Review question

# How could we integrate $e^{-x}\sin^n x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8134

## Solution

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 $I_4 = \frac{1}{85} \bigl(24 - 41 e^{-\frac{1}{2}\pi}\bigr).$