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2022 Paper 3 Q5
D: 1500.0 B: 1500.0
integration symmetric functions substitution even and odd functions hyperbolic functions continuous functions proof

  1. Show that \[ \int_{-a}^{a} \frac{1}{1+\mathrm{e}^x}\,\mathrm{d}x = a \quad \text{for all } a \geqslant 0. \]
  2. Explain why, if \(\mathrm{g}\) is a continuous function and \[ \int_0^a \mathrm{g}(x)\,\mathrm{d}x = 0 \quad \text{for all } a \geqslant 0, \] then \(\mathrm{g}(x) = 0\) for all \(x \geqslant 0\). Let \(\mathrm{f}\) be a continuous function with \(\mathrm{f}(x) \geqslant 0\) for all \(x\). Show that \[ \int_{-a}^{a} \frac{1}{1+\mathrm{f}(x)}\,\mathrm{d}x = a \quad \text{for all } a \geqslant 0 \] if and only if \[ \frac{1}{1+\mathrm{f}(x)} + \frac{1}{1+\mathrm{f}(-x)} - 1 = 0 \quad \text{for all } x \geqslant 0, \] and hence if and only if \(\mathrm{f}(x)\mathrm{f}(-x) = 1\) for all \(x\).
  3. Let \(\mathrm{f}\) be a continuous function such that, for all \(x\), \(\mathrm{f}(x) \geqslant 0\) and \(\mathrm{f}(x)\mathrm{f}(-x) = 1\). Show that, if \(\mathrm{h}\) is a continuous function with \(\mathrm{h}(x) = \mathrm{h}(-x)\) for all \(x\), then \[ \int_{-a}^{a} \frac{\mathrm{h}(x)}{1+\mathrm{f}(x)}\,\mathrm{d}x = \int_0^a \mathrm{h}(x)\,\mathrm{d}x\,. \]
  4. Hence find the exact value of \[ \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} \frac{\mathrm{e}^{-x}\cos x}{\cosh x}\,\mathrm{d}x\,. \]

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*}

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2006 Paper 3 Q2
D: 1700.0 B: 1513.8
integration trigonometric definite integral symmetry t-substitution even and odd functions algebraic manipulation parameter

Let \[ I = \int_{-\frac12 \pi}^{\frac12\pi} \frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} \, \d\theta \text{ and } J = \int_{-\frac12 \pi}^{\frac12\pi} \frac {\sec^2\theta}{1+\tan^2\theta\cos^22\alpha} \, \d\theta \] where \(0 < \alpha < \frac14\pi\,\).

  1. Show that \[ I = \int_{-\frac12 \pi}^{\frac12\pi} \frac {\cos^2\theta}{1+\sin\theta\sin2\alpha} \d\theta \] and hence that \[ \displaystyle 2I = \int_{-\frac12 \pi}^{\frac12\pi} \frac {2}{1+\tan^2\theta\cos^22\alpha} \, \d\theta \]
  2. Find \(J\).
  3. By considering \(I\sin^2 2\alpha +J\cos^2 2\alpha \), or otherwise, show that \(I =\frac12 \pi \sec^2\alpha\).
  4. Evaluate \(I\) in the case \(\frac14\pi < \alpha < \frac12\pi\).

Solution:

  1. \(\,\) \begin{align*} && I &= \int_{-\frac12 \pi}^{\frac12\pi} \frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} \, \d\theta \\ \phi = -\theta, \d \phi = - \d \theta: &&&= \int_{\phi=\frac12\pi}^{\phi=-\frac12\pi} \frac{\cos^2(-\phi)}{1-\sin(-\phi)\sin 2 \alpha} (-1) \d \phi \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \frac {\cos^2\phi}{1+\sin\phi\sin2\alpha} \d\phi \\ \\ && 2I &= \int_{-\frac12 \pi}^{\frac12\pi} \frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} \, \d\theta +\int_{-\frac12 \pi}^{\frac12\pi} \frac {\cos^2\theta}{1+\sin\theta\sin2\alpha} \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} +\frac {\cos^2\theta}{1+\sin\theta\sin2\alpha} \right) \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {2\cos^2\theta}{1-\sin^2\theta\sin^22\alpha} \right) \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {2\cos^2\theta}{1-(1-\cos^2\theta)(1-\cos^22\alpha)} \right) \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {2\cos^2\theta}{\cos^2\theta+\cos^22\alpha-\cos^2 \theta \cos^2 2 \alpha)} \right) \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {2}{1+\cos^22\alpha(\sec^2 \theta - 1))} \right) \, \d\theta \\ &&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( \frac {2}{1+\tan^2 \theta \cos^22\alpha} \right) \, \d\theta \\ \end{align*}
  2. \(\,\) \begin{align*} && J &= \int_{-\frac12 \pi}^{\frac12\pi} \frac {\sec^2\theta}{1+\tan^2\theta\cos^22\alpha} \, \d\theta \\ &&&= \left [\sec 2 \alpha \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) \right]_{-\frac12 \pi}^{\frac12\pi} \\ &&&= \sec(2\alpha)\pi = \frac{\pi}{\cos 2 \alpha} \end{align*}
  3. \(\,\) \begin{align*} && I\sin^2 2\alpha +J\cos^2 2\alpha &= \int_{-\frac12\pi}^{\frac12 \pi} \frac{\sin^2 2 \alpha+\cos^2 2 \alpha \sec^2 \theta}{1+\tan^2 \theta \cos^2 2\alpha} \d \theta \\ &&&= \int_{-\frac12\pi}^{\frac12 \pi} \frac{\sin^2 2 \alpha+\cos^2 2 \alpha (1 + \tan^2 \theta)}{1+\tan^2 \theta \cos^2 2\alpha} \d \theta \\ &&&= \pi \\ \\ \Rightarrow && I &= \frac{\pi - \pi \cos 2 \alpha}{\sin^2 2 \alpha} \\ &&&= \pi \frac{2\sin^2 \alpha}{4 \sin^2 \alpha \cos^2 \alpha} \\ &&&= \frac12 \pi \sec^2 \alpha \end{align*}
  4. If \(\frac14 \pi < \alpha < \frac12 \pi\) then our calculation for \(J\) is not correct. \begin{align*} && J &= \int_{-\frac12 \pi}^{\frac12\pi} \frac {\sec^2\theta}{1+\tan^2\theta\cos^22\alpha} \, \d\theta \\ &&&= \left [\sec 2 \alpha \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) \right]_{-\frac12 \pi}^{\frac12\pi} \\ &&&= \sec(2\alpha) \left ( \lim_{\theta \to \frac{\pi}{2}} \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) - \lim_{\theta \to -\frac{\pi}{2}} \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) \right) \\ &&&= \sec(2\alpha) \left ( \tan^{-1} \left ( \lim_{x\to -\infty} x \right) - \tan^{-1} \left ( \lim_{x\to \infty} x \right) \right) \\ &&&= -\pi \sec 2 \alpha \end{align*} Still using the same logic, we can say \begin{align*} && I &= \frac{\pi+\pi\cos 2 \alpha}{\sin^2 2 \alpha} \\ &&&= \pi \frac{2 \cos^2 \alpha}{4 \sin^2 \alpha \cos^2 \alpha}\\ &&&= \frac12 \pi \cosec^2 \alpha \end{align*}

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1996 Paper 2 Q8
D: 1600.0 B: 1485.5
second order differential equation functional equation complementary function particular integral trigonometric symmetry even and odd functions cos nθ

Suppose that \[{\rm f}''(x)+{\rm f}(-x)=x+3\cos 2x\] and \({\rm f}(0)=1\), \({\rm f}'(0)=-1\). If \({\rm g}(x)={\rm f}(x)+{\rm f}(-x)\), find \({\rm g}(0)\) and show that \({\rm g}'(0)=0\). Show that \[{\rm g}''(x)+{\rm g}(x)=6\cos 2x,\] and hence find \({\rm g}(x)\). Similarly, if \({\rm h}(x)={\rm f}(x)-{\rm f}(-x)\), find \({\rm h}(x)\) and show that \[{\rm f}(x)=2\cos x -\cos2x-x.\]

Solution: \begin{align*} && g(0) &= f(0)+f(-0) = 2f(0) = 2 \\ && g'(x) &= f'(x) - f'(-x) \\ && g'(0) &= f'(0) - f'(-0) = 0 \\ && g''(x) &= f''(x) +f''(-x) \\ \Rightarrow && g''(x) + g(x) &= f''(x) +f''(-x) + f(x) + f(-x) \\ &&&= f''(x)+ f(-x) +f''(-x) + f(x) \\ &&&= x + 3 \cos 2x + (-x + 3 \cos (-2x) ) \\ &&&= 6 \cos 2x \\ \end{align*} Considering the homogeneous part, we should expected a solution of the form \(g(x) = A \sin x + B \cos x\). Seeking an integrating factor of the form \(g(x) = C \cos 2x\) we see that \(-4C \cos 2x + C \cos 2x = 6 \cos 2x \Rightarrow -3C = 6 \Rightarrow C = -2\). Therefore the general solution is \begin{align*} && g(x) &= A\sin x + B \cos x - 2\cos 2x \\ && g(0) &= B - 2 = 2\\ && g'(0) &= A = 0 \\ \Rightarrow && g(x) &= 4\cos x - 2\cos 2x \\ \end{align*} \begin{align*} && h(0) &= f(0) - f(-0) = 0 \\ && h'(x) &= f'(x) + f'(-x) \\ && h'(0) &= f'(0) + f'(-0) = -2 \\ && h''(x) &= f''(x) - f''(-x) \\ \Rightarrow && h''(x) - h(x) &= f''(x) - f''(-x) -( f(x) - f(-x)) \\ &&&= f''(x) +f(-x)- (f''(-x) + f(x)) \\ &&&= x + 3\cos 2x - (-x + 3 \cos(-2x)) \\ &&&= 2x \end{align*} Considering the homogeneous part, we should expect a solution of the form \(Ae^x + Be^{-x}\). For a specific integral, we can take \(-2x\), ie \begin{align*} && h(x) &= Ae^x + Be^{-x} - 2x \\ && h(0) &= A+B =0 \\ && h'(0) &= A-B-2 =-2 \\ \Rightarrow && A &=B = 0 \\ \Rightarrow && h(x) &= -2x \end{align*} Therefore \(f(x) = \frac12(f(x) + f(-x)) + \frac12(f(x) -f(-x)) = 2\cos x - \cos 2x -x\)

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