Problems

Solution: \begin{align*} \cos 3\theta &= \cos 2\theta\cos\theta - \sin 2\theta \sin \theta \\ &= (2\cos^2\theta-1)\cos \theta - 2\cos \theta \sin^2 \theta \\ &= 2\cos^3\theta-\cos \theta - 2\cos \theta(1- \cos^2 \theta) \\ &= 4\cos^3 \theta - 3\cos \theta \end{align*} \begin{align*} 0 &= 24x^{3}-72x^{2}+66x-19 \\ &= 24(y+1)^3-72(y+1)^2+66(y+1)-19 \\ &= 24(y^3+3y^2+3y+1)-72(y^2+2y+1)+66(y+1)-19\\ &= 24y^3+(72-144+66)y+(24-72+66-19) \\ &= 24y^3-6y-1 \\ &= 24b'^3z^3 - 6b'z - 1 \\ &= \frac{2}{\sqrt{3}}(4 z^3 -3z) - 1 \\ \end{align*} Therefore if \(b = \sqrt{3}, a = 1\), we have: \(4z^3 - 3z = \frac{\sqrt{3}}{2}\) So if \(z = \cos \theta \Rightarrow \cos 3\theta = \frac{\sqrt{3}}2 \Rightarrow 3 \theta = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6} \Rightarrow \theta = \frac{\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}\). Since \(\frac{\cos x}{\sqrt{3}} < 1\) we only need to approximate the first part to 2 significant figures. Therefore: \begin{align*} \sqrt{3} &\approx 1 + \frac{1}{1 + \frac{1}{2+\frac11}} = \frac{7}{4} = 1.75\\ \cos \tfrac{\pi}{18} &\approx \cos \frac{1}{6} \approx 1 - \frac{1}{2} \frac{1}{6^2} = \frac{71}{72} \approx 1 - \frac{1}{70} = 1 - 0.014 = 0.986 \\ \frac{\cos \tfrac{\pi}{18}}{\sqrt{3}} & \approx \frac{.986}{1.75} = 0.57 \\ \end{align*} Final answers: \(1.57, 0.803, 0.629\). We wouldn't be able to solve \(x^3 + 1= 0\) using this method, as we would have 2 non-real roots

Solution: Let \(z = x^\alpha, \frac{\d z}{\d x}=\alpha x^{\alpha-1} \), then \begin{align*} \frac{\d y}{\d x} &= \frac{\d y}{\d z} \frac{\d z}{\d x} \\ &= \alpha x^{\alpha-1}\frac{\d y}{\d z} \\ \\ \frac{\d^2 y}{\d x^2} &= \frac{\d }{\d x} \left ( \alpha x^{\alpha-1}\frac{\d y}{\d z} \right) \\ &= \alpha (\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha x^{\alpha-1} \frac{\d ^2 y}{\d z^2} \frac{\d z}{\d x} \\ &= \alpha(\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha^2 x^{2\alpha-2} \frac{\d ^2y}{\d z^2} \end{align*} \begin{align*} && 0 &=x\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}-b\frac{\mathrm{d}y}{\mathrm{d}x}+x^{2b+1}y \\ &&&= x \left ( \alpha(\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha^2 x^{2\alpha-2} \frac{\d ^2y}{\d z^2}\right) - b \left ( \alpha x^{\alpha-1}\frac{\d y}{\d z} \right) + x^{2b+1}y \\ &&&= \alpha^2 x^{2\alpha-1} \frac{\d^2 y}{\d z^2} +\left (\alpha(\alpha-1)x^{\alpha-1}-b\alpha x^{\alpha-1} \right) \frac{\d y}{\d z} + x^{2b+1} y \\ \end{align*} If we set \(\alpha = b +1\) the middle term disappears, so we get \begin{align*} && 0 &= (b+1)^2 x^{2b+1} \frac{\d^2 y}{\d z^2} + x^{2b+1} y \\ \Rightarrow && 0 &= (b+1)^2 \frac{\d^2 y}{\d z^2} + y \\ \Rightarrow && y &= A \sin \left (\frac{z}{b+1} \right) + B \cos \left (\frac{z}{b+1} \right) \\ &&&= \boxed{A \sin \left (\frac{x^{b+1}}{b+1} \right) + B \cos \left (\frac{x^{b+1}}{b+1} \right)} \\ \\ \lim_{x \to 0}: && y &\to B \\ && \frac{\d y}{\d x} &= A x^b \cos\left (\frac{x^{b+1}}{b+1} \right) - B x^b \sin\left (\frac{x^{b+1}}{b+1} \right) \\ b>0: && \frac{\d y}{\d x} &\to 0 \\ \end{align*} So there are infinitely many different solutions with \(B = 1\) and \(A\) is anything it wants to be. If \(b = 0\) \(y' \to A\) so \(A =0 \) and unique. If \(b < 0\) \(x^b \to \infty\) so we need \(A = 0\), unique. However, we also need \(y' \to 0\), so we need to check \(y' = -x^b \sin \left ( \frac{x^{b+1}}{b+1}\right) \to 0\), \begin{align*} y' &= -x^b \sin \left ( \frac{x^{b+1}}{b+1}\right) \\ &\approx -x^b \left ( \frac{x^{b+1}}{b+1}\right) \\ &= - \frac{x^{2b+1}}{b+1} \end{align*} so we need \(2b+1>0 \Rightarrow b > -\frac12\). Therefore the solution is unique on \((-\frac12,0]\)

