24: Differentiation and Integration with Trig Functions © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

24: Differentiation and 24: Differentiation and Integration with Trig Integration with Trig

FunctionsFunctions

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

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Module C4

OCR

This presentation follows on from “Differentiating some

Trig Functions”

In C3 you learnt to differentiate using the following methods:

• the product rule

• the quotient rule

This presentation gives you some practice using the above methods with trig functions.

• inspection

You also learnt to integrate some compound functions using :

Reminder:

xy sin3A function such as is a product, BUT we don’t need the product rule.When we differentiate, a constant factor just “tags along” multiplying the answer to the 2nd factor.

xdx

dyxy cos3sin3 e.g.

However, the product rule will work even though you shouldn’t use it

xvu sin3 and

N.B.

0dx

du

Reminder:

xy sin3A function such as is a product, BUT we don’t need the product rule.When we differentiate, a constant factor just “tags along” multiplying the answer to the 2nd factor.

xdx

dyxy cos3sin3 e.g.

However, the product rule will work even though you shouldn’t use it

xvu sin3 and

so,

dx

dvu

dx

duv

dx

dy x

dx

dycos3 as

before.

0dx

dux

dx

dvcos

Product Rule

Use the product rule to differentiate the following.

1.

xxy sin42.

xey x cos

Exercise

Product Rule

xxxxdx

dycossin4 43

dx

dvu

dx

duv

dx

dy

34xdx

du x

dx

dvcos

)cossin4(3 xxxx

4xu xv sinLet and

Remove common factors:

1.

xxy sin4

Solutions:

Product Rule

)sin(cos xexedx

dy xx dx

dvu

dx

duv

dx

dy

xedx

du x

dx

dvsin

xeu xv cosLet and

2.

xey x cos

xexedx

dy xx sincos

)sin(cos xxedx

dy x

Product Rule or Chain Rule?

Decide how you would differentiate each of the following ( but don’t do them ):

xy 2sin

xy 3sin

xxy sinxy sin2

(a)(b)(c)(d)

Product rule

Chain rule

This is a simple function

For products we use the product rule and for functions of a function we use the chain rule.

3)(sin xy Chain rule

Product Rule or Chain Rule?Exercis

eDecide with a partner how you would differentiate the following ( then do them if you need the practice ):1.

2sin xy

2.

xxy sin 3.

xy 2sin

Write C for the Chain rule and P for the Product Rule

C

C or P

P


1.

xxy sin P xvxu sin

xdx

dv

dx

ducos1

xxxdx

dycossin

uyxu sin2 2coscos2 xu

du

dyx

dx

du

2cos2 xxdx

dy

2sin xy

2.

C

Solutions

dx

du

du

dy

dx

dy

dx

dvu

dx

duv

dx

dy


2sin uyxu

xudu

dyx

dx

dusin22cos

xxdx

dycossin2

3.

xy 2sin

xvxu sinsin

xdx

dvx

dx

ducoscos

xxxxdx

dycossincossin

xxdx

dycossin2

CEither

P

Or

dx

du

du

dy

dx

dy

dx

dvu

dx

duv

dx

dy

Quotient Rule

Use the quotient rule to differentiate the following. 1.

x

xy

sin

2.

4

cos

x

xy

Exercise

2vdx

dvu

dx

duv

dx

dy

v

uy

The quotient rule is

Quotient RuleSolutio

n:

2vdx

dvu

dx

duv

dx

dy

v

uy

x

xxxy

2sin

cossin

x

xy

sin

1.

1dx

du xdx

dvcos

andxu xv sin

Quotient Rule

xdx

dusin 34x

dx

dv

2vdx

dvu

dx

duv

dx

dy

v

uy

andxu cos 4xv v

uy

24

34

)(

cos4sin

x

xxxx

dx

dy

8

3 )cos4sin(

x

xxxx

dx

dy

Solution:

5x5

)cos4sin(

x

xxx

2.

4

cos

x

xy

Quotient Rule

2)(cos

)sin)((sin))(cos(cos

x

xxxx

dx

dy

v

uy xu sin xv cos

xdx

ducos x

dx

dvsin

x

xx2

22

cos

sincos

We can now differentiate the trig function

x

xxy

cos

sintan

by writing

xy tan

Quotient Rule

1sincos 22 xx

x

xx

dx

dy2

22

cos

sincos xy tanSo,

This answer can be simplified:

xcos

1 is defined as

xsecAlso,

xdx

dy 2sec xy tanSo,

xdx

dy2cos

1

Quotient Rule

Use the quotient rule ( or, for (a) and (b), the chain rule ) to find the derivatives with respect to x of

xy cosec(a) xy sec(b) xy cot(c)Before you check the solutions, look in your formula books to see the forms used for the answers. Try to get your answers into these forms.

