Find \[\int_{0}^{\frac{\pi}{4}} x \cos 4x \, dx\] Figure 3 shows part of the curve with equation \[ y = \sqrt{x} \sin 2x, \quad x > 0 \] The finite region R, shown shaded in Figure 3, is bounded by the curve, the x-axis and the line with equation \[ x = \frac{\pi}{4} \] The region R is rotated around \(2\pi\) radians about the x-axis to form a solid of revolution - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 5

To solve this integral, we will use the method of integration by parts. Let:

[ u = x \quad \Rightarrow \quad du = dx ]

[ dv = \cos 4x , dx \quad \Rightarrow \quad v = \frac{1}{4} \sin 4x ]

Using integration by parts, ( \int u , dv = uv - \int v , du ), we have:

[ \int x \cos 4x , dx = \left[ x \cdot \frac{1}{4} \sin 4x \right]_{0}^{\frac{\pi}{4}} - \int \frac{1}{4} \sin 4x , dx ]

Evaluating the first term:

[ \left[ x \cdot \frac{1}{4} \sin 4x \right]_{0}^{\frac{\pi}{4}} = \frac{\frac{\pi}{4}}{4} \sin(4 \cdot \frac{\pi}{4}) - 0 = \frac{\pi}{16} \cdot 1 = \frac{\pi}{16} ]

Now, we need to evaluate the integral ( \int \frac{1}{4} \sin 4x , dx ):

[ \int \frac{1}{4} \sin 4x , dx = -\frac{1}{16} \cos 4x + C ]

Substituting back, we find:

[ \int x \cos 4x , dx = \frac{\pi}{16} + \frac{1}{16} \cos 4x \bigg|_{0}^{\frac{\pi}{4}} = \frac{\pi}{16} - \left( \frac{1}{16} \cos(\pi) - \frac{1}{16} \cos(0) \right) = \frac{\pi}{16} - \left( \frac{1}{16} (-1) - \frac{1}{16} (1) \right) = \frac{\pi}{16} + \frac{1}{16} + \frac{1}{16} = \frac{\pi + 1}{16} ]