(y)\leq1\\ \Leftrightarrow&& 1\leq 2-F(y) \leq 2\\ \Leftrightarrow &&1 \leq (2-F(y))^2 \leq 4\\ \Leftrightarrow && 1 \geq \frac{1}{(2-F(y))^2} \geq \frac14 \\ \Leftrightarrow && 2 \geq \frac{2}{(2-F(y))^2} \geq\frac12 \end{align*} \begin{align*} && \mathbb{E}(Y) &= \int_a^b y G'(y) \d y \\ &&&= \int_a^b y F'(y) \underbrace{\frac{2}{(2-F(y))^2}}_{\in [\frac12, 2]} \d y \\ &&&\leq 2 \E[X] \\ &&&\geq \frac12 \E[X]\\ \\ && \E[Y^2] &\leq 2\E[X^2] \\ && \E[Y^2] &\geq \frac12\E[X^2] \\ \\ \Rightarrow && \var[Y] &= \E[Y^2]-\E[Y]^2 \\ &&& \leq 2 \E[X^2] - (\tfrac12\E[X])^2 \\ &&&= 2 \var[X] + \tfrac74(\E[X])^2 \end{align*}

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Solution: \begin{align*} && \int_1^a \frac{\ln x}{x} \d x &= \left [ \ln x \cdot \ln x\right ]_1^a - \int_1^a \frac{\ln x}{x} \d x \\ \Rightarrow && \int_1^a \frac{\ln x}{x} \d x &= \frac12 \left ( \ln a \right) ^2 \\ && \int_1^\infty \frac{\ln x}{x} \d x &= \lim_{a \to \infty} \frac12 (\ln a)^2 \\ &&&= \infty \end{align*} \begin{align*} && \pi \int_1^a \left ( \frac{\ln x}{x} \right)^2 \d x &= \pi \int_{u=0}^{u=\ln a} \left ( \frac{u}{e^u} \right)^2 e^u \d u \\ &&&= \pi \int_0^{\ln a} u^2 e^{-u} \d u \\ &&&= \pi \left [-u^2e^{-u} \right]_0^{\ln a} +\pi \int_0^{\ln a} 2u e^{-u} \d u \\ &&&= -\frac{\pi}{a} (\ln a)^2 + \pi \left [-2u e^{-u} \right]_0^{\ln a} + \pi \int_0^{\ln a} e^{-u} \d u \\ &&&= -\frac{\pi}{a} (\ln a)^2- \frac{2 \pi}{a}\ln a+\pi \left (1 - \frac{1}{a} \right) \\ \\ && \pi \int_1^{\infty} \left ( \frac{\ln x}{x} \right)^2 \d x &= \lim_{a \to \infty} \left ( -\frac{\pi}{a} (\ln a)^2- \frac{2 \pi}{a}\ln a+\pi \left (1 - \frac{1}{a} \right) \right) \\ &&&= \pi \end{align*}

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Solution: \begin{align*} && \int_0^1 \frac{x^4}{1+x^2} \d x &= \int_0^1 \frac{(x^2-1)(1+x^2)+1}{x^2+1} \d x\\ &&&= \int_0^1 \frac{1}{1+x^2} \d x -\int_0^1 (1-x^2) \d x \\ &&&= \left [\tan^{-1}x \right]_0^1 - \left [x - \tfrac13x^3 \right]_0^1 \\ &&&= \frac{\pi}{4} - \frac23 \end{align*}

