Unit 1

Mathematics – MATH 1111

Unit 1 1

UNIT 1 FURTHER DIFFERENTIATION AND INTEGRATION Unit Structure

1.0 Overview

1.1 Learning Objectives

1.2 Further Differentiation

1.2.1 Review

1.2.2 Logarithmic Differentiation

1.2.3 Differentiation of Inverse Trigonometric Functions

1.2.4 Parametric Differentiation

1.2.5 Taylor and Maclaurin Series

1.3 Further Integration

1.3.1 Integrals Involving Trigonometric Functions

1.3.2 Integrals Involving Inverse Trigonometric Functions

1.3.3 Integrals with a Quadratic in the Denominator

1.3.4 Integration by Parts

1.4 Summary

1.5 Supplementary Exercises

1.6 Answers to Activities and Supplementary Exercises


Unit 1 2

1.0 OVERVIEW

The objective of this Unit is to introduce further methods of differentiation and integration. The

main contents are as follows:

• Logarithmic differentiation.

• Differentiation of inverse trigonometric functions.

• Parametric differentiation.

• Taylor and Maclaurin series.

• Integrals involving trigonometric functions.

• Integrals involving inverse trigonometric functions.

• Integration by parts.

• General properties of the definite integral.

1.1 LEARNING OBJECTIVES By the end of this unit, you should be able to do the following: 1. Logarithmic differentiation.

2. Differentiation of inverse trigonometric functions.

3. Parametric differentiation.

4. Expansion of functions in power series

5. Integration of and involving inverse trigonometric functions.

6. Integration by parts.


Unit 1 3

1.2 FURTHER DIFFERENTIATION 1.2.1 Review Let n be a constant, u and v be functions of x.

Table 1.1: Differentiation

1 nxy = 1−= nnxdxdy

2 nxfy ))((= )())(( 1 xfxfndxdy n ′= −

dxdf

dxxdfxf ==′ )()(

3 vuy =

dxduv

dxdvu

dxdy

+=

2

2

2

2

2

2

2dx

udvdxdv

dxdu

dxvdu

dxyd

++=

Product rule for dxdy

4 vuy =

2vdxdvu

dxduv

dxdy −

=

Quotient Rule

5 dydx

dxdy 1= ,

dxdy

dydx 1= y is a function of x

only

6

dxdz

dzdy

dzdx

dzdy

dxdy

×==

Chain Rule given that

y and x are functions

of z

7 xdx

xd 1ln= , )(

)(1)(ln xfxfdx

xfd ′= Natural log

8 xx

edxed

=)( , )()(

)(

xfedx

de xfxf

′= , aadxad x

x

ln)(=

Exponential of

functions and a > 0

9 )cos()sin( xdx

xd= , )sin()cos( x

dxxd

−= , )(sec)tan( 2 xdx

xd=

10 )())(cos())(sin( xfxfdx

xfd ′= , )())(sin())(cos( xfxfdx

xfd ′−= ,

)())((sec))(tan( 2 xfxfdx

xfd ′=

Trigonometric

functions

11 dxdyyg

dxdy

dyydgyg

dxd )()()( ′== Implicit functions


Unit 1 4

Integration

1

⎪⎩

⎪⎨

⎧

−=+

−≠++=

+

∫1,ln

1,1

1

ncx

ncnx

dxx

n

n

cxdxx

+=∫ ln1

2

⎪⎪⎩

⎪⎪⎨

⎧

−=++

−≠+++

=+

+

∫1,ln1

1,)1()(

)(

1

ncbaxa

ncan

bax

dxbax

n

n

Can be applied to

power of a linear

expression in x only

3 dx

dxdvuuvdx

dxduv ∫∫ ⎥⎦

⎤⎢⎣⎡−=⎥⎦

⎤⎢⎣⎡

Integration by parts

4 cxfxfxdfdx

xfxf

+=≡∫∫ )(ln)()(

)()(' Function derivative

in numerator

5 Cxfdxxfxf

+=∫ )(2)()(' Function derivative

in numerator

6 cedxe xx +=∫ , ce

adxe baxbax += ++∫

1 , ca

adxax

x +=∫ ln

Exponentials

7 cxdxx +=∫ )sin()cos( , cxdxx +−=∫ )cos()sin( ,

cxdxx +=∫ )tan()(sec2

8 cxa

dxbax +=+∫ )sin(1)cos( , cbaxa

dxbax ++−=+∫ )cos(1)sin( ,

cbaxa

dxbax ++=+∫ )tan(1)(sec2

Trigonometry

Table 1.2

Tables 1.1-1.3 summarises all the basic formulae and rules that students need to know before

going deep through advanced differentiation and integration.


Unit 1 5

Common Mistakes

1 If xy 3= , then 13 −≠ xxdxdy

2 2

2

2

2

2

2

dxudv

dxvdu

dxyd

+≠ .

3 y

dyxd sin2

2

−= ; but,

ydx

yd sin12

2

−≠

4 C

xdx

xx +

+≠

+

∫ 122

1

5 C

xxdxx +

−≠∫ 2

cossin2

2 .

cx

edxex

x +≠∫ 2

22

Table 1.3

1.2.2 Logarithmic Differentiation

When a function consists of a number of factors it is often convenient to take logarithms before

differentiating. This will transform the problem of differentiating a product into that of

differentiating a sum.

A similar method can be applied to find the derivative of the function vuy = , where u and v are

functions of x.


Unit 1 6

Example 1

Find the derivative w.r.t. x of 2

2/3

2/12

)1()1()1(

+−+

xxx .

Step 1: Let 2

2/3

2/12

)1()1()1(

+−+

=x

xxy .

Then )1ln()1ln(2)1ln(ln 232

21 +−−++= xxxy .

Step 2: Differentiating both sides w.r.t. x, we have

)1(23

12

12

211

2 −−

−+

+=

xxxx

dxdy

y

)1)(1)(1(2

7732

23

+−++−+

=xxx

xxx

Step 3: Make dxdy subject of formula

=∴dxdy

)1)(1)(1(2773

2

23

+−++−+xxx

xxx .2

2/3

2/12

)1()1()1(

+−+

xxx

2/52/12

23

)1()1(2)1()773(

++−+−+

=xx

xxxx .

Example 2 Find 0),( >aadxd x .

Step 1: Let xay = .


Unit 1 7

Then on taking logs, we obtain axy lnln = .

Step 2: Now, differentiating w.r.t. x, we have

adxdy

yln1

=

.

Note: (i) 2ln22 xx

dxd

=

(ii) 19ln1919 tt

dtd −− −= .

Example 3

Differentiate xx xx )(lnsin + w.r.t. x.

Note: We can’t take logs directly since ])[(lnlnln])(lnln[ sinsin xxxx xxxx +≠+ .

We therefore do it in two parts, separately.

Step 1: Let xx xyxy )(ln, 2sin

1 == , so that dxdy

dxdyxx

dxd xx 21sin ])(ln[ +=+ .

