integrals - byju's...integrals. ˜ut these integrals are very similar geometrically. thus, y ˜...

INTEGRALS 287

v Just as a mountaineer climbs a mountain – because it is there, soa good mathematics student studies new material because

it is there. — JAMES B. BRISTOL v

7.1 IntroductionDifferential Calculus is centred on the concept of thederivative. The original motivation for the derivative wasthe problem of defining tangent lines to the graphs offunctions and calculating the slope of such lines. IntegralCalculus is motivated by the problem of defining andcalculating the area of the region bounded by the graph ofthe functions.

If a function f is differentiable in an interval I, i.e., itsderivative f ¢exists at each point of I, then a natural questionarises that given f ¢at each point of I, can we determinethe function? The functions that could possibly have givenfunction as a derivative are called anti derivatives (orprimitive) of the function. Further, the formula that givesall these anti derivatives is called the indefinite integral of the function and suchprocess of finding anti derivatives is called integration. Such type of problems arise inmany practical situations. For instance, if we know the instantaneous velocity of anobject at any instant, then there arises a natural question, i.e., can we determine theposition of the object at any instant? There are several such practical and theoreticalsituations where the process of integration is involved. The development of integralcalculus arises out of the efforts of solving the problems of the following types:(a) the problem of finding a function whenever its derivative is given,(b) the problem of finding the area bounded by the graph of a function under certain

conditions.These two problems lead to the two forms of the integrals, e.g., indefinite and

definite integrals, which together constitute the Integral Calculus.

Chapter 7INTEGRALS

G .W. Leibnitz(1646 -1716)

288 MAT�EMATICS

There is a connection, known as the Fundamental Theorem of Calculus, betweenindefinite integral and definite integral which makes the definite integral as a practicaltool for science and engineering. The definite integral is also used to solve many interestingproblems from various disciplines like economics, finance and probability.

In this Chapter, we shall confine ourselves to the study of indefinite and definiteintegrals and their elementary properties including some techniques of integration.

7.2 Integration as an Inverse Process of DifferentiationIntegration is the inverse process of differentiation. Instead of differentiating a function,we are given the derivative of a function and asked to find its primitive, i.e., the originalfunction. Such a process is called integration or anti differentiation.Let us consider the following examples:

�e know that (sin )d

xdx

�cos x ... (�)

�

( )�

d xdx

�x2 ... (2)

and ( )xde

dx�ex ... (�)

�e observe that in (�), the function cos x is the derived function of sin x. �e say

that sin x is an anti derivative (or an integral) of cos x. Similarly, in (2) and (�), �

�x

and

ex are the anti derivatives (or integrals) of x2 and ex, respectively. Again, we note thatfor any real number C, treated as constant function, its derivative is �ero and hence, wecan write (�), (2) and (�) as follows :

(sin �C) cos=d

x xdx

, �

2( �C)�

=d x xdx

and ( �C) =x xde e

dxThus, anti derivatives (or integrals) of the above cited functions are not unique.

Actually, there exist infinitely many anti derivatives of each of these functions whichcan be obtained by choosing C arbitrarily from the set of real numbers. For this reasonC is customarily referred to as arbitrary constant. In fact, C is the parameter byvarying which one gets different anti derivatives (or integrals) of the given function.

More generally, if there is a function F such that F ( ) � ( )d

x f xdx

, " x Î I (interval),

then for any arbitrary real number C, (also called constant of integration)

F ( ) �Cd

xdx

�f (x), x Î I

INTEGRALS 28�

Thus, �F � C, C Î R� denotes a family of anti derivatives of f.

Remark Functions with same derivatives differ by a constant. To show this, let g and hbe two functions having the same derivatives on an interval I.Consider the function f � g � h defined by f (x) � g (x) � h(x), " x Î I

Thendfdx

�f¢ � g¢ � h¢ giving f¢ (x) � g¢ (x) � h¢ (x) " x Î I

or f¢ (x) � �, " x Î I by hypothesis,i.e., the rate of change of f with respect to x is �ero on I and hence f is constant.

In view of the above remark, it is justified to infer that the family �F � C, C Î R�provides all possible anti derivatives of f.

�e introduce a new symbol, namely, ( )f x dxò which will represent the entire

class of anti derivatives read as the indefinite integral of f with respect to x.

Symbolically, we write ( ) �F ( ) �Cf x dx xò .

Notation Given that ( )dy f xdx

= , we write y � ( )f x dxò .

For the sake of convenience, we mention below the following symbols�terms�phraseswith their meanings as given in the Table (7.�).

Table 7.1

Symbols/Terms/Phrases Meaning

( )f x dxò Integral of f with respect to x

f (x) in ( )f x dxò Integrand

x in ( )f x dxò �ariable of integration

Integrate Find the integralAn integral of f A function F such that

F¢(x) � f (x)Integration The process of finding the integral

Constant of Integration Any real number C, considered asconstant function

2�� MAT�EMATICS

�e already know the formulae for the derivatives of many important functions.From these formulae, we can write down immediately the corresponding formulae(referred to as standard formulae) for the integrals of these functions, as listed belowwhich will be used to find integrals of other functions.

