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

How can we differentiate a product?

What techniques could we use to remove any occurrence of \(\cos x\) from the resulting expression?

Differentiating using the product rule gives \[\frac{d}{dx} (\cos x \sin^{n-1}x) = -\sin x\sin^{n-1}x +(n-1)\cos^2x \sin^{n-2}x,\] and then by using the identity \(\cos^2x = 1 - \sin^2x\) we get \[\begin{align*} \frac{d}{dx} (\cos x \sin^{n-1}x) &{}= -\sin^n x + (n-1)(1-\sin^2x)\sin^{n-2}x\\ &{}=(n-1)\sin^{n-2}x - n\sin^nx, \end{align*}\]

as required.

Writing \(s = \sin x\) and \(c = \cos x\) can be helpful. We have \[\frac{d}{dx} (cs^{n-1}) = -ss^{n-1} +(n-1)c^2s^{n-2}.\] Using \(c^2 = 1 - s^2\) we get \[\begin{align*} \frac{d}{dx} (cs^{n-1}) &{}=-s^n + (n-1)(1-s^2)s^{n-2}\\ &{}=(n-1)s^{n-2} - ns^n\\ &{}=(n-1)\sin^{n-2}x - n\sin^nx \end{align*}\]

as required.

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

Here we want to integrate by parts (our ‘product rule’ for integration).

The general formula for integration by parts is \[\int_a^b u \frac{dv}{dx} \, dx = \bigl[uv\bigr]_a^b - \int_a^b v\frac{du}{dx} \, dx.\]

We know how to integrate \(e^{-x}\) and differentiate \(\sin^n x\), so we take \(u=\sin^nx\) and \(\dfrac{dv}{dx} = e^{-x}\), giving \(v=-e^{-x}\). Then integration by parts yields \[I_n = \bigl[ -e^{-x} \sin^nx \bigr]_0^{\frac{1}{2}\pi} -\int_0^{\frac{1}{2}\pi} -e^{-x} n \sin^{n-1}x \cos x \, dx,\] which, after evaluating the limits, becomes \[\begin{equation} I_n = -e^{-\frac{1}{2}\pi} + n\int_0^{\frac{1}{2}\pi} e^{-x} \cos x \sin^{n-1}x \, dx. \label{eq:1} \end{equation}\]
  1. By using the results of (i) and (ii), or otherwise, show that \[(n^2 + 1) I_n = -e^{-\frac{1}{2}\pi} + n(n-1) I_{n-2}, \quad (n \ge 2).\]
Integrating the integral in \(\eqref{eq:1}\) by parts, taking \(u = \cos x\sin^{n-1}x\) and \(\dfrac{dv}{dx} = e^{-x}\) (and noticing that we’ve already worked out \(\dfrac{du}{dx}\) in part (i)) gives \[\begin{align*} \int_0^{\frac{1}{2}\pi} &e^{-x} \cos x \sin^{n-1}x \, dx\\ &\quad{}= \bigl[-e^{-x}\cos x \sin^{n-1}x \bigr]_0^{\frac{1}{2}\pi} + \int_0^{\frac{1}{2}\pi} e^{-x} \bigl((n-1) \sin^{n-2}x - n \sin^n x\bigr) \, dx\\ &\quad{}= (n-1)\int_0^{\frac{1}{2}\pi} e^{-x} \sin^{n-2}x \, dx -n \int_0^{\frac{1}{2}\pi} e^{-x} \sin^n x \, dx\\ &\quad{}= (n-1)I_{n-2} - nI_n \end{align*}\]

(The third line follows by evaluating the limits, which are both zero as long as \(n>1\); we are specifically told \(n\ge2\) here.)

Substituting this into \(\eqref{eq:1}\) now gives \[I_n = -e^{-\frac{1}{2}\pi} + n(n-1) I_{n-2} - n^2 I_n.\]

Rearranging gives \[\begin{equation} (n^2 + 1) I_n = -e^{-\frac{1}{2}\pi} + n(n-1) I_{n-2} \label{eq:2} \end{equation}\]

as required.

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

Rearranging \(\eqref{eq:2}\) gives \[I_n = \frac{1}{n^2 + 1} \bigl( -e^{-\frac{1}{2}\pi} + n(n-1) I_{n-2}\bigr).\]

So by setting \(n=4\) we get \[I_4 = \frac{1}{17} \bigl( -e^{-\frac{1}{2}\pi} + 12 I_2\bigr),\] and similarly, setting \(n=2\) gives \[I_2 = \frac{1}{5} \bigl( -e^{-\frac{1}{2}\pi} + 2 I_0\bigr).\]

By substituting the expression for \(I_2\) into that for \(I_4\) we get \[I_4 = \frac{1}{17} \left( -e^{-\frac{1}{2}\pi} + 12 \left(\frac{1}{5} \bigl( -e^{-\frac{1}{2}\pi} + 2 I_0\bigr)\right)\right),\] which rearranges to give \[I_4 = \frac{1}{85} \bigl(24 I_0 - 17e^{-\frac{1}{2}\pi}\bigr).\]

Now, by the definition of \(I_n\), \[\begin{align*} I_0 &{}=\int_0^{\frac{1}{2}\pi} e^{-x} \, dx \\ &{}= \bigl[-e^{-x}\bigr]_0^{\frac{1}{2}\pi} \\ &{}= 1 - e^{-\frac{1}{2}\pi}, \end{align*}\]

so the expression for \(I_{4}\) becomes \[I_4 = \frac{1}{85} \bigl(24 - 41 e^{-\frac{1}{2}\pi}\bigr).\]