Problems

Solution: \begin{align*} &&\int_0^{\frac14\pi} \sin (2x) \ln(\cos x)\, \d x &= \int_0^{\frac14 \pi} 2 \sin x \cos x \ln (\cos x) \d x \\ u = \cos \theta :&&&= \int_{u=1}^{u=\frac1{\sqrt2}} -2u \ln u \d u \\ &&&= \int_{\frac1{\sqrt{2}}}^1 2u \ln u \d u \\ &&&= \left [u^2 \ln u \right]_{\frac1{\sqrt{2}}}^1-\int_{\frac1{\sqrt{2}}}^1 u \d u \\ &&&= -\frac12 \ln \frac{1}{\sqrt{2}} - \l\frac12 - \frac14 \r \\ &&&= \frac14 (\ln 2 - 1) \end{align*} \begin{align*} && \int_0^{\frac14\pi} \cos (2x) \ln(\cos x)\, \d x &= \left [ \frac12 \sin 2x \ln (\cos x) \right]_0^{\frac14\pi}- \int_0^{\frac14\pi} \frac12 \sin 2x \frac{-\sin x}{\cos x} \d x \\ &&&=\frac12 \ln \frac{1}{\sqrt{2}}+\int_0^{\frac14\pi} \sin^2 x \d x \\ &&&= -\frac14 \ln 2 + \int_0^{\frac14\pi} \frac{1-\cos 2x }{2} \d x \\ &&&= -\frac14 \ln 2 +\frac{\pi}{8} -\frac{1}{4} \\ &&&= \frac18 (\pi - 2\ln 2 - 2) \\ &&&= \frac18 (\pi - \ln 4 - 2) \\ \end{align*} Notice that \(\cos x + \sin x = \sqrt{2} \cos (x -\frac{\pi}{4})\), so: \begin{align*} &&\int_{\frac14\pi}^{\frac12\pi} \big ( \cos(2x) + \sin (2x)\big) \ln \big( \cos x + \sin x\big)\d x &= \int_{\frac14\pi}^{\frac12\pi} \big ( \cos(2x) + \sin (2x)\big) \ln (\sqrt{2} \cos ( x - \frac{\pi}{4}) ) \d x \\ &&&= \int_{u=0}^{u=\frac{\pi}{4}} \l \cos(2u+\frac{\pi}{2})+\sin(2u+\frac{\pi}{2}) \r \cdot \l \frac12 \ln 2 +\ln (\cos u) \r \d x \\ &&&= \int_{0}^{\frac{\pi}{4}} \l -\sin 2u+\cos 2u \r \cdot \l \frac12 \ln 2 +\ln (\cos u) \r \d x \\ &&&= \frac14 \ln 2\left [ \cos 2u + \sin 2u \ \right]_{0}^{\frac{\pi}{4}} - \frac14(\ln2 - 1) + \frac18\pi - \frac14(\ln 2 +1) \\ &&&= \frac{\pi}{8}-\frac12 \ln 2 \end{align*}

Solution: \begin{align*} && T &= \int_{\frac13}^{\frac12} \frac{{\rm artanh}\, t}t \,\d t \\ && &=\int_{\frac13}^{\frac12} \frac{1}{2t}\ln \left ( \frac{1+t}{1-t} \right) \,\d t \\ u = \tfrac{1+t}{1-t}, t= \tfrac{u-1}{u+1}, \d t = \tfrac{2}{(u+1)^2} \d t &&&= \int_{u=2}^{u=3} \frac{1}{2t} \ln u \frac{2}{(u+1)^2} \d u \\ &&&= \int_2^3 \frac{u+1}{u-1} \ln u \frac{1}{(u+1)^2} \d u \\ &&&= \int_2^3 \frac{1}{u^2-1} \ln u \d u \end{align*} \begin{align*} && U&= \int _{\ln 2 }^{\ln 3 } \frac{u}{2\sinh u}\, \d u \\ v = e^u, \d v = e^u \d u &&&= \int_{v=2}^{v=3} \frac{\ln v}{v - \frac{1}{v}} \frac{1}{v} \d v \\ &&&= \int_2^3 \frac{1}{v^2-1} \ln v \d v \end{align*} \begin{align*} &&V &= - \int_{\frac13}^{\frac12} \frac{\ln v}{1-v^2} \,\d v \\ u = \tfrac1v, \d u = -\tfrac1{v^2} \d v &&&= -\int_{u=3}^{u=2} \frac{-\ln u}{1 - \frac{1}{u^2}} \frac{-1}{u^2} \d u \\ &&&= -\int_3^2 \frac{\ln u}{u^2-1} \d u \\ &&&= \int_2^3 \frac{1}{u^2-1} \ln u \d u \end{align*} \begin{align*} &&X&= \int _{\frac12\ln2}^{\frac12\ln3} \ln ({\coth x})\, \d x \\ u = \coth x, \d u =(1-u^2) \d x &&&= \int_{u = 3}^{u=2} \ln u \frac{1}{1-u^2} \d u \\ &&&= \int_2^3 \frac{\ln u}{u^2-1} \d u \end{align*} Therefore all integrals are equal to the same integral, namely \(\displaystyle \int_2^3 \frac{\ln u}{u^2-1} \d u\)

