91Ó°ÊÓ

First, understand that \(r= \cos 2 \theta\) represents a rose curve with 2 petals. This can be visualized by drawing the graph.

The boundaries of the integral for one petal are when r goes from 0 back to 0 as \(\theta\) progresses. We can find these boundaries by setting \(r= \cos 2 \theta = 0\) and solve for \(\theta\). This means that 2\(\theta\) equals any of \( \frac{\pi}{2}, \frac{3\pi}{2}\). So \(\theta\) is \(\frac{\pi}{4}, \frac{3\pi}{4}\). Consequently, the boundaries of integration are from \(\frac{\pi}{4}\) to \(\frac{3\pi}{4}\).

The area \(A\) of a region in a polar graph with boundaries \(\alpha\) and \(\beta\) for \(\theta\), and radial range r, is given by \( \frac{1}{2}\) the integral from \(\alpha\) to \(\beta\) of \(r^2 d \theta\). This gives us \( A= \frac{1}{2}\) \(\int_{{\frac{\pi}{4}}}^{{\frac{3\pi}{4}}}\) \(\cos^2 2\theta d\theta\). The trick to solve this integral is to use Power-Reduction Identity: \(\cos^2 x = \frac{1+ \cos 2x}{2}\). Therefore, the integral becomes \( \frac{1}{2}\) \(\int_{{\frac{\pi}{4}}}^{{\frac{3\pi}{4}}}\) \( ((\frac{1+ \cos 4\theta}{2}) \) \([\)d\(\theta]\). Using standard integral rules, the solution is \(\frac{1}{2}(\frac{1}{2} \theta + \frac{1}{8} \sin 4\theta)\) evaluated from \(\frac{\pi}{4}\) to \(\frac{3\pi}{4}\). The result gives us \(A = 0.3927\).