Step 2: Now, xxy ln.sinln 1 = and )(lnln.ln 2 xxy =

Differentiating w.r.t. x, we have

xxx

xdxdy

ycosln1sin1 1

1

⋅+⋅=

aadxdy x ln=∴


Unit 1 8

⎟⎠⎞

⎜⎝⎛ ⋅+⋅=∴ xx

xxx

dxdy x cosln1sinsin1

Also, )(lnln1.ln1.1 2

2

xxx

xdxdy

y+=

⎟⎠⎞

⎜⎝⎛ +=∴ )(lnln

ln1)(ln2 x

xx

dxdy x

Step 3: Hence

=+ ])(ln[ sin xx xxdxd

⎟⎠⎞

⎜⎝⎛ + xx

xxx x cos.lnsinsin ⎟

⎠⎞

⎜⎝⎛ ++ )(lnln

ln1)(ln x

xx x .

Activity 1

1. Differentiate w.r.t. x each of the following:

(i) 432 )4()3()2( +++ xxx ;

(ii) 11

+−

xx ;

(iii) 1

)1(12 +

+−x

xx ;

(iv) 0,92

>+ aa x ;

(v) xx xx cossin )(tan)(cot + .

2. Using the fact that )0(ln >= aaaadxd xx , deduce that C

aadxa

xx +=∫ ln

.


Unit 1 9

1.2.2 Differentiation of Inverse Trigonometric Functions

Firstly let us consider the differentiation of the following functions:

Differentiation Proof

1 [ ] )(cos)cot( 2 xecdx

xd−=

[ ]

)(cossin

1tansec

)tan(1

)cot(

222

2

xecxx

xdx

xd

dxxd

−=−=−

=

⎥⎦

⎤⎢⎣

⎡

=

2 [ ] )cot()(cos)(cos xxecdx

xecd−=

[ ]

)cot()(cossincos

sin1

sincos

)sin(1

)(cos

2 xxecxx

xxx

dxx

d

dxxecd

−=−=−

=

⎥⎦

⎤⎢⎣

⎡

=

3 Exercise: Prove that [ ] )tan()(sec)(sec xxdx

xd= .

Table 1.4

Principal values of inverse trigonometric functions )(sin 1 x− , )(cos 1 x− and )(tan 1 x− .

Remark: Inverse trigonometric functions are not the reciprocals of the trig functions, i.e.,

xx

xx

tan1tan,

sin1sin 11 ≠≠ −− , etc.

The principal values of x1sin− are values of x1sin −=θ for which the trigonometric function

)sin(θ=x is a one to one function such that it inverse exists. Similar interpretation can be made

for the other trigonometric inverse functions.


Unit 1 10

Principle values

1 11 ≤≤− x 2

sin2

1 πθπ≤=≤− − x

2 11 ≤≤− x πθ ≤=≤ − x1cos0

3 ∞<<∞− x2

tan2

1 πθπ<=<− − x

Table 1.5

Figure 1.1-1.3 demonstrate relationship between trigonometry and inverse trigonometry for

)(sin 1 x− , )(cos 1 x− and )(tan 1 x− .

Figure 1.1


Unit 1 11

Figure 1.2

Figure 1.3


Unit 1 12

We next consider the differentiation of some important inverse trigonometry functions:

Let a be a positive constant such that ax ≠

(i) ,1sin22

1

xaax

dxd

−=− ax <

(ii) ,1cos22

1

xaax

dxd

−

−=− ax <

(iii) 22

1 1tan1xaa

xadx

d+

=⎟⎠⎞

⎜⎝⎛ −

(iv) 2

1

11cotx

xdxd

+−=−

(v)

1>x

(vi)

1>x

Table 1.6

Example 4: Prove that 22

1 1sinxaa

xdxd

−=− .

Proof:

Step 1: Let axy 1sin −= , then yax sin=

Step 2: Differentiating x w.r.t to y, we have

2

2 1sin1cos ⎟⎠⎞

⎜⎝⎛−=−==

axayaya

dydx

Step 3: Usingdydx

dxdy 1= , we have

22

2

222

11

1

1xa

axa

a

ax

adxdy

−=

−=

⎟⎠⎞

⎜⎝⎛−

= where ayax <= sin .

,1

1se2

1

−=−

xxx

dd

,1

1cose2

1

−−=−

xxx

dd


Unit 1 13

Example 5: Prove that 221tan

xaa

ax

dxd

−=− .

Proof:

Step 1: Let axy 1tan −= , then yax tan=

Step 2: Differentiating x w.r.t to y, we have

)1()tan1(sec2

22 ⎟⎠⎞

⎜⎝⎛−=−==

axayaya

dydx

Step 3: Usingdydx

dxdy 1= , we have

22

2

222

)(

1

)1(

1xa

a

axaa

axa

dxdy

−=

−=

⎟⎠⎞

⎜⎝⎛−

= .

Example 6: Prove that 1,1

1cos2

1 >−

−=− x

xxxec

dxd .

Proof: Step 1: Let xecy 1cos −= , then yecx cos= Step 2: Differentiating x w.r.t to y, we have

11coscoscotcos 22 −−=−−=−= xxyecyecyyecdydx

Step 3: Usingdydx

dxdy 1= , we have

112 −

−=

xxdxdy where 1cos >= yecx


Unit 1 14

Example 7

If ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= −2

21

4)4sin

xxy , show that 24

4xdx

dy+−

= .

Solution: Note: The function is quite complicate and it’s preferable to use a substitution which allows us to apply the chain rule.

Step 1: Let ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= 2

2

4)4

xxu , so that uy 1sin −= , or yu sin= .

Then,

2

2

2

2

22

44

441

1sin1cos

xx

xx

uxydydu

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

−=

−=−==

Step 2:

Also, 2222

22

)4(16

)4()2)(4()2)(4(

xx

xxxxx

dxdu

+−

=+

−−−+=

Step 3:

Now, using the chain rule

dxdu

dydu

dxdu

dudy

dxdy

×⎟⎟⎠

⎞⎜⎜⎝

⎛=×= 1

2222 44

44

)4(16

xxx

xx

dxdy

+−

=++

−=


Unit 1 15

Example 8

If ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= −2

21

11cos

xxy , show that 21

2xdx

dy+

= .

Since the function is rather complicated, we’ll use a substitution and the chain rule.

Let, 2

2

11

xxu

+−

= , so that uy 1cos−= , or yu cos= .

Then,

22 1cos1sin uyydydu

−−=−−=−=

2

2

2

2

12

111

xx

xx

+−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

−−= .

Also, ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= 2

2

11

xx

dxd

dxdu

22

22

)1()2()1()2()1(

xxxxx

+−−−+

=

22 )1(4x

x+−

= .

Now, using the chain rule

dxdu

dydu

dxdu

dudy

dxdy

×⎟⎟⎠

⎞⎜⎜⎝

⎛=×= 1

22

2

)1(4

2)1(

xx

xx

+−

×+

−=

212x+

= .


Unit 1 16

Example 9

Show that if 222

1cos

xcbbcx

y−

⎟⎠⎞

⎜⎝⎛

=

−

, 1<x , c, b are constants, then 03)( 22222 =−′−′′− ycxycyxcb .

Note: It’s preferable to rearrange the function in order to avoid a complex quotient rule.

Step 1: ⎟⎠⎞

⎜⎝⎛=− −

bcxxcby 1222 cos)( .

