Problems

Solution:

1. Suppose \(u_0 = u = \sin^2 \theta\) then \begin{align*} && u_1 &= 4 u_0 (1-u_0) \\ &&&= 4 \sin^2 \theta ( 1- \sin^2 \theta) \\ &&&= 4 \sin^2 \theta \cos^2 \theta \\ &&&= (2 \sin \theta \cos \theta)^2 \\ &&&= (\sin 2 \theta)^2 = \sin^2 2 \theta \\ \\ && u_2 & = 4u_1 (1-u_1) \\ &&&= 4 \sin^2 2\theta \cos^2 2 \theta \\ &&&= \sin^2 4 \theta \end{align*} Claim: \(u_n = \sin^2 2^n \theta\). Proof: (By Induction) Base case is clear, suppose it's true for \(n=k\), then \begin{align*} && u_{k+1} &= 4u_k(1-u_k) \\ &&&= 4 \sin^2 2^k \theta(1-\sin^2 2^k \theta) \\ &&&= (2 \sin 2^k \theta \cos 2^k \theta)^2 \\ &&&= (\sin 2^{k+1} \theta)^2 \\ &&&= \sin^2 2^{k+1} \theta \end{align*} Therefore since it is true for \(n = 1\) and if it's true for \(n = k\) it is true for \(n=k+1\) it must be true for all \(k\).
2. Suppose \(v_n = \alpha u_n + \beta\) then \begin{align*} && (\alpha u_{n+1}+\beta) &= -p(\alpha u_n + \beta)^2 + q(\alpha u_n + \beta) + r \\ &&&= -p\alpha^2u_n^2+\alpha(q-2p\beta) u_n -p \beta^2 +q \beta+r \\ \Rightarrow && u_{n+1} &= u_n(q-2p\beta -p \alpha u_n) -(p\beta^2-(q-1)\beta-r) \end{align*} So if \(\alpha = \frac{4}{p}\) and \(q-2p\beta = 4\) ie \(\beta = \frac{q-4}{2p}\) then we also need the constant term to vanish, ie \begin{align*} 0 &&&= p\beta^2-(q-1)\beta+r \\ &&&= p \left (\frac{q-4}{2p} \right)^2 - (q-1) \frac{q-4}{2p} - r \\ \Rightarrow && 0 &= p(q-4)^2 -(q-1)(q-4)2p - 4p^2r \\ \Rightarrow && 0 &= (q-4)^2-2(q-1)(q-4)-4pr \\ &&&= q^2-8q+16-2q^2+10q-8-4pr \\ \Rightarrow && 4pr &= -q^2+2q+8 \end{align*} Suppose \(v_{n+1} = -v_n^2 + 2v_n +2\) then since \(4\cdot 1 \cdot 2 = 8\) and \(8 + 4 -4 = 8\) we can apply our method. \(v_n = 4u_n + \frac{-2}{2} = 4u_n -1 = 4\sin^2 (2^{n-1} \pi)-1\)

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*} && 0 &= x + 3\sqrt{x} - \frac12 \\ \sqrt{x} = y: && 0&= y^2 + 3y - \frac12 \\ \Rightarrow && y &= \frac{-3\pm\sqrt{3^2+2}}{2} \\ &&&= \frac{-3 \pm \sqrt{11}}{2} \\ y > 0: && x &= \left ( \frac{\sqrt{11}-3}{2} \right)^2 \end{align*}
2. • \begin{align*} && 0 &= x + 10\sqrt{x+2} - 22 \\ y = \sqrt{x+2}: && 0 &= y^2 - 2 + 10y - 22 \\ &&&= y^2 + 10y - 24 \\ &&&= (y-2)(y+12) \\ \Rightarrow && y &= 2, -12 \\ y > 0: && x &= 2 \end{align*}
   • Let \(y = \sqrt{2x^2-8x-3}\), so \begin{align*} && 0 &= x^2 - 4x +\sqrt{2x^2-8x-3} - 9 \\ && 0 &= \frac{y^2+3}{2} + y - 9 \\ &&&= \frac12 y^2 +y - \frac{15}{2} \\ &&&= \frac12 (y-3)(y+5) \\ \Rightarrow && y &= 3,-5 \\ y > 0: && 9 &= 2x^2-8x-3 \\ \Rightarrow && 0 &= 2x^2-8x-12 \\ &&&= 2(x^2-4x-6) \\ \Rightarrow && x &= 2 \pm \sqrt{10} \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*}

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:

1. \(\,\) \begin{align*} && y &= xv \\ && y' &= v + xv' \\ \Rightarrow && 0 &= x^2 v \left ( v + x\frac{\d v}{\d x} \right) +(x^2v^2) - 2x^2 \\ &&&= 2x^2v^2 + x^3 v \frac{\d v}{\d x} - 2x^2 \\ \Rightarrow && 0 &= xv \frac{\d v}{\d x} + 2v^2-2 \\ \\ \Rightarrow && \frac{v}{1-v^2} \frac{\d v}{\d x} &= \frac{2}{x} \\ \Rightarrow && \int \frac{v}{1-v^2} \d v &=2 \ln |x| \\ \Rightarrow && -\frac12\ln |1-v^2| &= 2\ln |x| + C \\ \Rightarrow && 4\ln |x| + \ln |1-v^2| &= K \\ \Rightarrow && x^4(1-v^2) &= K \\ \Rightarrow && x^2(x^2-y^2) &= K \end{align*}
2. \(\,\) \begin{align*} && 0 &= xv \left (v +x \frac{\d v}{\d x} \right) + 6x + 5xv \\ &&&= x^2 v \frac{\d v}{\d x} +xv^2 + 6x+5xv \\ \Rightarrow && 0 &= xv\frac{\d v}{\d x} +v^2 +5v+6 \\ \Rightarrow && -\int \frac{1}{x} \d x &=\int \frac{v}{v^2+5v+6} \d v \\ \Rightarrow && -\ln |x| &= \int \frac{v}{(v+2)(v+3)} \d v \\ &&&= \int \left (\frac{3}{v+3} - \frac{2}{v+2} \right) \d v \\ \Rightarrow && -\ln |x| &= 3\ln |v+3| - 2 \ln |v+2| + C\\ \Rightarrow && -\ln |x| &= \ln \frac{|v+3|^3}{|v+2|^2} + C \\ \Rightarrow && \frac{1}{|x|}|v+2|^2 &= A|v+3|^3 \\ \Rightarrow && \frac{1}{|x|}|\frac{y}{x} + 2|^2&= A|\frac{y}{x} + 3|^3 \\ \Rightarrow && \frac{1}{|x|^3} |y +2x|^2 &= \frac{A}{|x|^3}|y + 3x|^3 \\ \Rightarrow && (y+2x)^2 &= A|y+3x|^3 \end{align*}

Solution: \begin{align*} && t &= x + \sqrt{1+x^2} \\ &&\frac1t &= \frac{1}{x+\sqrt{1+x^2}} \\ &&&= \frac{\sqrt{1+x^2}-x}{1+x^2-1} \\ &&&= \sqrt{1+x^2}-x \\ \Rightarrow && x &=\frac12 \left ( t - \frac1t\right) \\ \Rightarrow && \d x &=\frac12 \left (1 + \frac1{t^2} \right)\d t \\ \\ \Rightarrow && \int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x &= \int_{t=1}^{t = \infty}f(t) \frac12\left (1 + \frac1{t^2} \right)\d t \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right)f(t) \d t \end{align*} \begin{align*} && I &= \int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \d x \\ &&&= \int_0^\infty \frac1 {(x+\sqrt{x^2+1})^2} \d x \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right) \frac{1}{t^2} \d t \\ &&&= \frac12 \left [-\frac1t-\frac13\frac1{t^3} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac43 = \frac23 \end{align*} \begin{align*} && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&x &= \tan \theta\\ && \d x &= \sec^2 \theta = (1+x^2) \d \theta\\ && \tan\theta &= \frac{s}{\sqrt{1-s^2}}\\ \Rightarrow && \tan^2 \theta &= \frac{s^2}{1-s^2} \\ \Rightarrow && \sin \theta &= \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \\ && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&&= \int_0^{\frac12 \pi} \frac{1}{\left (1+ \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \right )^3} \d \theta \\ &&&= \int_{x=0}^{x=\infty} \frac{1}{\left(1 + \frac{x}{\sqrt{1+x^2}} \right)^3} \frac{1}{1+x^2} \d x \\ &&&= \int_0^{\infty} \frac{\sqrt{1+x^2}}{(\sqrt{1+x^2}+x)^3} \d x \\ &&J_a &= \int_0^{\infty} \frac{\sqrt{1+x^2}+x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \frac23 \\ &&J_b &= \int_0^{\infty} \frac{\sqrt{1+x^2}-x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \int_0^{\infty} \frac{1}{(\sqrt{1+x^2}+x)^4} \d x\\ &&&= \frac12\int_1^{\infty} \left (1 +\frac1{t^2} \right)\frac{1}{t^4} \d t \\ &&&= \frac12 \left [-\frac13 t^{-3}-\frac15t^{-5} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac8{15} = \frac4{15} \\ \Rightarrow && J &= \frac12(J_a+J_b) = \frac7{15} \end{align*}

Solution: \begin{align*} &&z &= y^n \left( \frac{\d y}{\d x}\right)^{2} \\ \Rightarrow && \frac{\d z}{\d x} &= ny^{n-1}\left( \frac{\d y}{\d x}\right)^{3} + y^{n} \cdot 2 \left( \frac{\d y}{\d x}\right) \left( \frac{\d^2 y}{\d x^2}\right) \\ &&&= y^{n-1} \left( \frac{\d y}{\d x}\right) \left (n \left( \frac{\d y}{\d x}\right)^2 + 2y \frac{\d^2 y}{\d x^2} \right) \end{align*}

1. Let \(z = y (y')^2\), then \begin{align*} && \frac{\d z}{\d x} &= y' \sqrt{y} \\ &&&= \sqrt{z} \\ \Rightarrow && \int z^{-1/2} \d z &= x+C \\ \Rightarrow && 2\sqrt{z} &= x + C \\ x = 0, z=0: && C &= 0 \\ \Rightarrow && y(y')^2 &= \frac14 x^2 \\ \Rightarrow &&\sqrt{y} \frac{\d y}{\d x} &= \frac{1}{2}x\\ \Rightarrow && \int \sqrt{y} \d y &= \int \frac{1}{2}x\d x \\ \Rightarrow && \frac23y^{3/2} &=\frac14x^2 + K \\ x = 0, y = 1: && K &= \frac23 \\ \Rightarrow && y &= \left (\frac38 x^2 + 1 \right)^{2/3} \end{align*}
2. Let \(z = y^{-2} (y')^2\) \begin{align*} && \frac{\d z}{\d x} &= y^{-3} \frac{\d y}{\d x} \left (-2 \left( \frac{\d y}{\d x}\right) + 2y \frac{\d^2 y}{\d x^2} \right) \\ &&&= y^{-3} \frac{\d y}{\d x} 2y^2 \\ &&&= 2y^{-1}(y') = 2 \sqrt{z} \\ \Rightarrow && 2\sqrt{z} &= 2x + C \\ x = 0, z = 0: && C&= 0 \\ \Rightarrow && z &= x^2 \\ \Rightarrow && \frac{\d y}{\d x} &= xy \\ \Rightarrow && \ln |y| &= \frac12 x^2 + K \\ x =0 , y =1; && K &= 0 \\ \Rightarrow && y &= e^{\frac12 x^2} \end{align*}

Solution: \begin{align*} \int_0^1 \frac{x \e^x}{1+x} \, \d x &= \int_0^1 \frac{(x+1-1) \e^x}{1+x} \, \d x \\ &= \int_0^1 \left ( e^x -\frac{\e^x}{1+x} \right )\, \d x \\ &= \e-1-E \end{align*} \begin{align*} \int_0^1 \frac{x^2 \e^x}{1+x} \, \d x &= \int_0^1 \frac{(x^2+x-x) \e^x}{1+x} \, \d x \\ &= \int_0^1 \left ( xe^x -\frac{x\e^x}{1+x} \right )\, \d x \\ &= \left [xe^{x} \right]_0^1 - \int_0^1 e^x \, \d x -(\e-1-E) \\ &= \e-(\e-1)-(\e -1 -E) \\ &= 2-\e + E \end{align*}

1. Since \(\displaystyle u = \frac{1-x}{1+x},\frac{\d u}{\d x} = \frac{-(1+x)-(1-x)}{(1+x)^2}\), \begin{align*} && \int_0^1 \frac{\e^{\frac{1-x}{1+x}}}{1+x}\, \d x &= \int_{u=1}^{u=0} \frac{e^u}{1+x} \cdot \frac{(1+x)^2}{-2} \d u \\ &&&= \int_0^1 \frac{(1+x) e^u}{2} \d u \\ &&&= \int_0^1 \frac{\left ( 1 + \frac{1-u}{1+u} \right) e^u}{2}\, \d u \\ &&&= \frac12 \int_0^1 \left (e^u + \frac{e^{u}}{1+u} - \frac{ue^u}{1+u} \right) \, \d u \\ &&&= \frac12 \left( \e-1 + E - (\e - 1 - E) \right) \\ &&&= E \end{align*}
2. Since \(\displaystyle u = x^2-1, \d u = 2x \d x\)\begin{align*} \int_1^{\sqrt2} \frac {\e^{x^2}}x \, \d x &= \int_{u=0}^{u=1} \frac{e^{u+1}}{x} \frac{1}{2x} \d u \\ &= \int_0^1 \frac{e^{u+1}}{2(u+1)} \d u \\ &= \frac{\e}{2} E \\ &= \frac{E\e}{2} \end{align*}

