Problems

Solution: \begin{align*} && t &= x+ \sqrt{x^2+2bx+c} \\ && \frac{\d t}{\d x} &= 1 + \frac{x+b}{\sqrt{x^2+2bx+c}} \\ &&&= \frac{x + \sqrt{x^2+2bx+c} + b}{\sqrt{x^2+2bx+c}} \\ &&&= \frac{t+b}{t-x} \\ \Rightarrow && \frac{\d x}{\d t} &= \frac{t-x}{t+b} \\ \\ && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{t-x} \frac{t-x}{t+b} \d t \\ &&&= \int \frac{1}{t+b} \d t \\ &&&= \ln (t + b) +C \\ &&&= \ln \left (x + \sqrt{x^2+2bx+c} + b \right) + C \end{align*} If \(b^2 = c\) then we have \(x^2+2bx+b^2 = (x+b)^2\) so \(\sqrt{x^2+2bx+c^2} = x+b\) (if \(x+b>0\)), so \begin{align*} && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{x+b} \d x\\ &&&= \ln (x + b) + C \\ &&&= \ln(x+b) + \ln 2 + C' \\ &&&= \ln (2(x+b)) + C' \\ &&&= \ln \left(x + b + \sqrt{(x+b)^2} \right)+C'\\ &&&= \ln \left(x + b + \sqrt{x^2+2bx+c} \right)+C'\\ \end{align*} If \(x+b < 0\) then the antiderivative is \(\ln 0\). \begin{align*} && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= -\int \frac{1}{x+b} \d x\\ &&&= -\ln |x + b| + C \\ \end{align*} which are clearly different.

Solution: \begin{align*} && I &= \int_0^{\frac{1}{2}\pi} \frac{\sin 2x}{1+\sin^2x} \d x \\ &&&= \int_0^{\frac{1}{2}\pi} \frac{2 \sin x \cos x}{1+\sin^2x} \d x \\ &&&= \left [\ln (1 + \sin^2 x) \right]_0^{\pi/2} \\ &&&= \ln 2 \\ \\ && J &= \int_0^{\frac{1}{2}\pi} \frac{\sin x}{1+\sin^2x} \d x \\ &&&= \int_0^{\frac{1}{2}\pi} \frac{\sin x}{2-\cos^2x} \d x \\ &&&= \frac{1}{2\sqrt{2}}\int_0^{\frac{1}{2}\pi} \left ( \frac{\sin x}{\sqrt{2}-\cos x}+ \frac{\sin x}{\sqrt{2}+\cos x} \right) \d x \\ &&&= \frac{1}{2\sqrt{2}} \left [\ln (\sqrt{2}-\cos x) - \ln (\sqrt{2}+\cos x) \right]_0^{\pi/2} \\ &&&= \frac{1}{2\sqrt{2}} \left (-\ln(\sqrt{2}-1)+\ln(\sqrt{2}+1) \right) \\ &&&= \frac1{2\sqrt{2}} \ln \left (\frac{\sqrt{2}+1}{\sqrt{2}-1} \right)\\ &&&= \frac1{\sqrt{2}} \ln (\sqrt{2}+1) \end{align*} \begin{align*} && (1+\sqrt{2})^5 + (1-\sqrt{2})^5 &= 2(1+10\cdot2+5\cdot2^2) \\ &&&= 82 \\ && |(1-\sqrt{2})^5| & < 1 \\ && (1+\sqrt{2})^5 &< 83 < 99 \\ \\ && 1.4^2 &= 1.96 \\ &&&< 2 \\ \Rightarrow && 1.4 &<\sqrt{2} \\ \\ \Rightarrow && 2^{\sqrt{2}} &> 2^{1.4} \\ &&&=2^{7/5} \\ &&&= {128}^{1/5} \\ &&&>99^{1/5} \\ &&&>1+\sqrt{2} \end{align*} \begin{align*} && \ln 2 & > \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1) \\ \Leftrightarrow && \sqrt{2} \ln 2 &> \ln(\sqrt{2}+1) \\ \Leftrightarrow && 2^{\sqrt{2}} &> 1+\sqrt{2} \end{align*} which we have already shown, so the first integral is larger.

Solution:

1. 