Step 2: Differentiate both sides w.r.t. x:

222

2

222

222

2

1

)()(2

2

xcbc

bcxb

cxcbyxcb

xcy

−

−=

⎟⎠⎞

⎜⎝⎛−

−=−′+

−

−

Step 3: We now clear square roots before differentiating again. So, multiplying by 222 xcb −

throughout, we have

cyxcxcby

cxcbyyxc−=−−′

−=−′+−2222

2222

)()(

i.e., cxycyxcb −=−′− 2222 )(

Step 4: Differentiating w.r.t. x,

0)2()( 222222 =−′−−′+′′− ycyxcxcyyxcb ,

or, 03)( 22222 =−′−′′− ycxycyxcb .


Unit 1 17

Example 10

Show that if 1,1

sin2

1

<−

=−

xx

xy , then 0'3'')1( 2 =−−− yyxyx .

Here it’s better to rearrange the function to avoid using the quotient rule.

Thus, we write xxy 12 sin1 −=− .

Now differentiate both sides w.r.t. x:

2

2

2 11'1)2(

121

xyxx

xy

−=−+−

−

We now clear square roots before differentiating again. So, multiplying by 21 x− throughout,

we have

1')1( 2 =−+− yxxy

i.e., 1')1( 2 =−− xyyx

Differentiating w.r.t. x,

0)'()2(''')1( 2 =+−−+− yxyxyyx ,

or, 0'3'')1( 2 =−−− yxyyx .


Unit 1 18

Activity 2

1. Differentiate w.r.t. x:

(i) )43(sin 31 xx −− ;

(ii) ⎟⎟⎠

⎞⎜⎜⎝

⎛+−−

2

21

11sin

xx ;

(iii) ⎟⎠⎞

⎜⎝⎛−

xx2cos

sintan 1 .

2. If xxy 1tan −= show that

223 )1(')( xxyyxx ++=+ ,

and

0)1(2'2'')1( 2 =+−++ yyxyx .


Unit 1 19

1.2.4 Parametric Differentiation

Suppose that if x and y are given in terms of a parameter t, then the chain rule gives the

derivative of y w.r.t. x as shown in the table below:

Derivative of y w.r.t x Chain Rule

First (1st )

dtdx

dtdy

dxdt

dtdy

dxdy

=×=

[ ] [ ][ ]

dtdx

dtd

dxdt

dtd

dxd

KKK

=×=

Second (2nd)

dtdx

dtdxdyd

dxdt

dtdxdyd

dxdy

dxd

dxyd ⎥⎦

⎤⎢⎣⎡

==×⎥⎦⎤

⎢⎣⎡

=⎟⎠⎞

⎜⎝⎛=2

2

Third (3rd)

dtdx

dtdx

ydd

dxdt

dtdx

ydd

dxyd

dxd

dxyd ⎥

⎦

⎤⎢⎣

⎡

==×⎥⎦

⎤⎢⎣

⎡

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

2

2

2

2

2

3

3

M M nth

dtdx

dtdx

ydd

dxdt

dtdx

ydd

dxyd

dxd

dxyd n

n

n

n

n

n

n

n ⎥⎦

⎤⎢⎣

⎡

==×⎥⎦

⎤⎢⎣

⎡

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

−

−

−

− 1

1

1

1

1

1

Table 1.7

Example 11: Let tytx 5sin,5cos == . Then, since

tdtdxt

dtdy 5sin5,5cos5 −== ,

it follows that

tt

tdxdy 5cot

5sin55cos5

−=−

= .


Unit 1 20

Let’s now find the second derivative, 2

2

dxyd . We use the chain rule again:

⎟⎠⎞

⎜⎝⎛=

dxdy

dxd

dxyd2

2

dxdt

dxdy

dtd

×⎟⎠⎞

⎜⎝⎛=

dtdx

dxdy

dtd

⎟⎠⎞

⎜⎝⎛= .

Similarly, for the 3rd derivative we have

⎟⎟⎠

⎞⎜⎜⎝

⎛= 2

2

3

3

dxyd

dxd

dxyd

dxdt

dxyd

dtd

×⎟⎟⎠

⎞⎜⎜⎝

⎛= 2

2

dtdx

dxyd

dtd

⎟⎟⎠

⎞⎜⎜⎝

⎛= 2

2

.

Proceed likewise for the other higher derivatives.


Unit 1 21

So, returning to our example, we find

)5sin5()5cot(2

2

ttdtd

dxyd

−−=

tt

5sin55 cosec5 2

−=

t5 cosec3−= ;

and you can now easily show that

)5sin5()5 cosec( 33

3

ttdtd

dxyd

−−=

tt 5 cosec5cot3 4−= .

Activity 3

Find 2

2

dxyd if

(i) tytx 2tan,2sec == ;

(ii) ttyttx 3sinsin3,3coscos3 −=−= .


Unit 1 22

1.2.5 Taylor and Maclaurin Series We shall now briefly look at the representation or expansion of certain functions in power series.

A power series in x is simply an infinite series of the form

LL +++++++ rr xaxaxaxaxaa 4

43

32

210 ,

where the ia ’s are constants not all zero.

We assume that our functions are continuous, single-valued and have continuous derivatives up

to the nth order in a given interval.

Taylor Series

This is a representation of a function )(xf by a power series in )( ax − ; i.e., we are expanding

)(xf about the point ax = . Thus

K+−

+−

+−

+= )('''!3

)()(''!2

)()('!1

)()()(32

afaxafaxafaxafxf ( †)

An equivalent form of the series is obtained by putting hax += in ( †)

K++++=+ )('''!3

)(''!2

)('!1

)()(32

afhafhafhafhaf .


Unit 1 23

Maclaurin Series

This is a special case of Taylor’s series obtained by putting 0=a in ( †)

K++++= )0('''!3

)0(''!2

)0('!1

)0()(32

fxfxfxfxf ,

i.e., we are now expanding the function about the origin.

It is clear that not all functions can have series expansions as they or their derivatives may not

exist (i.e., they are infinite) at ax = or at 0=x . Thus xxx cot,ln,/1 do not have Maclaurin

series as they are all infinite at the origin. However, they can be expanded about some other

point.

Finally, we note that both Taylor and Maclaurin series do not generally converge for all values

of x, but only within a restricted range of values of x.

Let us first consider a few examples of Maclaurin series.

Example 12

xexf =)(

We have xexf =)( 1)0( =f

xexf =)(' 1)0(' =f

M M

xn exf =)()( 1)0()( =nf


Unit 1 24

Hence,

KK +++++++=!!4!3!2

1432

nxxxxxe

nx

This is known as the Exponential series and is valid for all values of x. Example 13 xxf sin)( =

We have xxf sin)( = 0)0( =f

xxf cos)(' = 1)0(' =f

xxf sin)('' −= 0)0('' =f

xxf cos)(''' −= 1)0(''' −=f

The values of the derivatives at 0=x form cycles of 1,0,1,0 − . Hence

KK +−

−++−+−=

−−

)!12()1(

!7!5!3sin

121753

nxxxxxx

nn

This is the sine series and is valid for all values of x (in radians). Similarly,

KK +−

−++−+−=

−−

!)22()1(

!6!4!21cos

221642

nxxxxx

nn

.