Solution: \begin{align*} && x &= \frac{1}{t^2-1} \\ && t &= \sqrt{\frac{x+1}{x}}\\ \Rightarrow && \frac{\d x}{\d t} &= \frac{-2t}{(t^2-1)^2} \\ \Rightarrow && I &= \int \frac{1}{\sqrt{x(x+1)}} \d x \\ &&&= \int \frac{1}{\sqrt{\frac1{t^2-1} \frac{t^2}{t^2-1}}} \cdot \frac{-2 t}{(t^2-1)^2} \d t \\ &&&= \int \frac{t^2-1}{t} \frac{-2t}{(t^2-1)^2} \d t \\ &&&= -\int \frac{2}{t^2-1} \d t \\ &&&= - \frac{2}{2 \cdot 1} \ln \left | \frac{t-1}{t+1} \right| +C \\ &&&= \ln \left | \frac{t+1}{t-1} \right| + C \\ &&&= \ln \left | \frac{\sqrt{\frac{x+1}{x}}+1}{\sqrt{\frac{x+1}{x}}-1} \right| + C \\ &&&= \ln \left | \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}-\sqrt{x}} \right| + C \\ &&&= 2\ln \left | \sqrt{x+1}+\sqrt{x}\right| + C \\ &&&= 2\ln \left ( \sqrt{x+1}+\sqrt{x}\right) + C \\ \end{align*} \begin{align*} && V&= \pi \int_{1/8}^{9/16} y^2 \d x \\ &&&= \pi \int_{1/8}^{9/16} \left ( \frac1x + \frac{1}{x+1} - \frac{2}{\sqrt{x(x+1)}}\right) \d x \\ &&&= \pi \left [ \ln x + \ln (x+1) - 4 \ln(\sqrt{x+1} + \sqrt{x}) \right]_{1/8}^{9/16} \\ &&&= \pi \left ( \ln \frac{9}{16} + \ln \frac{25}{16} - 4 \ln \left ( \frac54 + \frac34\right) \right) +\\ &&&\quad -\pi \left ( \ln \frac{1}{8} + \ln \frac{9}{8} - 4 \ln \left ( \frac1{2\sqrt{2}} + \frac3{2\sqrt{2}}\right) \right) \\ &&&= \pi \left ( 2 \ln 3 - 8 \ln 2 + 2 \ln 5 - 4\ln2 \right) - \pi \left ( -6 \ln 2 + 2\ln 3 - 2\ln 2\right) \\ &&&= \pi (2 \ln 5 - 4 \ln 2 ) \\ &&&= 2 \pi \ln \tfrac54 \end{align*}

Solution:

1. \(\,\) \begin{align*} && I &= \int_0^a \frac{f(x)}{f(x)+f(a-x)} \d x \\ u =a-x, \d u = - \d x: &&& \int_{u=a}^{u=0} \frac{f(a-u)}{f(a-u)+f(u)} (-1) \d u \\ &&&= \int_0^a \frac{f(a-u)}{f(u)+f(a-u)} \d u \\ &&&= \int_0^a \frac{f(a-x)}{f(x)+f(a-x)} \d x \\ \Rightarrow && 2 I &= \int_0^a \left ( \frac{f(x)}{f(x)+f(a-x)} + \frac{f(a-x)}{f(x)+f(a-x)} \right) \d x \\ &&&= \int_0^a 1 \d x \\ &&&= a \\ \Rightarrow && I &= \frac{a}{2} \end{align*} \begin{align*} && J &= \int_0^1 \frac{\ln (x+1)}{\ln (2+x-x^2)}\, \d x \\ &&&= \int_0^1 \frac{\ln (x+1)}{\ln((x+1)(2-x))} \d x \\ &&&= \int_0^1 \frac{\ln (x+1)}{\ln(x+1) + \ln ((1-x)+1)} \d x \\ &&&= \frac{1}{2} \tag{\(f(x) = \ln (x+1)\)} \\ \\ && K &= \int_0^{\frac\pi 2} \frac{\sin x } {\sin(x+\frac \pi 4 )} \, \d x \\ &&&= \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}} \\ &&&= \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x + \sin (\frac{\pi}{2}-x)} \d x\\ &&&= \frac{\pi}{2\sqrt{2}} \end{align*}
2. \(\,\) \begin{align*} &&I &= \int_{\frac12}^2 \frac{\sin x }{x(\sin x + \sin \frac1x)} \d x \\ u = 1/x, \d u = -1/x^2 \d x : &&&= \int_{u = 2}^{u=\frac12} \frac{\sin \frac1u}{\frac{1}{u}(\sin \frac1u + \sin u)} (-\frac{1}{u^2} ) \d u \\ &&&= \int_{\frac12}^2 \frac{\sin \frac1u}{u (\sin u + \sin \frac1u)} \d u \\ \Rightarrow && 2I &= \int_{\frac12}^2 \left ( \frac{\sin x }{x(\sin x + \sin \frac1x)} + \frac{\sin \frac1x }{x(\sin x + \sin \frac1x)}\right) \d x \\ &&&= \int_{\frac12}^2 \frac{1}{x} \d x\\ &&&= 2\ln2 \\ \Rightarrow && I &= \ln 2 \end{align*}

Solution:

1. \(\,\) \begin{align*} && \lim_{x \to 0} x^m(\ln x)^n &= \lim_{t \to \infty} (e^{-t})^m (\ln e^{-t})^n \\ &&&= \lim_{t \to \infty} e^{-mt} (-t)^n \\ &&&= (-1)^n \lim_{t \to \infty} e^{-mt} t^n = 0 \\ \\ && \lim_{x \to 0} x^x &= \lim_{x \to 0} e^{x \ln x} \\ &&&= \exp \left (\lim_{x \to 0} x \ln x \right) \\ &&&= \exp \left (0 \right) = 1 \end{align*}
2. \(\,\) \begin{align*} && I_{n} &= \int_0^1 x^m (\ln x)^n \d x \\ && I_{n+1} &= \int_0^1 x^m (\ln x)^{n+1} \d x \\ &&&= \left [\frac{x^{m+1}}{m+1} (\ln x)^{n+1} \right]_0^1 - \frac{1}{m+1} \int_0^1 x^{m+1} (n+1) (\ln x)^n \cdot x^{-1} \d x \\ &&&= 0 - \frac{1}{m+1} \lim_{x \to 0} \left (x^{m+1} (\ln x)^{n+1} \right) - \frac{n+1}{m+1} \int_0^1 x^m (\ln x)^n \d x \\ &&&= - \frac{n+1}{m+1} I_n \\ \\ && I_0 &= \int_0^1 x^m \d x = \frac{1}{m+1} \\ && I_1 &= -\frac{1}{(m+1)^2} \\ && I_2 &= \frac{2}{(m+1)^3} \\ && I_n &= (-1)^n \frac{n!}{(m+1)^{n+1}} \end{align*}
3. \(\,\) \begin{align*} && \int_0^1 x^x \d x &= \int_0^1 e^{x \ln x} \d x \\ &&&= \int_0^1 \sum_{k=0}^{\infty} \frac{x^k(\ln x)^k}{k!} \d x \\ &&&= \sum_{k=0}^{\infty} \int_0^1 \frac{x^k(\ln x)^k}{k!} \d x\\ &&&= \sum_{k=0}^{\infty} (-1)^k \frac{k!}{k!(k+1)^{k+1}}\\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^{k+1}}\\ &&&= 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \cdots \end{align*}

Solution:

1. 

   + - ↺
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: \begin{align*} && I &= \int_0^{\frac14\pi} \ln (1 + \tan \theta) \d \theta \\ \theta = \tfrac14\pi - \phi, \d \theta = -\d\phi: &&&= \int_0^{\frac14 \pi} \ln ( 1 + \tan (\tfrac14\pi - \phi)) \d \phi \\ &&&= \int_0^{\frac14 \pi} \ln \left ( 1 + \frac{1 - \tan \phi}{1+\tan \phi} \right) \d \phi \\ &&&= \int_0^{\frac14 \pi} \ln \left ( \frac{2}{1+\tan \phi} \right) \d \phi \\ &&&= \tfrac14 \pi \ln 2 - I \\ \Rightarrow && I &= \tfrac18\pi \ln 2 \end{align*} \begin{align*} && J &= \int_0^1 \frac{\ln(1+x)}{1+x^2} \d x \\ x= \tan \theta \d \theta, \d \theta = \frac{\d x}{1+x^2} &&&= \int_0^{\frac14 \pi} \ln(1 + \tan \theta) \d \theta \\ &&&= \tfrac18 \pi \ln 2 \end{align*} \begin{align*} && K &= \int_0^{\frac12 \pi} \ln \left ( \frac{1 + \sin x}{1 + \cos x} \right) \d x \\ y = \tfrac12\pi - x, \d y = -\d x: &&&= \int_0^{\frac12\pi} \ln \left ( \frac{1+\cos y}{1+\sin y}\right) \d y \\ &&&= -K \\ \Rightarrow && K &= 0 \end{align*}