6. (i) Find \[ \int x e^{x} \, dx \] (ii) Find \[ \int \frac{8}{(2x - 1)^{3}} \, dx, \quad x > \frac{1}{2} \] (iii) Given that \[ y = \frac{\pi}{6} \text{ at } x = 0, \] solve the differential equation \[ \frac{dy}{dx} = e^{x} \csc 2y \cos 2y \] - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 7

To solve this integral, we can use a substitution method. Let [ u = 2x - 1 ] such that [ du = 2 , dx ] or [ dx = \frac{du}{2} ].
Replacing in the integral gives:
[ \int \frac{8}{u^{3}} \cdot \frac{du}{2} = 4 \int u^{-3} , du ].
Integrating leads to:
[ 4 \cdot \left( -\frac{1}{2u^{2}} \right) + C = -\frac{2}{u^{2}} + C ]
Substituting back for [ u ]:
[ -\frac{2}{(2x - 1)^{2}} + C ].

To solve this differential equation, we start by separating variables.
Rearranging gives:
[ \frac{dy}{\csc 2y \cos 2y} = e^{x} , dx ]
Recall that [ \csc 2y \cos 2y = \frac{1}{\sin 2y} ]. Thus,
[ \sin 2y , dy = e^{x} , dx ]
Now, we integrate both sides:
[ -\cos 2y = e^{x} + C ].
Using the initial condition [ y = \frac{\pi}{6} \text{ when } x = 0 ]:
[ -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2} = e^{0} + C \Rightarrow C = -\frac{3}{2} ].
Thus, the solution becomes:
[ \cos 2y = -e^{x} - \frac{3}{2} ]
This solution represents the relationship between [ x ] and [ y ] in the defined equation.