6.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8c963567-d751-4898-b7a7-7095d90514f0-09_639_1179_246_386}
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\caption{Figure 3}
\end{figure}
Figure 3 shows a sketch of the curve with equation \(y = \frac { 2 \sin 2 x } { ( 1 + \cos x ) } , 0 \leqslant x \leqslant \frac { \pi } { 2 }\).
The finite region \(R\), shown shaded in Figure 3, is bounded by the curve and the \(x\)-axis.
The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 2 \sin 2 x } { ( 1 + \cos x ) }\).
| \(x\) | 0 | \(\frac { \pi } { 8 }\) | \(\frac { \pi } { 4 }\) | \(\frac { 3 \pi } { 8 }\) | \(\frac { \pi } { 2 }\) |
| \(y\) | 0 | | 1.17157 | 1.02280 | 0 |
- Complete the table above giving the missing value of \(y\) to 5 decimal places.
- Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(R\), giving your answer to 4 decimal places.
- Using the substitution \(u = 1 + \cos x\), or otherwise, show that
$$\int \frac { 2 \sin 2 x } { ( 1 + \cos x ) } d x = 4 \ln ( 1 + \cos x ) - 4 \cos x + k$$
where \(k\) is a constant.
- Hence calculate the error of the estimate in part (b), giving your answer to 2 significant figures.