Exercise

Quotient Rule

(a) Solution:

xxysin

1cosec

2)(sin

)(cos1)0(sin

x

xx

dx

dy

xx

x

sinsin

cos

xx

x

sin

1

sin

cos

xx coseccot )cotcosec( xx

Quotient Rule

xxycos

1sec

2)(cos

)sin(1)0(cos

x

xx

dx

dy

xx

x

coscos

sin

xx

x

cos

1

cos

sin

xx sectan )tansec( xx

(b) Solution:

Quotient Rule

x

xxysin

coscot

2)(sin

coscos)sin(sin

x

xxxx

dx

dy

x

xx2

22

sin

)cos(sin

x2sin

1

x2cosec

(c) Solution:

Integration

xcos xsin Cxsin

dxyy

xcos C

Since integration is the reverse of differentiation, for the trig ratios we have

Integration

xy cos

e.g. 1Radians!

The definite integral can give an area, so this result may seem surprising. However, the graph shows us why it is correct.

This part gives a negative integral

This part gives a positive integral

but the integral for the area below is negative.

The areas above and below the axis are equal, . . .

dxx

0cos xsin 0

0sinsin 0

Integration

xy cos

e.g. 1Radians!




How would you find the area?Ans: Find the integral from 0 to

and double it.2

dxx

0cos xsin 0

0sinsin 0

Integration

dxx

0sin2. xcos 0

0coscos

11 2

xy sin The area is above the axis, so the integral gives the entire area.

N.B. 00cos

Integration

Before we try to integrate compound functions, we need to be able to recognise them, and know the rule for differentiating them.

dx

du

du

dy

dx

dy )(xguwhere ,

the inner function.

If ,))(( xgfy

e.g. For xy 3sin

We saw that in words this says: differentiate the inner

function multiply by the derivative of the outer function

we get dx

dyx3cos3

x3cos3

Integration

Since indefinite integration is the reverse of differentiation, we get

dxx3cos3

integrate the outer function

Cx 3sin

So,

The rule is:

dxxcos 3 Cx

3sin

3If we divide C by 3, we get another constant, say C1, but we usually just write C.

Integration


dxx3cos3

integrate the outer function divide by the derivative of the inner

function

Cx 3sin

So,

The rule is:

Cx

3sin dxxcos 33

Integration


dxx3cos3


function

Cx 3sin

The rule is:

Cx

3sin

So, dxxcos 33

Integration

Cx 3sin


dxx3cos3


function

So,

The rule is:

Cx

3sin dxxcos 33

Tip: We can check the answer by differentiating it. We should get the function we wanted to integrate.

Integration

However, we can’t integrate all compound functions in this way.

dxx 2cose.g. 2sin xx2

CTHIS IS WRONG !

Let’s try the rule on another example:


function

Integration

However, we can’t integrate all compound functions in this way.

dxx 2cose.g. 2sin xx2

CTHIS IS WRONG !

The rule has given us a quotient, which, if we differentiate it, gives:

2

22

)2(

)(sin2)cos2(2

x

xxxx

. . . nothing like the function we wanted to integrate.

Let’s try the rule on another example:

2vdx

dvu

dx

duv

Integration

When we differentiate the inner function of the 1st example, we get 3, a constant.

The 2nd example gives 2x,which is a function of x.

Dividing by the 3 does NOT give a quotient

of the form ( since v is a function of x ). v

u

What is the important difference between

dxcos and dxcos ?x3 2x

Integration





u

So, the important difference is that the 1st example has an inner function that is linear; it differentiates to a constant. The 2nd is non-linear. We can’t integrate it.


dxcos and dxcos ?x3 2x

IntegrationExercises

Solutions:

dxx4sin1. Cx

4

4cos

1.

Find

dxx4sin 2. 2

03cos

dxx

2. 2

03cos

dxx2

03

3sin

x

3

0sin

3

)(3sin 2

3

1

Integration

We found earlier that

xdx

dyxy 2sectan

so, Cxdxx tansec2

You need to remember this exception to the general rule that we can only integrate directly if the inner function is linear.We also have, for example,

dxx2sec2 Cx

2

2tan

Integration

There are 3 more important trig integrals.

We have a function of a function . . .

but the inner function . . .

xsin dx2)(

e.g. Find dxx2sin

Integration

e.g. Find dxx2sin


is not linear.

dx2)( xsin

but the inner function . . .