Derivatives Integrals (Anti derivatives)

(i)�

�

nnd x

xdx n

+æ ö=ç ÷+è ø

��

C�

nn x

x dxn

+

= ++ò , n ¹ ��

�articularly, we note that

�d

xdx

= � Cdx x= +ò

(ii) sin cosd

x xdx

= � cos sin Cx dx x= +ò

(iii) �cos sind

x xdx

= � sin cos Cx dx – x= +ò

(iv) 2tan secd

x xdx

= � 2sec tan Cx dx x= +ò

(v) 2�cot cosecd

x xdx

= � 2cosec cot Cx dx – x= +ò

(vi) sec sec tand

x x xdx

= � sec tan sec Cx x dx x= +ò

(vii) �cosec cosec cotd

x x xdx

= � cosec cot �cosec Cx x dx x= +ò

(viii)��

2

�sin

�

dx

dx – x= �

��

2sin C

�

dxx

– x= +ò

(ix)��

2

��cos

�

dx

dx – x= �

��

2cos C

�

dx– x

– x= +ò

(x)��

2

�tan

�d

xdx x

=+ �

��2 tan C

�dx

xx

= ++ò

(xi)��

2

��cot

�d

xdx x

=+ �

��2 cot C

�dx

– xx

= ++ò

INTEGRALS 2��

(xii)��

2

�sec

�

dx

dx x x –= �

��

2sec C

�

dxx

x x –= +ò

(xiii)��

2

��cosec

�

dx

dx x x –= �

��

2cosec C

�

dx– x

x x –= +ò

(xiv) ( )x xde e

dx= � Cx xe dx e= +ò

(xv)�

log � �d

xdx x

= ��

log � � Cdx xx

= +ò

(xvi)x

xd aa

dx log aæ ö

=ç ÷è ø

� Cx

x aa dx

log a= +ò

ANote In practice, we normally do not mention the interval over which the variousfunctions are defined. �owever, in any specific problem one has to keep it in mind.

7.2.1 Geometrical interpretation of indefinite integral

Let f (x) � 2x. Then 2( ) Cf x dx x= +ò . For different values of C, we get differentintegrals. �ut these integrals are very similar geometrically.

Thus, y � x2 � C, where C is arbitrary constant, represents a family of integrals. �yassigning different values to C, we get different members of the family. These togetherconstitute the indefinite integral. In this case, each integral represents a parabola withits axis along y�axis.

Clearly, for C � �, we obtain y � x2, a parabola with its vertex on the origin. Thecurve y � x2 � � for C � � is obtained by shifting the parabola y � x2 one unit alongy�axis in positive direction. For C � � �, y � x2 � � is obtained by shifting the parabolay � x2 one unit along y�axis in the negative direction. Thus, for each positive value of C,each parabola of the family has its vertex on the positive side of the y�axis and fornegative values of C, each has its vertex along the negative side of the y�axis. Some ofthese have been shown in the Fig 7.�.

Let us consider the intersection of all these parabolas by a line x � a. In the Fig 7.�,we have taken a � �. The same is true when a � �. If the line x � a intersects theparabolas y � x2, y � x2 � �, y � x2 � 2, y � x2 � �, y � x2 � 2 at ��, ��, �2, ��, ��2 etc.,

respectively, then dydx

at these points equals 2a. This indicates that the tangents to the

curves at these points are parallel. Thus, 2C2 C F ( )x dx x x= + =ò (say), implies that

2�2 MAT�EMATICS

the tangents to all the curves y � FC (x), C Î R, at the points of intersection of thecurves by the line x � a, (a Î R), are parallel.

Further, the following equation (statement) ( ) F ( ) C (say)f x dx x y= + =ò ,

represents a family of curves. The different values of C will correspond to differentmembers of this family and these members can be obtained by shifting any one of thecurves parallel to itself. This is the geometrical interpretation of indefinite integral.

7.2.2 Some properties of indefinite integralIn this sub section, we shall derive some properties of indefinite integrals.

(I) The process of differentiation and integration are inverses of each other in thesense of the following results :

( )d

f x dxdx ò �f (x)

and ( )f x dx¢ò �f (x) � C, where C is any arbitrary constant.

Fig 7.1

INTEGRALS 2��

Proof Let F be any anti derivative of f, i.e.,

F( )d

xdx

�f (x)

Then ( )f x dxò �F(x) � C

Therefore ( )d

f x dxdx ò � F ( ) �C

dx

dx

� F ( ) � ( )d

x f xdx

Similarly, we note that

f ¢(x) � ( )d

f xdx

and hence ( )f x dx¢ò �f (x) � C

where C is arbitrary constant called constant of integration.(II) Two indefinite integrals with the same derivative lead to the same family of

curves and so they are equivalent.Proof Let f and g be two functions such that

( )d

f x dxdx ò � ( )

dg x dx

dx ò

or ( ) ( )d

f x dx – g x dxdx

é ùë ûò ò � �

�ence ( ) ( )f x dx – g x dxò ò � C, where C is any real number (� hy?)

or ( )f x dxò � ( ) Cg x dx +ò

So the families of curves � �( ) C , C Rf x dx + Îòand 2 2( ) C , C Rg x dx + Îò are identical.

