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Integration usingtrig identities ora trig substitution mc-TY-intusingtrig-2009-1 Some integrals involving trigonometric functions can be evaluated by using the trigonometric identities. These allow the integrand to be written in an alternative form which may be more amenable to integration. On occasions a trigonometric substitution will enable an integral to be evaluated.

Both of these topics are described in this unit.

In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial on this topic, you should be able to: •use trigonometric identities to integratesin2x,cos2x, and functions of the formsin3xcos4x. •integrate products of sines and cosines using a mixture of trigonometric identities and integration by substitution •use trigonometric substitutions to evaluate integrals

Contents

1.Introduction2

2.Integrals requiring the use of trigonometric identities 2

3.Integrals involving products of sines and cosines 3

4.Integrals which make use of a trigonometric substitution 5

www.mathcentre.ac.uk 1c?mathcentre 2009

1. IntroductionBy now you should be well aware of the important results that?

coskxdx=1 ksinkx+c? sinkxdx=-1kcoskx+c However, a little more care is needed when we wish to integrate more complicated trigonometric functions such as? sin

2xdx,?

sin3xcos2xdx, and so on. In case like these trigonometric identities can be used to write the integrand in an alternative form which can be integrated more readily. Sometimes, use of a trigonometric substitution enables an integral to be found. Such substitu- tions are described in Section 4.

2. Integrals requiring the use of trigonometric identities

The trigonometric identities we shall use in this section, or which are required to complete the

Exercises, are summarised here:

2sinAcosB= sin(A+B) + sin(A-B)

2cosAcosB= cos(A-B) + cos(A+B)

2sinAsinB= cos(A-B)-cos(A+B)

sin

2A+ cos2A= 1

cos2A= cos2A-sin2A = 2cos 2A-1 = 1-2sin2A sin2A= 2sinAcosA

1 + tan

2A= sec2A

Some commonly needed trigonometric identities

Example

Suppose we wish to find?

0 sin2xdx. The strategy is to use a trigonometric identity to rewrite the integrand in an alternative form which does not include powers ofsinx. The trigonometric identity we shall use here is one of the 'double angle" formulae: cos2A= 1-2sin2A

By rearranging this we can write

sin 2A=1

2(1-cos2A)

Notice that by using this identity we can convert an expression involvingsin2Ainto one which has no powers in. Therefore, our integral can be written? 0 sin2xdx=? 01

2(1-cos2x)dx

www.mathcentre.ac.uk 2c?mathcentre 2009 and this can be evaluated as follows: 01

2(1-cos2x)dx=?12?

x-12sin2x?? 0 =?1

2x-14sin2x?

0 2

Example

Suppose we wish to find?

sin3xcos2xdx. Note that the integrand is a product of the functionssin3xandcos2x. We can use the identity

2sinAcosB= sin(A+B)+sin(A-B)to express the integrand as the sum of two sine functions.

WithA= 3xandB= 2xwe have

sin3xcos2xdx=1 2? (sin5x+ sinx)dx 1 2? -15cos5x-cosx? +c =-1

10cos5x-12cosx+c

Exercises 1

Use the trigonometric identities stated on page 2 to find the following integrals.

1. (a)?

cos

2xdx(b)?

π/2

0 cos2xdx(c)? sin2xcos2xdx

2. (a)

π/3

π/62cos5xcos3xdx(b)?

(sin

2t+ cos2t)dt(c)?

sin7tsin4tdt.

3. Integrals involving products of sines and cosines

In this section we look at integrals of the form

sin mxcosnxdx. In the first example we see how to deal with integrals in whichmis odd.

Example

Suppose we wish to find?

sin

3xcos2xdx.

Study of the integrand, and the table of identities shows that there is no obvious identity which will help us here. However what we will do is rewrite the termsin3xassinxsin2x, and use the identitysin2x= 1-cos2x. The reason for doing this will become apparent. sin

3xcos2xdx=?

