Title: Default for 13.6 Integration of Trigonometric Functions
Course: HELM Workbooks 2023-2024

Integration of Trigonometric Functions

Integrals involving trigonometric functions are commonplace in engineering mathematics. This is especially true when modelling waves and alternating current circuits. When the root-mean-square (rms) value of a waveform, or signal is to be calculated, you will often find these results in an integral of the form \[\int\sin^2t \,\mathrm{d}t.\]

In this Section you will learn how such integrals can be evaluated.

Integration of trigonometric functions

Simple integrals involving trigonometric functions have already been dealt with in previous workbooks. See what you can remember:

Q: Exercise

You will need to refer to a Table of Trigonometric Identities to answer this question.

1. \({\displaystyle\int [value] \,\mathrm{d}x}=\)
2. \({\displaystyle\int_{[value]}^{[value]} [value] \,\mathrm{d}t}=\)
3. \({\displaystyle\int (\cos^2x + \sin^2x) \,\mathrm{d}x}=\)
Feedback: 1. We need to find the integral \(\displaystyle \int [value]\,\mathrm{d}x.\) Using \( \cos^2 x = \dfrac{1+ \cos 2x}{2}\) the integral can be written: \({\displaystyle\int \dfrac{1+ \cos 2x}{2} \,\mathrm{d}x}.\) Using the Key point we find \({\displaystyle\int [value]\,\mathrm{d}x= [value]}.\) 2. We need to find the integral \(\displaystyle{ \int_{[value]}^{[value]} [value] \,\mathrm{d}t}.\) In part (a) we found the the indefinite integral \( \displaystyle{ \int [value]\,\mathrm{d}x= [value]} .\) Hence \(\displaystyle{ \int_{[value]}^{[value]} [value] \,\mathrm{d}t= \left[ [value] \right]_{[value]}^{[value]}= [value]}.\) 3. We need to find the integral \(\displaystyle \int (\cos^2x + \sin^2x)\, \mathrm{d}x.\) Using \( \cos^2 x + \sin^2 x = 1\) the integral can be written: \(\displaystyle \int (\cos^2x + \sin^2x)\,\mathrm{d}x = \int 1 \,\mathrm{d}x= [value].\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Magnetic flux

The magnitude of the magnetic flux density on the axis of a solenoid, as in the Figure below, can be found by the integral: \[B= \int_{\beta_1}^{\beta_2} \frac{\mu_0 n I}{2} \sin\beta \, \mathrm{d}\beta\] where \(\mu_0\) is the permeability of free space (\(\approx 4\pi\times 10^{-7} \mbox{Hm}^{-1}\)), \(n\) is the number of turns and \(I\) is the current.

[Image: Solenoid and the angles defining its extent]
A solenoid and angles defining its extent.

Problem in words

Predict the magnetic flux in the middle of a long solenoid.

Mathematical statement of the problem

We assume that the solenoid is so long that \(\beta_1\approx 0\) and \(\beta_2\approx \pi\) so that \[B = {\displaystyle\int_{\beta_1}^{\beta_2}\dfrac{\mu_0 n I}{2}\sin \beta \, \mathrm{d}\beta}\approx {\displaystyle\int_0^{\pi}\dfrac{\mu_0 n I}{2}\sin\beta \, \mathrm{d}\beta}.\]

Mathematical analysis

The factor \(\dfrac{\mu_0 n I}{2}\) can be taken outside the integral, i.e. \[\begin{aligned} B &= \dfrac{\mu_0 n I}{2}{\displaystyle\int_0^{\pi}\sin\beta \, \mathrm{d}\beta} \\ &= \dfrac{\mu_0 n I}{2}\Bigg[ -\cos\beta\Bigg]_0^{\pi} \\ &=\dfrac{\mu_0 nI}{2}(-\cos\pi +\cos 0) \\ &=\dfrac{\mu_0 n I}{2}(-(-1)+1) = \mu_0 nI.\end{aligned}\]

Interpretation

The magnitude of the magnetic flux density at the midpoint of the axis of a long solenoid is predicted to be approximately \(\mu_0 nI\), i.e. proportional to the number of turns and proportional to the current flowing in the solenoid.

Orthogonality relations

In general two functions \(f(x), g(x)\) are said to be orthogonal to each other over an interval \(a\le x\le b\) if

\({\displaystyle\int_a^bf(x)g(x)\,\mathrm{d}x}=0\)

It follows from the previous Task that \(\sin x\) and \(\cos x\) are orthogonal to each other over the interval \(0\le x\le 2\pi\).