�ence, in this sense, ( ) and ( )f x dx g x dxò ò are equivalent.


A Note The equivalence of the families � �( ) �C ,Cf x dx Îò R and

2 2( ) �C ,Cg x dx Îò R is customarily expressed by writing ( ) � ( )f x dx g x dxò ò ,

without mentioning the parameter.

(III) ( ) � ( ) ( ) � ( )f x g x dx f x dx g x dx=ò ò òProof �y �roperty (I), we have

� ( ) � ( )�d

f x g x dxdx

é ùë ûò �f (x) � g (x) ... (�)

�n the otherhand, we find that

( ) � ( )d

f x dx g x dxdx

é ùë ûò ò � ( ) � ( )

d df x dx g x dx

dx dxò ò�f (x) � g (x) ... (2)

Thus, in view of �roperty (II), it follows by (�) and (2) that

( ) ( )f x g x dx+ò � ( ) ( )f x dx g x dx+ò ò .

(I�) For any real number k, ( ) ( )k f x dx k f x dx=ò ò

Proof �y the �roperty (I), ( ) ( )d

k f x dx k f xdx

=ò .

Also ( )d

k f x dxdx

é ùë ûò � ( ) � ( )

dk f x dx k f x

dx ò

Therefore, using the �roperty (II), we have ( ) ( )k f x dx k f x dx=ò ò .

(�) �roperties (III) and (I�) can be generalised to a finite number of functionsf�, f2, ..., fn and the real numbers, k�, k2, ..., kn giving

� � 2 2( ) ( ) ( )n nk f x k f x ... k f x dx+ + +ò� � � 2 2( ) ( ) ( )n nk f x dx k f x dx ... k f x dx+ + +ò ò ò .

To find an anti derivative of a given function, we search intuitively for a functionwhose derivative is the given function. The search for the requisite function for findingan anti derivative is known as integration by the method of inspection. �e illustrate itthrough some examples.

INTEGRALS 2��

Example 1 �rite an anti derivative for each of the following functions using themethod of inspection:

(i) cos 2x (ii) �x2 � �x� (iii)�x

, x ¹ �

Solution(i) �e look for a function whose derivative is cos 2x. Recall that

ddx

sin 2x �2 cos 2x

or cos 2x � �2

ddx

(sin 2x) ��

sin 22

dx

dxæ öç ÷è ø

Therefore, an anti derivative of cos 2x is �

sin 22

x .

(ii) �e look for a function whose derivative is �x2 � �x�. Note that

� �dx x

dx+ ��x2 � �x�.

Therefore, an anti derivative of �x2 � �x� is x� � x�.(iii) �e know that

� � �(log ) �and �log ( )� ( �) �

d dx , x – x – , x

dx x dx – x x= > = = <

Combining above, we get �

log �d

x , xdx x

= ¹

Therefore, �

logdx xx

=ò is one of the anti derivatives of �x

.

Example 2 Find the following integrals:

(i)�

2

�x –dx

xò (ii) 2�( �)x dx+ò (iii) ò

�2 �

( 2 � )+ xx e dxx

Solution(i) �e have

�2

2� –x –

dx x dx – x dxx

=ò ò ò (by �roperty �)


� � � 2 �

� 2C C� � 2 �

–x x–

–

+ +æ ö æ ö+ +ç ÷ ç ÷+ +è ø è ø

� C�, C2 are constants of integration

� 2 �

� 2C C2 �

–x x– –

–+ �

2

� 2�

�C C2x

–x

+

� 2 �

�C2x

x+ , where C � C� � C2 is another constant of integration.

ANote From now onwards, we shall write only one constant of integration in thefinal answer.

(ii) �e have2 2� �( �)x dx x dx dx+ = +ò ò ò

�

2�

�C

2 ��

xx

+

+ ++

� ��

C�

x x+ +

(iii) �e have � �2 2� �

( 2 ) 2x xx e – dx x dx e dx – dxx x

+ = +ò ò ò ò

�

��

22 �log �C

� �2

xxe x

+

++

��22

2 �log �C�

xx e x+


(i) (sin cos )x x dx+ò (ii) cosec (cosec cot )x x x dx+ò

(iii) 2

� sincos– x

dxxò


(sin cos ) sin cosx x dx x dx x dx+ = +ò ò ò� �cos sin Cx x+ +

INTEGRALS 2�7

(ii) �e have2(cosec (cosec �cot ) cosec cosec cotx x x dx x dx x x dx= +ò ò ò

� �cot cosec Cx – x +(iii) �e have

2 2 2

� sin � sincos cos cos– x x

dx dx – dxx x x

=ò ò ò

� 2sec tan secx dx – x x dxò ò� tan sec Cx – x +

Example 4 Find the anti derivative F of f defined by f (x) � �x� � �, where F (�) � �

Solution �ne anti derivative of f (x) is x� � �x since

�( � )d

x – xdx

��x� � �

Therefore, the anti derivative F is given by

F(x) �x� � �x � C, where C is constant.

Given that F(�) ��, which gives,

� �� ´ � � C or C � ��ence, the required anti derivative is the unique function F defined by

F(x) �x� � �x � �.

