Problems

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. \(r = a \sec \theta \pm b\). The points on \(r = a \sec \theta \Leftrightarrow r \cos \theta = a \Leftrightarrow x = a\) are points on the line \(x = a\). Therefore points on the curve \(r = a \sec \theta \pm b\) are points which are a distance \(b\) from the line \(x = a\) measured towards \(O\). So \(A\) is the origin and \(d = b\).
   

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2. 

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   The loop starts and ends when \(r = a \sec \theta - b = 0 \Rightarrow \cos \theta = \frac{a}{b}\), so when \(a = 1, b = 2\), this is \(-\frac{\pi}{3}\) to \(\frac{\pi}{3}\) \begin{align*} && A &= \frac12 \int r^2 \d \theta \\ &&&= \frac12 \int_{-\pi/3}^{\pi/3} \left ( \sec \theta - 2 \right)^2 \d \theta \\ &&&= \frac12 \int_{-\pi/3}^{\pi/3} \left (\sec^2 \theta - 4 \sec \theta + 4\right)\d \theta \\ &&&= \frac12 \left [ \tan \theta -4 \ln | \sec \theta + \tan \theta| + 4 \theta \right]_{-\pi/3}^{\pi/3} \\ &&&= \frac12 \left (\left (\tan \frac{\pi}3 - 4 \ln | \sec \frac{\pi}3 + \tan \frac{\pi}3 | + 4\left ( \frac{\pi}3 \right)\right) - \left (\tan \left (-\frac{\pi}3 \right) - 4 \ln | \sec \left (-\frac{\pi}3 \right)+ \tan\left ( -\frac{\pi}3 \right) | + 4\left ( -\frac{\pi}3 \right)\right) \right) \\ &&&= \frac12 \left ( 2\sqrt{3} - 4 \ln |2 + \sqrt{3}| + 4 \ln |2-\sqrt{3}| + \frac{8\pi}3 \right) \\ &&&= \sqrt{3} + 2\ln \frac{2-\sqrt{3}}{2+\sqrt{3}} + \frac{4\pi}3 \\ &&&= \sqrt{3} + 4 \ln (2 - \sqrt{3})+ \frac{4\pi}3 \end{align*}

Solution:

1. \(\,\) \begin{align*} && \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] - \int -\frac{x-2}{2-x} \d x \\ && \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] + \int 1 \d x \\ &&&= -(2-x) \ln (2-x) +(2-x) + C \end{align*}
2. \(\,\)
   

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3. \begin{align*} && \text{Area} &= \int_0^{\sqrt{3}} \ln | x^2 - 4 | \d x \\ &&&= \int_0^\sqrt{3} \ln(4-x^2) \d x \\ &&&= \int_0^\sqrt{3} \left ( \ln(2-x) + \ln (2+x) \right) \d x \\ &&&= \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_0^{\sqrt{3}} \\ &&&= \left ( -(2-\sqrt{3}) \ln (2-\sqrt{3}) +(2-\sqrt{3}) +(2+\sqrt{3})\ln(2+\sqrt{3})-(2+\sqrt{3}) \right) - \\ &&&\quad \quad \left (- 2\ln (2)+2 +2\ln(2)-2 \right) \\ &&&=\left ( -(2-\sqrt{3}) \ln \left ( \frac{1}{2+\sqrt{3}} \right) -2\sqrt{3} +(2+\sqrt{3})\ln(2+\sqrt{3}) \right) \\ &&&= 4\ln(2 + \sqrt{3}) - 2 \sqrt{3} \end{align*}
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   \begin{align*} && \text{Area} &= 2 \left ( \int_0^\sqrt{3} \ln (4-x^2) \d x - \lim_{t \to 2}\int_{\sqrt{3}}^t \ln(4-x^2) \d x \right) \\ &&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2}\int_{\sqrt{3}}^t \left ( \ln (2-x) + \ln (2+x) \right) \d x \\ &&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2} \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_{\sqrt{3}}^{t} \\ &&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} - 2 \lim_{t \to 2} \left(-(2-t) \ln (2-t) +(2-t) +(2+t)\ln(2+t)-(2+x) \right) \\ &&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} -2(4 \ln4-4) \\ &&&= 16 \ln(2 + \sqrt{3}) - 16 \ln 2 +8(1-\sqrt{3}) \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*}