This is also true over any interval \(\alpha\le x\le \alpha + 2\pi\) (e.g. \(\pi /2\le x\le 5\pi/2\), or \(-\pi\le x\le \pi\)).

More generally there is a whole set of orthogonality relations involving these trigonometric functions on intervals of length \(2\pi\) (i.e. over one period of both \(\sin x\) and \(\cos x\)). These relations are useful in connection with a widely used technique in engineering, known as Fourier analysis where we represent periodic functions in terms of an infinite series of sines and cosines called a Fourier series. (This subject is covered in another section.)

We shall demonstrate the orthogonality property \(I_{mn}={\displaystyle\int_0^{2\pi}\sin mx\sin nx \,\mathrm{d}x} = 0\) where \(m\) and \(n\) are integers such that \(m\ne n\). The secret is to use a trigonometric identity to convert the integrand into a form that can be readily integrated.

You may recall the identity \(\sin A\sin B\equiv\dfrac{1}{2}(\cos (A-B)-\cos (A+B))\)

It follows, putting \(A=mx\) and \(B=nx\) that provided \(m\ne n\)

\[\begin{eqnarray} I_{mn}&=&\dfrac{1}{2}{\displaystyle\int_0^{2\pi}[\cos (m-n)x-\cos (m+n)x]\,\mathrm{d}x}\\ &=&\dfrac{1}{2}\left[\frac{\sin(m-n)x}{(m-n)}-\dfrac{\sin (m+n)x}{(m+n)}\right]_0^{2\pi}\\ &=&0\end{eqnarray}\]

because \((m-n)\) and \((m+n)\) will be integers and \(\sin (\)integer\(\times 2\pi)=0\). Of course \(\sin 0 = 0\). Why does the case \(m=n\) have to be excluded from the analysis? (left to the reader to figure out!)

The corresponding orthogonality relation for cosines

\(J_{mn}={\displaystyle\int_0^{2\pi}\cos mx\cos nx\,\mathrm{d}x} = 0\) follows by use of a similar identity to that just used. Here again \(m\) and \(n\) are integers such that \(m\ne n\).

Example

Use the identity \(\sin A\cos B\equiv\dfrac{1}{2}(\sin (A+B)+\sin (A-B))\) to show that

\(K_{mn}={\displaystyle\int_0^{2\pi}\sin mx\cos nx\,\mathrm{d}x} = 0\)

\(m\) and \(n\) integers, \(m\ne n\).

Solution

\begin{eqnarray*} K_{mn}&=&\dfrac{1}{2}{\displaystyle\int_0^{2\pi}[\sin (m+n)x+\sin (m-n)x]\,\mathrm{d}x}\\ &=&\dfrac{1}{2}\left[-\frac{\cos (m+n)x}{(m+n)}-\dfrac{\cos (m-n)x}{(m-n)}\right]_0^{2\pi}\\ &=&-\dfrac{1}{2}\left[\frac{\cos (m+n)2\pi -1}{(m+n)}+\dfrac{\cos (m-n)2\pi -1}{(m-n)}\right]\\ &=& 0 \end{eqnarray*}

(recalling that \(\sin (\)integer\(\times 2\pi) = 1\)

Example

Derive the orthogonality relation \(K_{mn}={\displaystyle\int_0^{2\pi}\sin mx\cos nx\,\mathrm{d}x} = 0\)\(m\) and \(n\) integers, \(m= n\)

Hint: You will need to use a different trigonometric identity to that used in the previous Example.

Solution

\(K_{mn}={\displaystyle\int_0^{2\pi}\sin mx\cos mx\,\mathrm{d}x}\)

Putting \(m=n\ne 0\), and then using the identity \(\sin 2A\equiv 2\sin A\cos A\) we get

\begin{eqnarray*}K_{mm}&=&\int_0^{2\pi}\sin mx\cos mx\ \, \mathrm{d}x\\ &=&\frac{1}{2}\int_0^{2\pi}\sin 2mx\ \,\mathrm{d}x\\ &=&\frac{1}{2}\left[-\frac{\cos 2mx}{2m}\right]_0^{2\pi}=-\frac{1}{4m}(\cos 4 m\pi -\cos 0)=-\frac{1}{4m}(1-1) = 0 \end{eqnarray*}

Putting \(m=n=0\) gives \(K_{00} = \dfrac{1}{2}{\displaystyle\int_0^{2\pi}\sin 0\cos 0\,\mathrm{d}x} = 0\).