Remarks(i) �e see that if F is an anti derivative of f, then so is F � C, where C is any

constant. Thus, if we know one anti derivative F of a function f, we can writedown an infinite number of anti derivatives of f by adding any constant to Fexpressed by F(x) � C, C Î R. In applications, it is often necessary to satisfy anadditional condition which then determines a specific value of C giving uniqueanti derivative of the given function.

(ii) Sometimes, F is not expressible in terms of elementary functions vi�., polynomial,logarithmic, exponential, trigonometric functions and their inverses etc. �e are

therefore blocked for finding ( )f x dxò . For example, it is not possible to find2– xe dxò by inspection since we can not find a function whose derivative is

2– xe

2�8 MAT�EMATICS

(iii) � hen the variable of integration is denoted by a variable other than x, the integralformulae are modified accordingly. For instance

� ��

C C� � �y

y dy y+

= + = ++ò

7.2.3 Comparison between differentiation and integration�. �oth are operations on functions.2. �oth satisfy the property of linearity, i.e.,

(i) � � 2 2 � � 2 2( ) ( ) ( ) ( )d d d

k f x k f x k f x k f xdx dx dx

+ = +

(ii) � � 2 2 � � 2 2( ) ( ) ( ) ( )k f x k f x dx k f x dx k f x dx+ = +ò ò ò�ere k� and k2 are constants.

�. �e have already seen that all functions are not differentiable. Similarly, all functionsare not integrable. �e will learn more about nondifferentiable functions andnonintegrable functions in higher classes.

�. The derivative of a function, when it exists, is a unique function. The integral ofa function is not so. �owever, they are unique upto an additive constant, i.e., anytwo integrals of a function differ by a constant.

�. � hen a polynomial function � is differentiated, the result is a polynomial whosedegree is � less than the degree of �. � hen a polynomial function � is integrated,the result is a polynomial whose degree is � more than that of �.

�. �e can speak of the derivative at a point. �e never speak of the integral at apoint, we speak of the integral of a function over an interval on which the integralis defined as will be seen in Section 7.7.

7. The derivative of a function has a geometrical meaning, namely, the slope of thetangent to the corresponding curve at a point. Similarly, the indefinite integral ofa function represents geometrically, a family of curves placed parallel to eachother having parallel tangents at the points of intersection of the curves of thefamily with the lines orthogonal (perpendicular) to the axis representing the variableof integration.

8. The derivative is used for finding some physical quantities like the velocity of amoving particle, when the distance traversed at any time t is known. Similarly,the integral is used in calculating the distance traversed when the velocity at timet is known.

�. Differentiation is a process involving limits. So is integration, as will be seen inSection 7.7.

INTEGRALS 2��

��. The process of differentiation and integration are inverses of each other asdiscussed in Section 7.2.2 (i).

EXERCISE 7.1Find an anti derivative (or integral) of the following functions by the method of inspection.

1. sin 2x 2. cos �x 3. e2x

4. (ax � b)2 5. sin 2x � � e�x

Find the following integrals in Exercises � to 2�:

6. �(� � �) xe dxò 7. 22

�(�� )x dx

xò 8. 2( )ax bx c dx+ +ò

9. 2(2 )xx e dx+ò 10.2

�x – dx

xæ öç ÷è øò 11.

� 2

2

� �x x –dx

x+

ò

12.� � �x x

dxx

+ +ò 13.

� 2 ��

x x x – dxx –

- +ò 14. (� )– x x dxò

15. 2( � 2 �)x x x dx+ +ò 16. (2 �cos )xx – x e dx+ò17. 2(2 �sin � )x – x x dx+ò 18. sec (sec tan )x x x dx+ò

19.2

2

seccosec

xdx

xò 20. 2

2 ��sincos

xxò dx.

Choose the correct answer in Exercises 2� and 22.

21. The anti derivative of �

xx

æ ö+ç ÷è ø

equals

(A)� �� 2�

2 C�

x x+ + (�)2

2�2 �C

� 2x x+ +

(C)� �2 22

2 C�

x x+ + (D)� �2 2� �

C2 2

x x+ +

22. If ��

�( ) �

df x x

dx x= - such that f (2) � �. Then f (x) is

(A)�

�

� �2�8

xx

+ - (�)�

�

� �2�8

xx

+ +

(C)�

�� 2�

8x

x+ + (D)

��

� �2�8

xx

+ -

�� MAT�EMATICS

7.3 Methods of IntegrationIn previous section, we discussed integrals of those functions which were readilyobtainable from derivatives of some functions. It was based on inspection, i.e., on thesearch of a function F whose derivative is f which led us to the integral of f. �owever,this method, which depends on inspection, is not very suitable for many functions.�ence, we need to develop additional techniques or methods for finding the integralsby reducing them into standard forms. �rominent among them are methods based on:

�. Integration by Substitution2. Integration using �artial Fractions�. Integration by �arts

7.3.1 Integration by substitutionIn this section, we consider the method of integration by substitution.

The given integral ( )f x dxò can be transformed into another form by changingthe independent variable x to t by substituting x � g (t).

Consider I � ( )f x dxò

�ut x � g(t) so that dxdt

� g¢(t).

