Problems

Solution:

1. \begin{align*} && I &= \int_0^{\infty} \frac{\sqrt{x}-1}{\sqrt{x(x^3+1)}} \, dx \\ x = u^{-1}, \d x = -u^{-2} \d u: && &= \int_{u=\infty}^{u = 0} \frac{u^{-1/2} - 1}{\sqrt{u^{-1}(u^{-3}+1)}} (-u^{-2}) \d u \\ &&&= \int_0^\infty \frac{u^{-1/2} -1}{\sqrt{1+u^3}} \d u \\ &&&= \int_0^\infty \frac{1-\sqrt{u}}{\sqrt{u(u^3+1)}} \d u \\ &&&= -I \\ \Rightarrow && I &= 0 \end{align*}
2. \begin{align*} && I &= \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx \\ x = u^{-2}, \d x = -2u^{-3} \d u: && &= \int_{u = \infty}^{u = 0} \frac{1}{\sqrt{u^{-6}+1}} (-2u^{-3}) \d u \\ &&&= \int_0^\infty \frac{2}{\sqrt{1+u^6}} \d u \\ &&&= 2\int_0^{\infty} \frac{1}{\sqrt{x^6+1}} \, \d x \end{align*}
3. \begin{align*} && I &= \int_0^{\infty} \frac{x^r - 1}{\sqrt{x^s(x^p+1)}} \, dx \\ x = u^{-1}, \d x = - u^{-2} \d t: &&&= \int_{u=\infty}^{u = 0} \frac{u^{-r}-1}{\sqrt{u^{-s}(u^{-p}+1)}} (- u^{-2}) \d u \\ &&&= \int_0^\infty \frac{u^{-r}-1}{\sqrt{u^{4-s}(u^{-p}+1)}} \d u \\ &&&= \int_0^\infty \frac{1-u^{r}}{\sqrt{u^{4-s+2r}(u^{-p}+1)}} \d u \\ &&&= \int_0^\infty \frac{1-u^{r}}{\sqrt{u^{4-s+2r-p}(1+u^p)}} \d u \\ \end{align*} Therefore if \(4-s+2r-p = s\) or \(r = \frac{p}2+s-2\) we have \(I = -I\), ie \(I = 0\).
4. \begin{align*} && I &= \int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx \\ x = u^{-t}, \d x = -t u^{-t-1}:&&&= \int_{u = -\infty}^{u = 0} \frac{1}{\sqrt{u^{-pt}+1}} (-t u^{-t-1}) \d u \\ &&&= t \int_0^\infty \frac1{\sqrt{u^{-pt+2t+2}+u^{2t+2}}} \d u \end{align*} Therefore if \(\begin{cases} 2t+2 &= q \\ 2t+2 -pt &= 0 \end{cases}\), ie \(q = 2t+2, p = 2 + 2/t\) we will have found the \(p, q\) desired for any \(t\) (or \(k\)). [Alternatively] Let \(\displaystyle I(p) =\int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx\), then clearly \(I(p)\) is decreasing and as \(p \to \infty\) \(I(p) \to 1\), so our integral can take any values on \((1, \infty)\) and so for any positive value we can find two values with a given ratio. In particular given \(k\) and \(p\) we can find a suitable \(q\).

Solution:

1. \begin{align*} && b(p,q) &= \int_0^1 x^{p-1}(1-x)^{q-1}\, \d x \\ u = 1-x, \d u = -\d x && &= \int_{u=1}^{u = 0} (1-u)^{p-1}u^{q-1} (-1) \, \d u \\ &&&= \int_0^1 (1-u)^{p-1}u^{q-1} \d u \\ &&&= \int_0^1 u^{q-1}(1-u)^{p-1} \d u \\ &&&= b(q,p) \end{align*}
2. \begin{align*} b(p+1,q) + b(p,q+1) &= \int_0^1 x^p(1-x)^{q-1} \d x + \int_0^1 x^{p-1}(1-x)^{q} \d x \\ &= \int_0^1 \left (x^p(1-x)^{q-1} + x^{p-1}(1-x)^{q}\right) \d x \\ &= \int_0^1 x^{p-1}(1-x)^{q-1} \left (x + (1-x) \right) \d x \\ &= \int_0^1 x^{p-1}(1-x)^{q-1} \d x \\ &= b(p,q) \end{align*} Therefore \(b(p+1,q) = b(p,q) - b(p,q+1)\), in particular \(2b(p+1,p) = b(p+1,p)+b(p,p+1) = b(p,p) \Rightarrow b(p+1,p) = \frac12 b(p,p)\) as required.
3. \begin{align*} && b(p,q) &= \int_0^1 x^{p-1} (1-x)^{q-1} \d x \\ x = \sin^2 \theta, \d x = 2 \sin \theta \cos \theta \d \theta && &= \int_{u=0}^{u = \pi/2} \sin^{2p-2} \theta (1-\sin^2 \theta)^{q-1} \cdot 2 \sin \theta \cos \theta \d \theta \\ &&&= 2 \int_0^{\pi/2} \sin^{2p-1} \theta \cos^{2q-2} \cos \theta \d \theta \\ &&&= 2 \int_0^{\pi/2} \sin^{2p-1} \theta \cos^{2q-1} \theta \d \theta \end{align*} \begin{align*} b(p,p) &= 2\int_0^{\pi/2} (\sin \theta)^{2p-1}(\cos \theta)^{2p-1} \d \theta \\ &= 2 \int_0^{\pi/2} \left (\frac12 \sin 2\theta \right)^{2p-1} \d \theta \\ &= \frac1{2^{2p-1}} 2 \int_0^{\pi/2} (\sin 2 \theta)^{2p-1} \d \theta \\ &= \frac1{2^{2p-1}} 2 \int_{x=0}^{x=\pi} (\sin x)^{2p-1} 2 \d x\\ &= \frac1{2^{2p-1}} 2 \int_{x=0}^{x=\pi/2} (\sin x)^{2p-1} \d x\\ &= \frac1{2^{2p-1}} 2 \int_{0}^{\pi/2} (\sin x)^{2p-1} (\cos x)^{0} \d x\\ &= \frac1{2^{2p-1}} b(p,\tfrac12) \end{align*}
4. \begin{align*} &&b(p,q) &= \int_0^1 x^{p-1}(1-x)^{q-1} \d x \\ t = \frac{x}{1-x}, \d t = (1-x)^{-2} \d x &&&= \int_{t=0}^{t = \infty} \left ( \frac{t}{1+t} \right)^{p-1} \left ( 1-\frac{t}{1+t} \right)^{q+1} \d t\\ x = \frac{t}{1+t} && &=\int_0^\infty t^{p-1} (1+t)^{-(p-1)-(q+1)} \d t \\ &&&= \int_0^{\infty} \frac{t^{p-1}}{(1+t)^{p+q}} \d t \end{align*}
5. \begin{align*} I &= \int_0^\infty \frac{t^{3/2}}{(1+t)^6} \, dt \\ &= b( \tfrac52, \tfrac72) \\ &= b( \tfrac52, \tfrac52+1) \\ &= \tfrac12 b( \tfrac52, \tfrac52) \\ &= \frac12 \cdot \frac1{2^{4}} b(\tfrac52, \tfrac12) \\ &= \frac{1}{2^5} \cdot 2 \int_0^{\pi/2} (\sin \theta)^{4} \d \theta \\ &= \frac1{2^4} \int_0^{\pi/2}\left (\frac{1-\cos 2 \theta}{2} \right)^2 \d \theta \\ &= \frac1{2^6} \int_0^{\pi/2}\left (1 - 2 \cos 2 \theta + \cos^{2} 2 \theta \right) \d \theta \\ &= \frac1{2^6} \int_0^{\pi/2}\left (1 - 2 \cos 2 \theta + \frac{\cos 4 \theta + 1}{2} \right) \d \theta \\ &= \frac1{2^6} \left [\frac32 \theta - \sin 2 \theta + \frac18 \sin 4 \theta \right]_0^{\pi/2} \\ &= \frac1{2^6} \frac{3 \pi}{4} \\ &= \frac{3 \pi}{2^8} \end{align*}

Solution:

1. \begin{align*} \sqrt{x^2+1} &= |x|\sqrt{1+\frac{1}{x^2}} \\ &=|x| \left (1 + \frac12 \frac{1}{x^2} + \cdots \right) & \text{if } \left (\frac{1}{x^2} < 1 \right) \\ &= |x| + \frac12 \frac{1}{|x|} + \cdots \\ &\approx |x| + \frac{1}{2|x|} \end{align*}
   

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2. 1. \begin{align*} && \tan( g(x) + g(-x)) &= \tan \left ( \tan^{-1}(\sqrt{x^2+1}-x) + \tan^{-1}(\sqrt{x^2+1}+x) \right) \\ &&&= \frac{\sqrt{x^2+1}-x+\sqrt{x^2+1}+x}{1-1} \\ \Rightarrow && g(x) + g(-x) &\in \left \{\cdots, -\frac{\pi}{2}, \frac{\pi}{2}, \cdots \right\} \end{align*} But \(g(x), g(-x) > 0\) and \(g(x), g(-x) \in (-\frac{\pi}{2}, \frac{\pi}{2})\), therefore it must be \(\frac{\pi}{2}\).
   2. \begin{align*} && \tan(2(k(x) + k(-x))) &= \tan(\tan^{-1}x + \tan^{-1}(-x)) \\ &&&= 0 \\ \Rightarrow && k(x)+k(-x) &\in \left \{\cdots, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \cdots \right\} \\ \end{align*} But \(k(x) \in (-\frac{\pi}{4}, \frac{\pi}{4})\), therefore \(k(x) + k(-x) = 0\).
   3. Let \(t = \tan k(x)\). \begin{align*} && \tan \left ( \tan^{-1} \frac{1}{x} \right) &= \frac{2 \tan\left ( \frac12 \tan^{-1} \frac1x \right)}{ 1-\tan^2\left ( \frac12 \tan^{-1} \frac1x \right)} \\ \Rightarrow && \frac1x &= \frac{2t}{1-t^2} \\ \Rightarrow && 1-t^2 &= 2tx \\ \Rightarrow && 0 &= t^2+2tx - 1 \\ \Rightarrow && 0 &= (t+x)^2 - 1-x^2 \\ \Rightarrow && t &= -x \pm \sqrt{1+x^2} \end{align*} Since \(t > 0\), \(t = \sqrt{1+x^2}-x = f(x) = \tan g(x)\)
   4. 

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   5. \begin{align*} \int_0^1 k(x) \d x &= \int_0^1 \frac12 \tan^{-1} \left ( \frac1x \right) \d x \\ &= \left [ \frac{x}{2} \tan^{-1}\left ( \frac1x \right) \right]_0^1 - \int_0^1 \frac{x}{2} \frac{-1/x^2}{1+1/x^2} \d x \\ &= \left [ \frac{x}{2} \tan^{-1}\left ( \frac1x \right) \right]_0^1 + \frac14 \int_0^1 \frac{2x}{1+x^2} \d x \\ &= \frac12 \frac{\pi}{4} + \frac14 \ln(2) \\ &= \frac{\pi + \ln 4}{8}\end{align*} Therefore \(\displaystyle \int_{-1}^0 g(x) \d x = -\frac{\pi + \ln 4}{8}\)

Solution:

1. 

   + - ↺
   By symmetry, the centre of mass will lie on the \(y\) axis. Notice that a single slice (when revolved around the \(y\)-axis) has volume \(y \cdot \pi \cdot ((x+ \delta x)^2 - x^2) = 2 \pi x y \delta x\), and COM at height \(\frac12 y\) so we can conclude: \[ \overline{y} \sum_{\delta x} 2 \pi x y \delta x = \sum_{\delta x} \pi xy^2 \delta x\] \begin{align*} && \overline{y} \int_0^r 2xy \d x &= \int_0^r y^2 x \d x \\ \Rightarrow && \overline{y} 2\int_0^r \left (r - \frac{x^n}{r^{n-1}} \right)x \d x &= \int_0^r \left (r - \frac{x^n}{r^{n-1}} \right)^2 x \d x \\ \Rightarrow && \overline{y} \left [r \frac{x^2}{2} - \frac{1}{r^{n-1}} \frac{x^{n+2}}{n+2} \right]_0^r &= \left [r^2 \frac{x^2}{2} - \frac{2}{r^{n-2}} \frac{x^{n+2}}{n+2} + \frac{1}{r^{2n-2}} \frac{x^{2n+2}}{2n+2} \right]_0^r \\ \Rightarrow && 2\overline{y} \left (\frac{r^3}{2} - \frac{r^3}{n+2} \right) &= \left (\frac12 r^4 - \frac{2}{n+2}r^4 + \frac{1}{2n+2}r^4 \right) \\ \Rightarrow && \overline{y}r^3 \frac{n}{(n+2)} &= r^4\frac{(n+1)(n+2)-2\cdot2\cdot(n+1)+(n+2)}{2(n+1)(n+2)} \\ \Rightarrow && \overline{y} \frac{n}{(n+2)} &= r \left ( \frac{n^2}{2(n+1)(n+2)} \right) \\ \Rightarrow && \overline{y} &= \frac{nr}{2(n+1)} \\ &&&= r \left (1 -p^n \right) \end{align*} as required.
2. \begin{align*} && r^{n-1}y &= r^n - x^n \\ \frac{\d}{\d x}: && r^{n-1} \frac{\d y}{\d x} &= -n x^{n-1} \\ && \frac{\d y}{\d x} &= -np^{n-1} \end{align*} Therefor the normal has the equation: \begin{align*} && \frac{y-r(1-p^n)}{x-rp} &= \frac{1}{np^{n-1}} \\ \Rightarrow && Y &= \frac{-rp}{np^{n-1}} + r(1-p^n) \\ &&&= r \left (1 - p^n - \frac{1}{np^{n-2}} \right) \end{align*} If \(n = 4\) then \begin{align*} && Y &= r\left (1 - p^4 - \frac{1}{4p^{2}} \right) \\ \Rightarrow && \frac{\d Y}{\d p} &= r \left (-4p^3 + \frac{1}{2p^3} \right) \end{align*} Therefore there is a stationary point if \(p^6 = \frac18 \Rightarrow p =2^{-1/2}\). Clearly this will be a maximum (sketch or second derivative) therefore, \(Y = r(1 - \frac14 - \frac{2}{4}) = \frac14 r\)
3. The centre of mass of this shape can be found using this table: \begin{array}{|c|c|c|} \hline \text{} & \overline{y} & \text{mass} \\ \hline r^3y = r^4 - x^4 & \frac{2r}{5} & \frac{4\pi r^3}{6} = \frac23 \pi r^3\\ ry = -(r^2 - x^2) & -\frac{r}{3}& \frac{2 \pi r^3}{4}=\frac12\pi r^3 \\ \text{combined} & \frac{(\frac25 \cdot \frac23-\frac13 \cdot \frac12)r^4}{\frac76 r^3} = \frac3{35}r & \frac76 \pi r^3\\ \hline \end{array} Normals to the surface through points on the upper surface will meet the \(y\)-axis between \((-\infty, \frac14 r)\), and since \(p = 0 \to -\infty\) and \(p = 1 \to -\frac14 r\), so normals will pass through \((0, \frac3{35}r)\) from two different points. Normals to the surface through points on the lower surface will go through \(-r(1 - p^2 - \frac12) =- r(\frac12 -p^2)\) which ranges monotonically from \(\frac12 r \to -\frac12 r\) so there will be one point where the normal goes through \(\frac3{35}r\). Therefore there are three angles where the vector \(\overrightarrow{AB}\) is not vertical but the normal to the surfaces runs through the centre of mass (ie the the solid can rest in equilibrium)

Solution:

1. Note that \(\,\) \begin{align*} && (8+x^3)^{-1} &= \tfrac18(1 + \tfrac18x^3)^{-1} \\ &&&= \tfrac18( 1 - \left (\tfrac{x}{2} \right)^3 + \left (\tfrac{x}{2} \right)^6 -\left (\tfrac{x}{2} \right)^9 + \cdots ) \end{align*} So \begin{align*} && \int_0^1 \frac{x^m}{8+x^3} \d x &= \int_0^1 x^m \sum_{k=0}^{\infty} \frac{1}{2^3} \left ( - \frac{x}{2} \right)^{3k} \d x \\ &&&= \sum_{k=0}^{\infty} (-1)^k \frac{1}{2^{3(k+1)}} \int_0^1 x^{m+3k} \d x \\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \frac{1}{3k+m+1} \\ \end{align*}
2. Notice that \begin{align*} && S &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k+3} - \frac{2}{3k+2} + \frac{4}{3k+1} \right) \\ &&&= \int_0^1 \frac{x^2}{8+x^3} \d x - 2\int_0^1 \frac{x^1}{8+x^3}+4\int_0^1 \frac{x^0}{8+x^3} \d x \\ &&&= \int_0^1 \frac{x^2-2x+4}{x^3+8} \d x \\ &&&= \int_0^1 \frac{1}{x+2} \d x = \left[ \ln(x+2)\right]_0^1 \\ &&&= \ln 3 - \ln 2 = \ln \tfrac32 \end{align*}
3. Firstly, note that \begin{align*} && \frac{2k+1}{(3k+1)(3k+2)} &= \frac13 \left ( \frac{1}{3k+1} +\frac{1}{3k+2} \right) \end{align*} so \begin{align*} && S_2 &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{72(2k+1)}{(3k+1)(3k+2)} \right) \\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} 24 \left ( \frac{1}{3k+1} +\frac{1}{3k+2} \right) \\ &&&= 24 \left [ \int_0^1 \frac{x^0}{8+x^3} \d x + \int_0^1 \frac{x^1}{8+x^3} \d x \right] \\ &&&= 24 \left [ \int_0^1 \frac{1+x}{(x+2)(x^2-2x+4)} \d x \right] \\ &&&= 24 \left [ \int_0^1 \frac{x+8}{12(x^2-2x+4)}-\frac{1}{12(x+2)} \d x \right] \\ &&&= 2\left [ \int_0^1 \frac{x+8}{x^2-2x+4}-\frac{1}{x+2} \d x \right] \\ &&&= 2\left [ \int_0^1 \frac{x-1}{x^2-2x+4}+ \frac{9}{(x-1)^2+3}-\frac{1}{x+2} \d x \right] \\ &&&=2 \left [ \int_0^1 \frac12\ln(x^2-2x+4)+3\sqrt{3}\arctan \frac{x-1}{\sqrt{3}} -\ln(x+2) \right]_0^1 \\ &&&=2 \left ( \frac12 \ln3+3\sqrt3 \arctan 0 - \ln 3 \right) - 2\left (\frac12 \ln 4-3\sqrt3 \arctan \frac{1}{\sqrt3} - \ln 2 \right) \\ &&&=\sqrt3 \pi - \ln3 \end{align*}

Solution:

1. \(\,\) \begin{align*} && I &= \int_{-a}^a \frac{1}{1+e^x} \d x \\ &&&= \int_{-a}^a \frac{e^{-x}}{e^{-x}+1} \d x \\ &&&= \left [ -\ln(1 + e^{-x} ) \right ]_{-a}^a \\ &&&= \ln(1 + e^a) - \ln(1 + e^{-a}) \\ &&&= \ln \left ( \frac{1+e^a}{1+e^{-a}} \right) \\ &&&= \ln \left ( e^a \frac{1+e^a}{e^a+1} \right) \\ &&& = a \end{align*}
2. Suppose \(g\) is continuous and \(\int_0^a g(x) \d x = 0\) for all \(a \geq 0\) then \(g(x) = 0\) for all \(x\). Proof: Differentiate with respect to \(a\) to obtain \(g(a) = 0\) for all \(a\) as required. \begin{align*} && a &= \int_{-a}^a \frac{1}{1+ f(x)} \d x \\ \Leftrightarrow && 1 &= \frac{1}{1 + f(a)} + \frac{1}{1 + f(-a)} \tag{FTC} \\ \Leftrightarrow && (1+f(x))(1+f(-x)) &= 2+f(-x) + f(x) \\ \Leftrightarrow && f(x) f(-x) & = 1 \end{align*}
3. \(\,\) \begin{align*} && J &= \int_{-a}^a \frac{h(x)}{1 + f(x)} \d x \\ y = - x: &&&=\int_{-a}^a \frac{h(-y)}{1 + f(-y)} \d y \\ &&&= \int_{-a}^a \frac{h(y)}{1 + f(-y)} \d y \\ \Rightarrow && 2J &= \int_{-a}^a h(x) \left ( \frac{1}{1+f(x)} + \frac{1}{1+f(-x)} \right) \d x \\ &&&= \int_{-a}^a h(x) \d x \\ &&&= \int_{-a}^0 h(x) \d x + \int_0^a h(x) \d x\\ &&&= \int_0^a h(-x) \d x + \int_0^a h(x) \d x \\ &&&= 2 \int_0^a h(x) \d x \\ \Rightarrow && J &= \int_0^a h(x) \d x \end{align*}
4. First notice that \(h(x) = \cos x = h(-x)\). Also notice that if \(f(x) = e^{2x}\) then \(f(x)f(-x) = 1\) so \begin{align*} && K &= \int_{-\frac12 \pi}^{\frac12\pi} \frac{e^{-x}\cos x}{\cosh x} \d x \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \frac{2h(x)}{1+f(x)} \d x \\ &&&= 2 \int_0^{\frac12 \pi} h(x) \d x \\ &&&= 2 \int_0^{\frac12 \pi} \cos x \d x \\ &&&= 2 \end{align*}

Solution: \begin{questionparts} \item \(\,\) \begin{align*} && I_n &= \int_0^{\beta} (\sec x + \tan x)^n \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} \left ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( (\sec x + \tan x)^{2}+1\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( \sec^2 x + \tan^2 x + 2\sec x \tan x + 1\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( 2\sec x \tan x +2\sec^2 x \right) \, \d x \\ &&& = \left [\frac1n(\sec x + \tan x)^{n} \right]_0^{\beta} \\ &&&= \frac1n[(\sec \beta + \tan \beta)^n - 1] \end{align*} Notice that by AM-GM \(\tfrac12( ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}) \geq (\sec x + \tan x)^{n}\) with equality not holding most of the time. Integrating we obtain our result. \item \(\,\) \begin{align*} && J_n &= \int_0^{\beta} (\sec x \cos \beta + \tan x )^n \d x \\ && \tfrac12( J_{n+1} + J_{n-1}) &= \tfrac12 \int_0^{\beta} \left ( (\sec x \cos \beta + \tan x )^{n+1} +(\sec x \cos \beta + \tan x )^{n-1}\right ) \d x \\ && &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( (\sec x \cos \beta + \tan x )^{2} + \right ) \d x \\ && &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec^2 x \cos^2 \beta + \tan^2 x+ 2\sec x \tan x \cos \beta +1 \right ) \d x \\ && &= \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\tfrac12(\cos^2 \beta +1)\sec^2 x \right ) \d x \\ && &< \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\sec^2 x \right ) \d x \\ &&&= \left [\frac1n (\sec x \cos \beta + \tan x)^{n} \right]_0^{\beta} \\ &&&= \frac1n[ (1 + \tan \beta)^n - \cos^n \beta] \end{align*} But notice we can use the same AM-GM argument from before to show that \(J_n < \tfrac12( J_{n+1} + J_{n-1}) < \frac1n[ (1 + \tan \beta)^n - \cos^n \beta]\)

Solution:

1. \(\,\) \begin{align*} && x &= \frac1{1-u} \\ \Rightarrow && \d x &= \frac{1}{(1-u)^2} \d u \\ && I &= \int \frac{1}{x^{\frac32}(x-1)^{\frac12} } \d x \\ &&&= \int \frac1{(1-u)^{-\frac32}u^{\frac12}(1-u)^{-\frac12}} (1-u)^{-2} \d u \\ &&&= \int u^{-\frac12} \d u \\ &&&= 2\sqrt{u} + C \\ &&&= 2\sqrt{1-\frac{1}{x}} + C \end{align*}
2. \(\,\) \begin{align*} && J &= \int \frac{1}{(x-2)^{\frac32}(x+1)^{\frac12}} \d x \\ y = x+1: &&&= \int \frac{1}{(y-3)^{\frac32}y^{\frac12}} \d y \\ y = 9(3-u)^{-1}: &&&= \int \frac1{\left (9(3-u)^{-1}-3 \right)^{\frac32}3(3-u)^{-\frac12}} \frac{9}{(3-u)^2} \d u \\ &&&= \int \frac1{\left (3u(3-u)^{-1} \right)^{\frac32}3(3-u)^{-\frac12}} \frac{9}{(3-u)^2} \d u \\ &&&= \frac{1}{\sqrt3} \int u^{-\frac32} \d u \\ &&&= -\frac2{\sqrt3} u^{-\frac12} + C \\ &&&= -\frac2{\sqrt3} \sqrt{\frac{y}{3(y-3)}} + C \\ &&&= -\frac2{3} \sqrt{\frac{x+1}{x-2}} + C \\ \end{align*}
3. \(\,\) \begin{align*} && K &= \int_2^{\infty} \frac{1}{(x-1)(x-2)^{\frac12}(3x-2)^{\frac12}} \d x \\ y = x - 1: &&&=\int_{y=1}^{\infty} \frac{1}{y(y-1)^{\frac12}(3y+1)^{\frac12}} \d y \\ y = (1-u)^{-1}: &&&= \int_{u=0}^{u=1} \frac{1}{(1-u)^{-1}(u(1-u)^{-1})^{\frac12}((4-u)(1-u)^{-1})^{\frac12}} \frac{1}{(1-u)^2} \d u \\ &&&= \int_0^1 \frac{1}{u^{\frac12}(4-u)^{\frac12}} \d u \\ &&&= \int_0^1 \frac{1}{\sqrt{4-(u-2)^2}} \d u \\ &&&= \left [-\sin^{-1} \left ( \frac{2-u}{2} \right) \right]_0^1 \\ &&&= \sin^{-1} 1 - \sin^{-1} \tfrac12 \\ &&&= \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \end{align*}