Solution:

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1. \begin{align*} && 4x^{2}(x^{2}-5)&=(5x-2)(x^{2}-4) \\ && \frac{2x(x^2-5)}{x^2-4} &= \frac{5x-2}{2x} = \frac52 -\frac1x \end{align*} Therefore it will have two roots as the hyperbola will intersect our graph in two places. (In the upper left and lower right quadrants).
2. \begin{align*} && 4x^{2}(x^{2}-5)^{2}&=(x^{2}-4)^{2}(x^{2}+1) \\ && \frac{2x(x^2-5)}{x^2-4} &= \frac{(x^2-4)(x^2+1)}{2x} \end{align*}
   

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   No solutions \begin{align*} && 4z^{2}(z-5)^{2}&=(z-4)^{2}(z+1) \\ && \frac{2z(z^2-5)}{z^2-4} &= \frac{(z+5)(z-4)(z+1)}{(z-5)(z+4)} \end{align*}
   

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   5 solutions

Solution:

1. \begin{align*} \int_{1}^{\mathrm{e}}\frac{\ln x}{x^{2}}\,\mathrm{d}x &= \left [-\frac{\ln x}{x} \right]_1^e + \int_1^e \frac{1}{x^2} \, \d x \\ &= -\frac{1}{e} + \left [ -\frac{1}{x} \right]_1^e \\ &= 1 - \frac{2}{e} \end{align*}
2. \begin{align*} \int\frac{\cos x}{\sin x\sqrt{1+\sin x}}\,\mathrm{d}x &= \int \frac{2u}{(u^2-1)u} \d u \tag{\(u^2 = 1+\sin x\)} \\ &= \int \frac{1}{u-1} - \frac{1}{u+1} \d u \\ &= \ln(u-1) - \ln (u+1) + C \\ &= \ln \l \frac{u-1}{u+1} \r + C \\ &= \ln \l \frac{\sqrt{\sin x + 1} + 1}{\sqrt{\sin x + 1} -1} \r + C \end{align*}

Solution: \begin{align*} I &=\int_{1}^{2}\frac{(2-2x+x^{2})^{k}}{x^{k+1}}\,\mathrm{d}x \\ u = x-1 &, \quad \d u = \d x \\ &= \int_{u = 0}^{u=1} \frac{(u^2+1)^k}{(u+1)^{k+1}} \d u \\ &= \boxed{\int_0^1 \frac{(1+x^2)^k}{(1+x)^{k+1}} \d x} \\ x = \tan \theta &, \quad \d x = \sec^2 \theta \d \theta \\ &= \int_{\theta = 0}^{\theta = \pi/4} \frac{\sec^{2k+2} \theta }{(1 + \tan \theta)^{k+1}} \d \theta \\ &= \int_0^{\pi/4} \frac{\d \theta}{\cos^{2k+2} \theta (\frac{\sin \theta + \cos \theta}{\cos \theta})^{k+1}} \\ &= \int_0^{\pi/4} \frac{\d \theta}{\cos^{k+1} \theta ({\sin \theta + \cos \theta})^{k+1}} \\ &= \int_0^{\pi/4} \frac{\d \theta}{\cos^{k+1} \theta (\sqrt{2} \cos (\frac{\pi}{4} - \theta))^{k+1}} \\ I &= \boxed{ \int_0^{\pi/4} \frac{\d \theta}{(\sqrt{2}\cos \theta \cos (\frac{\pi}{4} - \theta))^{k+1}}} \\ \end{align*} Since \(f(\theta) = \cos \theta \cos (\frac{\pi}{4} - \theta)\) is symmetric about \(\frac{\pi}{8}\) this integral is twice the integral to \(\frac{\pi}{8}\). \(\tan 2 \theta = \frac{2\tan \theta}{1 - \tan^2 \theta} \Rightarrow 1 = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}} \Rightarrow \tan \frac{\pi}{8} = \sqrt{2}-1\). Therefore, using the same substitution we must have: \[ I=2\int_{0}^{\sqrt{2}-1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x \] Let \(u = x^2\), then \(\d u = 2 x\d x\) \begin{align*} \int_{1}^{\sqrt{2}}\left(\frac{2-2x^{2}+x^{4}}{x^{2}}\right)^{k}\frac{1}{x}\,\mathrm{d}x &= \int_{u = 1}^{u = 2} \l \frac{2-2u+u^2}{u}\r^k \frac{1}{2u} \d u \\ &= \frac12 I \\ &= \int_{0}^{\sqrt{2}-1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x \\ u = 1+x & \quad \d u = \d x \\ &= \int_1^{\sqrt{2}} \frac{(1+(u-1)^2)^k}{u^{k+1}} \d u \\ &= \int_{1}^{\sqrt{2}}\left(\frac{2-2u+u^{2}}{u}\right)^{k}\frac{1}{u}\,\mathrm{d}x \\ &= \int_{1}^{\sqrt{2}}\left(\frac{2-2x+x^{2}}{x}\right)^{k}\frac{1}{x}\,\mathrm{d}x \end{align*}

Solution: Let \(y = xv\) then \(y' = v + xv'\) and so \(x^2y' = x^2v + x^3v'\) Our differential equation is now: \begin{align*} && x^2 y' &= \frac{1+y^2}{(1+x^2)\frac{y}{x}} \\ \Rightarrow && xy' &= \frac{(1+y^2)}{(1+x^2)y} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x(1+x^2)} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x} - \frac{x}{1+x^2} \\ \Rightarrow && \frac12 \ln(1+y^2) &= \ln x - \frac12 \ln(1+x^2) + C \\ \Rightarrow && \frac12 \ln (1 + y^2) &= \frac12 \ln \l \frac{x^2}{1+x^2}\r + C \\ \Rightarrow && 1+y^2 &= \frac{Dx^2}{1+x^2} \\ \Rightarrow && D &= 4 \quad \quad: (x = 1, v = 1, y = 1) \\ \Rightarrow && 1 + x^2v^2&= \frac{4x^2}{1+x^2}\\ \Rightarrow && v^2 &= \frac{3x^2-1}{x^2(1+x^2)} \\ \Rightarrow && v &= \sqrt{\frac{3x^2-1}{x^2(1+x^2)}} \\ \end{align*} As \(x \to \infty\), \(v \to 0\)

Solution: When we apply the substitution \(x = \frac1{t}\), \(t\) runs from \(-1 \to -\infty\) as \(x\) goes from \(-1 \to 0\). Then it runs from \(\infty \to 1\) as \(x\) runs from \(0 \to 1\). So we would be able to show that: \[ \int_{-1}^{1}\frac{1}{(1+x^{2})^{2}}\,\mathrm{d}x = \int_{-1}^{-\infty}\frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t + \int_{\infty}^1 \frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t \] Let \(x = \tan u, \d x = \sec^2 u \d u\) \begin{align*} \int_{-1}^1 \frac1{(1+x^2)^2} \d x &= \int_{u = -\pi/4}^{u = \pi/4} \frac{\sec^2 u}{(1+\tan^2 u)^2} \d u \\ &= \int_{u = -\pi/4}^{u = \pi/4} \frac{1}{\sec^2 u} \d u \\ &= \int_{-\pi/4}^{\pi/4} \cos^2 u \d u \\ &= \int_{-\pi/4}^{\pi/4} \frac{1 + \cos 2 u}{2} \d u \\ &= \left [ \frac{2u + \sin 2u}{4} \right]_{-\pi/4}^{\pi/4} \\ &= \frac{\pi}{4} + \frac{1}{2} \end{align*} Let \(t = \tan u, \d t = \sec^2 u \d u\) \begin{align*} \int_{-1}^1 \frac{-t^2}{(1+t^2)^2} \d x &= \int_{u = -\pi/4}^{u = \pi/4} \frac{-\tan^2 u \sec^2 u}{(1+\tan^2 u)^2} \d u \\ &= -\int_{u = -\pi/4}^{u = \pi/4} \frac{\tan^2 u}{\sec^2 u} \d u \\ &= -\int_{-\pi/4}^{\pi/4} \sin^2 u \d u \\ &= -\int_{-\pi/4}^{\pi/4} \frac{1 - \cos 2 u}{2} \d u \\ &= -\left [ \frac{2u - \sin 2u}{4} \right]_{-\pi/4}^{\pi/4} \\ &= \frac{1}{2}-\frac{\pi}{4} \end{align*}