Figure 3 shows a sketch of the curve with equation $y = \frac{2 \sin 2x}{1 + \cos x}$, $0 \leq x \leq \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 8

The trapezium rule states:

[ \text{Area} \approx \frac{h}{2} \left( f(a) + 2\sum_{i=1}^{n-1} f(x_i) + f(b) \right) ] Where $h$ is the width of each interval and $x_i$ are the points on the x-axis.

In this case, $h = \frac{\frac{\pi}{2} - 0}{4} = \frac{\pi}{8}$, so:

[ \text{Area} \approx \frac{\pi/8}{2} \left( 0 + 2(1.17157 + 1.02280) + 1.53626 \right) ] Calculating this yields an approximate area of $0.73508$.

Let $u = 1 + \cos x$. Then, we have:

[ du = -\sin x : dx \text{, or } dx = -\frac{du}{\sin x} ] To express the integral in terms of $u$, we rewrite:

[ \int \frac{2 \sin 2x}{1 + \cos x} : dx = \int \frac{2(2\sin x \cos x)}{u} \cdot -\frac{du}{\sin x} ] This simplifies to:

[ = -\int \frac{4\cos x}{u} : du \text{, leading to}] [ 4\ln(u) + k - 4\cos x = 4\ln(1 + \cos x) - 4\cos x + k]

The actual area of the region can be computed with the integral:

[ \int\limits_0^{\frac{\pi}{2}} \frac{2\sin 2x}{1 + \cos x} : dx ] Using the value found in part (c), we can compare this with the trapezium estimate of $0.73508$. Evaluating gives an actual area of approximately $1.15043$. Therefore, the error estimate is: [ |1.15043 - 0.73508| \approx 0.41535 \text{, or } \approx 0.42 \text{ (to 2 significant figures)} ]