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   If \(y = \textrm{arcosh} x \), then \(\tanh\textrm{arcosh} x/2 = \sqrt{\frac{\cosh \textrm{arcosh} x-1}{\cosh \textrm{arcosh} x+1}} = \sqrt{\frac{x-1}{x+1}} = \frac{x-1}{\sqrt{x^2-1}}\)
2. \begin{align*} \int \textrm{arcosh} x \d x &= \left [x \textrm{arcosh} x \right] - \int \frac{x}{\sqrt{x^2-1}} \d x \\ &= x \textrm{arcosh} x - \sqrt{x^2-1}+C \\ \int \frac{x-1}{\sqrt{x^2-1}} &= \sqrt{x^2-1} - \textrm{arcosh} x +C \end{align*} Therefore \begin{align*} && \int_1^x \textrm{arcosh} t \d t &> \int_1^x \frac{t-1}{\sqrt{t^2-1}} \d t \\ \Rightarrow && x \textrm{arcosh} x - \sqrt{x^2-1} - 0 &> \sqrt{x^2-1} - \textrm{arcosh} x - 0 \\ \Rightarrow && (x+1) \textrm{arcosh} x &> 2\sqrt{x^2-1} \\ \Rightarrow && \textrm{arcosh} x & > 2\frac{\sqrt{x^2-1}}{x+1} \\ &&&= 2 \frac{\sqrt{x-1}}{\sqrt{x+1}} \\ &&&= 2 \frac{x-1}{\sqrt{x^2-1}} \end{align*}
3. Integrating both sides again, \begin{align*} && \int_1^x \textrm{arcosh} t \d t &> 2 \int_1^x \frac{t-1}{\sqrt{t^2-1}} \d t \\ \Rightarrow && x \textrm{arcosh} x - \sqrt{x^2-1} &> 2\left (\sqrt{x^2-1} - \textrm{arcosh}x \right) \\ \Rightarrow && (x+2)\textrm{arcosh} x &> 3 \sqrt{x^2-1} \\ \Rightarrow && \textrm{arcosh} x &> 3 \frac{\sqrt{x^2-1}}{x+2} \end{align*}

Solution: \begin{align*} && \cos^4 \theta - \sin^4 \theta &= (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) \\ &&&= \cos^2 \theta - \sin^2 \theta \\ &&&= \cos 2 \theta \\ \\ && 1&= (\cos^2 \theta + \sin^2 \theta)^2 \\ &&&= \cos^4 \theta + \sin^4 \theta + 2 \sin^2 \theta \cos^2 \theta \\ &&&= \cos^4 \theta + \sin^4 \theta + \frac12 ( \sin^2 2 \theta) \\ \Rightarrow && \cos^4 \theta + \sin^4 \theta &= 1 - \tfrac12 \sin^2 2 \theta \end{align*} \begin{align*} && I &= \int_{0}^{\pi/2} \cos^4 \theta \d \theta \\ && J &= \int_0^{\pi/2} \sin^4 \theta \d \theta \\ && I-J &= \int_0^{\pi/2} \cos 2 \theta \d \theta = 0 \\ && I+J &= \int_0^{\pi/2} (1- \frac12 \sin^2 2 \theta) \d \theta \\ &&&= \frac{\pi}{2} - \frac14 \int_0^{\pi} \sin^2 \theta \d \theta \\ &&&= \frac{\pi}{2} - \frac{\pi}{8} \\ &&&= \frac{3\pi}{8} \\ \Rightarrow && I=J &= \frac{3\pi}{16} \end{align*} \begin{align*} && \cos^6 \theta + \sin^6 \theta &= (\cos^2 \theta + \sin^2 \theta)(\cos^4 \theta - \cos^2 \theta \sin^2 \theta + \sin^4 \theta) \\ &&&= 1-\tfrac12 \sin^2 2\theta - \tfrac14 \sin^2 2 \theta \\ &&&= 1 - \tfrac34 \sin^2 2 \theta \\ %&& \cos^6 \theta - \sin^6 \theta &= (\cos^2 \theta - \sin^2 \theta)(\cos^4 \theta + \cos^2 \theta \sin^2 \theta + \sin^4 \theta) \\ %&&&= \cos 2 \theta (1 - \tfrac12 \sin^2 2 \theta + \tfrac14 \sin^2 2 \theta) \\ %&&&= \cos 2 \theta (1 - \tfrac14 \sin^2 2 \theta) \\ \end{align*} \begin{align*} && I &= \int_{0}^{\pi/2} \cos^6 \theta \d \theta \\ && J &= \int_0^{\pi/2} \sin^6 \theta \d \theta \\ && I-J &= 0 \\ && I+J &= \int_0^{\pi/2} (1 - \tfrac34 \sin^2 2 \theta) \d \theta \\ &&&= \frac{\pi}{2} - \frac{3\pi}{16} = \frac{5\pi}{16} \\ \Rightarrow && I = J &= \frac{5\pi}{32} \end{align*}

