Reduction formulae

4 Reduction formulae

You have seen earlier in this Workbook how to integrate $sin x$ and ${sin}^{2} x$ (which is ${(sin x)}^{2}$ ). Applications sometimes arise which involve integrating higher powers of $sin x$ or $cos x$ . It is possible, as we now show, to obtain a reduction formula to aid in this Task.

Task!

Given $I_{n} = \int {sin}^{n} (x) d x$ write down the integrals represented by $I_{2}, I_{3}, I_{10}$

$I_{2} = \int {sin}^{2} x d x I_{3} = \int {sin}^{3} x d x I_{10} = \int {sin}^{10} x d x$

To obtain a reduction formula for $I_{n}$ we write

${sin}^{n} x = {sin}^{n - 1} (x) sin x$

and use integration by parts.

Task!

In the notation used earlier in this Workbook for integration by parts (Key Point 5, page 31) put $f = {sin}^{n - 1} x$ and $g = sin x$ and evaluate $\frac{d f}{d x}$ and $\int g d x$ .

$\frac{d f}{d x} = (n - 1) {sin}^{n - 2} x cos x$ (using the chain rule of differentiation),

$\int g d x = \int sin x d x = - cos x$

Now use the integration by parts formula on $\int {sin}^{n - 1} x sin x d x$ . [Do not attempt to evaluate the second integral that you obtain.]

\begin{array}{rcl} \int {sin}^{n - 1} x sin x d x & = & {sin}^{n - 1} (x) \int g d x - \int \frac{d f}{d x} \int g d x \\ = & {sin}^{n - 1} (x) (- cos x) + (n - 1) \int {sin}^{n - 2} x {cos}^{2} x d x \end{array}

We now need to evaluate $\int {sin}^{n - 2} x {cos}^{2} x d x$ . Putting ${cos}^{2} x = 1 - {sin}^{2} x$ this integral becomes:

$\int {sin}^{n - 2} (x) d x - \int {sin}^{n} (x) d x$

But this is expressible as $I_{n - 2} - I_{n}$ so finally, using this and the result from the last Task we have $I_{n} = \int {sin}^{n - 1} (x) sin x d x = {sin}^{n - 1} (x) (- cos x) + (n - 1) (I_{n - 2} - I_{n})$

from which we get Key Point 9:

Key Point 9

Reduction Formula

Given $I_{n} = \int {sin}^{n} x d x$

$I_{n} = - \frac{1}{n} {sin}^{n - 1} (x) cos x + \frac{n - 1}{n} I_{n - 2}$

This is our reduction formula for $I_{n}$ . It enables us, for example, to evaluate $I_{6}$ in terms of $I_{4}$ , then $I_{4}$ in terms of $I_{2}$ and $I_{2}$ in terms of $I_{0}$ where

$I_{0} = \int {sin}^{0} x d x = \int 1 d x = x$ .

Task!

Use the reduction formula in Key Point 9 with $n = 2$ to find $I_{2}$ .

\begin{array}{rcl} I_{2} & = & - \frac{1}{2} [sin x cos x] + \frac{1}{2} I_{0} \\ = & - \frac{1}{2} [\frac{1}{2} sin 2 x] + \frac{x}{2} + c \\ i.e. \int {sin}^{2} x d x & = & - \frac{1}{4} sin 2 x + \frac{x}{2} + c \end{array}

as obtained earlier by a different technique.

Task!

Use the reduction formula in Key Point 9 to obtain $I_{6} = \int {sin}^{6} x d x$ .

Firstly obtain $I_{6}$ in terms of $I_{4}$ , then $I_{4}$ in terms of $I_{2}$ :

Using Key Point 9 with $n = 6$ gives $I_{6} = - \frac{1}{6} {sin}^{5} x cos x + \frac{5}{6} I_{4}$ .

Then, using Key Point 9 again with $n = 4$ , gives $I_{4} = - \frac{1}{4} {sin}^{3} x cos x + \frac{3}{4} I_{2}$ Now substitute for $I_{2}$ from the previous Task to obtain $I_{4}$ and hence $I_{6}$ .