1. \(\,\) \begin{align*} && I &= \int_0^1 x^3 \; \tan ^{-1} \left(\frac {1-x} {1+x} \right) \,\d x \\ &&&= \left [ \frac{x^4}{4}\tan ^{-1} \left(\frac {1-x} {1+x} \right) \right]_0^1 -\int_0^1 \frac{x^4}{4} \frac{1}{1 +\left(\frac {1-x} {1+x} \right) ^2 } \cdot \frac{-2}{(1+x)^2} \d x \\ &&&= \frac{1}{2} \int_0^1 \frac{x^4}{(1+x)^2+(1-x)^2} \d x \\ &&&= \frac{1}{4} \int_0^1 \frac{x^4}{1+x^2} \d x \\ &&&= \frac{\pi}{16} - \frac{1}{6} \end{align*}
2. \(\,\) \begin{align*} && J &= \int_0^1 \frac {(1-y)^3} {(1+y)^5} \; {{\tan}^{-1} y}\, \d y \\ &&&= \left [ \frac {(y(1+y^2)} {(1+y)^4} \tan^{-1}y \right]_0^1 - \int_0^1 \frac {(y(1+y^2)} {(1+y)^4} \frac{1}{1+y^2} \d y \\ &&&= \frac{\pi}{32} - \int_0^1 \frac{y}{(1+y)^4} \d y \\ &&&= \frac{\pi}{32} - \left[ - \frac{3y+1}{6(1+y)^3} \right]_0^1 \\ &&&= \frac{\pi}{32} +\frac{4}{6 \cdot 8} - \frac{1}{6} \\ &&&= \frac{\pi}{32} - \frac{1}{12} \end{align*}

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Solution: \begin{align*} && \int_{-1}^1 |x e^x |\d x &= \int_{-1}^0 |xe^x| \d x + \int_0^1 |xe^x| \d x \\ &&&= \int_{-1}^0 -xe^x \d x + \int_0^1 x \e^x \d x \\ &&&= -\int_{-1}^0 xe^x \d x + \int_0^1 x \e^x \d x \\ \\ && \int xe^x \d x &= xe^x - \int e^x \d x \\ &&&= xe^x - e^x \\ \\ \Rightarrow && \int_{-1}^1 |x e^x |\d x &= \left [ xe^x - e^x \right]_0^{-1}+ \left [ xe^x - e^x \right]_0^{1} \\ &&&= -e^{-1}-e^{-1} +e^{0} + e^1 - e^1 +e^0 \\ &&&= 2-2e^{-1} \end{align*}

1. \(\,\) \begin{align*} && I &= \int_0^4 | x^3-2x^2-x+2| \d x \\ &&&= \int_0^4 |(x-2)(x-1)(x+1)| \d x\\ &&&= \int_0^1( x^3-2x^2-x+2) \d x- \int_1^2 ( x^3-2x^2-x+2) \d x + \int_2^4 ( x^3-2x^2-x+2) \d x \\ &&&= \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_0^1 - \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_1^2 + \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_2^4 \\ &&&= 2 \left ( \frac14 - \frac23 -\frac12 + 2\right) - 2 \left ( \frac14 2^4 - \frac23 2^3 -\frac12 2^2 + 2 \cdot 2\right)+ \left ( \frac14 4^4 - \frac23 4^3 -\frac12 4^2 + 2 \cdot 4\right) \\ &&&= \frac{133}{6} \end{align*}
2. \(\,\) \begin{align*} && J &= \int_{-\pi}^\pi | \sin x + \cos x | \d x \\ &&&= \int_{-\pi}^{\pi} | \sqrt{2} \sin(x + \tfrac{\pi}{4})| \d x \\ &&&= 2\sqrt{2}\int_0^\pi \sin x \d x \\ &&&= 4\sqrt{2} \end{align*}

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Solution: \begin{align*} && g(\lambda) &= \int_0^1 \big(\f(x)-\lambda x \big)^{\!2} \,\d x \\ &&&= \int_0^1 \left ( f(x)^2 -2\lambda xf(x) + \lambda^2 x^2\right) \d x \\ &&&= \frac13\lambda^2 - 2\lambda \int_0^1 x f(x) \d x + \int_0^1 f(x)^2 \d x \\ \end{align*} Differentiating (or completing the square) it is clear the minimum occurs when \(\displaystyle \lambda = 3 \int_0^1 xf(x) \d x\) \begin{align*} && R^2 &= \int_0^1 (f(x) - \lambda x )^2 \d x \\ &&&= \frac13\lambda^2 - 2\lambda \int_0^1 x f(x) \d x + \int_0^1 f(x)^2 \d x \\ &&&= \frac13 \left (\lambda -3\int_0^1 xf(x) \d x \right)^2 -\frac13 \left ( 3\int_0^1 xf(x) \d x \right)^2+\int_0^1 f(x)^2 \d x \\ \end{align*} When \(\lambda = 3\int_0^1 xf(x) \d x \) clearly this is the desired result. \begin{align*} && \lambda &= 3\int_0^1 xf(x) \d x \\ &&&= 3\int_0^1 x \sin(\pi x /n) \d x \\ &&&= 3 \left [-x \frac{n}{\pi} \cos (\pi x /n) \right]_0^1 + \frac{3n}{\pi} \int_0^1 \cos(\pi x /n) \d x \\ &&&= -\frac{3n}{\pi}\cos(\pi/n) + \frac{3n}{\pi} \left [ \frac{n}{\pi} \sin(\pi x /n)\right]_0^1 \\ &&&= -\frac{3n}{\pi} \cos(\pi/n) + \frac{3n^2}{\pi^2} \sin(\pi /n) \\ \text{for large }n: &&&\approx -\frac{3n}{\pi}\left ( 1 - \frac12\frac{\pi^2}{n^2} + o(1/n^4)\right) + \frac{3n^2}{\pi^2} \left (\frac{\pi}{n} - \frac16 \frac{\pi^3}{n^3} +o(1/n^5) \right) \\ &&&= \left (\frac32 -\frac12\right)\frac{\pi}{n} + o(1/n^3) \\ &&&= \frac{\pi}{n} + o(1/n^2) \end{align*} Therefore for large \(n\), \(\lambda \approx \frac{\pi}n\) \begin{align*} && \int_0^1 \sin^2(\pi x/n) \d x &= \frac12\int_0^1(1- \cos(2\pi x/n)) \d x\\ &&&= \frac12\left ( 1 - \frac{n}{2\pi}\left[\sin(2\pi x/n) \right]_0^1 \right) \\ &&&= \frac12 -\frac{n}{4\pi}\sin(2\pi /n) \\ \\ && R^2 &= \frac12 -\frac{n}{4\pi}\sin(2\pi /n) - \frac13 \left ( \frac{\pi}{n}+o(1/n^2)\right)^2 \\ &&&= \frac12 - \left ( \frac{1}{2} -\frac16\frac{\pi}{n}+o(1/n^3) \right) - o(1/n^2) \\ &&& = \frac16 \frac{\pi}{n} + o(1/n^2) \\ &&&\to 0 \text{ as } n \to \infty \end{align*} We should expect these results as for \(n\) very large \(\sin(\pi x/n) \approx \frac{\pi }{n}x\) so the best linear approximation is likely to be \(\lambda = \frac{\pi}{n}\) and we should expect it to improve to the point that we cannot tell the difference, ie \(R^2 \to 0\)

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Solution: \begin{align*} && y &= \frac{ax+b}{cx+d} \\ &&&= \frac{\frac{a}{c}(cx+d) - \frac{da}{c} + b}{cx+d} \\ \Rightarrow && y' &= \left (b - \frac{da}{c} \right)(-1)(cx+d)^{-2} \cdot c \\ &&&= (ad-bc)(cx+d)^{-2} \end{align*} \begin{align*} && y &= \frac{x+1}{x+3} \\ && \frac{\d y}{\d x} &= \frac{2}{(x+3)^2} \\ \Rightarrow && I &= \int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x+1}{x+3}\right)\d x \\ &&&= \int_{y=1/3}^{y=1/2} \frac12 \ln y \, \d y \\ &&&= \frac12 \left [ y \ln y - y \right]_{1/3}^{1/2} \\ &&&= \frac12 \left ( \frac12\ln \frac12 - \frac12 - \frac13 \ln\frac13 + \frac13 \right) \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} \end{align*} \begin{align*} && J &= \int_0^1 \frac1{(x+3)^2} \ln \left ( \frac{x^2+3x+2}{(x+3)^2} \right) \d x \\ &&&= \int_0^1 \frac1{(x+3)^2} \left ( \ln \frac{x+1}{x+3} + \ln \frac{x+2}{x+3} \right) \d x \\ &&&= I + \int_0^1 \frac1{(x+3)^2} \ln \left ( \frac{x+2}{x+3} \right) \d x \\ &&&= I + \int_{y=2/3}^{y=3/4} \ln y\, \d y \\ &&&= I + \left [ y \ln y- y\right]_{2/3}^{3/4} \\ &&&= I + \left ( \frac34 \ln \frac34 - \frac34 - \frac23 \ln \frac23 + \frac23 \right) \\ &&&= I + \left ( \frac34 \ln 3 - \frac32 \ln 2- \frac1{12} - \frac23 \ln 2 + \frac23 \ln 3\right) \\ &&&= I + \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} + \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= \frac{19}{12} \ln 3 -\frac{29}{12}\ln 2 - \frac16 \end{align*} \begin{align*} && K &= \int_0^1 \frac{1}{(x+3)^2} \; \ln\left(\frac{x+1}{x+2}\right)\d x \\ &&&= \int_0^1 \frac{1}{(x+3)^2} \; \left ( \ln\left(\frac{x+1}{x+3}\right) - \ln \left ( \frac{x+3}{x+2} \right) \right)\d x \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} - \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= -\frac54 \ln 3 +\frac{23}{12} \ln 2 \end{align*}

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Solution:

1. Let \(u = 1-t, \d u = -\d t\), then: \begin{align*} \mathrm{I}(a,b) &= \int_0^1 t^a(1-t)^b \d t \\ &= \int_{u=1}^{u=0} -(1-u)^a u^b \d u \\ &= \int_0^1(1-u)^a u^b \d u \\ &= \mathrm{I}(b, a) \end{align*}
2. \begin{align*} \mathrm{I}(a+1,b)+\mathrm{I}(a,b+1) &= \int_0^1 t^{a+1}(1-t)^b + t^a(1-t)^{b+1} \d t \\ &= \int_0^1 (t+(1-t))t^a(1-t)^b \d t \\ &= \int_0^1 t^a(1-t)^b \d t \\ &= \mathrm{I}(a,b) \end{align*}
3. Integrating by parts with \(\frac{du}{dt} = t^a, v = (1-t)^{b}\)\begin{align*} \mathrm{I}(a,b) &= \int_0^1 t^a (1-t)^b \d t \\ &= \left [ \frac{t^{a+1}}{a+1} (1-t)^b \right ]_0^1 + \int_0^1 \frac{t^{a+1}}{a+1} b(1-t)^{b-1} \\ &= \frac{b}{a+1} \int_0^1 t^{a+1}(1-t)^{b-1} \d t \\ &= \frac{b}{a+1} \mathrm{I}(a+1,b-1) \end{align*} Claim: \(\mathrm{I}(a,b) = \frac{a!b!}{(a+b+1)!}\) Proof: Note that \(I(a,0) = \frac{1}{a+1}\) so the formula holds for this case. We will induct on \(b\). The base case is done. Suppose that for \(b = k\) our formula is true, ie: \(\mathrm{I}(a,k) = \frac{a!k!}{(a+k+1)!}\) for all \(a\) (and fixed \(k\)) \begin{align*} \mathrm{I}(a,k+1) &= \frac{k+1}{a+1} \mathrm{I}(a+1,k) \\ &= \frac{k+1}{a+1} \frac{(a+1)!k!}{(a+1+k+1)!} \\ &= \frac{a!(k+1)!}{(a+(k+1)+1)!} \end{align*} So the formula is true for \(b=k+1\). Therefore, since it is true if \(b=0\) and if \(b=k\) is true then \(b=k+1\) is true, it is true for all values of \(b\).