Note that the particular case \(m=n=1\) was considered earlier in this Section.

To obtain a reduction formula for \(I_n\) we write \(\sin^nx=\sin^{n-1}(x)\sin x\) and use integration by parts.

Example

In the notation used earlier in this Workbook for integration by parts put \(f=\sin ^{n-1}x\) and \(g=\sin x\) and evaluate \(\dfrac{\mathrm{d}f}{\mathrm{d}x}\) and \({\displaystyle\int g\,\mathrm{d}x}\).

Solution

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

\({\displaystyle\int g\,\mathrm{d}x} = {\displaystyle\int \sin x\,\mathrm{d}x}=-\cos x\)

Now use the integration by parts formula on \({\displaystyle\int\sin^{n-1}x\sin x\,\mathrm{d}x}\).

[Do not attempt to evaluate the second integral that you obtain.]

We now need to evaluate \({\displaystyle\int \sin^{n-2}x\cos^2x\,\mathrm{d}x}\).

Putting \(\cos^2x = 1-\sin^2 x\) this integral becomes: \({\displaystyle\int\sin^{n-2}(x)\,\mathrm{d}x}-{\displaystyle\int\sin^n(x)\,\mathrm{d}x}\)

But this is expressible as \(I_{n-2}-I_n\) so finally, using this and the result from the last Task we have \(I_n={\displaystyle\int\sin^{n-1}(x)\sin x\,\mathrm{d}x}=\sin^{n-1}(x)(-\cos x)+(n-1)(I_{n-2}-I_n)\) from which we get the following Key Point:

Key Point

Reduction Formula

\(I_n = {\displaystyle\int\sin^nx\,\mathrm{d}x}\) \(I_n = -\dfrac{1}{n}\sin^{n-1}(x)\cos x+\dfrac{n-1}{n}I_{n-2}\)

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 \(I_0={\displaystyle\int\sin^0x\,\mathrm{d}x}={\displaystyle\int 1\,\mathrm{d}x}=x\).

Example

Use the reduction formula from the Key Point with \(n=2\) to find \(I_2\).

Solution

\begin{eqnarray*}I_2&=&-\dfrac{1}{2}[\sin x\cos x]+\dfrac{1}{2}I_0\\ &=&-\dfrac{1}{2}\left[\dfrac{1}{2}\sin 2x\right] +\dfrac{x}{2}+c\\{\displaystyle\int\sin^2x\,\mathrm{d}x} &=& -\dfrac{1}{4}\sin 2x +\dfrac{x}{2}+c \end{eqnarray*}

as obtained earlier by a different technique.

Example

Use the reduction formula from the Key Point to obtain \(I_6={\displaystyle\int\sin^6x\,\mathrm{d}x}\) with \(n=2\) to find \(I_2\).

Solution

Use the reduction formula in the Key Point \(I_6={\displaystyle\int\sin^6x\,\mathrm{d}x}\).

Firstly obtain \(I_6\) in terms of \(I_4\), then \(I_4\) in terms of \(I_2\):

Using the Key Point with \(n=6\) gives

\(I_6 = -\dfrac{1}{6}\sin^5x\cos x+\dfrac{5}{6}I_4\).

Then, using the Key Point again with \(n=4\), gives

\(I_4 = -\dfrac{1}{4}\sin^3x\cos x+\dfrac{3}{4}I_2\)

Now substitute for \(I_2\) from the previous Task to obtain \(I_4\) and hence \(I_6\).

\(I_4 = -\dfrac{1}{4}\sin^3x\cos x-\dfrac{3}{16}\sin 2x +\dfrac{3}{8}x + c\)

therefore

\(I_ 6 = -\dfrac{1}{6}\sin^5x\cos x-\dfrac{5}{24}\sin^3x\cos x-\dfrac{5}{32}\sin2x+\dfrac{5}{16}x + c\)

Definite integrals can also be readily evaluated using the reduction formula in the Key Point. For example, \(I_n={\displaystyle\int_0^{\pi /2}\sin^nx\,\mathrm{d}x}\) so \(I_{n-2}={\displaystyle\int_0^{\pi /2}\sin ^{n-2}x\,\mathrm{d}x}\)

We obtain, immediately \(I_n=\dfrac{1}{n}\Bigg[-\sin^{n-1}(x)\cos x\Bigg]_0^{\pi /2}+\dfrac{n-1}{n}I_{n-2}\)

or, since \(\cos \dfrac{\pi}{2} = \sin 0 = 0,\) \(I_n=\frac{(n-1)}{n}I_{n-2}\)

