Problems

Filters
Clear Filters

1 problem found

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*}

View