�e write dx �g¢(t) dt

Thus I � ( ) ( ( )) ( )f x dx f g t g t dt= ¢ò òThis change of variable formula is one of the important tools available to us in the

name of integration by substitution. It is often important to guess what will be the usefulsubstitution. �sually, we make a substitution for a function whose derivative also occursin the integrand as illustrated in the following examples.

Example 5 Integrate the following functions w.r.t. x:(i) sin mx (ii) 2x sin (x2 � �)

(iii)� 2tan secx x

x(iv)

�

2

sin (tan )�

– xx+

Solution(i) � e know that derivative of mx is m. Thus, we make the substitution

mx � t so that mdx � dt.

Therefore, �

sin sinmx dx t dtm

=ò ò � � �m

cos t � C � � �m

cos mx � C

INTEGRALS ��

(ii) Derivative of x2 � � is 2x. Thus, we use the substitution x2 � � � t so that2x dx � dt.

Therefore, 22 sin ( �) sinx x dx t dt+ =ò ò � � cos t � C � � cos (x2 � �) � C

(iii) Derivative of x is �2� �

2 2

–x

x= . Thus, we use the substitution

�so that giving

2x t dx dt

x= = dx � 2t dt.

Thus,� 2 � 2tan sec 2 tan secx x t t t dt

dxtx

=ò ò � � 22 tan sect t dtòAgain, we make another substitution tan t � u so that sec2 t dt � du

Therefore, � 2 �2 tan sec 2t t dt u du=ò ò � �

2 C�u

+

� �2tan C

�t + (since u �tan t)

� �2tan C (since )

�x t x+ =

�ence,� 2tan secx x

dxxò � �2

tan C�

x +

Alternatively, make the substitution tan x t=

(iv) Derivative of �2

�tan

�– x

x=

+. Thus, we use the substitution

tan�� x � t so that 2�dx

x+ � dt.

Therefore , �

2

sin (tan ) sin�

– x dx t dtx

=+ò ò � � cos t � C � � cos(tan ��x) � C

Now, we discuss some important integrals involving trigonometric functions andtheir standard integrals using substitution technique. These will be used later withoutreference.

(i) ò tan = log sec + Cx dx x

�e havesin

tancos

xx dx dx

x=ò ò

��2 MAT�EMATICS

�ut cos x � t so that sin x dx � � dt

Then tan log C log cos Cdt

x dx – – t – xt

= = + = +ò òor tan log sec Cx dx x= +ò

(ii) òcot = log sin + Cx dx x

�e havecos

cotsin

xx dx dx

x=ò ò

�ut sin x � t so that cos x dx � dt

Then cotdt

x dxt

=ò ò � log Ct + � log sin Cx +

(iii) òsec = log sec + tan + Cx dx x x

�e havesec (sec tan )

secsec �tanx x x

x dx dxx x

+=ò ò

�ut sec x � tan x � t so that sec x (tan x � sec x) dx � dt

Therefore, sec log �C �log sec tan Cdtx dx t x xt

= = + +ò ò(iv) òcosec = log cosec – cot + Cx dx x x

�e havecosec (cosec cot )

cosec(cosec cot )

x x xx dx dx

x x+

=+ò ò

�ut cosec x � cot x � t so that � cosec x (cosec x � cot x) dx � dt

So cosec � �log �� log �cosec cot � Cdt

x dx t x xt

= = = + +ò ò

�2 2cosec cot

�log Ccosec cot

x xx x

-+

-

� log cosec cot Cx – x +


(i) � 2sin cosx x dxò (ii) sin

sin ( )x

dxx a+ò (iii)

�� tan

dxx+ò

INTEGRALS ��


� 2 2 2sin cos sin cos (sin )x x dx x x x dx=ò ò� 2 2(��cos ) cos (sin )x x x dxò

�ut t � cos x so that dt � � sin x dx

Therefore, 2 2sin cos (sin )x x x dxò � 2 2(�� )t t dt- ò

� � �

2 �( � ) C� �t t

– t t dt – –æ ö

= +ç ÷è ø

ò

� � �� cos cos C

� �– x x+ +

(ii) �ut x � a � t. Then dx � dt. Therefore

sin sin ( )sin ( ) sin

x t – adx dt

x a t=

+ò ò

� sin cos cos sin

sint a – t a

dttò

� cos �sin cota dt a t dtò ò� �(cos ) (sin ) log sin Ca t – a té ù+ë û

� �(cos ) ( ) (sin ) log sin ( ) Ca x a – a x aé ù+ + +ë û

� �cos cos (sin ) log sin ( ) C sinx a a a – a x a – a+ +

�ence, sin

sin ( )x

dxx a+ò � x cos a � sin a log �sin (x � a)� � C,

where, C � � C� sin a � a cos a, is another arbitrary constant.

(iii)cos

� tan cos sindx x dx

x x x=

+ +ò ò

� � (cos �sin �cos �sin )2 cos sin

x x x x dxx x+ò


� � � cos �sin2 2 cos sin

x xdx dx

x x+

+ò ò

� �C � cos sin

2 2 2 cos sinx x – x

dxx x

+ ++ò ... (�)

Now, consider cos sin

Icos sin

x – xdx

x x=

+ò�ut cos x � sin x � t so that (cos x � sin x) dx � dt

Therefore 2I log Cdt

tt

= = +ò � 2log cos sin Cx x+ +

�utting it in (�), we get

� 2C C�� log cos sin

� tan 2 2 2 2dx x

x xx

= + ++ò

� � 2C C�� log cos sin

2 2 2 2x

x x+ + +

� � 2C C�� log cos sin C C

2 2 2 2x

x x , æ ö+ + = +ç ÷è ø

EXERCISE 7.2Integrate the functions in Exercises � to �7:

1. 2

2�

xx+

2.2log x

x3.

�logx x x+

4. sin sin (cos )x x 5. sin ( ) cos ( )ax b ax b+ +

6. ax b+ 7. 2x x + 8. 2� 2x x+

9. 2(� 2) �x x x+ + + 10.�

x – x 11.�

xx +

, x � �

12.�

� ��( �)x – x 13.2

� �(2 � )x

x+ 14.�

(log )mx x, x � �, �¹m

15. 2� �x

– x16. 2 �xe + 17. 2x

x

e

INTEGRALS ��

18.�

2�

–tan xex+

19.2

2

��

x

x

e –e +

20.2 2

2 2

x – x

x – xe – ee e+

21. tan2 (2x � �) 22. sec2 (7 � �x) 23.�

2

sin

�

– x

– x

24.2cos �sin�cos �sin

x – xx x+ 25. 2 2

�cos (� tan )x – x 26.

cos xx

27. sin 2 cos 2x x 28.cos

� sinx

x+ 29. cot x log sin x

30.sin

� cosx

x+ 31. 2sin

� cos

x

x+ 32.�

� cot x+

33.�

� tan– x 34.tan

sin cosx

x x 35.2� log x

x+

36.2( �) logx x x

x+ +

37.� � �sin tan

�

–x x

x8+

Choose the correct answer in Exercises �8 and ��.

38.�

��

�� log ��

xex

x dxx

++ò equals

(A) ��x � x�� C (�) ��x � x�� C(C) (��x – x��)�� C (D) log (��x + x��) � C

39. 2 2 equalssin cos

dxx xò

(A) tan x � cot x � C (�) tan x � cot x �C(C) tan x cot x � C (D) tan x � cot 2x � C

7.3.2 Integration using trigonometric identities� hen the integrand involves some trigonometric functions, we use some known identitiesto find the integral as illustrated through the following example.

Example 7 Find (i) 2cos x dxò (ii) sin 2 cos �x x dxò (iii) �sin x dxò


Solution(i) Recall the identity cos 2x � 2 cos2 x � �, which gives

cos2 x � � cos 2

2x+

Therefore, 2cosò x dx � �

(��cos 2 )2

x dxò � � �

cos 22 2

dx x dx+ò ò

� �

sin 2 C2 �x

x+ +

(ii) Recall the identity sin x cos y � �2

�sin (x � y) � sin (x � y)� (� hy?)

Then sin 2 cos�ò x x dx � �

sin � sin2

é ùë ûò òx dx x dx

� � �

cos � cos C2 �

– x xé ù+ +ê úë û

� � �cos � cos C

�� 2– x x+ +

(iii) From the identity sin �x � � sin x � � sin� x, we find that

sin�x � �sin sin �

�x – x

Therefore, �sin x dxò � � �

sin sin ��

x dx – x dxò ò

� � �

� cos cos � C� �2

x x+ +

Alternatively, � 2sin sin sinx dx x x dx=ò ò � 2(��cos ) sinx x dxò�ut cos x � t so that � sin x dx � dt

Therefore, �sin x dxò � 2�– t dt- ò � �

2 C�t– dt t dt – t+ = + +ò ò

� ��cos cos C

�– x x+ +

Remark It can be shown using trigonometric identities that both answers are equivalent.

INTEGRALS ��7

EXERCISE 7.3Find the integrals of the functions in Exercises � to 22:

1. sin2 (2x � �) 2. sin �x cos �x 3. cos 2x cos �x cos �x4. sin� (2x + �) 5. sin� x cos� x 6. sin x sin 2x sin �x

7. sin �x sin 8x 8.� cos� cos

– xx+ 9.

cos� cos

xx+

10. sin� x 11. cos� 2x 12.2sin

� cosxx+

13. cos 2 cos 2cos cos

x –x –

aa

14.cos sin

� sin 2x – x

x+15. tan� 2x sec 2x

16. tan�x 17.� �

2 2

sin cossin cos

x xx x+

18.2

2

cos 2 2sincosx x

x+

19. �

�sin cosx x

20. 2cos 2

cos sin

x

x x+21. sin � � (cos x)

22.�

cos ( ) cos ( )x – a x – bChoose the correct answer in Exercises 2� and 2�.