(sinx·sin2x)cos2xdx sinx(1-cos2x)cos2xdx www.mathcentre.ac.uk 3c?mathcentre 2009 At this stage the substitutionu= cosx,du=-sinxdxenables us to rapidly complete the solution:

We find

sinx(1-cos2x)cos2xdx=-? (1-u2)u2du (u4-u2)du u5

5-u33+c

1

5cos5x-13cos3x+c

In the case whenmis even andnis odd we can proceed in a similar fashion, use the identity cos

2A= 1-sin2Aand the substitutionu= sinx.

Example

To find?

sin

4xcos3xdxwe write?

sin

4x(cos2x·cosx)dx. Using the identitycos2x= 1-sin2x

this becomes sin

4x(cos2x·cosx)dx=?

sin

4x(1-sin2x)cosxdx

(sin

4xcosx-sin6xcosx)dx

Then the substitutionu= sinx,du= cosxdxgives

(u4-u6)du=u5

5-u77+c

sin5x

5-sin7x7+c

In the case when bothmandnare even you should try using the double angle formulae, as in

Exercise 2 Q2 below.

Exercises 2

1. (a) Find?

cos

3xdx(b)?

cos

5xdx(c)?

sin

5xcos2xdx.

2. Evaluate

?sin2xcos2xdxby using the double angle formulae sin

2x=1-cos2x

2cos2x=1 + cos2x2

3. Using the double angle formulae twice find

sin

4xcos2xdx.

www.mathcentre.ac.uk 4c?mathcentre 2009

4. Integrals which make use of a trigonometric substitutionThere are several integrals which can be found by making a trigonometric substitution. Consider

the following example.

Example

Suppose we wish to find?1

1 +x2dx.

Let us see what happens when we make the substitutionx= tanθ. Our reason for doing this is that the integrand will then involve1

1 + tan2θand we have an identity

(1 + tan2A= sec2A) which will enable us to simplify this.

Withx= tanθ,dx

dθ= sec2θ, so thatdx= sec2θdθ. The integral becomes ?1

1 +x2dx=?11 + tan2θsec2θdθ

?1 sec2θsec2θdθ

1dθ

=θ+c = tan -1x+c So?1

1 +x2dx= tan-1x+c. This is an important standard result.

We can generalise this result to the integral

?1 a2+x2dx: We make the substitutionx=atanθ,dx=asec2θdθ. The integral becomes ?1 a2+a2tan2θasec2θdθ and using the identity1 + tan2θ= sec2θthis reduces to 1 a?

1dθ=1aθ+c

1 atan-1xa+c This is a standard result which you should be aware of and be prepared to look up when necessary.

Key Point

?1

1 +x2dx= tan-1x+c?1a2+x2dx=1atan-1xa+c

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ExampleSuppose we seek?1

4 + 9x2dx.

We proceed by first extracting a factor of 4 from the denominator: ?1

4 + 9x2dx=14?

11 +94x2dx

This is very close to the standard result in the previous keypoint except that the term9

4is not

really wanted. Let us observe the effect of making the substitutionu=3

2x, so thatu2=94x2.

Thendu=3

2dxand the integral becomes

1 4?

11 +94x2dx= =14?

11 +u2·23du

1 6?

11 +u2du

This can be finished off using the standard result, to give 1

6tan-1u+c=16tan-132x+c.

We now consider a similar example for which a sine substitution is appropriate.

Example

Suppose we wish to find?1

⎷a2-x2dx. The substitution we will use here is based upon the observations that in the denominator we have a terma2-x2, and that there is a trigonometric identity1-sin2A= cos2A(and hence (a2-a2sin2A=a2cos2A).

We tryx=asinθ, so thatx2=a2sin2θ. Thendx

dθ=acosθanddx=acosθdθ. The integral becomes ?1 ?1 ⎷a2cos2θacosθdθ ?1 acosθacosθdθquotesdbs_dbs17.pdfusesText_23

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