Problems

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Solution:

1. \(\,\) \begin{align*} && I_n &= \int_0^1 x^n \arctan x \d x \\ &&&= \left [ \frac{x^{n+1}}{n+1} \arctan x\right]_0^1 - \int_0^1 \frac{x^{n+1}}{n+1} \frac{1}{1+x^2} \d x \\ &&&= \frac{1}{n+1} \frac{\pi}{4} - \frac{1}{n+1} \int_0^1 \frac{x^{n+1}}{1+x^2}\d x \\ \Rightarrow && (n+1)I_n &= \frac{\pi}{4} - \int_0^1 \frac{x^{n+1}}{1+x^2}\d x \\ && I_0 &= \frac{\pi}{4} - \int_0^1 \frac{x}{1+x^2} \d x \\ &&&= \frac{\pi}{4} - \left [\frac12 \ln(1+x^2) \right]_0^1 \\ &&&= \frac{\pi}{4} - \frac12 \ln 2 \end{align*}
2. \(\,\) \begin{align*} && (n+3)I_{n+2} + (n+1)I_n &=\left ( \frac{\pi}{4} - \int_0^1 \frac{x^{n+3}}{1+x^2} \d x \right)+ \left (\frac{\pi}{4} - \int_0^1 \frac{x^{n+1}}{1+x^2} \d x \right) \\ &&&=\frac{\pi}{2}+ \int_0^1 \frac{x^{n+1}+x^{n+3}}{1+x^2} \d x \\ &&&=\frac{\pi}{2}+ \int_0^1 x^{n+1} \d x \\ &&&= \frac{\pi}{2} + \frac{1}{n+2} \\ && 3I_2 + I_0 &= \frac{\pi}{2} + \frac{1}{2} \\ \Rightarrow && 3I_2 &=\frac{\pi}{4} + \frac12 \ln 2 + \frac12 \\ && 5I_4 + 3I_2 &= \frac{\pi}{2} + \frac14 \\ \Rightarrow && 5I_4 &= \frac{\pi}{2} + \frac14 - \left ( \frac{\pi}{4} + \frac12 \ln 2 + \frac12\right) \\ &&&= \frac{\pi}4-\frac12 \ln 2-\frac14 \\ \Rightarrow && I_4 &= \frac15 \left (\frac{\pi}4-\frac12 \ln 2-\frac14 \right) \\ &&&= \frac1{20} \left (\pi - 2\ln 2 -1 \right) \end{align*}
3. Claim: \[ (4n+1) I_{4n} =\frac{\pi}4-\frac12 \ln 2 - \frac12 \sum_{r=1}^{2n} (-1)^r \frac 1 {r} \,, \] Proof: Base case we have just shown above Assume true for \(n = k\), consider \(n = k+1\), then \begin{align*} && (4(k+1)+1) I_{4(k+1)} &= \frac{\pi}{2} + \frac{1}{4(k+1)} - (4k+3)I_{4k+2} \\ &&&= \frac{\pi}{2} + \frac{1}{4(k+1)} - \left (\frac{\pi}{2} + \frac{1}{2(2k+1)} - (4k+1)I_{4k} \right)\\ &&&= (4k+1)I_{4k} - \frac12 \left ( \frac{1}{2k+2} - \frac{1}{2k+1}\right) \\ &&&= \frac{\pi}4-\frac12 \ln 2 - \frac12 \sum_{r=1}^{2k} (-1)^r \frac 1 {r} - \frac12 \left ( \frac{1}{2k+2} - \frac{1}{2k+1}\right)\\ &&&= \frac{\pi}4-\frac12 \ln 2 - \frac12 \sum_{r=1}^{2(k+1)} (-1)^r \frac 1 {r} \\ \end{align*} as required.