23.2 2

2 2

sin cos is equal tosin cos

x x dxx x-

ò(A) tan x � cot x � C (�) tan x � cosec x � C(C) � tan x � cot x � C (D) tan x � sec x � C

24. 2

(� ) equalscos ( )

x

x

e x dxe x+

ò(A) � cot (exx) � C (�) tan (xex) � C(C) tan (ex) � C (D) cot (ex) � C

7.4 Integrals of Some Particular FunctionsIn this section, we mention below some important formulae of integrals and apply themfor integrating many other related standard integrals:

(1) ò 2 21 –= log + C

2 +–dx x a

a x ax a


(2) ò 2 21 += log + C

2 ––dx a x

a a xa x

(3) ò – 12 2

1 tan Cdx x= +

a ax + a

(4) ò 2 22 2

= log + – + C–

dxx x a

x a

(5) ò – 12 2

= sin + C–

dx xaa x

(6) ò 2 22 2

= log + + + C+

dxx x a

x a

�e now prove the above results:

(1) �e have 2 2

� �( ) ( )x – a x ax – a

=+

� � ( ) �( ) � � �

2 ( ) ( ) 2x a x – a

–a x – a x a a x – a x a

é ù+ é ù=ê ú ê ú+ +ë ûë û

Therefore, 2 2

�2

dx dx dx–

a x – a x ax – aé ù

= ê ú+ë ûò ò ò

� �

log ( )� log ( )� C2

| x – a – | x aa

+ +

� �

log C2

x – aa x a

++

(2) In view of (�) above, we have

2 2

� � ( ) ( )2 ( ) ( )�

a x a xa a x a xa x

é ù+ + -= ê ú+ -ë û

� � � �

2a a x a xé ù+ê ú- +ë û

INTEGRALS ��

Therefore, 2 2�dx

a xò � �2

dx dxa a x a x

é ù+ê ú- +ë ûò ò

� �

� log � � log � �� C2

a x a xa

- - + + +

� �

log C2

a xa a x

++

-

ANote The technique used in (�) will be explained in Section 7.�.

(3) �ut x � a tan q. Then dx � a sec2 q dq.

Therefore, 2 2

dxx a+ò �

2

2 2 2

� �

�

sectan

a da a+ò

� �� C tan C– x

da a a a

= + = +ò(4) Let x � a secq. Then dx � a secq tanq d q.

Therefore,2 2

dx

x a-ò �

2 2 2

sec�tan� �

sec �

a d

a a-ò

� �sec� � log sec��tan��Cd =ò

�2

�2log � Cx x

–a a

+ +

� 2 2�log log Cx x – a a+ - +

� 2 2log �Cx x – a+ , where C � C� � log �a�

(5) Let x � a sinq. Then dx � a cosq dq.

Therefore, 2 2

dx

a x-ò �

2 2 2

� �

�

cos

sin

a d

a – aò

� ��C �sin C– xd

a+ò

(6) Let x � a tan q. Then dx � a sec2q dq.

Therefore,2 2

dx

x a+ò �

2

2 2 2

� �

�

sec

tan

a d

a a+ò

� �� sec� ��log (sec tan ) Cd + +ò


�2

�2log � Cx xa a

+ + +

�2

�log log Cx x a | a |2+ + - +

�2log Cx x a2+ + + , where C � C� � log �a�

Applying these standard formulae, we now obtain some more formulae whichare useful from applications point of view and can be applied directly to evaluateother integrals.

(7) To find the integral 2dx

ax bx c+ +ò , we write

ax2 � bx � c � 2 2

222 �

b c b c ba x x a x –

a a a a a

é ùæ öé ù æ ö+ + = + +ê úç ÷ç ÷ê úë û è øê úè øë û

Now, put 2b

x ta

+ = so that dx � dt and writing 2

22�

c b– k

a a= ± . �e find the

integral reduced to the form 2 2

� dta t k±ò depending upon the sign of

2

2�c b–a a

æ öç ÷è ø

and hence can be evaluated.

(8) To find the integral of the type2

dx

ax bx c+ +ò , proceeding as in (7), we

obtain the integral using the standard formulae.

(9) To find the integral of the type 2

px qdx

ax bx c+

+ +ò , where p, q, a, b, c are

constants, we are to find real numbers A, � such that

2� �A ( ) ��A (2 ) ��d

px q ax bx c ax bdx

+ + +

To determine A and �, we equate from both sides the coefficients of x and theconstant terms. A and � are thus obtained and hence the integral is reduced toone of the known forms.

INTEGRALS ��

(10) For the evaluation of the integral of the type2

( )px q dx

ax bx c

+

+ +ò , we proceed

as in (�) and transform the integral into known standard forms.Let us illustrate the above methods by some examples.


(i) 2 ��dx

x -ò (ii)22

dx

x x-ò

Solution

(i) �e have 2 2 2�� dx dx

x x –=

-ò ò � �

log C8 �

x –x

1+

+�by 7.� (�)�

(ii)2 22 � �

=-

ò òdx dx

x x – x –

�ut x � � � t. Then dx � dt.