$I_{4} = - \frac{1}{4} {sin}^{3} x cos x - \frac{3}{16} sin 2 x + \frac{3}{8} x +$ constant

$∴ I_{6} = - \frac{1}{6} {sin}^{5} x cos x - \frac{5}{24} {sin}^{3} x cos x - \frac{5}{32} sin 2 x + \frac{5}{16} x + constant$

Definite integrals can also be readily evaluated using the reduction formula in Key Point 9. For example,

$I_{n} = \int_{0}^{π ∕ 2} {sin}^{n} x d x$ so $I_{n - 2} = \int_{0}^{π ∕ 2} {sin}^{n - 2} x d x$

We obtain, immediately

$I_{n} = \frac{1}{n} {[- {sin}^{n - 1} (x) cos x]}_{0}^{π ∕ 2} + \frac{n - 1}{n} I_{n - 2}$

or, since $cos \frac{π}{2} = sin 0 = 0,$ $I_{n} = \frac{(n - 1)}{n} I_{n - 2}$

This simple easy-to-use formula is well known and is called Wallis’ formula .

Key Point 10

Reduction Formula - Wallis’ Formula

Given $I_{n} = \int_{0}^{π ∕ 2} {sin}^{n} x d x$ or $I_{n} = \int_{0}^{π ∕ 2} {cos}^{n} x d x$

$I_{n} = \frac{(n - 1)}{n} I_{n - 2}$

Task!

If $I_{n} = \int_{0}^{π ∕ 2} {sin}^{n} x d x$ calculate $I_{1}$ and then use Wallis’ formula, without further integration, to obtain $I_{3}$ and $I_{5}$ .

$I_{1} = \int_{0}^{π ∕ 2} sin x d x = {[- cos x]}_{0}^{π ∕ 2} = 1$

Then using Wallis’ formula with $n = 3$ and $n = 5$ respectively

$I_{3} = \int_{0}^{π ∕ 2} {sin}^{3} x d x = \frac{2}{3} I_{1} = \frac{2}{3} \times 1 = \frac{2}{3}$

$I_{5} = \int_{0}^{π ∕ 2} {sin}^{5} x d x = \frac{4}{5} I_{3} = \frac{4}{5} \times \frac{2}{3} = \frac{8}{15}$

Task!

The total power $P$ of an antenna is given by

$P = \int_{0}^{π} \frac{η L^{2} I^{2} π}{4 λ^{2}} {sin}^{3} θ d θ$

where $η, λ, I$ are constants as is the length $L$ of antenna. Using the reduction formula for $\int {sin}^{n} x d x$ in Key Point 9, obtain $P$ .

Ignoring the constants for the moment, consider

$I_{3} = \int_{0}^{π} {sin}^{3} θ d θ$ which we will reduce to $I_{1}$ and evaluate.

$I_{1} = \int_{0}^{π} sin θ d θ = {[- cos θ]}_{0}^{π} = 2$

so by the reduction formula with $n = 3$

$I_{3} = \frac{1}{3} {[- {sin}^{2} x cos x]}_{0}^{π} + \frac{2}{3} I_{1} = 0 + \frac{2}{3} \times 2 = \frac{4}{3}$

We now consider the actual integral with all the constants.

Hence $P = \frac{η L^{2} I^{2} π}{4 λ^{2}} \int_{0}^{π} {sin}^{3} θ d θ = \frac{η L^{2} I^{2} π}{4 λ^{2}} \times \frac{4}{3}$ , so $P = η \frac{L^{2} I^{2} π}{3 λ^{2}}$ .

A similar reduction formula to that in Key Point 9 can be obtained for $\int {cos}^{n} x d x$ (see Exercise 5 at the end of this Workbook). In particular if

$J_{n} = \int_{0}^{π ∕ 2} {cos}^{n} x d x$ then $J_{n} = \frac{(n - 1)}{n} J_{n - 2}$

i.e. Wallis’ formula is the same for ${cos}^{n} x$ as for ${sin}^{n} x$ .

4 Reduction formulae

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