1

HKDSE Mathematics

Indefinite Integration

By Leon Lee

1. Introduction

Integration, either definite or indefinite, is a large topic in HKDSE Extended Module 2. With the

introduction of integrations by substitution and by parts, which were not required in HKCEE

Additional Mathematics, to the syllabus, a much wider variety of questions can be set compared

with the old syllabus. Therefore, students should pay much more attention to this topic, and try to

master this topic as well as possible for good results in M2 examination.

This note will demonstrate the techniques in solving problems involving indefinite integration as

detailed as possible. It will start from the basic problems, and gradually to the hardest problems

which involve advanced techniques in integration.

2. What is Indefinite Integration?

→ Indefinite integration can be considered the ‘reverse’ process of differentiation.

→ In differentiation, we find the derivative of a function.

In indefinite integration, we find the primitive function of a function.

→ In simpler words, if the derivative of )(xF is )(xf , then )(xF is a primitive function of )(xf ,

i.e. )()( xfxFdx

d= .

→ The primitive function of a function is not unique in nature.

Note that )(]2)([]1)([)( xfxFdx

dxF

dx

dxF

dx

d=+=+= .

Then )(xF , 1)( +xF and 2)( +xF are all primitive functions of )(xf .

When we replace 1 or 2 by any other real constants, sayπ , e , 2 or 10− , we still get the

same result.

→ From the above results, we see that there are infinitely many primitive functions for any

integrable function.

For convenience, if )()( xfxFdx

d= , then we write ∫ += CxFdxxf )()( .

→ The constant C is called the constant of integration, and it is

arbitrary in nature.

→ The sign ∫ is called the integral sign, and )(xf is called

the integrand.

Quick Example:

As x

x

dx

d

4

1

2=

, we have C

xdx

x+=∫

24

1.

F(x)+C f (x)

differentiation

indefinite integration

Figure 1 A diagrammatic representation of

differentiation and indefinite integration

2

3. Methods of Indefinite Integration

The methods of indefinite integration will be introduced below.

3.1. Elementary Integration

The following shows some fundamental indefinite integration results, which can be obtained

directly from differentiation.

(1) constant a is where, kCkxkdx∫ +=

(2) ∫ −≠+

+

=

+

1 where,1

1

nCn

xdxx

n

n

(3) ∫ ≠+= 0 where,ln1

xCxdxx

(4) 0 where, ≠+=∫ aCa

edxe

ax

ax

From (4), when 1=a , we have

(5) ∫ += Cedxexx .

The following gives the proof of (3). The remaining is left to readers as an exercise.

Proof:

When 0>x , xx lnln = . We havex

xdx

d 1)(ln = .

When 0<x , )ln(ln xx −= . We havexx

xdx

d 1)1(

1)][ln( =−

−

=− .

Combining the results, we have Cxdxx

+=∫ ln1

.

Furthermore, we have the following properties. They can also be easily proved by employing

differentiation.

(6) ∫ ∫= constant a is where,)()( kdxxfkdxxkf

(7) [ ]∫ ∫ ∫±=± dxxgdxxfdxxgxf )()()()(

Using (7) repeatedly, we have

(8) [ ] ∫∫∫ ∫ ±±±=±±± dxxfdxxfdxxfdxxfxfxfnn

)(...)()()(...)()(2121

Important!

is the absolute sign. It is defined as:

<−

≥=

0 when ,

0 when ,

xx

xx

x

3

Example 1

Find dxx

x∫

−

4

2 1. (HKDSE Sample Paper)

Solution:

Cx

xxxx

Cx

xxxx

dxxx

xxx

dxxx

xxx

dxxx

xCx

xCx

xCxdxx

x

+−−+−=

+−

+−⋅+⋅−=

+−+−=

+−+−=

−+

−+

−+

−+=

−

−

−

∫

∫

∫∫

3

3

69

3369

4258

4

258

43

24

3

2

224

2

324

1

42

4

2

3

1ln42

3

2

9

3ln4

36

64

9

)4

64(

1464

11)(

1)(

1)()(

1

Note: Always add the constant C after indefinite integration!

To make it simple, you should add it whenever the integral sign ∫ disappears.

Example 2

Find (a) ( ) dxx∫ +

2

1 (b) ∫ dxx10 .

Solution:

(a) ( )∫ + dxx2

1

Cxxx

Cxxx

dxxx

dxxx

+++=

++⋅+=

++=

++=

∫

∫

2

32

2

32

2

1

3

4

2

3

22

2

)12(

)12(

(b) We first let constant. a is where,10 aeaxx =

Solving, we have 10ln=a .

C

Ce

dxe

dx

x

x

x

x

+=

+=

= ∫

∫

10ln

10

10ln

10

10ln

10ln

10ln

10ln

ln10ln

=

=

=

a

axx

eaxx

(Binomial Theorem)

4

3.2. Integration by Substitution

The above formulae are very limited in usage and cannot deal with most integrals like

dxx

x

∫−

+

1

1 or dx

x

x

∫+1

2

2

. We need a new method called ‘integration by substitution’ to deal

with these integrals.

Let )( xgu = . Then, we have ∫ ∫= duufdxxgxgf )()('))(( .

The proof is given in the appendix of this note on p.46.

This method can be regarded as the ‘reverse’ of the chain rule in differentiation.

Refer to the following illustration to see how we can apply this method to find integrals:

Consider dx

x

x

∫+1

2

2

. We take 1)( 2+== xxgu . Then, we have xxg' 2)( = .

Cx

Cu

duu

dxxgxg

xdxxg

dxx

x

++=

+=

=

=

⋅=

+

∫

∫

∫∫

−

12

2

)(')(

1

2)(

1

1

2

2

2

1

2

1

2

As seen from the above example, we can see that we express the integral in the form

∫ dxxgxgf )('))(( and transform it into the form ∫ duuf )( for simpler calculation.

However, we usually use the following way:

Let 12+= xu , then xdxdu 2= , i.e. du

xdx

2

1= .

Cx

Cu

duu

duxu

xdx

x

x

++=

+=

=

⋅=

+

∫

∫∫

−

12

2

2

12

1

2

2

2

1

2

1

2

We use a suitable substitution )(xgu = at the beginning. We differentiate it with respect to x

(w.r.t. x, in short) to obtain dxxgdu )('= . By rearranging terms, we get duxg

dx)('

1= . We put

it into the original integral and express the whole integral in terms of the new variable u only.

Finally, we find the integral, and express the result in terms of the original variable x.

We express dx in

terms of du.

Here, x

xf1

)( = and 1)( 2+= xxg .

5

Example 3 (2012 DSE)

(a) Find ∫+

dxx

x 1.

(b) Using the substitution 12−= xu , find ∫

−

dxx

x

12

3

.

Solution:

(a) ∫∫ ++=+=+

Cxxdxx

dxx

xln)

11(

1

(b) Let 12−= xu , then xdxdu 2= , i.e. du

xdx

2

1= .

2

1 where,'1ln

2

1

2

1

1ln2

1)1(

2

1

ln2

1

2

1

1

2

1

1

22

22

2

3

−=+−+=

+−+−=

++=

+=

−∫∫

CC'Cxx

Cxx

Cuu

duu

udx

x

x

*Think About*:

(1) Why did we not to include 2

1− in final answer and use a new constant 'C instead?

(2) Why and how is 2x changed into 1+u ?

Example 4

Suppose 0>x .

Find (a) ∫+

dxebax (b) ∫

++

+ dxebaxcbxax

2

)2( (c) ∫ dxex

x

2

2

1 (d) ∫ dx

x

xln.

Solution:

(a) ∫∫ +=+=

+

++

Ca

ebaxde

adxe

bax

baxbax)(

1

(b) Cecbxaxdedxebaxcbxaxcbxaxcbxax+=++=+

++++++

∫∫222

)()2(2

(c) Cex

dedxex

xxx +−=

−= ∫∫

222

22

12

2

11

(d) Cx

xxddxx

x+== ∫∫ 2

)(ln)(lnln

ln 2

(by (a)) Always remember:

Express your final answer

in terms of the original

variable!

6

Example 5

Find ∫ + dxxx20132 )1( .

Solution:

Let 1+= xu , then dxdu = .

Cxxx

Cuuu

duuuu

duuuu

duuu

dxxx

++

++

−+

=

++−=

+−=

+−=

−=

+

∫

∫

∫

∫

2014

)1(

2015

)1(2

2016

)1(

20142015

2

2016

)2(

)12(

)1(

)1(

201420152016

201420152016

201320142015

20132

20132

20132

Example 6

Find (a) ∫−

+dx

x

x

1

1 (b) ∫

−

dxx

x

1

2

.

Solution:

(a) ∫∫−

+−=

−

+dx

x

xdx

x

x

1

21

1

1

Cxx

xdx

dx

dxxx

x

+−+=

−−

+=

−+

−

−=

∫ ∫

∫

1ln2

)1(1

12

1

2

1

1

(b) Let 1−= xu , then dxdu = .

∫−

dxx

x

1

2

'1ln22

)1(

1ln)1(22

)1(

ln22

)1

2(

12

)1(

2

2

2

2

2

Cxxx

Cxxx

Cuuu

duu

u

duu

uu

duu

u

+−++−

=

+−+−+−

=

+++=

++=

++=

+=

∫

∫

∫

We can write ∫ dx1 simply as

∫ dx , omitting the constant 1.

7

Example 7 (Long Division & Partial Fractions)

(a) Let 23

)()(

23

18622

23456

+−

+≡

+−

++−−−−

xx

xgxf

xx

xxxxxx, where )(xf and )(xg are

two polynomials with 4)(deg ≤xf and 2)(deg <xg . Find )(xf and )(xg .

(b) Let 2123

)(2

−

+

−

≡

+− x

B

x

A

xx

xg, where A and B are constants. Find A and B.

(c) Hence, find ∫+−

++−−−−dx

xx

xxxxxx

23

186

2

23456

.

Solution:

(a) Here, we perform long division.

43 2 1

1

8124

1 8 124

6 93

1 8 6 53

462

1 8 6 1 3 2

23 1

1 8 6 1 1 1 12 3 1

+++

+

+−+

++−+

+−+

++−−+

+−+

++−−−+

+−

++−−−−+−

Then, we have 23

1432

23

186

2

234

2

23456

+−

++++=

+−

++−−−−

xx

xxxx

xx

xxxxxx

.

Thus, xxxxxf 432)( 234+++≡ and 1)( ≡xg .

(b) From (a), 1)( ≡xg .

)1()2(1

2123

12

−+−≡

−

+

−

≡

+−

xBxA

x

B

x

A

xx

Put 1=x , we have )11()21(1 −+−= BA , i.e. 1−=A .

Put 2=x , we have )12()22(1 −+−= BA , i.e. 1=B .

(c) From above, we have1

1

2

1432

23

186 234

2

23456

−

−

−

++++=

+−

++−−−−

xx

xxxx

xx

xxxxxx

Cx

xxx

xx

Cxxxxxx

dxxx

xxxxdxxx

xxxxxx

+−

−++++=

+−−−++++=

−−

−++++=

+−

++−−−−∫∫

1

2ln2

25

1ln2ln225

1

1

2

1432

23

186

23

45

23

45

234

2

23456

Points to note:

1. The notation ‘deg’ is used to refer to the

degree of a polynomial. For instance, let

12)( 25+−= xxxh , then )(deg xh is

equal to 5.

2. This type of long division without

variables written is known as the method of

detached coefficients.

3. The method employed in (b), i.e. to break

an algebraic fraction into sum of fractions

with denominators of smaller degrees, is

called resolving an algebraic fraction into

partial fractions.

8

Example 8 (Reduction Formula)

Let nxxy )(ln= , where 0>x .

(a) Find dx

dy.

(b) Let ∫= dxxIn

n)(ln , where 0≥n .

Using the result in (a), or otherwise, show that 1

)(ln−

−=n

n

nnIxxI

(c) Hence, find 4I .

Solution:

(a) nnnn xxnxx

xnxdx

dy)(ln)(ln)1()(ln

1)(ln 11

+=⋅+⋅⋅=−−

(b) Integrating both sides w.r.t. x, we have

1

1

1

1

)(ln

)(ln)(ln)(ln

)(ln)(ln)(ln

)(ln)(ln

−

−

−

−

−=

−=

+=

+=

∫∫

∫∫

∫∫

n

n

n

nnn

nnn

nn

nIxxI

dxxnxxdxx

dxxdxxnxx

dxxdxxny

(c) Using the result in (b) repeatedly, we have

[ ][ ]

[ ]

Cxxxxxxxxx

dxxxxxxxxxx

Ixxxxxxxx

Ixxxxxx

Ixxxx

IxxI

++−+−=

+−+−=

−−+−=

−+−=

−−=

−=

∫24ln24)(ln12)(ln4)(ln

)(ln24ln24)(ln12)(ln4)(ln

ln24)(ln12)(ln4)(ln

2)(ln12)(ln4)(ln

3)(ln4)(ln

4)(ln

234

0234

0

234

1

234

2

34

3

4

4

Notes:

(1) The method employed in this question is to establish a reduction formula, which

reduces an integral from a higher to lower power, with the same form. A reduction

formula can be used repeatedly to lower the power of the integral, until it can be

integrated easily. In this way, the integrals in certain forms, no matter how high the

power is, can be found.

(2) We can find out a reduction formula by differentiation. However, we usually use a

technique called integration by parts to find them out. This will be introduced in

Section 3.5. (Refer to Example 34 for details)

9

3.3. Integration of Trigonometric Functions

Trigonometric functions can be differentiated. Similarly, they can also be integrated.

3.3.1. Basic Integration of Trigonometric Functions

From differentiation, we have the following basic results.

(1) ∫ +−= Cxxdx cossin

(3) ∫ += Cxxdx tansec2

(5) ∫ += Cxxdxx sectansec

For other integrals of trigonometric functions, different trigonometric identities are useful.

Example 9

Find (a) ∫ dxx

2sin

2 (b) ∫ xdx2tan .

Solution:

(a) Cxxxdxxxdxdxdxx

dxx

+−=−=−=−

= ∫∫∫∫∫ sin2

1

2

1cos

2

1

2

1cos

2

1

2

1

2

cos1

2sin

2

(b) Cxxdxxxdx +−=−= ∫∫ tan)1(sectan 22

(Can you find ∫ dxx

2cos

2 and ∫ xdx2cot ? Give them a try!)

3.3.2. Integration of Trigonometric Functions with Method of Substitution

The method of substitution can also be used in finding integrals of trigonometric functions.

Example 10

Find ∫ xdxx cos3sin .

Solution:

[ ]

Cxx

xxdxxd

dxxx

dxxxxx

xdxx

+−−=

⋅+⋅=

+=

−++=

∫∫

∫

∫

∫

2cos4

14cos

8

1

)2(2sin2

1

2

1)4(4sin

4

1

2

1

)2sin4(sin2

1

)3sin()3sin(2

1

cos3sin

(2) ∫ += Cxxdx sincos

(4) ∫ +−= Cxxdx cotcsc2

(6) ∫ +−= Cxxdxx csccotcsc

Point to note:

Generally, we have the following formulas

for constants a and b, with 0≠a :

Cbaxa

dxbax

Cbaxa

dxbax

++=+

++−=+

∫

∫

)sin(1

)cos(

)cos(1

)sin(

They can be proven easily by differentiation

or method of substitution, with substitution

baxu += . They are widely accepted

and are not required to be proven again

in your calculations.

10

Example 11

(a) Prove that x

x

x

2sin

2cos1tan

−

= .

(b) Hence, or otherwise, find ∫−

dxx

x

2sin

2cos1.

Solution:

(a) LHStancos

sin

cossin

sin

2

2cos1

cossin2

2

2sin

2cos1RHS

2

====

−

⋅=

−

= x

x

x

xx

xx

xxx

x

Thus, we have x

x

x

2sin

2cos1tan

−

= .

(b) Cxxdx

dxx

xxdxdx

x

x+−=−===

−

∫ ∫∫∫ cosln)(coscos

1

cos

sintan

2sin

2cos1

Example 12

Find ∫ xdxsec .

Solution:

∫ xdxsec

∫+

+= dx

xx

xxx

tansec

)tan(secsec

∫+

+

=

xx

xxd

tansec

)tan(sec (Here, xxu tansec += and dxxxxdu )tan(secsec += .)

Cxx ++= tansecln

Notes:

(1) Students are strongly recommended to recite the substitution used in Example 12, as well

as the way to find this integral.

(2) Alternatively, you can find the integral using partial fractions as follows:

∫ ∫∫∫∫−

=

−

=== duu

xdx

dxx

xdx

xxdx

222 1

1)(sin

sin1

1

cos

cos

cos

1sec (for xu sin= )

We can let u

B

u

A

u −

+

+

≡

− 111

1

2. Solving the simultaneous equations set up by

comparing like terms, we can get 2

1=A and

2

1=B . (Refer to Example 7 for details)

Then, Cx

xCuudu

udu

uxdx +

−

+=+−−+=

−

+

+

= ∫∫∫sin1

sin1ln

2

11ln

2

11ln

2

1

1

1

2

1

1

1

2

1sec .

(3) Can you find ∫ xdxcsc ? Give it a try! (Hint: Let xxu cotcsc += )

(or Cx +secln )

11

Example 13 (Subsidiary Angle)

(a) Let2

0π

θ << . By differentiating )tanln(sec θθ + , show that ∫ ++= Cd )tanln(secsec θθθθ ,

where C is a constant.

(b) Let )cos(sincos αθθθ +=− r , where r and α are constants with 0>r and 22

π

α

π

<<− .

Find r and α .

(c) Hence, find ∫−

θθθd

sincos

1.

Solution:

(a) θθθθθθ

θθθ

sec)sectan(sectansec

1)tanln(sec 2

=+

+

=+

d

d

Integrating both sides w.r.t. θ , we have

Cd ++=∫ )tanln(secsec θθθθ

(b) θααθαθθθ sincossincos)cos(sincos rrr −=+=−

Comparing like terms, we have

−=−

=

)2...(....................1cos

..(1).................... 1sin

α

α

r

r

1tan:)2(

)1(=α

We get 4

π

α = .

2222222 11)cos(sin:)2()1( +=++ ααr

We get 2=r .

(c) From (b), we have

+=−4

cos2sincosπ

θθθ .

C

d

dd

+

++

+=

+

+=

+

=−

∫

∫∫

4tan

4secln

2

2

44sec

2

1

4cos

1

2

1

sincos

1

πθ

πθ

πθ

πθ

θπ

θ

θθθ

Notes:

(1) The form )sin( α±xr or )cos( α±xr is called the subsidiary angle form. It expresses the sum

(or difference) of two sine (or cosine) functions (with any non-zero coefficients) as a product

of a non-zero real number and a sine (or cosine) function.

(2) Subsidiary angle form is NOT required in HKDSE M2 Examination. However, question

format as in Example 13 is possible to appear in the examination.

12

Example 14

Find (a) ∫ dxx

x)sin(ln (b) ∫ dxxx )(cossecsin 2 (c) ∫ ++ dxxxx )1cot()1csc( 22 .

Solution:

(a) ∫ dxx

x)sin(ln

Cx

xdx

+−=

= ∫)cos(ln

)(ln)sin(ln

(b) ∫ dxxx )(cossecsin 2

Cx

xdx

+−=

−= ∫)tan(cos

)(cos)(cossec2

(c) ∫ ++ dxxxx )1cot()1csc( 22

Cx

xdxx

++−=

+++= ∫

)1csc(2

1

)1()1cot()1csc(2

1

2

222

3.3.3. Integration of Special Types of Trigonometric Functions

The integrals of some trigonometric functions can be found easily by certain methods.

(Here, p and q are positive integers.)

Type A: ∫ xdxxqp

cossin

Example 15

Find (a) ∫ xdxx cossin 5 (b) ∫ xdxx36 cossin (c) ∫ xdxx

46 cossin .

Solution:

(a) Cx

xxdxdxx +== ∫∫ 6

sin)(sinsincossin

6

55

(b) ∫ xdxx36 cossin

Cxx

xxdxxd

xdxx

xdxxx

+−=

−=

−=

⋅⋅=

∫ ∫

∫

∫

9

sin

7

sin

)(sinsin)(sinsin

)(sin)sin1(sin

coscossin

97

86

26

26

13

(c) We first express xx46

cossin as a sum of nxcos , where n is an integer.

( )

)10cos8cos26cos34cos82cos26(512

1

6cos2

110cos

2

18cos6cos22cos24cos42cos33

256

1

)8cos2cos8cos4cos2cos44cos42cos33(256

1

)2cos1(8cos4cos43256

1

)2cos1(2

8cos14cos21

128

1

)2cos1)(4cos4cos21(128

1

)2cos1(2

4cos1

32

1

)2cos1(2sin32

1

2

2cos12sin

2

1

sin)cos(sincossin

2

2

4

4

2446

xxxxx

xxxxxxx

xxxxxxx

xxx

x

x

x

xxx

x

x

xx

x

x

xxxxx

−++−−=

−−+++−−=

−++−−=

−+−=

−

++−=

−+−=

−

−=

−=

−

=

=

Cxxxxxx

Cxxxxxx

dxxxxxxxdxx

+−++−−=

+

−++−−=

−++−−= ∫∫

10sin5120

18sin

2048

16sin

1024

14sin

256

12sin

512

1

256

3

10sin10

18sin

4

16sin

2

14sin22sin6

512

1

)10cos8cos26cos34cos82cos26(512

1cossin 46

Notes:

(1) If at least one of p or q is odd, we use the substitution xu sin= or xu cos= , in order to

convert the integral into a form only with the substitution used. In Examples 13(a) and 13(b),

we used xu sin= as the substitution and the whole integral was expressed in terms of xsin .

(2) If both p and q are even, we use the identities2

2cos1sin

2 x

x

−

= and2

2cos1cos

2 x

x

+=

repeatedly to lower the power of the integral. The integral is eventually expressed as a sum of

sine and cosine functions, and can then be integrated easily.

14

Type B: ∫ xdxxqp

sectan

Example 16

Find (a) ∫ xdxx25 sectan (b) ∫ xdxx

42 sectan (c) ∫ xdxx53 sectan .

Solution:

(a) ∫ xdxx25 sectan

Cx

xxd

+=

= ∫

6

tan

)(tantan

6

5

(b) ∫ xdxx42 sectan

Cxx

xxdxxd

xdxx

xdxxx

++=

+=

+=

⋅=

∫∫

∫

∫

3

tan

5

tan

)(tantan)(tantan

)(tan)1(tantan

secsectan

35

24

22

222

(c) ∫ xdxx53 sectan

Cxx

xxdxxd

xxdx

xdxxxx

+−=

−=

−=

⋅=

∫∫

∫

∫

5

sec

7

sec

)(secsec)(secsec

)(secsec)1(sec

tansecsectan

57

46

42

42

Notes:

(1) If q is even, we use the substitution xu tan= and convert the whole integral in terms of xtan .

Besides this, xn

sec (where n is an even number) can be expressed in terms of xtan using the

relation xx22

tan1sec += .

(2) If both p and q are odd, we use the substitution xu sec= and convert the whole integral in

terms of xsec . Besides this, xn

tan (where n is an even number) can be expressed in terms of

xsec using the relation 1sectan22

−= xx .

(3) It is much more difficult to find the integral if p is even and q is odd. A new technique called

‘integration by parts’ is required, and it will be introduced in Section 3.5.

(Refer to Example 33 on p. 33 for details.)

(4) Integrals in the form ∫ xdxxqp

csccot are found in similar way.

15

Type C: ∫ xdxn

sin , ∫ xdxn

cos , etc.

The techniques employed are similar to those used in finding Type A and Type B integrals.

Example 17

(a) Find (i) ∫ xdx4

sin (ii) ∫ xdx5cos (iii) ∫ xdx

4tan (iv) ∫ xdx5

cot .

(b) By differentiating xxn

cossin1− , show that ∫∫

−−−

+−= xdxn

nxx

nxdx

nnn 21sin

1cossin

1sin ,

where 2≥n . Hence, find ∫ xdx4

sin .

Solution:

(a) (i) ∫ xdx4

sin

1

2

2

4sin32

12sin

4

1

8

3

)4cos2cos43(8

1

2

4cos12cos21

4

1

)2cos2cos21(4

1

2

2cos1

Cxxx

dxxx

dxx

x

dxxx

dxx

++−=

+−=

++−=

+−=

−=

∫

∫

∫

∫

(ii) ∫ xdx5cos

Cxxx

xdxx

xdxx

++−=

+−=

−=

∫

∫

53

42

22

sin5

1sin

3

2sin

)(sin)sinsin21(

cos)sin1(

(iii) ∫ xdx4tan

Cxxx

dxxxxd

xdxxdxx

xdxx

++−=

−−=

−=

−=

∫∫

∫∫

∫

tan3

tan

)1(sec)(tantan

tantansec

tan)1(sec

3

22

222

22

We try to express the integral

as a sum of cosine functions.

We try to express the integral

in terms of xsin .

16

(iv) ∫ xdx5

cot

Cxxx

xdx

xx

dxx

xxxdxxd

xdxxdxxxdxx

xdxx

+++−=

++−=

++−=

+−=

⋅−=

∫

∫∫∫

∫∫∫

∫

sinlncsc4

csc

)(sinsin

1csc

4

csc

sin

cos)(csccsc2)(csccsc

cotcotcsc2cotcsc

cot)1(csc

2

4

2

4

3

24

22

(b) Let xxyn

cossin1−

= .

xnxn

xxnx

xxnx

xxnxxxdx

dy

nn

nn

nn

nn

sinsin)1(

)sin1(sin)1(sin

cossin)1(sin

))(cos)(sin1)((cos)sin(sin

2

22

22

21

−−=

−−+−=

−+−=

−+−=

−

−

−

−−

Integrating both sides w.r.t. x, we have

∫∫

∫∫

∫∫

−−

−−

−

−+−=

−−=

−−=

xdxn

nxx

nxdx

xxdxnxdxn

xdxnxdxny

nnn

nnn

nn

21

12

2

sin1

cossin1

sin

cossinsin)1(sin

sinsin)1(

2

3

03

234

8

3cossin

8

3cossin

4

1

)(sin2

1cossin

2

1

4

3cossin

4

1

sin4

3cossin

4

1sin

Cxxxxx

dxxxxxx

xdxxxxdx

++−−=

+−+−=

+−=

∫

∫∫

An integral may take up different forms. We obtain different results for the same integral ∫ xdx4

sin .

If we subtract the result in (a)(i) from that in (b), we get:

0

2cos2sin16

12sin

4

12sin

16

32cos2sin

16

12sin

16

1

2cos2sin16

12sin

4

12sin

16

3

2

2cos12sin

2

1

4

1

4sin32

12sin

4

1

8

3

8

3cossin

8

3cossin

4

1 3

=

−+−+−=

−+−

−

−=

+−−

+−−

xxxxxxx

xxxx

x

x

xxxxxxxx

It follows that C1 = C2. There is no difference between the two integrals, and they are solely equivalent.

17

3.4. Integration by Trigonometric Substitution

Integration of trigonometric functions can be helpful in finding certain indefinite integrals. Before

learning how to do so, we have to learn about inverse trigonometric functions.

3.4.1. Inverse Trigonometric Functions

We have come across with a lot of trigonometric functions, say, xy sin= . To express x in terms of y,

we define yx1

sin−

= (or yx arcsin= ), where y1sin − is read as ‘arcsine y’.

(a) Principal Values

Consider xy sin= . For2

2=y , we may have ,...

4

9,

4

3,

4

πππ

=x (in fact, 4

)1(π

πn

nx −+= ,

where n is an integer). A function should give one and only one output for every input (e.g.

for yx1

sin−

= , if x is a function of y, then for any value of y, there is one and only one

corresponding value of x). To make sure the relationship, we define an interval called

‘principal values’ to restrict the value of x.

Refer to the following illustration:

Let xy1

sin−

= . (Note that x and y is exchanged here, unlike the above example)

For arcsine functions, the principal values are

−

2,

2

ππ

(i.e. 22

ππ

≤≤− x ).

When 2

1=x , we have

6

π

=y , i.e. 62

1sin

1 π

=− .

When 1−=x , we have 2

π

−=y , i.e. 2

)1(sin 1 π

−=−− .

Note: Our calculators are user-friendly. When you enter inverse trigonometric functions, it will

automatically show the principal value as the answer.

The principal values of different functions are shown in the table below:

Function Principal Values

x1

sin−

−

2,

2

ππ

x1

cos− [ ]π,0

x1

tan−

−

2,

2

ππ

x1

csc−

−−2

,π

π or

2,0π

x1

sec−

−−2

,π

π or

2,0π

x1

cot− ),0( π

Note: 02

cot =π

.

Point to note:

[ ] refers to closed interval (i.e. including the

limits) while ( ) refers to open interval (i.e.

excluding the limits).

Examples:

(1)

−∈

2,

2

ππ

x means 22

ππ

≤≤− x .

(2)

−∈

2,

2

ππ

x means 22

ππ

<<− x .

(3)

−∈

2,

2

ππ

x means 22

ππ

≤<− x .

18

Example 18

Find (a) 75.0sin1− (b)

4cos

1 π− (c) 3.3cot1− , in radian measures, correct to 3 decimal places.

Solution:

(a) 848.075.0sin1

≈−

(i.e. 75.0848.0sin ≈ )

(b) 667.04

cos1

≈−π

(i.e. 4

667.0cosπ

≈ )

(c) We cannot enter ‘ 1cot

− ’ directly into our calculators.

Therefore, we use the following method.

294.0

3.3

1tan3.3cot

11

≈

= −−

(i.e. 3.3294.0cot ≈ )

(b) Graphs of Inverse Trigonometric Functions

The following shows the graphs of the six trigonometric functions and those of their inverse

functions for comparison:

Let 3.3cot1−

=x .

Then, we have

3.3cot =x

3.3

1tan

3.3

1tan

3.3tan

1

1−=

=

=

x

x

x

i.e.

= −−

3.3

1tan3.3cot

11

x

y

O

y=sin x

y=sin−1

x

1 −1 x

y

O

x

y

O

y=cos x

x

y

O

y=cos−1

x

1

π

2

π

−1

19

Figure 2 The graphs of the six trigonometric functions and their inverse functions

x

y

O

y=tan x

x

y

O

y=tan−1

x

x

y

O

x

y

O

x

y

O

y=csc x

x

y

O

y=csc−1

x

y=sec x x

y

O

1 −1

y=sec−1

x

2

π

1 −1

2

π

−

y=cot x

x

y

O

y=cot−1

x 2

π

π−

20

3.4.2. Basic Integration by Trigonometric Substitution

Trigonometric substitutions are useful in finding integrals involving 22xa − , 22

xa + and 22ax − ,

where a is a constant. The following examples will illustrate how we use the substitution to find

these integrals.

Type A: Integrals involving 22xa −

We use the substitution θsinax = , where 22

πθ

π<<− .

Example 19

Find ∫−

dx

x2

25

1.

Solution:

Let θsin5=x , then θθddx cos5= .

Cx

C

d

d

d

ddx

x

+=

+=

=

=

=

⋅

−

=

−

−

∫

∫

∫

∫∫

5sin

cos

cos

cos5

cos5

cos5)sin5(25

1

25

1

1

2

22

θ

θ

θθ

θ

θ

θ

θ

θθ

θ

Notes:

(1) In the range 22

πθ

π<<− , θcos is positive. As a result, θθ coscos

2= .

(Normally, we do not have to specify the range of θ in our calculations.)

(2) We can also use the substitution θcos5=x as follows:

Let θcos5=x , then θθddx sin5−= .

Cx

C

d

d

d

ddx

x

+−=

+−=

−=

−=

−=

−⋅

−

=

−

−

∫

∫

∫

∫∫

5cos

sin

sin

sin5

sin5

)sin5()cos5(25

1

25

1

1

2

22

θ

θ

θθ

θ

θ

θ

θ

θθ

θ

21

Example 20

Find ∫ − dxx2

9 .

Solution:

Let θsin3=x , then θθddx cos3= .

C

d

d

d

ddxx

++=

+=

=

−=

⋅−=−

∫

∫

∫

∫∫

θθ

θθ

θθ

θθθ

θθθ

2sin4

9

2

9

)2cos1(2

9

cos9

cossin19

cos3)sin3(99

2

2

22

Now, we need to express the result in terms of x.

3sin

1 x−

=θ

θθθ cossin22sin =

[We construct a right-angled triangle as shown on the right such that3

sinx

=θ .]

[Length of adjacent side = 222

93 xx −=− ]

[Then, we have3

9cos

2x−

=θ .]

9

92

3

9

322sin

22xxxx −

=

−

=θ

Cxxx

Cxxx

dxx

+−+=

+

−+=−

−

−

∫

21

2

12

92

1

3sin

2

9

9

92

4

9

3sin

2

99

Notes:

(1) Sometimes, we may need to further perform integration of trigonometric functions after using

a trigonometric substitution.

(2) The steps of finding θcos (enclosed by [ ]) can be skipped, as long as the right-angled triangle

(with the lengths of three sides clearly indicated) is clearly drawn.

θ

x 3

29 x−

22

Type B: Integrals involving 22xa +

We use the substitution θtanax = , where 22

πθ

π<<− .

Example 21

It is given that Cxxxdx ++=∫ tanseclnsec . (Refer to Example 12 for details)

Find (a) ∫+

dxx 3

1

2 (b) ∫

+

dx

x 3

1

2

.

Solution:

(a) Let θtan3=x , then θθddx2

sec3= .

1

1

1

2

2

2

2

2

2

2

3tan

3

3

3

3

sec

sec

3

3

)1(tan3

sec3

3)tan3(

sec3

3

1

Cx

C

d

d

ddxx

+=

+=

=

+

=

+

=

+

−

∫

∫

∫∫

θ

θθ

θ

θθ

θ

θ

θ

θ

(b) Let θtan3=x , then θθddx2

sec3= .

3ln''' where,''3ln

'3ln3ln

'33

3ln

'tansecln

sec

sec

sec

1tan3

sec3

3)tan3(

sec3

3

1

2

2

2

2

2

2

2

2

2

−=+++=

+−++=

+++

=

++=

=

=

+⋅

=

+

=

+

∫

∫

∫

∫∫

CCCxx

Cxx

Cxx

C

d

d

d

ddx

x

θθ

θθ

θθ

θ

θ

θ

θ

θ

θ

θ

Note:

As 22

πθ

π<<− , θθ secsec

2= , but not θsec− .

θ

x

3

32+x

23

Example 22

Find ∫+

dx

x

x

42

3

by (a) a trigonometric substitution (b) an algebraic substitution.

Solution:

(a) Let θtan2=x , then θθddx2

sec2= .

Cxx

Cxx

Cxx

C

d

d

d

ddx

x

x

+−+=

++−+=

+

+−

+=

+−=

−=

=

⋅=

⋅+

=+

∫

∫

∫

∫∫

)8(43

1

44)4(3

1

2

48

2

4

3

8

sec8sec3

8

)(sec)1(sec8

sectan8

sec2sec2

tan8

sec24)tan2(

)tan2(

4

22

22

3

2

23

2

3

2

3

2

2

3

2

2

3

2

3

θθ

θθ

θθθ

θθ

θ

θ

θθ

θ

θ

(b) Let 42+= xu , then xdxdu 2= .

Cxx

Cxx

Cuu

duuu

du

u

u

duxu

xdx

x

x

+−+=

++−+=

+−=

−=

⋅−

=

⋅=

+

∫

∫

∫∫

−

)8(43

1

)4(4)4(3

1

43

1

)4(2

1

4

2

1

2

1

4

22

2

1

22

3

2

2

1

2

3

2

1

2

1

2

1

3

2

3

Notes:

An integral can be found by different methods. In this example, we used trigonometric substitution

and algebraic substitution, and we still arrived at the same result.

θ

2

42+x

x

24

Type C: Integrals involving 22ax −

We use the substitution θsecax = , where 2

0π

θ << or2

πθπ −<<− .

Example 23

It is given that Cxxxdx ++=∫ tanseclnsec . (Refer to Example 12 for details)

Find (a) ∫−

dx

xx 9

1

2

(b) ∫−

dx

x 9

1

2

.

Solution:

(a) Let θsec3=x , then θθθ ddx tansec3= .

1

1

2

22

3sec

3

1

3

1

tan

tan

3

1

tan3

tan

9)sec3(sec3

tansec3

9

1

Cx

C

d

d

ddx

xx

+=

+=

=

=

−

=

−

−

∫

∫

∫∫

θ

θθ

θ

θ

θ

θ

θ

θθ

θθ

(b) Let θsec3=x , then θθθ ddx tansec3= .

3ln''' where,''9ln

'3

9

3ln

'tansecln

sec

tan

tansec

9)sec3(

tansec3

9

1

2

2

2

22

−=+−+=

+−

+=

++=

=

=

−

=

−

∫

∫

∫∫

CCCxx

Cxx

C

d

d

ddx

x

θθ

θθ

θ

θ

θθ

θ

θ

θθ

Notes:

In the range 2

0π

θ << or2

πθπ −<<− , θθ tantan

2= , but not θtan− .

θ

3

x 9

2−x

25

3.4.3. Variations of Integration by Trigonometric Substitution

Example 24

Find ∫+

dxx

x

2sin1

cos.

Solution:

Let θtansin =x , then θθdxdx2

seccos = .

Cx

C

d

ddxx

x

+=

+=

=

+

=

+

−

∫

∫∫

)(sintan

tan1

sec

sin1

cos

1

2

2

2

θ

θ

θθ

θ

Example 25 (Method of Completing the Square)

Find ∫−+−

dx

xx 2012

1

2

.

Solution:

We first express 20122

−+− xx in the form of 22 )( bxa −− , where a and b are constants.

22

2

22

2

22

)6(4

)6(16

2

1220

2

1212

20)12(2012

−−=

−−=

+−

+−−=

−−−=−+−

x

x

xx

xxxx

Let θsin46 =−x , then θθddx cos4= .

Cx

C

d

d

dx

x

dx

xx

+−

=

+=

=

−

=

−−

=

−+−

−

∫

∫

∫∫

4

6sin

cos

cos

)sin4(4

cos4

)6(4

1

2012

1

1

22

222

θ

θθ

θ

θ

θ

θ

Notes:

The method of completing the square is used in converting quadratic expressions into the form

required for using trigonometric substitutions. We usually use this method when we encounter a

quadratic denominator, while the numerator is a constant.

26

3.5. Integration by Parts

3.5.1. Introduction

There are still some integrals that cannot be found using the above techniques, e.g. ∫ xdxx sin ,

∫ xdxx ln and ∫ xdxex

cos , etc. which are mostly in the form of the products of two functions.

Here, we need a new technique called integration by parts to deal with these integrals.

Let )(xfu = and )(xgv = be two differentiable functions.

Then, we have ∫∫ −= vduuvudv .

The proof is amazingly simple with the help of the product rule of differentiation.

Proof:

'')( vuuvuvdx

d+=

Integrating both sides w.r.t. x, we have

∫∫

∫∫+=

+=

vduudv

dxvudxuvuv ''

i.e. ∫∫ −= vduuvudv .

The following example illustrates how we can use integration by parts to find an integral.

Example 26

Find ∫ xdxln .

Solution:

Consider ∫ xdxln .

Take xu ln= and xv = .

We have dxx

du1

= and dxdv = .

Cxxx

dxxx

dxx

xxx

vduuv

udvxdx

+−=

−=

⋅−=

−=

=

∫

∫

∫

∫∫

ln

ln

1ln

ln

In fact, it can be written in a simpler way:

Cxxx

dxxx

dxx

xxx

xxdxxxdx

+−=

−=

⋅−=

−=

∫

∫

∫∫

ln

ln

1ln

)(lnlnln

27

3.5.2. Common Integrals involving Integration by Parts

There are many integrals that can be found using integration by parts. Some of them are listed here.

Type A: ∫ xdxxn

sin or ∫ xdxxn

cos

Example 27

Find (a) ∫ xdxx sin (b) ∫ xdxx cos (c) ∫ dxx

x2

sin22 .

Solution:

(a) ∫ xdxx sin

Cxxx

xdxxx

xxd

++−=

+−=

−=

∫

∫

sincos

coscos

)(cos

(b) ∫ xdxx cos

Cxxx

xdxxx

xxd

++=

−=

=

∫

∫

cossin

sinsin

)(sin

(c) ∫ dxx

x2

sin22

( )

Cxxxxxx

xdxxxxx

xxdxxx

xdxx

xdxxdxx

dxxx

++−−=

+−=

+−=

−=

−=

−=

∫

∫

∫

∫∫

∫

sincossin2

1

6

sinsin2

1

6

)(sin2

1sin

2

1

6

)(sin2

1

6

cos2

1

2

1

cos12

1

2

3

2

3

22

3

2

3

22

2

Notes:

Integration by parts can be used repeatedly to lower the power of some integrals. In DSE M2

examination, however, you are required to use this technique at most twice only to find an integral.

Here, xu = , xv cos= ,

dxdu = and xdxdv sin−= .

Here, xu = , xv sin= ,

dxdu = and xdxdv cos= .

Use integration by parts

repeatedly here (as in part

(a)) to find ∫ xdxx sin .

28

Type B: ∫ xdxxn

ln

Example 28

Find ∫ xdxxn

ln for (a) 1−≠n (b) 1−=n .

Solution:

(a) ∫ xdxxn

ln

Cxn

xxn

dxxn

xxn

xdxn

xxn

xxdn

nn

nn

nn

n

+

+

−

+

=

+

−

+

=

+

−

+

=

+

=

++

+

++

+

∫

∫

∫

1

2

1

1

11

1

)1(

1ln

1

1

1

1ln

1

1

)(ln1

1ln

1

1

)(ln1

1

(b) ∫−

xdxx ln1

Cx

xxd

dxx

x

+=

=

=

∫

∫

2

)(ln

)(lnln

ln

2

*Think About*:

Can we find the integral by expressing ∫ xdxxn

ln as ∫−+ )ln(ln1

xxxdxxnn ?

Type C: ∫ dxexxn

Example 29

Find (a) ∫ dxxex (b) ∫ dxex

x22 .

Solution:

(a) ∫ dxxex

Cexe

dxexe

exd

xx

xx

x

+−=

−=

=

∫

∫ )(

29

(b) ∫ dxexx22

Cexeex

dxexeex

exdex

dxxeex

xdeex

edx

xxx

xxx

xx

xx

xx

x

++−=

+−=

−=

−=

−=

=

∫

∫

∫

∫

∫

2222

2222

222

222

2222

22

4

1

2

1

2

1

2

1

2

1

2

1

)(2

1

2

1

2

1

)(2

1

2

1

)(2

1

Type D: ∫ bxdxeax

sin or ∫ bxdxeax

cos

Example 30

Find (a) ∫ xdxex

sin (b) ∫ bxdxeax

cos , where 0, ≠ba .

Solution:

(a) ∫ xdxex

sin

∫

∫

∫

∫

∫

∫

−+−=

−+−=

+−=

+−=

+−=

−=

xdxexexe

exdxexe

xdexe

xdxexe

exdxe

xde

xxx

xxx

xx

xx

xx

x

sinsincos

)(sinsincos

)(sincos

coscos

)(coscos

)(cos

Rearranging terms, we have

CCCxexexdxe

Cxexexdxe

xxx

xxx

2

1' where,'sin

2

1cos

2

1sin

sincossin2

=++−=

++−=

∫

∫

Note: You can try to find the integral by first expressing ∫ xdxex

sin as ∫ )(sin x

exd .

You will get the same result at last!

Use integration by parts

repeatedly here to find

∫ dxxex2 .

30

(b) ∫ bxdxeax

cos

∫

∫

∫

∫

∫

∫

−+=

−+=

+=

+=

−=

=

bxdxea

bbxe

a

bbxe

a

bxdea

bbxe

a

bbxe

a

ebxda

bbxe

a

bxdxea

bbxe

a

bxdea

bxea

ebxda

axaxax

axaxax

axax

axax

axax

ax

cossincos1

)(sinsincos1

)(sincos1

sincos1

)(cos1

cos1

)(cos1

2

2

2

22

2

Rearranging terms, we have

Cba

aCCbxe

ba

bbxe

ba

abxdxe

Cbxea

bbxe

abxdxe

a

ba

axaxax

axaxax

22

2

2222

22

22

' where,'sincoscos

sincos1

cos

+

=+

+

+

+

=

++=

+

∫

∫

3.5.3. Other Examples Involving Integration by Parts

Example 31

(a) Let xy 1sin

−

= . By consideringdy

dx, or otherwise, show that

21

1

xdx

dy

−

= .

(b) Hence, or otherwise, find (i) ∫−

xdx1

sin (ii) ∫−

dxx21 )(sin .

Solution:

(a) From xy 1sin

−

= , we have yx sin= .

Then, we have ydy

dxcos= .

By inverse function rule of differentiation, we also have ydy

dx

dx

dy

cos

11

=

=

−

.

Also note that 221sin1cos xyy −=−= .

Thus, we have2

1

1

xdx

dy

−

= .

31

Notes:

(1) For xy 1sin

−

= , 22

ππ

≤≤− y . In this range, ycos is positive, and hence

yy2

sin1cos −= , but not y2

sin1−− .

(2) The result can also be arrived at by finding ∫−

dx

x2

1

1 using the substitution θsin=x .

(Refer to Example 19 for details)

(b) (i) There are two methods for finding this integral.

Method 1:

Cxxx

xd

x

xx

dx

x

xxx

xxdxxxdx

+−+=

−

−

+=

−

−=

−=

−

−

−

−−−

∫

∫

∫∫

21

2

2

1

2

1

111

1sin

)1(1

1

2

1sin

1sin

)(sinsinsin

Method 2:

Let xu1

sin−

= . Then, we have ux sin= , i.e. ududx cos= .

Cxxx

Cxxx

Cuuu

uduuu

uud

uduuxdx

+−+=

++=

++=

−=

=

=

−

−−−

−

∫

∫

∫∫

21

111

1

1sin

)cos(sin)sin(sin)(sin

cossin

sinsin

)(sin

cossin

(ii) Again, there are two methods for this integral.

Method 1:

Cxxxxx

dx

x

xxxxx

xdxxxxx

xxdxx

dx

x

xxxx

xxdxxdxx

+−−+=

−

−−−+=

−−−+=

−+=

−

−=

−=

−−

−−

−−−

−−

−

−

−−−

∫

∫

∫

∫

∫∫

21sin2)(sin

1

121sin2)(sin

)(sin121sin2)(sin

)1(sin2)(sin

1

sin2)(sin

)(sin)(sin)(sin

2121

2

2

2121

122121

2121

2

1

21

212121

By (a), we have

2

1

1

1)(sin

x

xdx

d

−

=−

.

Reminders:

(1) xx =− )sin(sin 1

(2) We find )cos(sin 1x

− by

constructing the triangle as

shown:

21 x−

x

1

x1

sin−

32

Method 2:

We also let xu1

sin−

= . Then, we have ux sin= , i.e. ududx cos= .

Cxxxxx

Cxxxxx

Cuuuuu

uduuuuu

uuduu

uduuuu

uuduu

udu

uduudxx

+−−+=

+−+=

+−+=

−+=

+=

−=

−=

=

=

−−

−−−−−

−

∫

∫

∫

∫

∫

∫∫

21sin2)(sin

)sin(sin2)cos(sinsin2)sin(sin)(sin

sin2cos2sin

cos2cos2sin

)(cos2sin

sin2sin

)(sinsin

)(sin

cos)(sin

2121

111121

2

2

2

2

22

2

221

Example 32

Find (a) ∫ xdxex

cossin

for π<< x0 (b) ∫ dxx)sin(ln for 0>x .

Solution:

(a) Let xu sin= , then xdxx

du cossin2

1= .

Cex

Ceue

dueue

eud

duue

uduexdxe

x

uu

uu

u

u

ux

+−=

+−=

−=

=

=

⋅=

∫

∫

∫

∫∫

sin

sin

)1sin(2

22

22

)(2

2

2cos

(b) Let xu ln= , then dxx

du1

= .

∫

∫∫=

=

udue

uduxdxx

u

sin

sin)sin(ln

Repeating the steps in Example 30, we have Cueueudueuuu

++−=∫ sin2

1cos

2

1sin .

Cxxxx

Cxexedxxxx

+−=

++−=∫

)cos(ln2

1)sin(ln

2

1

)sin(ln2

1)cos(ln

2

1)sin(ln lnln

33

Example 33 (Finding ∫ xdxxqp

sectan for even p and odd q)

It is given that Cxxxdx ++=∫ tanseclnsec . (Refer to Example 12 for details)

Find (a) ∫ xdx3

sec (b) ∫ xdxx sectan2 (c) ∫ xdxx

32sectan (d) ∫ − dxxx 1

22 .

Solution:

(a) ∫ xdx3

sec

∫

∫

∫

∫∫

∫

−++=

++−=

++=

+=

+=

xdxxxxx

xxxxdxx

xxxxd

xdxxdxx

dxxx

3

2

2

sectanseclntansec

tansecln)(tansectansec

tansecln)(sectan

sectansec

)1(tansec

Rearranging terms, we have

2

3

1

3

tansecln2

1tansec

2

1sec

tanseclntansecsec2

Cxxxxxdx

Cxxxxxdx

+++=

+++=

∫

∫

(b) ∫ xdxx sectan2

3

3

3

2

tansecln2

1tansec

2

1

tanseclntansecln2

1tansec

2

1

secsec

sec)1(sec

Cxxxx

Cxxxxxx

xdxxdx

xdxx

++−=

++−++=

−=

−=

∫∫

∫

Alternatively,

∫ xdxx sectan2

∫

∫∫

∫

∫

∫

−+−=

−−=

−=

−=

=

xdxxxxxx

xdxxdxxxx

xdxxx

xxdxx

xxd

2

2

3

tansectanseclntansec

sectansectansec

sectansec

)(tansectansec

)(sectan

34

Rearranging terms, we have

3

2

2

tansecln2

1tansec

2

1sectan

tanseclntansecsectan2

Cxxxxxdxx

cxxxxxdxx

++−=

++−=

∫

∫

(c) ∫ xdxx32

sectan

∫

∫∫

∫∫

∫∫

∫

−+−+−=

+−−=

+−−=

−=

−=

xdxxxxxxxxxx

xdxxxxxdxxxx

xxdxxxxdxx

xxdxxd

xxdx

323

2323

33

3

2

sectan3tansecln2

1tansec

2

1tansectansec

sectantansecsectan3tansec

)(sectantansec)(sectantansec

)(tansec)(tansec

)(tansec)1(sec

Rearranging terms, we have

5

332

4

332

tansecln8

1tansec

8

1tansec

4

1sectan

tansecln2

1tansec

2

1tansecsectan4

Cxxxxxxxdxx

Cxxxxxxxdxx

++−−=

++−−=

∫

∫

(d) Let θsec=x , then θθθ ddx tansec= .

5

2223

5

3

32

2222

1ln8

11

8

11

4

1

tansecln8

1tansec

8

1tansec

4

1

sectan

tansec1secsec1

Cxxxxxx

C

d

ddxxx

+−+−−−−=

++−−=

=

⋅−=−

∫

∫∫

θθθθθθ

θθθ

θθθθθ

Notes:

(1) The technique of integration by parts followed by rearranging terms is very useful in finding

integrals in the form ∫ xdxxqp

sectan , where p is even and q is odd.

(2) Here is a challenging problem: find ∫ +− dxxx 2042 . Try it out!

[Hint: Make good use of part (a) of Example 33 by using a suitable substitution!]

(For answers, please read the appendix of this note on p.46.)

12−x x

1

(by (b))

θ

35

Example 34 (Reduction Formula)

(a) Prove the reduction formulas in Examples 8 and 17, i.e.

(i) ∫∫−

−= dxxnxxdxxnnn 1)(ln)(ln)(ln for 0≥n ;

(ii) ∫∫−−

−+−= xdx

n

nxx

nxdx

nnn 21sin

1cossin

1sin for 2≥n .

(b) Let ∫+

=nmn

x

dxI

)1(, where m and n are positive integers.

Show that nnmnI

mn

mn

xmn

xI

1

)1(1

−+

+

=+

. Hence, or otherwise, find ∫+

32 )1(x

dx.

Solution:

(a) (i) ∫ dxxn)(ln

∫

∫

∫

−

−

−=

⋅−=

−=

dxxnxx

dxx

xxnxx

xxdxx

nn

nn

nn

1

1

)(ln)(ln

1)(ln)(ln

)(ln)(ln

(ii) ∫ xdxn

sin

∫∫

∫

∫

∫

∫

−−−+−=

−−+−=

−+−=

+−=

−=

−−

−−

−−

−−

−

xdxnxdxnxx

dxxxnxx

xdxxnxx

xxdxx

xxd

nnn

nn

nn

nn

n

sin)1(sin)1(cossin

)sin1(sin)1(cossin

cossin)1(cossin

)(sincoscossin

)(cossin

21

221

221

11

1

Rearranging terms, we have

∫∫

∫∫−−

−−

−+−=

−+−=+−

xdxn

nxx

nxdx

xdxnxxxdxn

nnn

nnn

21

21

sin1

cossin1

sin

sin)1(cossinsin)11(

Notes:

(1) Rearranging terms is a very commonly used technique used in proving reduction formula using

integration by parts.

(2) In proving reduction formulas involving trigonometric functions, the trigonometric identities

1cossin22

=+ xx , xx22

sec1tan =+ and xx22

csc1cot =+ are helpful on most occasions.

36

(b) nI

∫

∫

∫

+

+

−

++

+=

++

+=

+−

+=

dxx

xmn

x

x

dxx

mxxn

x

x

xxd

x

x

nm

m

nm

nm

m

nm

nmnm

1

1

)1

)1()1(

)1(

(

)1(

)1(

1

)1(

1

11

)1(

)1(

1

)1(

1

)1(

+

++

−+

+

=

+

−

+

++

+

= ∫∫

nnnm

nmnm

m

nm

mnImnIx

x

dxx

mndxx

xmn

x

x

Rearranging terms, we have

nnmn

nnmn

Imn

mn

xmn

xI

Imnx

xmnI

1

)1(

)1()1(

1

1

−+

+

=

−+

+

=

+

+

First, we take 2=m and 2=n .

∫∫+

−+

+

=

+222232 )1()2)(2(

1)2)(2(

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

dx

x

x

x

dx

For ∫+

22 )1(x

dx, take 2=m and 1=n .

∫

∫∫

++

++

+=

+

−+

++

+=

+

18

3

)1(8

3

)1(4

1)1)(2(

1)1)(2(

)1)(1)(2(4

3

)1(4)1(

2222

222232

x

dx

x

x

x

x

x

dx

x

x

x

x

x

dx

For ∫+1

2x

dx, let θtan=x , then θθddx

2sec= .

1

1

1

2

2

2

tan

1tan

sec

1

Cx

C

d

d

x

dx

+=

+=

=

+

=

+

−

∫

∫∫

θ

θ

θ

θθ

Thus, we have Cxx

x

x

x

x

dx++

+

+

+

=

+

−

∫1

22232tan

8

3

)1(8

3

)1(4)1(.

Note: The integral ∫+

32 )1(x

dx can also be found by letting θtan=x at the very beginning.

The reduction formula can even be found by differentiation. Try them out!

37

3.6 Integration of Miscellaneous Functions

Some functions can be integrated with a combination of the above techniques.

3.6.1. Finding Integrals in the Form ∫++

++dx

FExDx

CBxAx

2

2

We follow the following steps:

(In the following, all capital letters represent constants.)

(1) Express ∫++

++dx

FExDx

CBxAx

2

2

as ∫∫++

++ dx

FExDx

IHxGdx

2.

(2) Express ∫++

+dx

FExDx

IHx

2 as ∫ ∫

++

+++

++

dxFExDx

KFExDxd

FExDx

J

2

2

2)( .

(3) Find ∫Gdx , ∫ ++

++

)( 2

2FExDxd

FExDx

J and ∫

++

dxFExDx

K

2 one by one.

The integral is then found.

Example 35

Find ∫ ++

−−dx

xx

xx

52

5432

2

.

Solution:

∫

∫∫

∫∫∫

∫∫∫

++

−++−=

+++

−++

++

−=

++

−

++

+−=

++

+−

++

++=

++

−−

dxx

xxx

dxxx

xxdxx

x

dxxx

dxxx

xdx

dxxx

xdx

xx

xxdx

xx

xx

4)1(

110)52ln(53

4)12(

110)52(

52

153

52

110

52

)22(53

52

2010

52

)52(3

52

543

2

2

2

2

2

22

22

2

2

2

For ∫++

dxx 4)1(

12

, let θtan21=+x , then θθddx2

sec2= .

1

1

1

2

2

2

2

1tan

2

1

2

1

2

1

4)tan2(

sec2

4)1(

1

Cx

C

d

ddxx

++

=

+=

=

+

=

++

−

∫

∫∫

θ

θ

θθ

θ

Thus, we have ∫ ++

−++−=

++

−−−

Cx

xxxdxxx

xx

2

1tan5)52ln(53

52

543 12

2

2

.

This step can be replaced by

long division.

x

x

xx

number real allfor 0

4)1(

52

2

2

>

++=

++

Hence, absolute sign is

not required here for

)52ln( 2 ++ xx .

38

Example 36

(a) Let 111

1

23+−

++

+

≡

+ xx

CBx

x

A

x, where A, B and C are constants. Find A, B and C.

(b) Hence, find ∫+

dxx 1

1

3.

Solution:

(a) From the question, we have111

1

23+−

++

+

≡

+ xx

CBx

x

A

x.

1

)()()(

)1)(1(

)1)(()1(

1

1

3

2

2

2

3

+

++−+++≡

+−+

++++−≡

+

x

CAxACBxBA

xxx

xCBxxxA

x

Hence, we have )()()(1 2CAxACBxBA ++−+++≡ .

Comparing the coefficients of like terms, we have

=+

=−+

=+

1

0

0

CA

ACB

BA

.

Solving the above system of linear equations, we have 3

2 and

3

1 ,

3

1=−== CBA .

(b) ∫∫∫+−

−−

+

=

+

dxxx

xdx

xdx

x 1

2

3

1

1

1

3

1

1

1

23

∫∫

∫∫

∫∫

+

−

++−+−

−+=

−+

+−

++−

−−+=

+−

−−+

+=

dx

x

xxdxx

x

dx

xx

dxxx

xx

dxxx

xxd

x

4

3

2

1

1

2

1)1(

1

1

6

11ln

3

1

2

11

2

1

3

6

1

1

12

6

11ln

3

1

1

42

6

1)1(

1

1

3

1

2

2

2

22

2

2

2

For ∫+

−

dx

x4

3

2

1

1

2, let θtan

2

3

2

1=−x , then θθddx

2sec

2

3= .

1

1

12

2

2

3

12tan

3

2

3

2

3

2

4

3tan

2

3

sec

2

3

4

3

2

1

1C

xCdddx

x

+

−=+==

+

=

+

−

−

∫∫∫ θθθ

θ

θ

Thus, we have Cx

xxxdxx

+

−++−−+=

+−

∫3

12tan

3

1)1ln(

6

11ln

3

1

1

1 12

3.

39

3.6.2. General t-substitution for integrals involving trigonometric functions

There are some integrals that involve trigonometric functions in the denominator and/or numerator

in the integrand, e.g. ∫+

dxxsin1

1 and ∫

++

dxxx 1cossin

1, etc. To find these integrals, we can

use a special method which makes the whole integral be in terms of the variable2

tanx

t = .

Before talking about how we can use this method, we first see how we can express the three basic

trigonometric functions, xsin , xcos and xtan , in terms of 2

tanx

.

Example 37

Let 2

tanx

t = . Express xsin , xcos and xtan in terms of t.

Solution:

2

2

2

2

1

2

2tan1

2tan2

2sec

2tan2

2cos

2cos

2sin

2

2cos

2sin2sin

t

t

x

x

x

x

x

x

x

xxx

+

=

+

=

=

=

=

Notes:

You may also consider constructing a right-angled triangle as shown after finding xtan . In this

case, xsin and xcos can also be found. However, this is NOT considered a good method, as you

can only find xsin and xcos for π<< x0 in this case. Using trigonometric identities would

be a better method as the result is true for all real values of x.

There are also other methods to find out the way to express the three functions

in terms of t. Interested readers can search for more information on the internet.

2

2

2

2

2

2

2

1

1

1

)1(2

1

2tan1

2

1

2sec

2

12

cos2cos

t

t

t

t

x

x

xx

+

−=

+

+−=

−

+

=

−=

−=

2

2

1

2

2tan1

2tan2

tan

t

t

x

x

x

−

=

−

=

40

Example 38

Let ∫+

= dxx

Isin1

1. Using the substitution

2tan

xt = , or otherwise, find I.

Solution:

Let 2

tanx

t = .

dtt

dtx

dx

dxx

dt

2

2

2

1

2

2tan1

2

2sec

2

1

+

=

+

=

=

Cx

Ct

tdt

dttt

dtt

t

t

dxx

I

+

+

−=

+

+

−=

+

+

=

++

=

+

⋅

+

+

=

+

=

∫

∫

∫

∫

12

tan

2

1

2

)1()1(

12

12

2

1

2

1

21

1

sin1

1

2

2

2

2

Notes:

(1) Readers should pay attention to the way we express dx in terms of t.

(2) This integral can be especially found in an easier way:

Cx

xd

x

dxx

x

dxxx

dxxxxx

I

+

+

−=+

+

=

+

=

+

=

++

=

∫

∫∫∫

12

tan

2)1

2(tan

)12

(tan

2

)12

(tan

2sec

)2

cos2

(sin

1

2cos

2cos

2sin2

2sin

1

2

2

2

222

(3) You may also consider:

'sectancos

)(cossec

cos

sin1

sin1

sin1

sin1

12

2

22 ∫ ∫∫∫∫ +−=+=−

=

−

−=

+

Cxxx

xdxdxdx

x

xdx

x

xdx

x

*Think about*:

Can you show that the two answers obtained differ by a constant (1, in this case)?

(by Example 37)

41

Example 39

Find ∫+−

dxxx 5cos3sin2

1.

Solution:

Again, we let 2

tanx

t = , then dtt

dx2

1

2

+

= . (Refer to Example 38)

∫

∫

∫

∫

∫

∫∫

+

+

=

+

+

=

−+

++

=

++=

+++−=

+⋅

+

+

−−

+

=+−

dt

t

dt

t

dt

tt

dttt

dtttt

dtt

t

t

t

tdx

xx

16

3

4

1

1

4

1

4

3

4

14

1

4

11

16

1

2

14

1

124

1

55334

2

1

2

51

13

1

22

1

5cos3sin2

1

2

2

2

2

22

2

2

2

2

Let θtan4

3

4

1=+t , then θθddt

2sec

4

3= .

C

x

Ct

C

d

ddxxx

++

=

++

=

+=

=

+

⋅=

+−

−

−

∫

∫∫

3

12

tan4

tan3

1

3

14tan

3

1

3

1

3

1

16

3tan

4

3

sec

4

3

4

1

5cos3sin2

1

1

1

2

2

θ

θ

θ

θ

θ

Notes:

It is common that we have to use a further substitution to find the integral obtained by letting

2tan

xt = , and the techniques in Section 3.6.1 are useful in most situations.

42

4. Applications of Indefinite Integration

Indefinite integration is useful in both mathematics and daily life applications.

A. Applications in Coordinate Geometry

It is known that )(' xf represents the slope of the tangent of a point (x,y) to the curve )(xfy = .

By integrating )(' xf with respect to x, it is expected that we can obtain the original curve

)(xfy = .

However, is it the case? No!

Integrating )(' xf w.r.t. x, we have

CxfCydxxf +=+=∫ )()('

We observe that there is an extra constant C. In fact, what

we have obtained is a family of curves, which can be

obtained by translating the curve of )(xfy = upwards or

downwards.

We can further observe that all these curves have the same

)(' xf , i.e. the slope of tangent to all curves at point (x,y) is

the same.

In order to find out the equation of a specific curve instead

of a family of curves, we need the coordinates of a specific

point to find out the constant C. This restricts the possibility

of curves to only one curve, but not the family of curves as

mentioned.

Similarly, suppose )('' xf is given, we have

1)(')('' Cxfdxxf +=∫

[ ]211

)()(' CxCxfdxCxf ++=+∫

To find out )(xf from )('' xf , we need to integrate twice.

We also need the coordinates of two points on the curve.

By substituting them into21

)( CxCxfy ++= , we can find out the constants 1

C and 2

C .

Finally, we can find out the original function )(xf .

Refer to the examples on the next page.

x

y

O

Figure 3 The slope of tangent to the

curve at point (x,y) is equal to )(' xf .

Slope = )(' xf

x

y

O

Figure 4 These curves, obtained by

translating )(xfy = upwards or

downwards, have the same )(' xf .

43

Example 40

The slope of any point (x,y) on a curve Q is 2

1 x

x

+

. It is known that Q passes through the origin.

Find the equation of Q.

Solution:

From the question, we have 2

1 x

x

dx

dy

+

= .

Integrating both sides w.r.t. x, we have

Cx

xdx

dxx

xy

++=

+

+

=

+

=

∫

∫

2

2

2

2

1

)1(1

1

2

1

1

Put (0,0) into Cxy ++=2

1 , we have

1

010 2

−=

++=

C

C

Thus, the equation of Q is 11 2−+= xy .

Example 41

For curve Y, it is given that xedx

yd 2

2

2

= . If Y passes through points A(0,1) and B(1,1), find the

equation of curve Y.

Solution:

From xedx

yd 2

2

2

= , by integration, we have 1

22

2

1Cedxe

dx

dyxx

+== ∫ .

By further integration, we have 21

2

1

2

4

1

2

1CxCedxCey xx ++=

+= ∫

Put (0,1) and (1,1) into 21

2

4

1CxCey

x

++= , we get 4

3

2=C and 2

21

4

11 eCC −=+ .

Solving the above system of linear equations, we get 4

12

1

eC

−= and

4

3

2=C .

Thus, the equation of curve Y is 4

3

4

1

4

12

2+

−+= x

eey

x .

44

B. Applications in Physics

Consider an object travelling in a straight line:

We define some terms as follows:

Displacement (s) is the distance of an object from its original position, with the consideration of

direction.

Velocity (v) is the derivative of displacement with respect to time (t).

Acceleration (a) is the derivative of velocity with respect to time.

If we take right hand side as positive direction, If PQ = 3 m, then the displacement of P is 3 m. On

the contrary, if we take left hand side as positive direction, then the displacement of P is −3 m.

Note that 2

2

dt

sd

dt

dva == .

By integrating a or v with respect to t, we can obtain v and s respectively.

Example 42

The velocity of a particle travelling along a straight line is given by ttv 25)( −= , where )(tv is the

velocity (in m/s) and t is time (in second). It is known that the particle passes through its original

position after 4 seconds. Take right hand side as positive direction.

(a) Find the displacement of the particle after t seconds.

(b) What is the displacement of the particle when it is momentarily at rest?

Solution:

(a) Let the displacement of the particle after t seconds be )(ts m.

Then Cttdttdttvts +−=−== ∫∫25)25()()( .

When 4=t , 0)( =ts .

4

0)4()4(52

−=

=+−

C

C

Thus, the displacement of the particle is ( 452

−+− tt ) m.

(b) The particle is momentarily at rest when 0)( =tv , i.e. 2

5=t .

The displacement of particle = m 4

94

2

55

2

52

=−

+

−

P Q

45

C. Other Applications

Other than in physics, in many situations, if the rate of change is given, we can find out the original

quantity by integrating the rate with respect to time.

Example 43

The figure on the right shows an inverted circular cone with base radius 30

cm and height 40 cm. At the beginning, the cone was full of water. A small

hole was then drilled at the apex of the cone and water is flowing out of

the cone at a rate of )12( +t cm3/s, where t is time in second.

(a) Express the volume of water remaining in the cone after t seconds.

(b) After how many seconds will the cone be empty?

Correct your answer to the nearest second.

Solution:

(a) Let the volume of water remaining in the cone be V cm3.

Then, we have

Ctt

dttV

tdt

dV

+−−=

+−=

+−=

∫2

)12(

)12(

When 0=t , we have 32 cm 12000)40()30(3

1ππ ==V .

Thus, the volume of water remaining in the cone after t seconds is )12000( 2tt −−π cm

3.

(b) When the cone is empty, 0=V .

194

)0 rejected,(2

1480001or

2

1480001

)1(2

)12000)(1(4)1(1

012000

012000

2

2

2

≈

>+−−++−

=

−−±−=

=−+

=−−

t

t

tt

tt

Q

ππ

π

π

π

Thus, the required time is 194 seconds.

More applications of integration will be found in the topic ‘Definite integrals’.

The writer would like to thank Mr. Yue Kwok Choy for his kind reading and checking of this

article.

-END-

Figure 5 An inverted circular cone

Negative sign is added

since the volume of

water is decreasing.

46

Appendix 1

Proof of Method of Substitution in Integration

To prove:

Let )( xgu = . Then, we have ∫ ∫= duufdxxgxgf )()('))(( .

Consider the function ))(( xgFy = , where ))(( xgF is a primitive function of ))(( xgf .

Then )('))(()(' ))((' xgxgfxgxgFdx

dy== .

Integrating both sides w.r.t. x, we have dxxgxgfy ∫= )(' ))(( .

But )()(' ufuFdu

dy == .

Integrating both sides w.r.t. u, we have ∫= duufy )( .

Then, we have ∫ ∫= duufdxxgxgf )()('))(( .

This completes this proof.

Appendix 2

Finding ∫ +− dxxx 2042

∫∫ +−=+− dxxdxxx222 4)2(204

Now let θtan42 =−x , then θθddx2

sec4= .

C'xxxxxx

Cxxxxxx

Cxxxx

d

d

dxxdxxx

+−++−++−−=

+−

++−

+

−

+−=

+++=

=

⋅+=

+−=+−

∫

∫

∫∫

2204ln8204)2(2

1

4

2

4

204ln8

4

2

4

2048

33(a)) Example(by tansecln8tansec8

sec16

sec44)tan4(

4)2(204

22

22

3

222

222

θθ

θθθ