6. (a) Use integration by parts to show that
$$\int _ { 0 } ^ { \frac { \pi } { 4 } } x \sec ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 4 } \pi - \frac { 1 } { 2 } \ln 2 .$$
\begin{figure}[h]
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\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{a1b078fe-96e3-4d62-bf0d-415294ba022f-5_841_1202_459_434}
\end{figure}
The finite region \(R\), bounded by the equation \(y = x ^ { \frac { 1 } { 2 } } \sec x\), the line \(x = \frac { \pi } { 4 }\) and the \(x\)-axis is shown in Fig. 1. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
(b) Find the volume of the solid of revolution generated.
(c) Find the gradient of the curve with equation \(y = x ^ { \frac { 1 } { 2 } } \sec x\) at the point where \(x = \frac { \pi } { 4 }\).