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Solution:

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4. Note that \(\frac{x+x^4}{1+x^2+x^4}\) is continuous, and nicely behaved on \((-\infty, \infty)\) so we can see that \(\lim_{x \to \infty} h_1(x) = \pi + \frac{\pi}{4} = \frac{5\pi}{4}\). \(\frac{4x-x^3}{1-x^4}\) on the other hand has asymptotes at \(\pm 1\). So as as \(x \to 1\), \(h_1(x) \to \pi + \frac{\pi}{2} = \frac{3\pi}{2}\). Then as \(x \to \infty\) we increase by another \(\frac{\pi}{2}\), so \(\lim_{x \to \infty} h_2(x) = 2\pi\)

An alternative way to think about the last two parts is to consider \(h\) as giving the (continuous) argument (shifted by \(\pi\)) of \((1-t^2)+it\) (blue), \((1+t^2+t^4)+i(t+t^4)\) (orange) or \((1-t^4)+i(4t-t^3)\) (green). We can see the orange line never wraps around the origin, so the argument is always easy to find. The blue does one full circuit, from \(-\pi\) to \(\pi\) (or \(0\) to \(2\pi\) in our world. And the green line also does a full \(2\pi\) loop.

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Solution:

1. Claim: \(\tan S_n = \frac n {n+1}\) Proof: (By Induction) Base case: (\(n=1\)): \begin{align*} && \tan \left ( \sum_{m=1}^1 \arctan \left ( \frac{1}{2m^2} \right) \right) &= \tan \left ( \arctan \left ( \frac{1}{2} \right) \right) \\ &&&= \frac12 = \frac{1}{1+1} \end{align*} Therefore the base case is true. Inductive step: Suppose our statement is true for some \(n = k\), ie \begin{align*} && \frac{k}{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) \right) \\ \Rightarrow && \tan S_{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) + \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right) \\ &&&= \frac{\tan S_k + \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)}{1-\tan S_k \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)} \\ &&&= \frac{\frac{k}{k+1} + \frac{1}{2(k+1)^2}}{1-\frac{k}{k+1} \frac{1}{2(k+1)^2}} \\ &&&= \frac{2k(k+1)^2+(k+1)}{2(k+1)^3-k} \\ &&&= \frac{k+1}{(k+1)+1} \end{align*} Therefore it is true for \(n=k+1\). Conclusion: Therefore by the principle of mathematical induction since our statement is true for \(n=1\) and if it is true for \(n=k\) it is true for \(n=k+1\) it is true for all \(n\geq1\) Since \(S_n < \frac12 \pi\) for all \(n\), we must have \(\arctan \frac{n}{n+1} = S_n\)
2. \(\tan (2\alpha_n) = \frac{4n^2}{4n^4-1} = \frac{2n^2+2n^2}{(2n^2)(2n^2)-1} = \frac{\frac{1}{2n^2}+\frac{1}{2n^2}}{1-\frac{1}{2n^2}\frac{1}{2n^2}} \Rightarrow \tan (\alpha_n) = \arctan \frac{1}{2n^2}\). In particular \(\displaystyle \sum_{n=1}^{N} \alpha_n = \arctan \frac{n}{n+1} \Rightarrow \sum_{n=1}^{\infty} \alpha_n \to \arctan 1 = \frac{\pi}{4} \)

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Solution:

1. \begin{align*} && I &= \int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1} \d x \\ u = 1/x, \d u = -1/x^2 \d x: &&&= \int_{u=b}^{u=1/b} \frac{1/u \ln(1/u)}{(a^2+u^{-2})(a^2u^{-2}+1)} (- \frac{1}{u^2}) \d u \\ &&&= \int_{1/b}^b \frac{-u\ln u}{(a^2u^2+1)(a^2+u^2)} \d u \\ &&&= -I \\ \Rightarrow && I &= 0 \end{align*}
2. \(\,\) \begin{align*} && I &= \int_{1/b}^b \frac{\arctan x}{x} \d x \\ u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=b}^{u=1/b} \frac{\arctan \frac1u}{\frac1u} \frac{-1}{u^2} \d u \\ &&&= \int_{1/b}^b \frac{\arctan \frac1u}{u} \d u \\ \Rightarrow && 2I &= \int_{1/b}^b \frac{\arctan x + \arctan \frac1x}{x} \d x \\ &&&= \int_{1/b}^b \frac{\frac{\pi}2}{x} \d x \\ &&&= \pi \ln b \\ \Rightarrow && I &= \frac{\pi}{2} \ln b \end{align*}
3. \(\,\) \begin{align*} && I_a &= \int_0^{\infty} \frac{1}{(a^2+x^2)^2} \d x \\ u = a/x, \d x = -a/u^2 \d u:&&&= \int_{u=0}^{u=\infty} \frac{1}{\left (a^2+\frac{a^2}{u^2} \right)^2} \frac{a}{u^2} \d u \\ &&&= \frac1{a^3}\int_0^{\infty} \frac{1}{(u+1/u)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{u^2}{(u^2+1)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{u^2+1-1}{(u^2+1)^2} \d u \\ &&&= \frac{1}{a^3} \int_0^{\infty} \frac{1}{(u^2+1)} - \frac{1}{(u^2+1)^2} \d u \\ &&&= \frac1{a^3} \frac{\pi}{2} - \frac{1}{a^3} I_1 \\ \Rightarrow && 2I_1 &= \frac{\pi}{2} \\ \Rightarrow && I_1 &= \frac{\pi}{4} \\ \Rightarrow && I_a &= \frac{\pi}{4a^3} \end{align*}

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Solution: Let \(t = \tan \frac12 x\), then \begin{align*} \frac{\d t}{\d x} &= \tfrac12 \sec^2 \tfrac12 t \\ &= \tfrac12 (1 + \tan^2 \tfrac12 ) \\ &= \tfrac12 (1 + t^2) \\ \\ \sin x &= 2 \sin \tfrac12 x \cos \tfrac12 \\ &= \frac{2 \frac{\sin \tfrac12 x}{ \cos \tfrac12x}}{\frac{1}{\cos^2 \tfrac12 x}} \\ &= \frac{2 \tan \tfrac12 x}{\sec^2 \tfrac12 } \\ &= \frac{2t }{1+t^2} \end{align*} Now consider \begin{align*} t = \tan \tfrac12 x: && \int_0^{\frac12\pi} \frac 1{1+a \sin x}\, \d x &= \int_{t=0}^{t = 1} \frac{1}{1 + a \frac{2t}{1+t^2}} \frac{2}{1+t^2} \d t \\ &&&= \int_0^1 \frac{2}{1+2at+t^2} \d t \\ &&&= \int_0^1 \frac{2}{(t+a)^2 + 1-a^2} \d t \\ (1-a^2) > 0: &&&= \left [ \frac{2}{\sqrt{1-a^2}} \arctan \frac{t+a}{\sqrt{1-a^2}} \right]_0^1 \\ &&&= \frac{2}{\sqrt{1-a^2}} \left ( \arctan \frac{1+a}{\sqrt{1-a^2}} - \arctan \frac{a}{\sqrt{1-a^2}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\frac{1+a}{\sqrt{1-a^2}}-\frac{a}{\sqrt{1-a^2}}}{1+\frac{1+a}{\sqrt{1-a^2}}\frac{a}{\sqrt{1-a^2}}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\frac{1}{\sqrt{1-a^2}}}{\frac{1+a}{1-a^2}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\sqrt{1-a}}{\sqrt{1+a}} \right) \end{align*} as required. Let \[ I_n = \int_0^{\frac12\pi} \frac{ \sin ^nx}{2+\sin x} \, \d x \qquad \quad (n\ge0). \] and consider \begin{align*} I_{n+1} + 2I_n &= \int_0^{\frac12\pi} \frac{ \sin ^{n+1}x+2\sin^{n} x}{2+\sin x} \, \d x \\ &= \int_0^{\frac12\pi} \frac{ \sin^n x (2 + \sin x)}{2+\sin x} \, \d x \\ &= \int_0^{\frac12\pi} \sin^n x \d x \end{align*} Therefore we can compute \begin{align*} I_0 &= \int_0^{\pi/2} \frac{1}{2 + \sin x} \d x \\ &= \frac12 \int_0^{\pi/2} \frac{1}{1 + \frac12 \sin x} \d x \\ &= \frac{1}{\sqrt{3/4}} \arctan \frac{\sqrt{1/2}}{\sqrt{3/2}} \\ &= \frac{2}{\sqrt{3}} \arctan \frac{1}{\sqrt{3}} \\ &= \frac{\pi}{3\sqrt{3}} \\ \\ I_1 &= \int_0^{\pi/2} 1 \d x - 2 I_0 \\ &= \frac{\pi}{2} - \frac{2\pi}{3\sqrt{3}} \\ I_2 &= \int_0^{\pi/2} \sin x \d x - 2I_1 \\ &= 1 - \pi + \frac{4\pi}{3\sqrt{3}} \\ I_3 &= \int_0^{\pi/2} \sin^2 x \d x - 2I_2 \\ &= \frac12 \int_0^{\pi/2} \sin^2 + \cos^2 x \d x - 2I_2 \\ &= \frac{\pi}{4} - 2 + 2\pi - \frac{8\pi}{3\sqrt{3}} \\ &= -2 + \frac{9\pi}{4} - \frac{8\pi}{3\sqrt{3}} \end{align*}

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Solution:

1. Given \(\frac{\d}{\d x} (\arctan x) \leq 1\) we must have \(\frac{\d}{ \d x} (x-\arctan x) \geq 0\) for \(x \geq 0\), but since \( 0 - \arctan 0 = 0\) this means that \(x - \arctan x \geq 0\), ie \( \arctan x \geq x\) for \(x \geq 0\) \(g(x) = x^{-1} \tan x \Rightarrow g'(x) = -x^{-2}\tan x +x^{-1} \sec^2 x\). If we can show \(f(x) = x \sec ^2 x - \tan x\) is positive that would be great. However \(f'(x) = x 2 \tan x \sec^2 x \geq 0\) and \(f(0) = 0\) so \(f(x)\) is positive and \(g'(x)\) is positive and hence increasing, therefore \(g(x) \geq g(\arctan x) \Rightarrow \frac{\tan x}{x} \geq \frac{x}{\arctan x}\) from which the result follows.

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Solution:

1. \begin{align*} && \tan (A+B) &= \frac{\tan A + \tan B}{1-\tan A \tan B}\\ &&&= \frac{\tan \arctan \frac12 + \tan \arctan \frac13}{1-\tan \arctan \frac12 \tan \arctan \frac13}\\ &&&= \frac{\frac12+\frac13}{1-\frac16} \\ &&&= \frac{3+2}{5} \\ &&&= 1 \\ \Rightarrow && A+B &= \frac{\pi}{4} + n \pi \end{align*} but since \(A,B\) are acute \(0 < A+B < \pi\), so \(A+B = \frac{\pi}{4}\) \begin{align*} && 1 &= \tan \frac{\pi}{4} \\ &&&= \tan \left ( \arctan {\frac1 p} + \arctan {\frac1 q}\right) \\ &&&= \frac{\frac1p + \frac1q}{1-\frac1{pq}} \\ &&&= \frac{q+p}{pq-1} \\ \Rightarrow && pq-1 &= q+p \\ \Rightarrow && 0 &= pq-q-p-q \\ &&&= (p-1)(q-1)-2 \\ \Rightarrow && 2 &= (p-1)(q-1) \end{align*} But \(p\),\(q\) are integers, so \(p-1 \in \{-2,-1,1,1\} \Rightarrow p \in \{-1,0,2,3\}\) but we cannot have \(p= 0\), so we must have \((p,q) = (2,3), (3,2)\)
2. \begin{align*} && 1 &= \tan \frac{\pi}{4} \\ &&&= \tan \left ( \arctan {\frac1 r} + \arctan \frac s {s+t} \right) \\ &&&= \frac{\frac1r + \frac{s}{s+t}}{1-\frac{s}{r(s+t)}} \\ &&&= \frac{s+t+sr}{r(s+t)-s} \\ \Rightarrow && rs+rt-s &= s+t + sr \\ \Rightarrow && 0 &= rt-2s-t \\ &&2s&= t(r-1) \end{align*} Since \((s,t) =1\), we must have \(t \mid 2\), so \( t = 1,2\) and \(r = 2s+1\) or \(r=s+1\) respectively.