Solution:

1. \(\,\) \begin{align*} && y &= 2x^3-bx^2+cx \\ \Rightarrow && y' &= 6x^2-2bx+c \end{align*} We must have \(p, q\) are the roots of this equation, ie \(\frac13b = p+q, \frac16c = pq\)
2. The point of inflection will be at \(\frac{b}6\) The graph will have rotational symmetry of \(180^{\circ}\) about the point of inflection.
   

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3. \begin{align*} && m-n &= 2(p^3-q^3)-b(p^2-q^2)+c(p-q) \\ &&&= (p-q)(2(p^2+qp+q^2)-b(p+q)+c) \\ &&&= (p-q)(2(p^2+qp+q^2)-3(p+q)^2+6pq) \\ &&&= (p-q)(-p^2-q^2+2pq) \\ &&&= (q-p)^3 \end{align*}
4. The area of \(R\) is \begin{align*} A &= \frac12 bh \\ &= \frac12 (q-p)(m-n) = \frac12(q-p)^4 \end{align*} as required.

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*}

Solution:

1. \(\,\) \begin{align*} && y &= \ln (x + \sqrt{3+x^2}) \\ \Rightarrow && y' &= \frac{1}{x + \sqrt{3+x^2}} \cdot \left (1 + \frac{x}{\sqrt{3+x^2}} \right) \\ &&&= \frac{1}{\sqrt{3+x^2}} \\ \\ && y &= x\sqrt{3+x^2} \\ && y' &= \sqrt{3+x^2} + \frac{x^2}{\sqrt{3+x^2}} \\ &&&= 2\sqrt{3+x^2} - \frac{3}{\sqrt{3+x^2}} \\ \\ \Rightarrow && \sqrt{3+x^2} &= \frac12(x \sqrt{3+x^2})' + \frac32(\ln(x+\sqrt{3+x^2})' \\ \Rightarrow && \int \sqrt{3+x^2}\, \d x &= \frac12x\sqrt{3+x^2} + \frac32 \ln (x+\sqrt{3+x^2}) + C \end{align*}
2. \(\,\) \begin{align*} && 3 \left ( \frac{\d y}{\d x} \right)^2 + 2x \frac{\d y}{\d x} &= 1 \\ && \frac{\d y}{\d x} &= \frac{-x \pm \sqrt{x^2+3} }3 \\ \Rightarrow && y &= -\frac{x^2}{6} \pm \frac16x\sqrt{3+x^2} \pm \frac12 \ln (x+\sqrt{3+x^2}) + C \\ y = 0, x = 1: && 0 &= -\frac16 \pm \frac13 \pm \frac12 \ln 3 \\ \Rightarrow && y &= -\frac{x^2}{6} \pm \frac12x\sqrt{3+x^2} \pm \frac32 \ln (x+\sqrt{3+x^2}) + \frac16 \mp \frac13 \mp \frac12 \ln 3 \end{align*}

Solution:

1. 

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2. Since \(f(x)\) is continuous, \(a -bx\) joins \((2k, \ln k)\) and \((4k ,0)\). ie it has a gradient of \(\frac{-\ln k}{2k}\) and is zero at \(4k\), hence \(\displaystyle b = -\frac{\ln k}{2k}, a = 2\ln k\). The \(3\) sections have areas \(\int_1^k \ln x \d x = k \ln k -k +1\), \(k \ln k, k \ln k\). Therefore \begin{align*} &&1&= 3k\ln k - k +1 \\ \Rightarrow &&0 &= k(3\ln k - 1) \\ \Rightarrow &&\ln k &= \frac13 \\ \Rightarrow &&k &= e^{1/3} \\ && a &= \frac23 \\ && b&= -\frac16e^{-1/3} \end{align*}
3. Clearly \(1 > k \ln k > \frac{1}{3}\), therefore the median must lie between \(k\) and \(2k\). So we need, \(\frac12\) to be the area of the rectangle + the triangle, ie: \begin{align*} && \frac12 &= k \ln k + (2k-M) \ln k \\ &&&= \frac13 k + \frac13 (2k - M) \\ \Rightarrow && M &= 3k - \frac32 \\ \Rightarrow && M &= 3e^{1/3} - \frac32 \end{align*}

