8. (a) Find \( \int_{0}^{\frac{\pi}{4}} x \cos 4x \, dx \)\n\nFigure 3 shows part of the curve with equation \( y = \sqrt{x} \sin 2x, \; x > 0 \)\n\nThe finite region \( R \), shown shaded in Figure 3, is bounded by the curve, the x-axis and the line with equation \( x = \frac{\pi}{4} \)\n\nThe region \( R \) is rotated through \( 2\pi \) radians about the x-axis to form a solid of revolution.\n(b) Find the exact value of the volume of this solid of revolution, giving your answer in its simplest form.\n(Solutions based entirely on graphical or numerical methods are not acceptable.) - Edexcel - A-Level Maths Pure - Question 1 - 2017 - Paper 9

To find the volume of the solid of revolution generated by rotating the region ( R ) around the x-axis, we apply the disk method: \n\n[ V = \pi \int_{0}^{\frac{\pi}{4}} (\sqrt{x} \sin 2x)^2 , dx ]\n[ = \pi \int_{0}^{\frac{\pi}{4}} x \sin^2 2x , dx ]\n\nUsing the identity ( \sin^2 2x = \frac{1 - \cos 4x}{2} ) gives us:\n\n[ = \pi \int_{0}^{\frac{\pi}{4}} x \cdot \frac{1 - \cos 4x}{2} , dx ]\n[ = \frac{\pi}{2} \left( \int_{0}^{\frac{\pi}{4}} x , dx - \int_{0}^{\frac{\pi}{4}} x \cos 4x , dx \right) ]\n\nCalculating the first integral, we have:\n[ \int_{0}^{\frac{\pi}{4}} x , dx = \left[ \frac{x^2}{2} \right]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \left( \frac{\pi^2}{16} \right) = \frac{\pi^2}{32} ]\n\nCombining this with the result from part (a):\n[ V = \frac{\pi}{2} \left( \frac{\pi^2}{32} - \frac{\pi - 1}{16} \right) = \frac{\pi^3}{64} - \frac{\pi^2 - 2\pi + 2}{32} ]\n[ = \frac{\pi^3 - 2\pi^2 + 4}{64} ] which is the final volume.