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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: \begin{align*} && \tan \left (\arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right) \right) &= \frac{\frac1{a+b}+\frac1{a+c}}{1-\frac{1}{(a+b)(a+c)}} \\ &&&= \frac{a+c+a+b}{(a+b)(a+c)-1} \\ &&&= \frac{2a+b+c}{a^2+ab+ac+bc-1} \\ &&&= \frac{2a+b+c}{2a^2+ab+ac} \\ &&&= \frac{1}{a} \\ &&&= \tan \arctan \frac1a\\ \Rightarrow && \arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right) &= \arctan \frac{1}{a} + n \pi \end{align*} Since \(\arctan x \in (-\frac{\pi}{2}, \frac{\pi}{2})\) the LHS \(\in (0, \pi)\) so \(n = 0\). \begin{align*} a=p+q, b = s, c = t:&& \arctan \! \!\left(\!\frac1 {p+q+s}\!\right) + \arctan \! \!\left(\!\frac 1{p+q+t}\!\right) &= \arctan \left ( \frac{1}{p+q} \right) \\ a=p+r, b= u, c = v && \arctan \! \!\left(\!\frac 1 {p+r+u}\!\right) + \arctan \! \!\left(\!\frac1 {p+r+v}\!\right) &= \arctan \! \!\left(\!\frac1 {p+r}\!\right) \\ a = p, b = q, c = r:&& \arctan \left ( \frac{1}{p+q} \right) +\arctan \! \!\left(\!\frac1 {p+r}\!\right) &= \arctan \left ( \frac1p \right) \end{align*} and the result follows. Taking \(p = 7\) we need to solve \[ \begin{cases} q+s &= 6 \\ q+t &= 14 \\ r+u &= 75 \\ r+v &= 180 \end{cases} \] also satisfying \(qr = 50\) etc, so say \(q = 1, r = 50, s = 5, v=25\)

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Solution: \begin{align*} && v^2 &= u^2 + 2as \\ (\uparrow): && 0 &= (u \sin \theta)^2 - 2gh \\ \Rightarrow && h &= \frac{u^2 \sin^2 \theta}{2g} \end{align*} To avoid hitting the ceiling \begin{align*} && \frac9{10}d &>\frac{u^2 \sin^2 \theta}{2g} \\ \Rightarrow && \frac{gd}{u^2}&>\frac{5\sin^2 \theta}{9} \\ \end{align*} In order to pass through \(P\) we need \begin{align*} && d &= u \cos \theta t \\ && \frac12 d &= u \sin \theta t - \frac12 g t^2 \\ \Rightarrow && \frac12 &= \tan \theta - \frac12 \frac{g d}{u^2} \sec^2 \theta \\ &&&<\tan \theta - \frac12 \frac{5 \sin^2 \theta}{9\cos^2 \theta} \\ \Rightarrow && 0 & >5 \tan^2 \theta -18\tan \theta +9 \\ &&&= (5\tan \theta - 3)(\tan \theta - 3) \\ \\ \Rightarrow && \tan \theta &\in \left ( \frac35, 3\right) \\ && \theta &= \left (\arctan \tfrac35, \arctan 3 \right) \end{align*}

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Solution: If \(\theta = \frac{\pi}{3}\) then \(\cos \theta = \frac12, \sin \theta = \frac{\sqrt{3}}{2}\) and clearly the equation is satisfied. We can also solve this equation using the harmonic formulae, namely: \begin{align*} && 5 &= 4 \cos \theta + 2\sqrt{3} \sin \theta \\ &&&= \sqrt{4^2+2^2 \cdot 3} \cos \left (\theta -\tan^{-1} \left (\frac{2\sqrt{3}}{4}\right) \right) \\ \Rightarrow && \frac{5}{\sqrt{28}} &= \cos \left ( \frac{\pi}{3} - \tan^{-1} \left (\frac{\sqrt{3}}{2}\right) \right) \\ \Rightarrow && \frac{\pi}{3} &= \arccos\left( \frac{5}{\sqrt{28}}\right) + \arctan \left (\frac{\sqrt{3}}{2}\right) \end{align*} From which the result follows. Similarly, notice that \(3 \cos \theta + 4 \sin \theta = \frac{7}{\sqrt{2}}\) is clearly solved by \(\frac{\pi}{4}\), but also writing it in harmonic form, we have \begin{align*} &&\frac{7}{\sqrt{2}} &= 5 \sin \left (\theta + \tan^{-1} \left ( \frac{3}{4} \right) \right) \\ \Rightarrow && \frac{7\sqrt{2}}{10} &= \sin \left ( \frac{\pi}{4} + \tan^{-1} \left ( \frac{3}{4} \right) \right) \\ \Rightarrow && \frac{\pi}{4} &= \arcsin \left ( \frac{7\sqrt{2}}{10} \right) - \arctan \left ( \frac{3}{4} \right) \end{align*} as required.

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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*}