However, we can use a trig formula to convert the function into one that we

can integrate.


Integration

Which double angle formula can we use to change the function so that it can be

integrated?

e.g. Find dxx2sin

ANS: AA 2sin212cos )2( b

Rearranging the formula: AA 2sin212cos

)2cos1(sin 212 AA

So, dxxdxx )2cos1(sin 212

2

2sin x2

1 x C

Integration

The previous example is an important application of a double angle formula.

The next 2 are also important. Try them yourself.

Exercise

1. Find dxx2cos

2. Find dxxx cossin

IntegrationExercis

e1. Find dxx2cos

Solution:

1cos22cos 2 AA

AA 221 cos)2cos1(

dxxdxx )2cos1(cos 212So,

Cx

x

2

2sin21

Integration

2. Find dxxx cossin

AAA cossin22sin Solution: AAA cossin2sin2

1

dxxdxxx 2sincossin 21So,

Cx

2

2cos

2

1

Cx

4

2cos

Integration

SUMMARY

A2cosThe rearrangements of the double angle formulae for are

)2cos1(cos 212 AA

)2cos1(sin 212 AA

They are important in integration so you should either memorise them or be able to obtain them very quickly.

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

The Product Rule

SUMMARY

dx

dvu

dx

duv

dx

dy

Otherwise use the product rule:If ,uvy where u and v are both functions

of x

To differentiate a product:

Check if it is possible to multiply out. If so, do it and differentiate each term.


For products we use the product rule and for functions of a function we use the chain rule.

xy 2sin

xy 3sin

xxy sinxy sin2

(a)(b)(c)(d)

Product rule

Chain rule

This is a simple function

3)(sin xy Chain rule

For example,

Quotient Rule

2vdx

dvu

dx

duv

dx

dy

v

uy

The quotient rule is

Quotient Rule

))(cos(cos

)sin)((sin))(cos(cos

xx

xxxx

dx

dy

v

uy xu sin xv cos

xdx

ducos x

dx

dvsin

x

xx2

22

cos

sincos

With the quotient rule we can differentiate the trig function

x

xxy

cos

sintan

by writing

xy tan

Quotient Rule

1sincos 22 xx

x

xx

dx

dy2

22

cos

sincos xy tanSo,

This answer can be simplified:

xcos

1 is defined as

xsecAlso,

xdx

dy 2sec xy tanSo,

xdx

dy2cos

1

Integration

xcos xsin Cxsin

dxyy

xcos C

Since integration is the reverse of differentiation, for the trig ratios we have

Integration

xy cos

dxx

0cos

e.g. 1Radians! xsin

0

0sinsin 0




To find the area, find the integral from 0 to

and double it.

2

Integration

Before we try to integrate compound functions, we need to be able to recognise them, and know the rule for differentiating them.

dx

du

du

dy

dx

dy )(xguwhere ,

the inner function.

If ,))(( xgfy

e.g. For xy 3sin

We saw that in words this says: differentiate the inner

function multiply by the derivative of the outer function

we get dx

dyx3cos3

x3cos3

Integration

Cx 3sin


dxx3cos3


function

So,

The rule is:

Cx

3sin dxxcos 33

Tip: We can check the answer by differentiating it. We should get the function we wanted to integrate.

Integration





u

So, the important difference is that the 1st example has an inner function that is linear; it differentiates to a constant. The 2nd is non-linear. We can’t integrate it.


dxx3cos and dxx 2cos ?

Integration

We found earlier that

xdx

dyxy 2sectan

so, Cxdxx tansec2

You need to remember this exception to the general rule that we can only integrate directly if the inner function is linear.We also have, for example,

dxx2sec2 Cx

2

2tan

Integration

e.g. Find dxx2sin


dx2)( xsin

but the inner function is not linear

However, we can use a trig formula to covert the function into one that we can

integrate.


Integration

Which double angle formula can we use to change the function so that it can be

integrated?

e.g. Find dxx2sin

ANS: AA 2sin212cos )2( b

Rearranging the formula: AA 2sin212cos

)2cos1(sin 212 AA

So, dxxdxx )2cos1(sin 212

2

2sin x2

1 x C

Integration

SUMMARY

A2cosThe rearrangements of the double angle formulae for are

)2cos1(cos 212 AA

)2cos1(sin 212 AA

They are important in integration so you should either memorise them or be able to obtain them very quickly.

Documents

24: Differentiation and Integration with Trig Functions © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules