Show that \(\dfrac{d}{dx} (\cos x \sin^{n-1}x) = (n-1) \sin^{n-2}x - n \sin^{n}x\).

Given that \(I_n = \displaystyle\int_0^{\frac{1}{2}\pi} e^{-x} \sin^n x \, dx\), show that \[I_n = -e^{-\frac{1}{2}\pi} + n \int_0^{\frac{1}{2}\pi} e^{-x} \cos x \sin^{n-1}x \, dx, \quad (n \ge 1).\]

By using the results of (i) and (ii), or otherwise, show that \[(n^2 + 1) I_n = -e^{-\frac{1}{2}\pi} + n(n-1) I_{n-2}, \quad (n \ge 2).\]

Show that \(I_4 = \frac{1}{85} (24 - 41 e^{-\frac{1}{2}\pi})\).