Solution:

1. \begin{align*} && t(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ u = v^{-1}, \d u = -v^{-2} \d v&&&= \int_{v = \infty}^{v = 1/x} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\ &&&= \int_{1/x}^\infty \frac{1}{1+v^2} \d v \\ \\ \Rightarrow && t(x) + t(1/x) &= \int_0^x \frac{1}{1+u^2} \d u + \int_0^{1/x} \frac{1}{1+u^2} \d u \\ &&&= \int_{1/x}^{\infty} \frac{1}{1+u^2} \d u + \int_0^{1/x} \frac{1}{1+u^2} \d u \\ &&&= \int_0^{\infty} \frac{1}{1+u^2} \d u \\ &&&= \frac12 p \\ \\ \Rightarrow && t(1) +t(1/1) = 2t(1) &= \frac12 p \end{align*}
2. \(\,\) \begin{align*} && y &= \frac{u}{\sqrt{1+u^2}} \\ \Rightarrow && y^2 &= \frac{u^2}{1+u^2} \\ &&&= 1-\frac{1}{1+u^2} \\ \Rightarrow && 1+u^2 &= \frac{1}{1-y^2} \\ \Rightarrow && u &= \frac{y}{\sqrt{1-y^2}} \\ \\ && \frac{\d u}{\d y} &= \frac{\sqrt{1-y^2} + y^2(1-y^2)^{-1/2}}{1-y^2} \\ &&&= \frac{1}{(1-y^2)^{3/2}} \\ \\ && t(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ &&&= \int_0^{y = x/\sqrt{1+x^2}} \frac{1}{1 + \frac{y^2}{1-y^2}} \frac{1}{(1-y^2)^{3/2}} \d y \\ &&&= \int_0^{x/\sqrt{1+x^2}} \frac{1-y^2}{(1-y^2)^{3/2}} \d y \\ &&&= \int_0^{x/\sqrt{1+x^2}} \frac{1}{(1-y^2)^{1/2}} \d y \\ &&&= s\left ( \frac{x}{\sqrt{1+x^2}} \right) \\ \\ \Rightarrow && s\left ( \frac{1}{\sqrt{2}} \right) &= t(1) = \frac14p \end{align*}
3. \(\,\) \begin{align*} && z &= \frac{u + \frac1{\sqrt{3}}}{1- \frac{1}{\sqrt{3}} u}\\ \Rightarrow && z - \frac{z}{\sqrt{3}}u &= u + \frac{1}{\sqrt{3}} \\ \Rightarrow && u &= \frac{z-\frac{1}{\sqrt{3}}}{1 + \frac{z}{\sqrt{3}}} \\ \\ \Rightarrow && \frac{\d u}{\d z} &= \frac{\sqrt{3}(\sqrt{3}+z ) -(\sqrt{3}z-1)}{\left (\sqrt{3}+z \right)^2} \\ &&&= \frac{4}{(\sqrt{3}+z)^2} \\ \\ \Rightarrow && t \left ( \frac{1}{\sqrt{3}} \right) &= \int_0^{1/\sqrt{3}} \frac{1}{1+u^2} \d u \\ &&&= \int_{z=1/\sqrt{3}}^{z=\sqrt{3}} \frac{1}{1 + \left ( \frac{\sqrt{3}z-1}{\sqrt{3}+z}\right)^2} \frac{4}{(\sqrt{3}+z)^2} \d z\\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{4}{(\sqrt{3}+z)^2+(\sqrt{3}z-1)^2} \d z \\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{4}{4+4z^2} \d z \\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \d z \end{align*} Notice that \(t(1/\sqrt{3})+t(\sqrt{3}) = \frac12p\) and also notice that \(t(1/\sqrt{3}) + t(1/\sqrt{3}) =t(\sqrt{3})\) so \(3t(1/\sqrt{3}) = \frac12p\)

Solution:

1. Firstly, we consider the probability that all darts lie within a distance \(s\) from the centre, ie \begin{align*} \mathbb{P}(\text{all darts within }s) &= \prod_{k=1}^s \mathbb{P}(\text{dart within }s) \\ &= \left ( \frac{\pi s^2}{\pi} \right)^n \\ &= s^{2n} \end{align*} Therefore the pdf is \(2ns^{2n-1}\), and the expected area is \(\int_{s=0}^1 \pi s^2 \cdot 2n s^{2n-1} \d s = 2n \pi \frac{1}{2n+2} = \frac{n}{n+1} \pi\). \begin{align*} \mathbb{P}(\text{n-1 within }s) &= \underbrace{s^{2n}}_{\text{all within }s} + \underbrace{ns^{2n-2}(1-s^2)}_{\text{all but 1 within }s}\\ &= ns^{2n-2}-(n-1)s^{2n} \end{align*} Therefore the pdf is \(n(2n-2)s^{2n-3} - 2n(n-1)s^{2n-1} = 2n(n-1)(s^{2n-3}-s^{2n-1})\) and the expected area is: \begin{align*} \int \pi s^2 \cdot2n(n-1)(s^{2n-3}-s^{2n-1})\d s &= 2n(n-1) \pi \left ( \frac{1}{2n} - \frac{1}{2n+2} \right) \\ &= n(n-1)\pi \frac{2}{n(n+1)} \\ &= \frac{n-1}{n+1} \pi \end{align*}
2. Now consider a square of side-length \(s\), we must have \(\mathbb{P}(\text{all darts within square}) = \left ( \frac{s^2}{4} \right)^n\) and therefore the pdf is \(n \frac{s^{n-1}}{4^n}\). Therefore the expected area is \(\displaystyle \int_0^2 s^2 \cdot n \frac{s^{n-1}}{4^n} \d s = \frac{n}{n+1} \frac{2^{2n+1}}{2^{2n}} = \frac{4n}{n+1}\)
3. It is clearly larger as the square dartboard contains all of the circular dartboard, and there will be some probability that the darts land outside the circular dartboard, making the circle much larger.

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*}

Solution:

1. \(\,\)
   

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   Notice that they are equal at \(1\) when \(x = \pi/2\), but this is a local minimum for \(\csc x\) whereas \(2x/\pi\) is increasing so there is a second intersection. Notice that \(\csc x = \frac{2x}{\pi} \Leftrightarrow x \sin x = \frac{\pi}{2}\) therefore our intersections are also the roots of \(x \sin x = \frac{\pi}{2}\) and the larger one is greater than \(\pi/2\) \begin{align*} && I &= \int_{\pi/2}^{\pi} \Bigl| x \sin x - \frac{\pi}{2} \Bigr| \d x \\ &&&= \int_{\pi/2}^{\alpha} \left ( x \sin x - \frac{\pi}{2} \right )\d x +\int_{\alpha}^{\pi} \left ( \frac{\pi}{2} -x \sin x \right) \d x \\ &&&= \left ( \pi - 2\alpha + \frac{\pi}{2}\right) \frac{\pi}{2} + \int_{\pi/2}^{\alpha} x \sin x\d x -\int_{\alpha}^{\pi} x \sin x \d x \\ &&&= \frac{3\pi^2}{4} - \alpha \pi + \left [-x \cos x \right]_{\pi/2}^{\alpha}+\left[x \cos x \right]_{\alpha}^{\pi} + \int_{\pi/2}^{\alpha} \cos x \d x - \int_{\alpha}^{\pi} \cos x \d x \\ &&&= \frac{3\pi^2}{4} - \alpha \pi -\alpha \cos \alpha -\pi -\alpha \cos \alpha+ \sin \alpha - 1+\sin \alpha \\ &&&= 2\sin \alpha + \frac{3\pi^2}{4} - \alpha \pi - 2\alpha \cos \alpha - 1 \end{align*}
2. \(\,\)
   

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   \begin{align*} && A &= \int_0^{\ln 2} ||e^x-1|-1| \d x \\ &&&= \int_0^{\ln 2} |e^x-2| \d x \\ &&&=\int_0^{\ln 2} (2-e^x) \d x \\ &&&= 2 \ln 2 - \left [e^x \right]_0^{\ln 2} \\ &&&= \ln 4 - (2-1) = \ln 4 - 1 \end{align*}