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Solution: \begin{align*} && u_n &= \int_0^{\tfrac12 \pi} \sin^{n} t \, \d t \\ && &= \int_0^{\tfrac12 \pi} \sin t \sin^{n-1} t \, \d t \\ && &= \left [ -\cos t \sin^{n-1} t \right]_0^{\tfrac12 \pi} + \int_0^{\tfrac12 \pi} \cos t (n-1) \sin^{n-2} t \cos t \d t \\ && &= 0 + (n-1)\int_0^{\tfrac12 \pi} \cos^2 t \sin^{n-2} t \d t \\ && &= (n-1) \int_0^{\tfrac12 \pi}(1-\sin^2 t) \sin^{n-2} t \d t \\ && &= (n-1)u_{n-2} - (n-1)u_n \\ \Rightarrow && n u_n &= (n-1)u_{n-2} \\ \end{align*} Mutplying both sides by \(u_{n-1}\) we obtain \(nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2}\). Therefore \(nu_nu_{n-1}\) is constant, ie is equal to \(\displaystyle u_1u_0 = \int_0^{\tfrac12 \pi} \sin^{1} t \, \d t \int_0^{\tfrac12 \pi} \sin^{0} t \, \d t = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}\)

Since \(0 < \sin t < 1\) for \(t \in (0, \tfrac{\pi}{2})\) we must have \(0 < \sin^n t < \sin^{n-1} t\), in particular \(0 < u_n < u_{n-1}\) Therefore \begin{align*} && nu_{n}u_{n-1} &= \tfrac{1}{2}\pi \\ \Rightarrow && nu_n u_n &< \tfrac{1}{2}\pi \tag{\(u_n < u_{n-1}\)} \\ \Rightarrow && nu_{n-1} u_{n-1} &> \tfrac{1}{2}\pi \tag{\(u_n < u_{n-1}\)} \\ \Rightarrow && nu_n^2 &< \tfrac12 \pi < n u_{n-1}^2 \end{align*} However we also have \(\tfrac12 \pi < (n+1)u_n^2\) (by considering the next inequality), so \(\left ( \frac{n}{n+1}\right) \tfrac12 \pi < n u_n^2 < \tfrac12 \pi\) but since as \(n \to \infty\) the right hand bound is constant and the left hand bound tends to \(\tfrac12 \pi\) therefore \(n u_n^2 \to \tfrac12 \pi\)

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Solution: \begin{align*} && P(x) &= (x-a)^mQ(x) \\ \Rightarrow && P'(x) &= m(x-a)^{m-1}Q(x) + (x-a)^mQ'(x) \\ &&&= (x-a)^{m-1}(\underbrace{mQ(x) + (x-a)Q'(x)}_{\text{a polynomial}}) \\ &&&= (x-a)^{m-1}R(x) \end{align*} Therefore \(P^{(r)}(a) = 0\) for \(1 \leq r \leq m-1\) since each time we differentiate we will have a factor of \((x-a)^{m-r}\) which is zero when we evaluate at \(x = a\). If \(P_n(x) = \frac{\d^n}{\d x^n}(x^2-1)^n\) then we are differentiating a degree \(2n\) polynomial \(n\) times. Each time we differentiate we reduce the degree by \(1\), therefore the degree of \(P_n\) is \(n\). \begin{align*} && \int_{-1}^1 x^mP_n(x) \d x &= \left [x^m \underbrace{\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= -\left [mx^{m-1} \underbrace{\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1+ \int_{-1}^1 m(m-1)x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= m(m-1)\int_{-1}^1 x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&& \cdots \\ &&&= (-1)^m m!\int_{-1}^1 \frac{\d^{n-m}}{\d x^{n-m}} \left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 \end{align*} If \(n = m\), we have \begin{align*} && \int_{-1}^1 x^n P_n(x) \d x&= (-1)^nn! \int_{-1}^1 (x^2-1)^n \d x \\ && &= (-1)^{2n}n! \cdot 2\int_{0}^1 (1-x^2)^n \d x \\ x = \sin \theta, \d x = \cos \theta \d \theta: &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n} \theta \cdot \cos \theta \d \theta \\ &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n+1} \theta \d \theta \\ &&&= 2 \cdot n!\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \\ &&&= \frac{(2^{2n+1})(n!)^{3}}{(2n+1)!} \\ \end{align*}

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Solution: \begin{align*} && \int_0^x \mathrm{sech}\, t \d t &= \int_0^x \frac{2}{e^t+e^{-t}} \d t \\ &&&= \int_0^x \frac{2e^t}{e^{2t}+1} \d t \\ &&&= \left [2 \arctan e^t \right ]_0^x \\ &&&= 2\tan^{-1}e^x- \frac{\pi}{2} \end{align*} \begin{align*} && I_n &= \int_0^x \mathrm{sech}^n\, t \d t \\ &&&= \int_0^x \mathrm{sech}^{n-2}\, t \mathrm{sech}^2\, t \d t \\ &&&= \left [ \mathrm{sech}^{n-2}\, t \cdot \tanh t\right ]_0^x - \int_0^x (n-2) \mathrm{sech}^{n-3} \, t \cdot (- \tanh t \mathrm{sech}\, t) \tanh t \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t \tanh^2 t \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t (1-\mathrm{sech}^2 \, t) \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2}-(n-2)I_n \\ \Rightarrow && (n-1)I_n &= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \\ \Rightarrow && I_n &= \frac{1}{n-1} \left ( \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \right) \\ \end{align*} \begin{align*} I_1 &= 2\tan^{-1}e^x- \frac{\pi}{2} \\ I_3 &= \frac12 \left ( \mathrm{sech}\, x \cdot \tanh x+ 2\tan^{-1}e^x- \frac{\pi}{2}\right) \\ &= \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \\ I_5 &= \frac14 \left (\mathrm{sech}^3\, x \cdot \tanh x + 3 \left ( \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \right) \right) \\ &= \frac14 \mathrm{sech}^3\, x \cdot \tanh x +\frac38 \mathrm{sech}\, x \cdot \tanh x+\frac34 \tan^{-1}e^x- \frac{3\pi}{16} \\ \\ I_2 &= \tanh x \\ I_4 &= \frac13 \left ( \mathrm{sech}^2 x\tanh x +2\tanh x\right) \\ &= \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \\ I_6 &= \frac15 \left ( \mathrm{sech}^4 x \tanh x+4\left ( \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \right) \right) \\ &= \frac15 \mathrm{sech}^4 x \tanh x + \frac4{15}\mathrm{sech}^2 x\tanh x + \frac{8}{15} \tanh x \end{align*}

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Solution:

1. \begin{align*} \int_{-\pi}^\pi |\sin x | \d x &= \int_{-\pi}^{0} - \sin x \d x + \int_0^\pi \sin x \d x \\ &= \left [\cos x \right]_{-\pi}^{0} +[-\cos x]_0^{\pi} \\ &= 1-(-1)+(1)-(-1) \\ &= 4 \end{align*}
2. \begin{align*} \int_{-\pi}^\pi \sin | x | \d x &= \int_{-\pi}^0 - \sin x \d x + \int_0^\pi \sin x \d x \\ &= 4 \end{align*}
3. \begin{align*} \int_{-\pi}^\pi x \sin x \d x &= \left [ -x \cos x \right]_{-\pi}^\pi + \int_{-\pi}^{\pi} \cos x \d x \\ &= \pi -(-\pi) + \left [\sin x \right]_{-\pi}^\pi \\ &= 2\pi \end{align*}
4. \begin{align*} \int_{-\pi}^{\pi} x^{10} \sin x \d x &\underbrace{=}_{x^{10}\sin x \text{ is odd}} 0 \end{align*}

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Solution:

1. \begin{align*} u = \frac{\pi}{2} - x :&& \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \int_{\frac12\pi}^0 \ln (\cos u) (- 1)\d u \\ &&&= \int_0^{\frac12 \pi} \ln (\cos x) \d x \\ \Rightarrow && 2 \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \int_0^{\tfrac12 \pi} \ln (\sin x) \d x +\int_0^{\tfrac12 \pi} \ln (\cos x) \d x \\ &&&= \int_0^{\tfrac12 \pi}\left (\ln (\sin x)+ \ln (\cos x) \right) \d x \\ &&&= \int_0^{\frac12 \pi} \ln \left (\frac12 \sin 2x \right) \d x \\ &&&= \int_0^{\frac12 \pi} \left ( \ln \left (\sin 2x \right) - \ln 2 \right)\d x \\ &&&= \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x - \frac{\pi}{2} \ln 2\\ \Rightarrow && \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \frac12 \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x - \frac{\pi}{4} \ln 2 \end{align*} \begin{align*} u = 2x, \d u = 2 \d x && \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x &= \int_0^{\pi} \ln (\sin u) \frac12 \d u \\ &&&= \frac12 \int_0^{\pi} \ln (\sin u) \d u \\ &&&=\frac12 \left ( \int_0^{\pi/2} \ln (\sin u) \d u + \int_{\pi/2}^{\pi} \ln (\sin u) \d u \right)\\ &&&= \int_0^{\pi/2} \ln (\sin u) \d u \\ \Rightarrow && I &= \frac12 I - \frac14 \pi \ln 2 \\ \Rightarrow && I &= -\frac12 \pi \ln 2 \end{align*}
2. \begin{align*} && \ln u &< u & \quad (u \geq 1)\\ \underbrace{\Rightarrow}_{u = \sqrt{x}} && \ln \sqrt{x} &< \sqrt{x} \\ \Rightarrow && \frac12 \ln x &< \sqrt{x} \\ \Rightarrow && \frac{\ln x}{x} &< \frac{2\sqrt{x}}{x} \\ &&&= \frac{2}{\sqrt{x}} \\ &&&\to 0 & (x \to \infty) \\ && x \ln x &= \frac{\ln 1/y}{y} \\ &&&= -\frac{\ln y}{y} \\ &&&\to 0 & (y \to \infty, x \to 0) \end{align*}
3. \begin{align*} \int_{0}^{\frac{1}{2}\pi}x\cot x\,\mathrm{d}x &= \left [ x \ln(\sin x) \right]_0^{\pi/2} - \int_0^{\pi/2} \ln (\sin x) \d x \\ &= \left ( \frac{\pi}{2} \ln 1 - \lim_{x \to 0} x \ln (\sin x) \right) - \left ( -\frac12 \pi \ln 2 \right) \\ &= \frac12 \pi \ln 2 \end{align*}

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Solution:

1. \begin{align*} kI_k &= \int_0^\theta \cos^k x k\cos kx \d x \\ &= \left [\cos^k x \sin kx \right]_0^\theta - \int_0^\theta k \cos^{k-1} x \sin x \sin k x \d x \\ &= \cos^k \theta \sin k \theta - k\int_0^\theta \cos^{k-1} x \sin x \sin k x \d x \\ &= \cos^k \theta \sin k \theta + k\int_0^\theta \cos^{k-1} x (\cos (k-1) x - \cos k x \cos x) \d x \\ &= \cos^k\theta \sin k \theta + k I_{k-1} - kI_k \\ \end{align*} So \(2kI_k = kI_{k-1} + \cos^k \sin k \theta\)
2. \begin{align*} && \sum_{r=0}^m \binom{m}{r} \cos 2r \theta &= \textrm{Re} \left ( \sum_{r=0}^m \binom{m}{r} \exp(i 2r \theta)\right) \\ &&&= \textrm{Re} \left ( \sum_{r=0}^m \binom{m}{r} \exp(i 2 \theta)^r\right) \\ &&&= \textrm{Re} \left ( \left (1+e^{2i \theta} \right)^m\right) \\ &&&= \textrm{Re} \left (\left (e^{i\theta}(e^{i \theta} +e^{-i\theta})\right)^m\right) \\ &&&= \textrm{Re} \left (\left (e^{i\theta}2\cos \theta\right)^m\right) \\ &&&= \textrm{Re} \left (2^m \cos^m \theta e^{im\theta}\right) \\ &&&= 2^m \cos^m \theta \cos m\theta \\ \end{align*}
3. \begin{align*} \sum_{r=1}^m \binom{m}{r}\frac{\sin 2r \theta}{2r} &= \int_0^\theta \sum_{r=1}^m \binom{m}{r}\cos 2r x\d x \\ &= \int_0^\theta \left (2^m \cos^m x\cos mx- 1 \right ) \d x\\ &= 2^m I_m - \theta \\ &= \frac{2^{m-1}}{m}\cos^m \theta \sin m \theta + 2^{m-1} I_{m-1} - \theta \\ &= \frac{2^{m-1}}{m}\cos^m \theta \sin m \theta + \frac{2^{m-2}}{m-1}\cos^m \theta \sin m \theta+2^{m-2} I_{m-2} - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta + I_0 - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta + \int_0^\theta 1 \cdot 1 \d \theta - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta \\ &= \sum_{r=0}^{m} \frac{2^{r-1}}{r} \cos^{r} \theta \sin r \theta \end{align*}

