Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 heta ext{sin} heta, ewline y = ext{sec} heta, ewline 0 < heta < rac{eta}{2}$ The point P(k, 8) lies on C, where k is a constant - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 2

The area A of the region R can be found via integration:

$A = \int_0^{\beta} \left(\text{sec}^2\theta + \tan\theta \text{sec}^2\theta \right) d\theta.$

We compute this from the boundaries, where the area function rac{1}{2} encompasses contributions from each segment of R.

Using integration by parts for the area formula derived:

$A = [\tan\theta - \text{ln}\left|\text{sec}\theta + \tan\theta\right|]_0^{\frac{\pi}{4}}$ as the limits for integration.

Evaluating these integral boundaries yields the area R in precise numerical form:

1. Calculate at upper bound: $\tan\left(\frac{\pi}{4}\right) - \text{ln}\left|\sec\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right)\right|$
2. Substitute at $0$: $\tan(0) - \text{ln}|\text{sec}(0) + \tan(0)|$ Completing this yields the exact value of the area for R.