This is the cosine series and is valid for all values of x (in radians). Likewise we have


Unit 1 25

22

,315

17152

3tan

753 π<<

π−++++= xxxxxx K ;

11,)1(432

)1ln(1432

≤<−+−

++−+−=+−

xn

xxxxxxnn

KK .

We shall now expand about some other point. Thus, we shall be finding the Taylor series of the function about the given point.

Example 14

Find 5/xe in powers of )5( −x .

Here we are expanding the function 5/xe about the point 5=x , i.e., finding its Taylor expansion

about 5=x . We’ll use ( †).

So, 5/)( xexf = ef =)5(

5/51)(' xexf = ef 5

1)5(' =

5/251)('' xexf = ef 25

1)5('' = .

⎥⎦

⎤⎢⎣

⎡+

−−

++−

+−+=∴−

− KK!)1(

)5(5

1!2

)5(251)5(

511

1

1

25/

nxxxee

n

nx .

The series converges for all x.


Unit 1 26

Example 15

Find the Taylor expansion of xln about 3=x up to and including the term in 4x .

xxf ln)( = 3ln)3( =f

xxf /1)(' = 3/1)3(' =f

2/1)('' xxf −= 9/1)3('' −=f

3/2)(''' xxf = 27/2)3(''' =f

4/6)( xxf iv −= 27/2)3( −=ivf .

Hence, using ( †), we obtain

K+−

−−

+−

−−+=!4

)3(272

!3)3(

272

!2)3(

91)3(

313lnln

432 xxxxx

K+−

−−

+−

−−

+=324

)3(81

)3(18

)3(3

)3(3ln432 xxxx .

Example 16 If 21 )(sin xy −= , prove that

02)1(2

22 =−−−

dxdyx

dxydx .

Hence using Maclaurin’s expansion, prove that the first two non-zero terms in the expansion of 21 )(sin x− are

42

31 xx + .


Unit 1 27

Solution: 21 )(sin xy −=

Differentiating y w.r.t x

2

1

1

1)(sin2x

xdxdy

−= − or y

dxdyx 21 2 =−

Again, differentiating w.r.t x

dxdyyxx

dxdy

dxydx 2/12/122

22 )2()1(

211 −− =−−+− or

dxdyxy

dxdyx

dxydx 22/12

22 1)1( −=−− −

2)1(2

22 =−−

dxdyx

dxydx [Proven]

Again, differentiating w.r.t x, we have

0)2()1(2

2

2

2

3

32 =⎥

⎦

⎤⎢⎣

⎡+−⎥

⎦

⎤⎢⎣

⎡−+−

dxdy

dxydxx

dxyd

dxydx or 03)1( 2

2

3

32 =−−−

dxdy

dxydx

dxydx

Again, differentiating w.r.t x, we have

033)1(2 2

2

3

3

2

2

4

42

3

3

=−⎥⎦

⎤⎢⎣

⎡+−⎥

⎦

⎤⎢⎣

⎡−+−

dxyd

dxydx

dxyd

dxydx

dxydx or

04)1(5 2

2

4

42

3

3=−−+−

dxyd

dxydx

dxydx

Evaluation of y and its derivatives:

When x = 0, then y = 0, 0=dxdy , 2

2

2=

dxyd , 0

3

3=

dxyd , 8

4

4=

dxyd

Using Maclaurin’s Expansion, 21 )(sin xy −= = L+′′′′+′′′+′′+′+ yxyxyxyxy

!4!3!2)0()0(

432

= L+++++ )8(24

)0(6

)2(2

00432 xxx

= L++3

42 xx [Proven]


Unit 1 28

Example 17

By using the Maclaurin’s expansion or otherwise show that the first three terms in the expansion

of ))2tan()2ln(sec( xx + in powers of x are

34

342

53 xxx ++ .

Solution: ))2tan()2ln(sec( xxy += , 00 ==xy

)2sec(2))2tan()2(sec(

)2(sec2)2tan()2sec(2 2x

xxxxx

xy

=+

+=

∂∂ , 2

0=

∂∂

=xxy

xdxdyxx

xx

xy 2tan22tan2sec4))2sec(2(2

2==

∂∂

=∂∂ , 0

02

2=

∂∂

=xxy

dxdyx

xyx

dxdyx

dxyd

x

xdxdy

xy 42tan8]2sec4[]2tan2[

]2tan2[22

2

2

3

3+

∂∂

=+=∂

∂=

∂∂ , 8

03

3=

∂∂

=xxy

2

23

2

22

2

2

2

232

2

2

2

222

2

2

4

4

4]2tan32[]16[]2tan8[

4]2tan32[]2tan32[]2tan8[

4]2sec2tan32[]2tan8[

dxydx

xy

xyx

xy

dxydx

xyx

xyx

xy

dxydxx

xyx

xy

xy

+∂∂

+∂∂

+∂∂

=

+∂∂

+∂∂

+∂∂

=

+∂∂

+∂∂

=∂∂

00

4

4=

∂∂

=xxy

3

33

2

222

3

32

2

22

3

3

5

5

4]2tan32[]2sec2tan192[

16]2sec2tan32[]2tan8[

dxydx

xyxx

xy

xyxx

xyx

xy

xy

+∂∂

+∂∂

+∂∂

+∂∂

+∂∂

=∂∂

1600

5

5=

∂∂

=xxy

Using Maclaurin’s Expansion, ))2tan()2ln(sec( xxy +=


Unit 1 29

= L+′′′′+′′′′+′′′+′′+′+ yxyxyxyxyxy!5!4!3!2

)0()0(5432

= L++++++!5

1600!3

802053 xxx

= L+++3

43

4253 xxx [Proven]

Activity 4 1. Show that

(i) 2/2/,72061

245

211sec 642 π<<π−++++= xxxxx K ;

(ii) 11,76.4.2

5.3.154.2

3.132

1sin753

1 <<−++++=− xxxxxx K ;

(iii) 11,12

)1(753

tan121753

1 ≤≤−+−

−++−+−=

−−− x

nxxxxxx

nn

KK .

2. Expand xcos about the point 3/π=x .


Unit 1 30

1.3 FURTHER INTEGRATION 1.3.1 Integrals Involving Trigonometric Functions

1 ;sincos;cossin ∫∫ +=+−= CxdxxCxdxx 2 ∫ ∫ +=+−== CxCxdx

xxdxx seclncosln

cossintan

3 Cxdxxxdxx +==∫ ∫ sinln

sincoscot

4 CxxCxdxx +−=+=∫ ]cot cosec[ln)2/tan(ln cosec

5

CxxCxdxx ++=+⎟⎠⎞

⎜⎝⎛ +π

=∫ ]tanln[sec24

tanlnsec

Table 1.8

Prove: CxxCxdxx +−=+=∫ ]cot cosec[ln)2/tan(ln cosec

Method: Use of the method of substitution to prove the first part

Step 1: Let 2

tan xt = . Then using the identityA

AA 2tan1tan22tan

−= and putting

2xA = , we get

22 1

2

2tan1

2tan2

tantt

x

x

x−

=−

= .

Step 2: We then deduce [construct the usual right-angled triangle] that

2

22

2

2

2 11sec,

21 cosec,

11cos,

12sin

ttx

ttx

ttx

ttx

+−

=+

=+−

=+

= .

Step 3: Also, dxtdxxdt )1(21

2sec

21 22 +== , so that 21

2t

dtdx+

= . Hence


Unit 1 31

Cttdt

tdt

ttdxx +==

++

= ∫∫∫ ln1

22

1 cosec 2

2

∴ Cxdxx +=∫ )2/tan(ln cosec .

Method: To be able to prove the second part we need another form for this integral which is

deduced by trigonometrical manipulation.

2cos

2sin2

2sin2

ln

2cos

2sin

ln2

tanln2

xx

x

x

xx

== [on multiplying top & bottom by 2/sin2 x ]

⎟⎠⎞

⎜⎝⎛ −

=x

xsin

cos1ln

]cot cosec[ln xx −=

Thus, we have shown that

CxxCxdxx +−=+=∫ ]cot cosec[ln)2/tan(ln cosec

Prove: CxxCxdxx ++=+⎟⎠⎞

⎜⎝⎛ +π

=∫ ]tanln[sec24

tanlnsec

Hint: This can be found by the same substitution as above, but it is perhaps more instructive to

deduce it from

]cot cosec[ln2

tanln cosec xxxdxx −==∫ ,

by the substitution φ+π

=2

x .

Then φ=φ= sec cosec, xddx and φ−= tancot x

∴ ]tanln[sec24

tanlnsec φ+φ=⎟⎠⎞

⎜⎝⎛ φ

+π

=φφ∫ d ,

or, CxxCxdxx ++=+⎟⎠⎞

⎜⎝⎛ +π

=∫ ]tanln[sec24

tanlnsec .


Unit 1 32

Products of sines and/or cosines of multiple angles may be integrated by parts. It is however,

easier to use the following trig identities, known as the Factor Formulae, to simplify the

integrand before integrating.

Factor Formulae

])cos()[cos(coscos 21 xbaxbabxax −++=

])sin()[sin(cossin 21 xbaxbabxax −++=

])cos()[cos(sinsin 21 xbaxbabxax +−−=

Table 1.9

Example 18

(i) ∫∫ += dxxxdxxx )4cos8(cos2cos6cos 21

Cxx++=

84sin

168sin .

(ii) ∫∫ += dxxxdxxx )3sin7(sin2cos5sin 21

Cxx+−−=

63cos

147cos

(iii) ∫∫ −+= dxxxdxxx )]4sin(10[sin7cos3sin 21

Cxx++−=

84cos

2010cos [Recall α−=α− sin)sin( ]

(iv) ∫∫ −= dxxxdxxx )7cos(cos4sin3sin 21 [Recall α=α− cos)cos( ]

Cxx +−= ]7sin[sin 71

21 .

Integrals of Products of Sines and/or Cosines of Multiple Angles


Unit 1 33

We’ll consider those integrals where m and n are integers and at least one of them is odd. The

case where m and n are both even will be dealt with later in Unit 4.

m n Substitution to be used

odd even xu sin=

even odd xu cos=

odd odd xu sin= or xu cos=

Table 1.10 Example 19

(a) ∫ dxx3sin

Here 3,0 == nm . Since n is odd, we put xu cos= , so that dxxdu sin−= .

Now,

)()1(sin)cos1(sin)(sinsin 2223 duudxxxdxxxdxx −−=−== ∫∫∫∫

.coscos3

31

331

Cxx

Cuu

+−=

+−=

(b) ∫ dxx5cos

Here 0,5 == nm . Since m is odd, we put xu sin= , so that dxxdu cos= .

Now,

∫∫∫∫ −=−== duudxxxdxxxdxx 222245 )1(cos)sin1(cos)(coscos

.sinsinsin

)21(

5513

32

5513

32

42

Cxxx

Cuuu

duuu

++−=

++−=

+−= ∫

Integrals of the form ∫ dxxx nm sincos .


Unit 1 34

Example 20

∫ dxxx 65 cossin

Here 5,6 == nm . Since n is odd, we put xu cos= , so that dxxdu sin−= .

∴ ∫∫ −= duuxdxxx 6465 sincossin

∫ −−= duux 622 )cos1(

∫ −−= duuu 622 )1(

∫ +−−= duuuu )2( 1086

Cuuu +−+−= 11/9/27/ 1197

Cxxx+−+−=

11cos

9cos2

7cos 1197

.

Example 21

∫ dxxx 43 sincos

Here, 4,3 == nm . Since m is odd, we put xu sin= , so that dxxdu cos= .

duuxdxxx ∫∫ = 4243 cossincos

duuu 42 )1(∫ −=

Cuu+−=

75

75

Cxx+−=

7sin

5sin 75

.


Unit 1 35

Activity 5

Find the following integrals:

(i) ∫ dxxx 2sin3sin ;

(ii) ∫ dxxx 2cos4cos ;

(iii) ∫ dxxx 5cos3sin ;

(iv) ∫ dxxx 32 cossin ;

(v) ∫ dxxx 3cos3sin 53 .

1.3.2 Integrals Involving Inverse Trigonometric Functions Consider the following standard integrals:

Integration Substitution used

Caxdx

xa+=

−−∫ 1

22sin1

θsinax =

Caxdx

xa+=

−

− −∫ 1

22cos1

θcosax =

stan

dard

inte

gral

s

Cax

adx

xa+=

+−∫ 1

22 tan11

θtanax =

Presence of 22 Xa − in the integrand θsinaX =

Hint Presence of 22 Xa + in the integrand.

θ= tanaX

Table 1.11


Unit 1 36

We now apply these standard integrals in the following examples:

Example 22: [Direct use of the standard integrals for easier problems]

(a) Cxx

dxx

dx+=

−=

−−∫∫ 3

sin)3(9

1

222;

(b)

Cx

x

dxx

dx+=

−=

−−∫∫ 5

sin)5(5

1

222.

(c) Cxx

dxx

dx+=

+=

+−∫∫ 2

tan21

)2(41

222 ;

(d)

Cxx

dxx

dx+=

+=

+−∫∫ 7

tan7

1)7(7

1222 .

Example 23: [Considering harder problems]

Find ∫− 249 xdx .

Solution:

Step 1: We write )(449 2492 xx −=− and thus

2492

492 2)(449 xxx −=−=− .

Step 2: Then,

∫∫ ∫−

=−

=− 2

492

492 2

1249 x

dxx

dxx

dx .

Step 3: Using the standard integral, with 4/92 =a , so that 2/3=a , we have

CxCxx

dxx

dx+=+=

−==

−−−∫∫ 3

2sin21

2/3sin

21

21

4911

2492

.

Example 24:


Unit 1 37

Evaluate ∫−

3/1

0 275 xdx .

Solution:

Step 1: ∫∫∫∫−

=−

=−

=−

3/1

0 2

3/1

0 2

3/1

0 2

3/1

0 2 )7/5(71

)7/5(7)7/5(775 xdx

xdx

xdx

xdx

Thus, if we consider 7/52 =a , then 7/5=a .

Step 2: Hence,

∫ ⎥

⎦

⎤⎢⎣

⎡=

−−3/1

0

3/1

0

1

2 7/5sin

71

75x

xdx

⎥⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛= −− 0sin

7/53/1sin

71 11

153235.0= [RADIAN mode]

Example 25:

Evaluate the following integral: ∫−+−

2/11

42 55324

1 dxxx

Solution:

Step 1: ∫−+−

2/11

42 55324

1 dxxx

= ∫−−

2/11

42)4(49

1

xdx

Step 2: Let )4(32

−= xu , then dudx23

= and thus

∫−−

2/11

42)4(49

1

x= du

u∫

−

1

021

123.

31

Step 3: Hence, ∫−+−

2/11

42 55324

1 dxxx

= ]101 )(sin21 u− =

4π


Unit 1 38

Example 26:

Find ∫ + 294 xdx .

Solution: Step 1:

∫∫∫ +=

+=

+ )9/4(91

)9/4(994 222 xdx

xdx

xdx ,

which is of the standard integral with 3/2=a . Step 2: Hence

Cxx

dxx

dx+=

+=

+−∫∫ 3/2

tan3/2

191

)9/4(91

941

22

Cx+= −

23tan

61 1 .

Example 27:

Find ∫ +−

2

02 17123 xx

dx , giving your answer to 4 decimal places.

Solution:

Step 1: ∫ +−

2

02 17123 xx

dx ∫∫⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=+−

=2

02

2

2

02

]35)2[(

31

]3/174[3x

dxxx

dx

Step 2: ∫ +−

2

02 17123 xx

dx 0.2576

352tan

53

31

2

0

1 =⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛−

= − x


Unit 1 39

Activity 6

Find the following integrals, using both of the methods given:

(i) ∫− 254 xdx ;

(ii) ∫−

1

0 267 tdt ;

(iii) ∫ + 238 xdx ;

(iv) ∫ +

2

1 2 54zdz .

Example 28:

Find dxx∫ − 24 .

Solution:

Step 1: This contains a term of the form 22 Xa − , with xXa == ,2

Let θ= sin2x such that θθ ddx cos2= .

Step 2: Thus, θθθ−=− ∫∫ ddxx cos2sin444 22


Unit 1 40

∫ θθθ= dcos2cos2

∫ θθ= d2cos4

∫ θθ+= d)2cos1(2

C+θ+θ= 2sin2 .

Step 3:

We need to revert back to terms of x. From our substitution θ= sin2x , it follows that 2

sin x=θ ,

so that 2

sin 1 x−=θ , and 4/1cos 2x−=θ , and therefore

4/14/1)2/(2cossin22sin 22 xxxx −=−=θθ=θ .

Hence,

∴ Cxxxdxx +−+=− −∫ 4/12

sin24 212 .


Unit 1 41

Example 29:

Find ∫+

3

3 29 xxdx .

Solution:

Step 1: Here the presence of the form 22 Xa + suggests that we put θ= tan3x .

Step 2: Then θθ= ddx 2sec3 and limits becomes

x )3/(tan 1 x−=θ

3 6/π

3 4/π

Step 3:

∴ ∫∫π

π θ+θ

θθ=

+

4/

6/ 2

23

3 2 tan99tan3

sec3

9

d

xxdx ∫

π

πθ

θθθ

=4/

6/

2

sec3tansec d

∫π

πθ

θθ

=4/

6/ tansec

31 d

∫π

πθθ=

4/

6/ cosec

31 d

4/6/)]2/ln[tan(

31 π

πθ=

)]12/tan(ln)8/tan([ln31

π−π=

145195.0= . [Radian mode]


Unit 1 42

Example 30:

Find dxx

x∫

− 249 .

Solution:

Step 1: 222322 )(2)4/9(449 xxx −=−=− , which now involves the form 22 Xa − ; so

we let θ= sin23x .

Step 2: θθθ

=θθθθ

=−

∴ ∫∫∫ dddxx

xsin

cos3cossincos349 2

23

23

2

∫ ∫∫ θθ−θθ=θθθ−

= ddd sin3 cosec3sinsin13

2

C+θ+θ−θ= cos3]cot cosec[ln3

Cxx

x+−+⎟

⎟⎠

⎞⎜⎜⎝

⎛ −−= 2

2

49493ln3 .

1.3.3 Integrals with a Quadratic in the Denominator

In this unit we shall consider only integrals of the form

Type Procedures

(i) ∫ ++ cbxax

dx2 If the quadratic is factorisable, then use partial fractions. If not,

complete the square and use the appropriate standard integral result or

the relevant substitution

(ii) ∫

++ cbxax

dx2

Whether the quadratic is factorisable or not, complete the square and

use the appropriate standard integral result or the relevant substitution.

Table 1.12


Unit 1 43

Example 31:

(a) Find ∫ −− 2076 2 xxdx .

Quadratic is factorisable; so we use partial fractions:

4352)43)(52(1

20761 23

3232

2 +−

−=

+−=

−− xxxxxx.

)]43/()52ln[(43522076 23

1233

232

2 Cxxdxxxxx

dx++−=

+−

−=

−−∴ ∫∫

(b) Find dxxx

dx∫ +− 1342

Quadratic is not factorisable; so we complete the square:

9)2(134 22 +−=+− xxx

∫∫ −+=

+−∴ 22 )2(9134 x

dxdxxx

dx [which is of the form ,22∫ + Xadx ]

Cx+

−= −

32tan

31 1 . [Alternatively, let θ=− tan3)2(x ]


Unit 1 44

(c) Find ∫−+ 232 xx

dx .

Complete the square even though quadratic is factorisable:

⎟⎠⎞

⎜⎝⎛ −−=−+ 2

612 )(

3625332 xxx .

∫∫∫−−

=−−

=−+

∴2

61

36252

61

36252 )(3

1

])([332 x

dx

x

dx

xx

dx ,

which is of the form ∫− 22 Xa

dx . Hence,

Cx

xx

dx+

−=

−+−∫ 6/5

6/1sin3

1

321

2

Cx+

−= −

516sin

31 1 .

Alternatively, we could have used the substitution θ=− sin)( 6

561x .

We shall now consider some integrals which, though not of types (i) and (ii) above, can still be

worked out by an appropriate trig substitution because they “contain” the standard integrals.


Unit 1 45

Example 32:

dxxx

x∫

−− 265

On completing the square we have

dxx

xdxxx

x∫∫

−−=

−− 225

412 )(65

,

in which we have the form 25

2122 ,, −==− xXaXa .

So, here we make the substitution

θ=− sin21

25x

θθ=θ+=∴ ddxx cos,sin 21

21

25

θ=θ−=−− cossin)( 212

41

412

25

41 x .

θθθθ+

=−−

∴ ∫∫ ddxxx

x coscos

sin

6521

21

21

25

2

∫ θθ+= d)sin5(21

C+θ−θ= )cos5(21 .

The answer must be given in terms of x. From our substitution θ=− sin21

25x , we have

52sin −=θ x , so that )52(sin 1 −=θ − x , and 2)52(1cos −−=θ x . The final answer is then

Cxxdxxx

x+−−−−=

−−−∫ ])52(1)52(sin5[

21

6521

2.


Unit 1 46

Example 33:

Find ∫+− 1342 xx

dx .

∫∫−+

=+− 22 )2(9134 x

dxdxxx

dx . [On completing the square.]

Here we have the form 2,3,22 −==+ xXaXa . So, let θ=− tan32x . The answer is

Cxxx ++−+− ]1342ln[ 2 .

However, we’ll see a neater way of doing this integral when we study Hyperbolic Functions!


Unit 1 47

Activity 7


(i) ∫ −+ 62 2 xxdx ;

(ii) ∫ ++ 875 2 xxdx ;

(iii) ∫−− 22 xx

dx ;

(iv) ∫ +− )34)(2( xxdx ;

(v) ∫ −

2/1

0 )1( yydy ;

(vi) ∫−+

2/1

0 2443 uu

du ;

(vii) ∫−

− ++

4/1

2 2 12 zzdz ;

(viii) dxxa

xa

∫−0 44

, [Let 2xu = ] ;

(ix) ∫−

32

2 2 3xx

dx , [Let xu /1= ] ;

(x) ∫ −−

3

2 2 123 ssds [Watch Out !]


Unit 1 48

Here we use the substitution )tan( anglehalft = , i.e., 2

tan mxt = .

Refer to the method for integrating cosec x.

Then dxtmdt )1(2

2+= ; 212sin

ttmx

+= , and 2

2

11cos

ttmx

+−

= .

Example 34:

Find ∫ + xdx

3cos45.

Here, 3=m . Let 2

3tan xt = . Then dxtdt )1(23 2+= , and 2

2

113cos

ttx

+−

= .

∫∫∫ +=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

+

+=

+∴ 2

2

2

2

932

1145

)1(32

3cos45 tdt

tt

dtt

xdx

Ct += − )3/(tan92 1

Cx+⎟

⎠⎞

⎜⎝⎛= −

23tan

31tan

92 1 .

Note: If the integrand involves xx 22 cos,sin or x2tan , we may use the simpler substitution

xu tan= , in which case

)1/(1cos),1/(sin),1/( 222222 uxuuxududx +=+=+= .

Integrals of the Form ∫ + bmxadx

sinand ∫ + bmxa

dxcos


Unit 1 49

Example 35:

( )

( ) .tan5/6tan301

5/6tan301

65

tan65sec

sin6cos5

1

12

2

2

22

Cx

Cuu

du

dxx

xxx

dx

+=

+=+

=

+=

+

−

−∫

∫∫

[Letting xu tan= ]

Activity 8


(i) ∫ + xdx

5cos23;

(ii) ∫ θ+θ

21cos53

d ;

(iii) ∫ φ+φ

3sin1d ;

(iv) ∫ + xdx

2sin32;

(v) ∫ + xdx

2cos43;

(vi) dxx

x∫ + 2

2

cos34sin ;

(vii) dxxxxx

∫π

++4/

0 22

22

sin4cos5sin2cos3 ;

(viii) dxx

x∫ + 2

2

cos32tan ;

(ix) ∫π

+

2/

0 22 sin4cos3 xxdx ;

(x) dxx

x∫π

+

3/

0 2

2

sin21sin .


Unit 1 50

1.3.4 Integration by Parts Recall the product rule of differentiation

dxdUV

dxdVUVU

dxd

+=)( .

Now integrate both sides w.r.t. x

dxdxdUVdx

dxdVUdxUV

dxd

∫∫∫ +=)( ,

i.e., dxdxdUVdx

dxdVUUV ∫∫ +=

or, dxdxdUVVUdx

dxdVU ∫∫ −= ,

or, dUVVUdVU ∫∫ −=

which is the well known formula used to integrate products of functions. We’ll now illustrate the method by a few examples. Example 36:

Find ∫ dxex x32

Here we choose 2xU = and dxedV x3= .

Then dxxdU 2= and xeV 331= .


Unit 1 51

dxxeexdxex xxx 23313

31232 ∫∫ −=∴

dxexex xx ∫−= 33232

31

We now need to find dxex x∫ 3 .

This time we choose xU = and dxedV x3= , so that dxdU = and xeV 331= .

dxeexdxex xxx ∫∫ −=∴ 3313

313

xx exe 3913

31 −= .

Hence,

=∫ dxex x32 (3232

31 −xex Cexe xx +− )3

913

31

Cxxe x ++−= ]269[ 23271 .

The above example shows that we sometimes have to use the method more than once to reach

the result. In the next example we shall see that the integral we start out with appears again in

the process.

Example 37:

Find dxxe x 3cos2∫ .

Let dxxeI x 3cos2∫= .

Choose xU 3cos= and dxedV x2= , so that dxxdU 3sin3−= and xeV 221= .


Unit 1 52

Then, dxxeexI xx 3sin33cos 2212

21 ∫ −⋅−=

dxxexe xx 3sin3cos 2232

21 ∫+=

Now, with xU 3sin= and dxedV x2= [ xeVdxxdU 221,3cos3 ==∴ ],

dxxeexdxxe xxx 3cos33sin3sin 2212

212 ∫∫ ⋅−=

dxxexe xx 3cos3sin 2232

21 ∫−= [See, it appears again!]

Ixe x232

21 3sin −=

]3sin[3cos 232

21

232

21 IxexeI xx −+=∴

We now collect all the I on the left

xexeI xx 3sin3cos 2432

21

413 +=∴

or, CxxeI x ++= ]3sin33cos2[2131 .

Hence,

Cxxedxxe xx ++=∫ ]3sin33cos2[3cos 21312 .


Unit 1 53

We shall integrate by parts.

Example 38:

∫∫ ⋅= −− dxxdxx 1sinsin 11

Here we choose xU 1sin −= and dxdV ⋅= 1 , so that xVdxx

dU =−

= ,1

12

.

dxxxxxdx

xxxxdxx ∫∫∫

−

−+=

−−=∴ −−−

2211

2

11

12sin

1sinsin

Cxxx +−+= − 21 1sin .

Example 39:

∫ − dxx1tan

Choose xU 1tan −= and dxdV ⋅= 1 , so that xVdxx

dU =+

= ,1

12 .

dxx

xxxdxx ∫∫ +−=∴ −−

211

1tantan

Cxxx ++−= − )1ln(tan 2211 .

Activity 9

Integrals of the Inverse Trigonometry


Unit 1 54


(i) dxxe x 2sin3∫ ;

(ii) dxxx 21sin∫ − ;

(iii) dxex x23∫ ;

(iv) ∫ − dxxx 1tan ;

(v) ∫ dxx)(lnsin [Hint: Let xz ln= , then integrate by parts.] ;

(vi) dxxx

∫2

1

ln .

1.4 SUMMARY

Having studied carefully this unit and done all the activities therein, you should now be familiar

with all the techniques of differentiation and integration presented. Integration is an art and only

lots of practice will enable you to be good at it. You should at a glance decide which method of

integration is the most appropriate. The following supplementary exercises will help you

consolidate what you have learnt so far.


Unit 1 55

1.5 SUPPLEMENTARY EXERCISES

1. Differentiate the following functions:

(i) 105 )2/()13( xx −+ ;

(ii) nxx ]1[ 2++ ;

(iii) )cos1ln(tan 3 xxa x −+ ;

(iv) )]sin2/()sin21[(sin 1 xx ++− ;

(v) )1/()1(tan 1 xx +−− ;

(vi) )/(lncot)ln/(cot 11 xxxx −− + ;

(vii) xxx )/1( + .

2. If yyx 1tan2 −−= , find the value of 22 / dxyd when 1=y .

3. Find dxdy / if ])sin[( 2yxy += .

4. Find 22 / dxyd when

(i) θ=θ= 33 sin,cos ayax ;

(ii) θ=θ−θ= sin],coscot[ln 21 ayax .

5. If ]cos[sin 211 xy −= , prove that

]1'[tan'' 2−= yyy

If 2/2/ π<<π− y , obtain the expansion of y in ascending powers of x as far as and

including the term in 2x , and show that the coefficient of 3x is zero.

6. Show that, for small values of x,


Unit 1 56

K+++++=+π 4324

2 42)(sec cxbxaxxx ,

and determine a, b and c. Expand )(sec2 42 x++ π in ascending powers of x as far as

the term involving 3x .

7. Find the following integrals:

(i) dxx

x∫ ++

+2)1(4

74 ;

(ii) ∫ φ+φ

sin1d ;

(iii) dxx

xx∫

−

−

2

1

1

sin ;

(iv) ∫++

1

0 2 1)1( xxdx ;

(v) )(,))((

bxabxxa

dxxa

b>>

−−∫ [Let θ−=− 2cos)( baxa ] ;

(vi) ∫−+ 123 2 xxx

dx [Let ux /1= ] ;

(vii) ∫−

dxx

e x

4

/1

, [Hint: 2

/1

24

/1 1x

exx

e xx −−

= ];

(viii) ∫π

θθ+θ2/

0)sin1ln(cos d .


Unit 1 57

1.6 ANSWERS TO ACTIVITIES AND SUPPLEMENTARY EXERCISES

Activity 1

(i) )72529()4()3)(2( 232 +++++ xxxxx ;

(ii) 2/32/1 )1()1(1

+− xx;

(iii) 22

23

)1(1217

+−−+−−

xxxxx ;

(iv) aax x ln2 92 + ;

(v) ]tanlnsincosec[)(tan]seccotln[cos)(cot cossin xxx xxxxx xx −+− .

Activity 2

1. (i) 21

3x−

;

(ii) 212x+

− ;

(iii) )3sinsin1(2

3coscos3xxxx

−− .


Unit 1 58

Activity 3

(i) t2cot− ;

(ii) tt cosec2sec331 .

Activity 4

KK +⎟⎠⎞

⎜⎝⎛ π

−⎟⎠⎞

⎜⎝⎛ π

+π

++⎟⎠⎞

⎜⎝⎛ π

−+⎟⎠⎞

⎜⎝⎛ π

−−⎟⎠⎞

⎜⎝⎛ π

−−=r

xr

r

xxxx3!

23cos

3!3.23

3!2.21

323

21cos

32

.

Activity 5

(i) Cxx +− 5sinsin 101

21 ;

(ii) Cxx ++ 6sin2sin 121

41 ;

(iii) Cxx +− 8cos2cos 161

41 ;

(iv) Cxx +− 5513

31 sinsin ;

(v) Cxx ++− 3cos3cos 82416

181 .


Unit 1 59

Activity 6

(i) Cx +⎟⎟⎠

⎞⎜⎜⎝

⎛−

25sin

51 1 ;

(ii) 0.483039 ;

(iii) ( ) Cx +− 8/3tan62

1 1 ;

(iv) 0.0740874.

Activity 7

(i) Cxx

+⎟⎠⎞

⎜⎝⎛

+−232ln

71 ;

(ii) Cx+⎟

⎠

⎞⎜⎝

⎛ +−

111710tan

1112 1 ;

(iii) Cx+⎟

⎠⎞

⎜⎝⎛ +−

312sin 1 ;

(iv) Cx+⎟

⎠⎞

⎜⎝⎛ −−

513sin

31 1 ;

(v) 2/π ;

(vi) 12/π ;

(vii) 7tan7

2 1− ;


Unit 1 60

(viii) 4/π ; (ix) 36

π ;

(x) )5/7ln(41 .

Activity 8

(i) Cx+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

25tan

51tan

552 1 ;

(ii) x

xC

41

41

tan2

tan2ln

−

+;

(iii) C+φ+

−]tan1[3

2

23

;

(iv) Cx +⎟⎟⎠

⎞⎜⎜⎝

⎛− tan25tan

101 1 ;

(v) Cx +⎟⎟⎠

⎞⎜⎜⎝

⎛− tan73tan

211 1 ;

(vi) Cxx+⎟⎟

⎠

⎞⎜⎜⎝

⎛+− −

7tan2tan

67

31 ;

(vii) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

π −

52tan

51

41 ;

(viii) Cxx ++⎟⎟⎠

⎞⎜⎜⎝

⎛− − tan

21tan

52tan

25

21 1 ;

(ix) 34

π ;


Unit 1 61

(x) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

π − 3tan3

132

1 1 .

Activity 9

(i) Cxxe x +− )2cos22sin3(3131 ;

(ii) Cxxx +−+− ]1sin[ 421221 ;

(iii) Cxxxe x +−+− ]3664[ 23281 ;

(iv) Cxxx +−+ − ]tan)1[( 1221 ;

(v) Cxxx +− ]lncosln[sin21 ;

(vi) 4242ln22 +− .

Supplementary Exercises

1. (i) 114 )2/()38()13(5 xxx −++ ;

(ii) 22 1/]1[ xxxn n +++ ;

(iii) )cos1/(sincos3]tanln[sec 322 xxxxaxa x −++ ;

(iv) )sin2/(3 x+ ;

(v) 212/1 x−− ;


Unit 1 62

(vi) 0 ;

(vii) )]1/()1()/1[ln()/1( 22 +−+++ xxxxxx x .

2. 27/4− .

3. ])cos[()(21

])cos[()(22

2

yxyxyxyx++−

++ .

4. (i) θθ= cosecsec'' 431ay ;

(ii) θθ= sinsec'' 41ay .

5. )3( 261 xy −π= .

6. 324592

472;3/64,3/40,8 xxxcba +++=== .

7. (i) Cxxx +++++ − ]2/)1[(tan)52ln(2 1232 ;

(ii) C+φ+

−2/tan1

2 ;

(iii) Cxxx +−− −12 sin1 ;

(iv) )21ln(2 2/1 +− ; (v) 2/)( π+ ba ;

(vi) Cxx

+⎟⎠⎞

⎜⎝⎛ −−

21cos 1 ; (vii) Ce

xxx +⎟

⎠⎞

⎜⎝⎛ ++ − /1

2 221 ; (viii) 14ln − .

Unit 1

Documents

x x ax

dx ln x x

ln x x ln x sin x

x sin x sin x

ln ln x x

ln x cos x dx x

f x dx f x

f x dx x dx f x y