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2003 Paper 2 Q2
D: 1600.0 B: 1500.0
trigonometry harmonic form R-alpha method arctan arccos arcsin exact values identity manipulation

Write down a value of \(\theta\,\) in the interval \(\frac{1}{4}\pi< \theta <\frac{1}{2}\pi\) that satisfies the equation \[ 4\cos\theta+ 2\sqrt3\, \sin\theta = 5 \;. \] Hence, or otherwise, show that \[ \pi=3\arccos(5/\sqrt{28}) + 3\arctan(\sqrt3/2)\;. \] Show that \[ \pi=4\arcsin(7\sqrt2/10) - 4\arctan(3/4)\;. \]

Solution: If \(\theta = \frac{\pi}{3}\) then \(\cos \theta = \frac12, \sin \theta = \frac{\sqrt{3}}{2}\) and clearly the equation is satisfied. We can also solve this equation using the harmonic formulae, namely: \begin{align*} && 5 &= 4 \cos \theta + 2\sqrt{3} \sin \theta \\ &&&= \sqrt{4^2+2^2 \cdot 3} \cos \left (\theta -\tan^{-1} \left (\frac{2\sqrt{3}}{4}\right) \right) \\ \Rightarrow && \frac{5}{\sqrt{28}} &= \cos \left ( \frac{\pi}{3} - \tan^{-1} \left (\frac{\sqrt{3}}{2}\right) \right) \\ \Rightarrow && \frac{\pi}{3} &= \arccos\left( \frac{5}{\sqrt{28}}\right) + \arctan \left (\frac{\sqrt{3}}{2}\right) \end{align*} From which the result follows. Similarly, notice that \(3 \cos \theta + 4 \sin \theta = \frac{7}{\sqrt{2}}\) is clearly solved by \(\frac{\pi}{4}\), but also writing it in harmonic form, we have \begin{align*} &&\frac{7}{\sqrt{2}} &= 5 \sin \left (\theta + \tan^{-1} \left ( \frac{3}{4} \right) \right) \\ \Rightarrow && \frac{7\sqrt{2}}{10} &= \sin \left ( \frac{\pi}{4} + \tan^{-1} \left ( \frac{3}{4} \right) \right) \\ \Rightarrow && \frac{\pi}{4} &= \arcsin \left ( \frac{7\sqrt{2}}{10} \right) - \arctan \left ( \frac{3}{4} \right) \end{align*} as required.

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2000 Paper 1 Q8
D: 1484.0 B: 1484.0
integration absolute value integration by parts polynomial factorisation trigonometric piecewise functions harmonic form

Show that \[ \int_{-1}^1 \vert \, x\e^x \,\vert \d x =- \int_{-1}^0 x\e^x \d x + \int_0^1 x\e^x \d x \] and hence evaluate the integral. Evaluate the following integrals:

  1. \(\displaystyle \int_0^4 \vert\, x^3-2x^2-x+2 \,\vert \, \d x\,;\)
  2. \(\displaystyle \int_{-\pi}^\pi \vert\, \sin x +\cos x \,\vert \; \d x\,.\)

Solution: \begin{align*} && \int_{-1}^1 |x e^x |\d x &= \int_{-1}^0 |xe^x| \d x + \int_0^1 |xe^x| \d x \\ &&&= \int_{-1}^0 -xe^x \d x + \int_0^1 x \e^x \d x \\ &&&= -\int_{-1}^0 xe^x \d x + \int_0^1 x \e^x \d x \\ \\ && \int xe^x \d x &= xe^x - \int e^x \d x \\ &&&= xe^x - e^x \\ \\ \Rightarrow && \int_{-1}^1 |x e^x |\d x &= \left [ xe^x - e^x \right]_0^{-1}+ \left [ xe^x - e^x \right]_0^{1} \\ &&&= -e^{-1}-e^{-1} +e^{0} + e^1 - e^1 +e^0 \\ &&&= 2-2e^{-1} \end{align*}

  1. \(\,\) \begin{align*} && I &= \int_0^4 | x^3-2x^2-x+2| \d x \\ &&&= \int_0^4 |(x-2)(x-1)(x+1)| \d x\\ &&&= \int_0^1( x^3-2x^2-x+2) \d x- \int_1^2 ( x^3-2x^2-x+2) \d x + \int_2^4 ( x^3-2x^2-x+2) \d x \\ &&&= \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_0^1 - \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_1^2 + \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_2^4 \\ &&&= 2 \left ( \frac14 - \frac23 -\frac12 + 2\right) - 2 \left ( \frac14 2^4 - \frac23 2^3 -\frac12 2^2 + 2 \cdot 2\right)+ \left ( \frac14 4^4 - \frac23 4^3 -\frac12 4^2 + 2 \cdot 4\right) \\ &&&= \frac{133}{6} \end{align*}
  2. \(\,\) \begin{align*} && J &= \int_{-\pi}^\pi | \sin x + \cos x | \d x \\ &&&= \int_{-\pi}^{\pi} | \sqrt{2} \sin(x + \tfrac{\pi}{4})| \d x \\ &&&= 2\sqrt{2}\int_0^\pi \sin x \d x \\ &&&= 4\sqrt{2} \end{align*}

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