Solution: Since \(\int f(x) \, dx = 1\), and \(f(x)\) is a triangle with base \(b-a\), it must have height \(\frac{2}{b-a}\) in order to have the desired area. Since \(g(a) = 0, g(c) = \frac{2}{b-a}\), \(g(x) = A(x-a)\) and \(\frac{2}{b-a} = A (c-a) \Rightarrow g(x) = \frac{2(x-a)}{(b-a)(c-a)}\) as required. Similarly, \(h(x) = B(x-b)\) and \(\frac{2}{b-a} = B(c-b) \Rightarrow h(x) = \frac{2(b-x)}{(b-a)(b-c)}\) The mean of the distribution will be: \begin{align*} \int_a^b xf(x) \, dx &= \int_a^c xg(x) \, dx + \int_c^b xh(x) \, dx \\ &= \frac{2}{(b-a)(c-a)} \int_a^c x(x-a) dx + \frac{2}{(b-a)(b-c)} \int_c^b x(b-x) \, dx \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \left [ \frac{x^3}{3} - a\frac{x^2}{2} \right ]_a^c + \frac{1}{b-c} \left [ b\frac{x^2}{2} - \frac{x^3}{3} \right ]_c^b\r \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \l \frac{c^3}{3} - a\frac{c^2}{2} - \frac{a^3}{3} + \frac{a^3}{2} \r + \frac{1}{b-c} \l \frac{b^3}{2} - \frac{b^3}{3} - \frac{bc^2}{2} + \frac{c^3}{3} \r \r \\ &= \frac{2}{(b-a)} \l \l \frac{c^2+ac+a^2}{3} - \frac{a(a+c)}{2} \r +\l \frac{b(b+c)}{2} - \frac{b^2+bc+c^2}{3} \r\r \\ &= \frac{2}{(b-a)} \l \frac{2c^2+2ac+2a^2}{6} - \frac{3a^2+3ac}{6} + \frac{3b^2+3bc}{6} - \frac{2b^2+2bc+2c^2}{6} \r \\ &= \frac{2}{(b-a)} \l \frac{-a^2+b^2-ac+bc}{6} \r \\ &= \frac{a+b+c}{3} \\ \end{align*} The median \(M\) satisfies: \begin{align*} && \int_a^M f(x) \, dx &= \frac12 \\ \end{align*} The left hand triangle will have area: \(\frac{c-a}{b-a}\) which will be \(\geq \frac12\) if \(c \geq \frac{a+b}{2}\). In this case we need \begin{align*} && \frac{(M-a)^2}{(b-a)(c-a)} &= \frac12 \\ \Rightarrow && M &= a + \sqrt{\frac12 (b-a)(c-a)} \end{align*} Otherwise, we need: \begin{align*} && \frac{(b-M)^2}{(b-a)(b-c)} &= \frac12 \\ \Rightarrow && M &= b - \sqrt{\frac12 (b-a)(b-c)} \end{align*} These are consistent, if \(c = \frac{b+a}{2}\)

Solution:

1. If \(f(x) = \sin nx\) then \(f'(x) = n \cos n x\) and so \begin{align*} && LHS &= \int_0^\pi \sin^2 n x \d x \\ &&&= \left [ \frac{x+\frac1{2n}\sin 2n x}{2} \right ]_0^{\pi} \\ &&&= \frac{\pi}{2} \\ \\ && RHS &= \int_0^{\pi} n^2 \cos^2 n x \d x \\ &&&= n^2 \left [ \frac{\frac{1}{2n}\sin 2n x + x}{2} \right]_0^{\pi} \\ &&&= n^2\frac{\pi}{2} \geq LHS \end{align*} [\(f(0) = 0, f(\pi) \neq 0\)] Suppose \(f(x) = x\) then \(f'(x) = 1\) and \(LHS = \frac{\pi^3}{3} > \pi = RHS\). [\(f(0) \neq 0, f(\pi) = 0\)] Suppose \(f(x) = \pi - x\) then \(f'(x) = -1\) and \(LHS = \frac{\pi^3}{3} > \pi = RHS\)
2. Suppose \(f(x) = x(\pi - x)\) then \(f'(x) = \pi - 2x\) and so \begin{align*} && \int_0^\pi x^2(\pi-x)^2 \d x &\leq \int_0^\pi (\pi-2x)^2 \d x \\ \Leftrightarrow && \left [\pi^2 \frac{x^3}{3} - 2\pi \frac{x^4}{4} + \frac{x^5}{5} \right]_0^{\pi} &\leq \left [ \pi^2x - 4\pi \frac{x^2}{2} + \frac{4x^3}{3} \right]_0^{\pi} \\ \Leftrightarrow && \pi^5 \left (\frac13 - \frac12+\frac15 \right) &\leq \pi^3 \left ( 1 - 2+\frac43 \right) \\ \Leftrightarrow && \pi^2 \frac{1}{30} &\leq \frac13 \\ \Leftrightarrow && \pi^2 &\leq 10 \end{align*} Suppose \(f(x) = p\sin \tfrac12 x + q \cos \tfrac12 x + r\), so \(f(0) = q + r\) and \(f(\pi) = p + r\), so say \(p = q = 1, r = -1\) \begin{align*} && LHS &= \int_0^{\pi} \left ( \sin \tfrac12 x + \cos \tfrac12 x-1\right)^2 \d x \\ &&&=\int_0^\pi \left ( \sin^2 \tfrac12 x + \cos^2 \tfrac12 x+1-2\sin \tfrac12 x - 2\cos \tfrac12 x+ \sin x \right)\\ &&&= \left [2x + 4\cos \tfrac12 x - 4\sin \tfrac12 x - \cos x \right]_0^{\pi} \\ &&&= \left ( 2\pi -4+1 \right) - \left ( 4-1 \right) \\ &&&= 2\pi -6\\ \\ && RHS&= \int_0^{\pi} \left ( \tfrac12 \cos \tfrac12 x -\tfrac12 \sin \tfrac12 x\right)^2 \d x \\ &&&= \int_0^{\pi} \left ( \tfrac14 \cos^2 \tfrac12 x +\tfrac14 \sin^2 \tfrac12 x-\tfrac14 \sin x\right) \d x \\ &&&= \frac{\pi}{4} - \frac12 \\ \Rightarrow && 2\pi -6 &\leq \frac{\pi}{4} - \frac12 \\ \Rightarrow && \frac{7\pi}{4} &\leq \frac{11}{2} \\ \Rightarrow && \pi &\leq \frac{22}{7} \end{align*} \(22^2/7^2 = 484/49 < 10\) therefore \(\pi \leq \frac{22}{7}\) is the better estimate.