Solution: If \(f''(x) = kf(x)f'(x)\) then we can see \begin{align*} && I &= \int [\f'(x)]^2 \,[\f(x)]^n \d x \\ &&&= \int f'(x) \cdot f'(x) [f(x)]^n \d x \\ &&&= \left[ f'(x) \cdot \frac{[f(x)]^{n+1}}{n+1} \right] - \int f''(x) \frac{[f(x)]^{n+1}}{n+1} \d x \\ &&&= \frac{1}{n+1} \left (f'(x) [f(x)]^{n+1} - \int kf'(x) [f(x)]^{n+2} \d x \right) \\ &&&= \frac{1}{n+1} \left (f'(x) [f(x)]^{n+1} - k \frac{[f(x)]^{n+3}}{n+3} \right) +C\\ &&&= \frac{[f(x)]^{n+1}}{n+1} \left ( f'(x) - \frac{k[f(x)]^2}{n+3} \right) + C \end{align*}

1. If \(f(x) = \tan x, f'(x) = \sec^2 x, f''(x) = 2 \sec^2 x \tan x = 2 \cdot f(x) \cdot f'(x)\), so \(\tan\) satisfies the conditions for the theorem. \begin{align*} && I &= \int \sec^4 x \tan^n x \d x \\ &&&= \left [\sec^2x \cdot \frac{\tan^{n+1} x}{n+1} \right] - \int 2 \sec^2 x \tan x \cdot \frac{\tan^{n+1} x}{n+1} \d x \\ &&&= \left [\sec^2x \cdot \frac{\tan^{n+1} x}{n+1} \right] - 2 \cdot \frac{\tan^{n+3} x}{(n+1)(n+3)} \\ \end{align*} So \begin{align*} && I &= \int \frac{\sin^4 x}{\cos^8 x} \d x \\ &&&= \int \tan^4 x \sec^4 x \d x \\ &&&= \int [\sec^2 x]^2 [\tan x]^4 \d x \\ &&&= \frac{\tan^{5}x}{5} \left ( \sec^2 x - \frac{2 \tan^2 x}{7} \right) + C \end{align*}
2. \begin{align*} && I &= \int \sec^2x\, (\sec x + \tan x)^6\,\d x \\ &&&= \int (\sec x (\sec x + \tan x))^2 \cdot (\sec x + \tan x )^4 \d x \\ &&&= \frac{(\sec x + \tan x)^5}{5} \left ( \sec x (\sec x + \tan x) - \frac{(\sec x + \tan x)^2}{7} \right) + C \end{align*}

Solution:

1. \begin{align*} && 0 &= \frac{\d u}{\d x} - \left ( \frac{x+2}{x+1} \right)u \\ \Rightarrow && \int \frac1u \d u &= \int 1 + \frac1{x+1} \d x \\ \Rightarrow && \ln |u| &= x + \ln |x+1| + C \\ \Rightarrow && u &= A(x+1)e^x \end{align*}
2. If \(y = ze^{-x}\), \(y' = (z'-z)e^{-x}\), \(y'' = (z''-2z'+z)e^{-x}\) \begin{align*} && 0 &= (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y \\ y = ze^{-x}: && 0 &= (x+1) \left ( \frac{\d^2 z}{\d x^2} - 2\frac{\d z}{\d x} +z\right)e^{-x} +x \left ( \frac{\d z}{\d x} -z\right)e^{-x} - ze^{-x} \\ &&&= (x+1) \frac{\d^2 z}{\d x^2} -(x+2)\frac{\d z}{\d x} \\ \Rightarrow && \frac{\d}{\d x} \left ( \frac{\d z}{\d x}\right) &= \left ( \frac{x+2}{x+1}\right) \frac{\d z}{\d x} \end{align*} Therefore \(\frac{\d z}{\d x} = A(x+1)e^x \) and so \begin{align*} z &= A \int (x+1)e^{x} \d x \\ &= A \left ( \left [ (x+1)e^x\right] - \int e^x \d x \right) \\ &= A(x+1)e^x - Ae^x + B \\ y &= Ax + Be^{-x} \end{align*}
3. We have found the complementary solution. To find a particular integral consider \(y = ax^2 + bx + c\), then \(y' = 2ax+b, y'' = 2a\) and we have \begin{align*} && x^2+2x+1 &= 2a(x+1) + x(2ax+b) - (ax^2+bx+c) \\ \Rightarrow && x^2+2x+1 &= ax^2+ 2ax + 2a-c \\ \Rightarrow && a = 1, &c=1 \end{align*} so the general solution should be \[ y = Ax + Be^{-x} + x^2+1 \]

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 \(\displaystyle \int_2^3 \frac{\ln u}{u^2-1} \d u\)