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Solution: Consider \(\frac12 x(t)^2\) then differentiating we obtain \(x(t)x'(t) = x(t)y(t)\). Also note that \(x(0) = \int_0^0 y(u) \d u = 0\) Therefore, \begin{align*} \int_0^1 x(t)y(t) \d t &= \left [ \frac12 x(t)^2 \right]_0^1 \\ &= \frac12[x(1)]^2 \end{align*} \begin{align*} && 0 &= y'' + (y^2-1)y' + y \\ \Rightarrow && 0 &= \int_0^1 \l y'' + (y^2-1)y' + y \r \d t \\ &&&= \left [y'(t) + \frac13y^3-y+x \right]_0^1 \\ &&&= x(1) \end{align*} Therefore \(x(1) = 0\). \begin{align*} && 0 &= xy'' + (y^2-1)y' x+ yx \\ \Rightarrow && 0 &= \int_0^1 \l xy'' + (y^2-1)y'x + xy \r \d t \\ &&&= \left [ x y' +(\frac13 y^3-y)x \right]_0^1 - \int_0^1 yy'+\frac13y^4-y^2 \d t \\ &&&= 0 - \frac13 \int_0^1 [y(t)]^4 \d t - \int_0^1 [y(t)]^2 \d t \\ \Rightarrow && \int_0^1 [y(t)]^2 \d t &= \frac13 \int_0^1 [y(t)]^4 \d t \end{align*} \begin{align*} && 0 &= yy'' + (y^2-1)y' y+ y^2 \\ \Rightarrow && 0 &= \int_0^1 \l yy'' + (y^2-1)y'y + y^2 \r \d t \\ &&&= \left [ y y' +(\frac14 y^4-\frac12y^2) \right]_0^1 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\ &&&= 0 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\ \Rightarrow && \int_0^1 [y'(t)]^2 \d t &= \int_0^1 [y(t)]^2 \d t \end{align*}

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Solution: Integrating by parts we obtain: \begin{align*} \int x \ln x \, \d x &= [\frac12 x^2 \ln x] - \int \frac12x^2 \cdot \frac1x \d x \\ &= \frac12 x^2 \ln x - \frac14 x^2 + C \end{align*}

We should have: \begin{align*} \int_0^1 x \ln \frac{1}{x} \d x &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ \left [ -\frac12 x^2 \ln x + \frac14 x^2 \right]_0^1 &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ \frac{1}{4} &=\lim_{n \to \infty} \frac{1}{n^2} \sum_{i=1}^n \l i \ln n - i \ln i \r \\ &= \lim_{n \to \infty} \frac{1}{n^2}\l \frac{n(n+1)}{2} \ln n - \sum_{i=1}^n i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{i=1}^n i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{i=1}^n \sum_{k=1}^i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{k=0}^{n-1} \sum_{i=0}^k \ln (n-i) \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{k=0}^{n-1} \ln \frac{n!}{(n-k)!}\r \\ \end{align*}