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*}

Solution:

1. \begin{align*} && \int \frac{\sinh x}{\cosh 2x} \d x &= \int \frac{\sinh x}{2\cosh^2 x -1} \d x \\ u = \cosh x, \d u = \sinh x \d x &&&= \int \frac{1}{2u^2 -1} \d u \\ &&&= \int\frac12 \left ( \frac{1}{\sqrt{2}u-1}-\frac{1}{\sqrt{2}u+1} \right) \d u \\ &&&= \frac1{2\sqrt{2}} \left (\ln (\sqrt{2}u-1) - \ln(\sqrt{2}u+1) \right) + C \\ &&&= \frac{1}{2\sqrt{2}} \ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) + C \end{align*}
2. \begin{align*} && \int \frac{\cosh x}{\cosh 2x} \d x &= \int \frac{\cosh x}{1+2\sinh^2 x} \d x \\ u = \sinh x && &= \int \frac{1}{1+2u^2} \d u \\ &&&=\frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2}u) + C \\ &&&= \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) + C \end{align*}
3. \begin{align*} u = e^x : && \int_0^1 \frac{1}{1+u^4} \d u &= \int_{x=-\infty}^{x=0} \frac{1}{1+e^{4x}}e^{x} \d x \\ &&&= \int_{-\infty}^{0} \frac{e^{-x}}{e^{2x}+e^{-2x}} \d x \\ &&&= \int_{-\infty}^{0} \frac{\cosh x - \sinh x}{2\cosh 2x } \d x \\ &&&= \frac12 \int_{-\infty}^{0} \frac{\cosh x}{\cosh 2x} \d x - \frac12 \int_{-\infty}^{0} \frac{\sinh x}{\cosh 2x} \\ &&&= \frac12 \left [\frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) \right]_{-\infty}^{0}-\frac12 \left [ \frac{1}{2\sqrt{2}}\ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) \right]_{-\infty}^{0} \\ &&&= 0 - \frac1{2\sqrt{2}} \frac{-\pi}{2}-\left (\frac1{4\sqrt{2}} \ln \left (\frac{\sqrt{2}-1}{\sqrt{2}+1} \right) - 0 \right) \\ &&&= \frac{\pi - \ln((\sqrt{2}-1)^2)}{4\sqrt{2}} \\ &&&= \frac{\pi + 2 \ln(1+\sqrt{2})}{4\sqrt{2}} \end{align*}

Solution:

1. \begin{align*} && I - I_1 &= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) \d x - \int_0^1 \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) - \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left (-y^2-2yy' \tan x - y^2 \tan^2 x \right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2(1+ \tan^2 x )\right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2 \sec^2 x\right) \d x\\ &&&= \int_0^1 -\frac{\d}{\d x} \left (y^2 \tan x \right) \d x\\ &&&= \left [-y^2 \tan x \right]_0^1 \\ &&&= 0 \\ \\ \Rightarrow && I &= I_1 = \int_0^1 \left ( y' + y \tan x \right)^2 d x \geq 0 \end{align*} The only way \(I_0 = 0\) is is \(y' + y \tan x =0\), so \begin{align*} && \frac{\d y}{\d x} &= - y \tan x \\ \Rightarrow && \int \frac{1}{y} &= \int -\tan x \d x \\ \Rightarrow && \ln |y| &= \ln |\cos x| + C \\ \Rightarrow && y &= A \cos x \\ \Rightarrow && A &= 0 \Rightarrow y = 0 \end{align*}
2. Let \(J_1 = \int_0^1 (y'+ay\tan ax)^2 \d x\), then \begin{align*} && J-J_1 &= \int_0^1 \left ( \left ( y' \right)^2 - a^2y^2 \right) - \left ( y' + ya \tan ax \right)^2 \d x\\ &&&= \int_0^1 \left (-a^2y^2-2yy' a \tan a x-y^2a^2 \tan^2 ax \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2(1+\tan^2 ax) \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2\sec^2 ax \right) \d x \\ &&&= \left [ - a y^2 \tan a x \right]_0^1 = 0 \end{align*} This is true if \(a < \frac{\pi}{2}\), since otherwise we might care about the order of the zero for \(y\) at \(x = 1\). Consider \(y = \cos \frac{\pi}{2} x\), then \(y' = -\frac{\pi}{2} \sin^2\frac{\pi}{2} x\) and \begin{align*} && \int_0^1 \frac{\pi^2}{4} \left (\sin^2 \frac{\pi}{2}x - \cos^2 \frac{\pi}{2} x \right) \d x &= -\frac{\pi^2}{4} \int_0^1 \cos(\pi x) \d x \\ &&&= 0 \end{align*}

Solution:

1. \begin{align*} u = \tan x, \d u = \sec^2 x \d x: &&\int_0^{\pi/4} \tan^n x \sec^2 x \d x &= \int_0^1 u^n \d u \\ &&&= \frac{1}{n+1} \end{align*} \begin{align*} u = \sec x, \d u = \sec x \tan x \d x: &&\int_0^{\pi/4} \sec^n x \tan x \d x &= \int_{u=1}^{u=\sqrt{2}} u^{n-1} \d u \\ &&&= \left [ \frac{u^n}{n}\right] \\ &&&= \frac{(\sqrt{2})^n - 1}n \end{align*}
2. \begin{align*} &&\int_0^{\frac14\pi} x \sec ^4 x \tan x \d x &= \left [x \frac{1}{4} \sec^4 x \right]_0^{\frac14\pi} - \frac14 \int_0^{\frac14\pi} \sec^4 x \d x \\ &&&= \frac{\pi}{4} - \frac14 \int_0^{\frac14\pi} \sec^2 x(1+ \tan^2 x) \d x \\ &&&= \frac{\pi}{4} - \frac14 \left [ \tan x+ \frac13 \tan^3 x \right] _0^{\frac14\pi} \\ &&&= \frac{\pi}{4} - \frac{1}{3} \end{align*} \begin{align*} \int_0^{\frac14\pi} \!\! x^2 \sec ^2 \! x\, \tan x \, \d x &= \left [x^2 \frac12 \tan^2 x \right]_0^{\frac14\pi} - \int_0^{\frac14\pi} x \tan^2 x \d x\\ &= \frac{\pi^2}{32} - \int_0^{\frac14\pi} x (\sec^2 x - 1) \d x\\ &= \frac{\pi^2}{32} - \left [x (-x + \tan x) \right]_0^{\frac14\pi} + \int_0^{\frac14\pi}-x + \tan x \d x \\ &= \frac{\pi^2}{32} - \frac{\pi}{4} (-\frac{\pi}{4} + 1) -\frac{\pi^2}{32} + \left [ -\ln \cos x \right]_0^{\pi/4} \\ &= \frac{\pi^2}{16}- \frac{\pi}{4} + \frac12 \ln 2 \end{align*}

Solution:

1. Let \(y = ux\), then \(\frac{\d y}{\d x} = x\frac{\d u}{\d x} = u\) and the differential equation becomes, \begin{align*} && xu' + u &= \frac{1}{u} +u \\ \Rightarrow && u' &= \frac{1}{ux} \\ \Rightarrow && u u' &= \frac1{x} \\ \Rightarrow && \frac12 u^2 &= \ln x + C \\ (x,y) = (1,2): && \frac12 4 &= C \\ \Rightarrow && \frac12 \frac{y^2}{x^2} &= \ln x + 2 \\ \Rightarrow && y^2 &= x^2 \l 2\ln x + 4 \r \\ \Rightarrow && y &= x \sqrt{4 + 2 \ln x} \quad (x > e^{-2}) \end{align*}
2. Let \(y = ux^2\) then \begin{align*} && \frac{\d y}{\d x} &= \frac{x^2}{y} + \frac{2y}{x} \\ \Rightarrow && u' x^2 + 2x u &= \frac{1}{u} + 2x u \\ \Rightarrow && u' u &= \frac{1}{x^2} \\ \Rightarrow && \frac12 u^2 &= -\frac{1}{x} + C \\ (x,y) = (1,2): && 2 &= C - 1 \\ \Rightarrow && \frac12 \frac{y^2}{x^4} &= 3 - \frac{1}{x} \\ \Rightarrow && y &= x\sqrt{2(3x^2-x)}, \quad (x > \frac13) \end{align*}

Solution:

1. \(\,\) \begin{align*} u = 1-x, \d u = -\d x && \int_0^1 x^{n-1}(1-x)^n \d x &= \int_{u=1}^{u=0} (1-u)^{n-1}u^n (-1) \d u \\ &&&= \int_0^1 u^n (1-u)^{n-1} \d u \\ &&&= \int_0^1 x^n (1-x)^{n-1} \d x \\ \\ \Rightarrow && 2\int_0^1 x^{n-1}(1-x)^n \d x &= \int_0^1 \left ( x^{n-1}(1-x)^n + x^{n}(1-x)^{n-1} \right)\d x \\ &&&= \int_0^1x^{n-1}(1-x)^{n-1} \left ( (1-x) + x \right) \d x\\ &&&= I_{n-1} \\ \\ && I_n &= \left [x^n \cdot (-1)\frac{(1-x)^{n+1}}{n+1}\right]_0^1 + \int_0^1 n x^{n-1} \frac{(1-x)^{n+1}}{n+1} \d x\\ &&&= \frac{n}{n+1} \int_0^1 x^{n-1} (1-x)^{n+1} \d x \\ \\ && I_n &= \frac{n}{n+1} \int_0^1 x^{n-1} (1-x)^{n+1} \d x \\ &&&= \frac{n}{n+1} \int_0^1 \left ( x^{n-1} (1-x)^{n} - x^n(1-x)^n \right) \d x \\ &&&= \frac{n}{n+1} \left (\frac12 I_{n-1} - I_n \right) \\ \Rightarrow && I_n \cdot \left ( \frac{2n+1}{n+1} \right) &= \frac{n}{2(n+1)} I_{n-1}\\ \Rightarrow && I_n &= \frac{n}{2(2n+1)} I_{n-1} \end{align*}
2. \(\,\) \begin{align*} && I_0 &= \int_0^1 1 \d x = 1 \\ \Rightarrow && I_1 &= \frac{1}{2 \cdot 3} \\ && I_n &= \frac{n}{2(2n+1)} \cdot \frac{n-1}{2(2n-1)}\cdot \frac{n-2}{2(2n-3)} \cdots \frac{1}{2 \cdot 3} \\ &&&= \frac{n!}{2^n (2n+1)(2n-1)(2n-3) \cdots 3} \\ &&&= \frac{n! (2n)(2n-2)\cdots 2}{2^n (2n+1)!} \\ &&&= \frac{(n!)^2 2^n}{2^n(2n+1)!} \\ &&&= \frac{(n!)^2}{(2n+1)^2} \end{align*}
3. \(\,\) \begin{align*} && I_{\frac12} &= \int_0^1 \sqrt{x(1-x)} \d x\\ x = \sin^2 \theta, \d x = 2 \sin \theta \cos \theta \d \theta: &&&= \int_{\theta =0}^{\theta = \frac{\pi}{2}} \sin \theta \cos \theta 2 \sin \theta \cos \theta \d \theta \\ &&&= \frac12 \int_0^{\pi/2} \sin^2 2 \theta \d \theta \\ &&&= \frac12 \int_0^{\pi/2} \frac{1-\cos 2 \theta}{2} \d \theta \\ &&&= \frac14 \left [\theta - \frac12 \sin 2 \theta \right]_0^{\pi/2} \\ &&&= \frac{\pi}{8} \\ \\ && I_{\frac32} &= \frac{3/2}{2 \cdot ( 2 \cdot \frac32 + 1)} I_{\frac12} \\ &&&= \frac{3}{4 \cdot 4} \frac{\pi}{8} \\ &&&= \frac{3 \pi}{128} \end{align*}