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

Reduction Formula - Wallis' Formula

Wallis' Formula

Given \(I_n={\displaystyle\int_0^{\pi /2}\sin ^nx\,\mathrm{d}x}\) or \(I_n={\displaystyle\int_0^{\pi /2}\cos^n x\,\mathrm{d}x}\)

\(\phantom{---}I_n=\dfrac{(n-1)}{n}I_{n-2}\)

Example

The total power \(P\) of an antenna is given by \(P={\displaystyle\int_0^{\pi}\frac{\eta L^2I^2\pi}{4\lambda^2}\sin^3\theta\phantom{.}\mathrm{d}\theta}\)

where \(\eta,\lambda, I\) are constants as is the length \(L\) of antenna.

Use the reduction formula for \({\displaystyle\int\sin^nx\,\mathrm{d}x}\) to obtain \(P\).

Solution

Ignoring the constants for the moment, consider

\(I_3 = {\displaystyle\int_0^{\pi}\sin^3\theta\,\mathrm{d}\theta}\) which we will reduce to \(I_1\) and evaluate.

\(I_1 = {\displaystyle\int_0^{\pi}\sin\theta\,\mathrm{d}\theta} = \Bigg[-\cos\theta\Bigg]_0^{\pi} = 2\) so by the reduction formula with \(n = 3\) \(I_3 = \dfrac{1}{3}\Bigg[-\sin^2x\cos x\Bigg]_0^{\pi}+\dfrac{2}{3}I_1=0+\dfrac{2}{3}\times 2 = \dfrac{4}{3}\)

We now consider the actual integral with all the constants.

Hence \(P = \dfrac{\eta L^2I^2\pi}{4\lambda^2}{\displaystyle\int_0^{\pi}\sin^3\theta\,\mathrm{d}\theta} = \dfrac{\eta L^2I^2\pi}{4\lambda^2}\times\dfrac{4}{3}\), so \(P=\eta\dfrac{L^2I^2\pi}{3\lambda^2}\).

A similar reduction formula to that in Key Point 9 can be obtained for \({\displaystyle\int\cos^nx\,\mathrm{d}x}\)

(see the Exercises at the end of this Workbook).

In particular if \(J_n={\displaystyle\int_0^{\pi /2}\cos^nx\,\mathrm{d}x}\) then \(J_n=\dfrac{(n-1)}{n}J_{n-2}\)

i.e. Wallis' formula is the same for \(\cos^nx\) as for \(\sin^n x\).

Harder trigonometric integrals

The following seemingly innocent integrals are examples, important in engineering, of trigonometric integrals that cannot be evaluated as indefinite integrals:

- \({\displaystyle\int\sin (x^2)\,\mathrm{d}x}\phantom{--}\) and \(\phantom{--}{\displaystyle\int\cos (x^2)\,\mathrm{d}x}\phantom{---}\)These are called Fresnel integrals.
- \({\displaystyle\int\frac{\sin x}{x}\,\mathrm{d}x}\phantom{---}\)This is called the Sine integral.

Definite integrals of this type, which are what normally arise in applications, have to be evaluated by approximate numerical methods.

Fresnel integrals with limits arise in wave and antenna theory and the Sine integral with limits in filter theory. It is useful sometimes to be able to visualize the definite integral. For example consider

\(F(t) = {\displaystyle\int_0^t\dfrac{\sin x}{x}\,\mathrm{d}x}, t>0\)

Clearly, \(F(0)={\displaystyle\int_0^0\dfrac{\sin x}{x}\,\mathrm{d}x} = 0\).

Recall the graph of \(\dfrac{\sin x}{x}\) against \(x\)

[Image: The sine of x over x as x increases]

For any positive value of \(t\), \(F(t)\) is the shaded area shown (the area interpretation of a definite integral was covered earlier in this Workbook). As \(t\) increases from \(0\) to \(\pi\), it follows that \(F(t)\) increases from \(0\) to a maximum value \(F(\pi)={\displaystyle\int_0^{\pi}\frac{\sin x}{x}\,\mathrm{d}x}\) whose value could be determined numerically (it is actually about 1.85). As \(t\) further increases from \(\pi\) to \(2\pi\) the value of \(F(t)\) will decrease to a local minimum at \(2\pi\) because the \(\dfrac{\sin x}{x}\) curve is below the \(x\)-axis between \(\pi\) and \(2\pi\). Note that the area below the curve is considered to be negative in this application. Continuing to argue in this way we can obtain the shape of the \(F(t)\) graph in the following Figure: (can you see why the oscillations decrease in amplitude?)

[Image]

The result \({\displaystyle\int_0^{\infty}\frac{\sin x}{x}\,\mathrm{d}x} = \dfrac{\pi}{2}\) is clearly illustrated in the graph (you are not expected to know how this result is obtained). Methods for solving such problems are dealt with in a later Section.

Q: Exercise

The mean square value of a function \(f(t)\) over the interval \(t=a\) to \(t=b\) is defined to be \[\dfrac{1}{b-a}\int_a^b (f(t))^2 \,\mathrm{d}t\]

Find the mean square value of \(f(t)=\sin t\) over the interval \(t=[value]\) to \(t = [value]\).
Feedback: We use the trigonometric identity \[\sin^2 \theta = \frac{1 - \cos 2 \theta}{2}.\] The mean square value is then \[\begin{aligned} \frac{1}{[value] - ([value])} \int_{[value]}^{[value]} \sin^2 t \,\mathrm{d}t &= [value] \int_{[value]}^{[value]} \frac{1 - \cos 2t}{2} \,\mathrm{d}t \\ &= [value] \left[[value]\right]_{[value]}^{[value]} \\ &= [value] \left[[value] - \left([value]\right)\right] \\ &= [value]. \end{aligned}\]
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

Use the identity \( \sin(A+B)+\sin(A-B)= 2\sin A \cos B \) to find the following integral:

\({\displaystyle\int [value] \,\mathrm{d}x}=\)

You will need to refer to a Table of Trigonometric Identities to answer this question.
Feedback: We can rewrite the given trigonometric identity as \[\sin A \cos B = \frac{\sin(A + B) + \sin(A - B)}{2}.\] Using this, we can rewrite the integrand setting \(A = [value]\) and \(B = [value]\). Then we evaluate the integral referring to the table. \[\begin{aligned} \int [value] \,\mathrm{d}x &= \int \left([value]\right) \,\mathrm{d}x \\ &= [value].\end{aligned}\]
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

Find \({\displaystyle\int ([value]) \,\mathrm{d}x}=\)

You will need to refer to a Table of Trigonometric Identities to answer this question.
Feedback: We use the following trigonometric identity to rewrite the integrad: \[\tan^2 \theta + 1 = \sec^2 \theta\] Note that \(\displaystyle \frac{d}{dx} \tan x = \sec^2 x\), so \(\displaystyle \int \sec^2 x \,\mathrm{d}x = \tan x + c\). Rewriting then gives \[\begin{aligned} \int ([value]) \,\mathrm{d}x &= \int \sec^2 x \,\mathrm{d}x \\ &= [value]. \end{aligned}\]
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Exercise

Write down the following integrals:

1. \(\displaystyle \int [value] \, \mathrm{d}x = \)
2. \(\displaystyle \int [value] \, \mathrm{d}x = \)
3. \(\displaystyle \int [value] \, \mathrm{d}x = \)
4. \(\displaystyle \int [value] \, \mathrm{d}x = \)
Feedback: 1. We simply use the integration table to get \[\int [value] \,\mathrm{d}x = [value]. \] 2. Again, we use the integration table to get \[\int [value] \,\mathrm{d}x = [value].\] 3. This is also in the table, which gives \[\int [value] \,\mathrm{d}x = [value].\] 4. The last one is in the table too, giving \[\int [value] \,\mathrm{d}x = [value].\]

Exercise

Use the trigonometric identity \(\displaystyle \sin^2 \theta \equiv \frac{1}{2}(1 - \cos \theta)\) to express the integral \(\displaystyle \int [value] \,\mathrm{d}x\) in an alternative form and hence evaluate it.

1. First use the identity to rewrite the integrand.
\(\displaystyle \int [value] \,\mathrm{d}x = \int \) \(\mathrm{d}x\)
2. Now evaluate the integral.
\(\displaystyle \int [value] \,\mathrm{d}x = \)
Feedback: We are asked to evaluate \(\displaystyle \int [value] \,\mathrm{d}x\) using the given trigonometric identity. 1. First, we rewrite the integrand using \(\displaystyle \sin^2 \theta \equiv \frac{1}{2}(1 - \cos \theta)\). This gives \[\int [value] \,\mathrm{d}x = \int [value] \,\mathrm{d}x.\] 2. Now we evaluate the integral, using the table. \[\int [value] \,\mathrm{d}x = \int [value] \,\mathrm{d}x = [value]\]

Exercise

Use the trigonometric identity \(\displaystyle \sin 2\theta \equiv 2 \sin \theta \cos \theta\) to express the integral \(\displaystyle \int [value] \,\mathrm{d}x\) in an alternative form and hence evaluate it.

1. First use the identity to rewrite the integrand.
\(\displaystyle \int [value] \,\mathrm{d}x = \int \) \(\mathrm{d}x\)
2. Now evaluate the integral.
\(\displaystyle \int [value] \,\mathrm{d}x = \)

This result is one example of what are called orthogonality relations.
Feedback: We are asked to evaluate \(\displaystyle \int [value] \,\mathrm{d}x\) using the given trigonometric identity. 1. First, we rewrite the integrand using \(\displaystyle \sin 2\theta \equiv 2 \sin \theta \cos \theta\). This gives \[\int [value] \,\mathrm{d}x = \int [value] \,\mathrm{d}x.\] 2. Now we evaluate the integral, using the table. \[\int [value] \,\mathrm{d}x = \int [value] \,\mathrm{d}x = [value]\]

The basic rules from which these results can be derived are summarised here:

Key Point

\({\displaystyle\int \sin (kx) \,\mathrm{d}x} = -\dfrac{\cos (kx)}{k} + c\)

\({\displaystyle\int \cos (kx) \,\mathrm{d}x} = \dfrac{\sin (kx)}{k} + c\)

In engineering applications it is often necessary to integrate functions involving powers of the trigonometric functions such as \({\displaystyle\int \sin^2x \,\mathrm{d}x}\) or \({\displaystyle\int \cos^2\omega t \,\mathrm{d}t}\). Note that these integrals cannot be obtained directly from the formulas in the Key Point above. However, by making use of trigonometric identities, the integrands can be re-written in an alternative form. It is often not clear which identities are useful and each case needs to be considered individually. Experience and practice are essential. Work through the following question.

Exercise

Given \(\displaystyle \int \sin^n (x) \,\mathrm{d}x\) write down the integrals represented by \(I_2\), \(I_{[value]}\) and \(I_{[value]}\).

1. \(\displaystyle I_2 = \int\) \(\mathrm{d}x\)
2. \(\displaystyle I_{[value]} = \int\) \(\mathrm{d}x\)
3. \(\displaystyle I_{[value]} = \int\) \(\mathrm{d}x\)
Feedback: We simply substitute the corresponding value of \(n\) in the formula. 1. \(\displaystyle I_2 = \int [value] \,\mathrm{d}x\) 2. \(\displaystyle I_{[value]} = \int [value] \,\mathrm{d}x\) 3. \(\displaystyle I_{[value]} = \int [value] \,\mathrm{d}x\)

Exercise

If \(\displaystyle I_n = \int_0^{\pi/2} \sin^2 x \,\mathrm{d}x\), calculate \(I_1\) and then use Wallis' formula, without further integration, to obtain \(I_3\) and \(I_5\).

Write all of your answers exactly, as integers or fractions.

1. First calculate \(I_1\).
\(\displaystyle \int_0^{\pi/2} [value] \,\mathrm{d}x = \)
2. Now use Wallis' formula to find \(I_3\).
\(\displaystyle \int_0^{\pi/2} [value] \,\mathrm{d}x = \)
3. Lastly, use Wallis' formula to find \(I_5\).
\(\displaystyle \int_0^{\pi/2} [value] \,\mathrm{d}x = \)
Feedback: We are given \(\displaystyle I_n = \int_0^{\pi/2} \sin^2 x \,\mathrm{d}x\). 1. This is a basic definite trigonometric integral: \[ I_1 = \int_0^{\pi/2} [value] \,\mathrm{d}x = \bigg[-\cos x\bigg]_0^{\pi/2} = [value] \] 2. Recall Wallis' formula: \[ I_n = \frac{n-1}{n} I_{n-2} \] We substitute \(n=3\) and \(I_1 = [value]\) in the formula. \[ I_3 = \int_0^{\pi/2} [value] \,\mathrm{d}x = \frac{3 - 1}{3} \cdot [value] = [value] \] 3. Again, we substitute \(n = 5\) and \(I_3 = [value]\) in Wallis' formula: \[ I_5 = \int_0^{\pi/2} [value] \,\mathrm{d}x = \frac{5 - 1}{5} \cdot [value] = [value] \]

Reduction formulae

You have seen earlier in this Workbook how to integrate \(\sin x\) and \(\sin^2x\) (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.