Therefore,22

dx

x x-ò �

2�

dt

– tò � �sin ( ) C– t + �by 7.� (�)�

� �sin ( ��) C– x +

Example 9 Find the following integrals :

(i) 2 � ��dx

x x- +ò (ii) 2� �� dx

x x+ -ò (iii) 2� 2

dx

x x-ò

Solution(i) �e have x2 � �x + �� x2 � �x � �2 � �2 � �� (x � �)2 � �

So,� ��dx

x x2 - +ò � 2 2

�

� 2dx

x – +ò

Let x � � �t. Then dx � dt

Therefore,� ��dx

x x2 - +ò � �

2 2

�tan C

2 22–dt t

t= +

+ò �by 7.� (�)�

� �� tan C

2 2– x –

+


(ii) The given integral is of the form 7.� (7). �e write the denominator of the integrand,

2� �� x x –+ �2 ��

�� x

x –æ ö+ç ÷è ø

�

2 2�� 7�

� �x –

é ùæ ö æ ö+ê úç ÷ ç ÷è ø è øê úë û

(completing the square)

Thus� ��

dxx x2 + -ò � 2 2

�� 7

� �

dx

xæ ö æ ö+ -ç ÷ ç ÷è ø è ø

ò

�ut ��

x t+ = . Then dx � dt.

Therefore,� ��

dxx x2 + -ò � 2

2

�� 7

�

dt

t æ ö- ç ÷è ø

ò

� �

�7� �log C

�7 �7� 2

� �

t –

t+

´ ´ +�by 7.� (i)�

� �

�� 7� � �log C

�� 7�7� �

x –

x

++

+ +

� ��

log C�7 � ��

xx

-+

+

� �� 2 � �

log C log�7 � �7 �

xx

-+ +

+

�� 2

log C�7 �

xx

-+

+ , where C � ��

C log�7 �

+

INTEGRALS ��

(iii) �e have 2 2� 2 �

�

dx dx

xx x x –2

=æ ö-ç ÷è ø

ò ò

�2 2

��

� �

dx

x – –æ ö æ öç ÷ ç ÷è ø è ø

ò (completing the square)

�ut ��

x – t= . Then dx � dt.

Therefore,� 2

dx

x x2 -ò �

22

��

�

dt

t – æ öç ÷è ø

ò

�2

2� �log C

��t t – æ ö+ +ç ÷

è ø�by 7.� (�)�

� 2� � 2log C

� ��x

x – x –+ +


(i)2

2 � �x

dxx x2

++ +ò (ii) 2

�

� �

x dxx �x

+

-ò

Solution(i) �sing the formula 7.� (�), we express

x � 2 � 2A 2 � � �d

x xdx

+ + + � A (� �) �x + +

Equating the coefficients of x and the constant terms from both sides, we get

�A � � and �A � � � 2 or A � ��

and � � �2

.

Therefore,2

2 � �x

x x2

++ +ò �

� � � �� 22 � � 2 � �

x dxdx

x x x x2 2

++

+ + + +ò ò

� � 2� �

I I� 2

+ (say) ... (�)


In I�, put 2x2 � �x � � � t, so that (�x � �) dx � dt

Therefore, I� � �log Cdt

tt

= +ò

� 2�log �2 � �� Cx x+ + + ... (2)

and I2 � 22

��22 � � �2

dx dxx x x x

=+ + + +

ò ò

� 2 2

�2 � �

2 2

dx

xæ ö æ ö+ +ç ÷ ç ÷è ø è ø

ò

�ut �2

x t+ = , so that dx � dt, we get

I2 � 22

�2 �

2

dt

t æ ö+ ç ÷è ø

ò � �2

�tan 2 C

�22

– t +´

�by 7.� (�)�

� �2

�tan 2 �C

2– xæ ö+ç ÷

è ø � �

2tan 2 � �C– x + ... (�)

�sing (2) and (�) in (�), we get

2 �2 � �log 2 � � tan 2 � C

� 22 � �–x

dx x x xx x2

+= + + + + +

+ +ò

where, C � � 2C C� 2

+

(ii) This integral is of the form given in 7.� (��). Let us express

x � � � 2A (� � ) ��d

– x – xdx

� A (� � � 2x) � �

Equating the coefficients of x and the constant terms from both sides, we get

� 2A � � and � � A � � � �, i.e., A � �2

– and � � �

INTEGRALS ��

Therefore,2

�

� �

xdx

x x

+

- -ò � 2 2

� 2�2 � � � �

– – x dx dx–

x x x x+

- - - -ò ò

��2

– I� � I2 ... (�)

In I�, put � � �x � x2 � t, so that (� � � 2x) dx � dt.

Therefore, I�� 2

� 2

� �

– x dx dttx x

-=

- -ò ò � �2 Ct +

� 2�2 � � C– x – x + ... (2)

Now consider I2 � 2 2� � � ( 2)

dx dx

x x – x=

- - +ò ò

�ut x � 2 � t, so that dx � dt.

Therefore, I2 ��

22 2sin �C

��–dt t

t=

-ò �by 7.� (�)�

� �2

2sin C

�– x +

+ ... (�)

Substituting (2) and (�) in (�), we obtain

2 �

2

� 2�� sin C

�� –x x

– x x– x – x

+ += +ò , where �

2C

C C2

–=

EXERCISE 7.4Integrate the functions in Exercises � to 2�.

1.2

�

��

xx +

2.2

�

� �x+3. 2

�

2 �– x +

4.2

�

� 2�– x5. �

�� 2

xx+

6.2

��x

x-

7. 2

�

�

x –

x –8.

2

� �

x

x a+9.

2

2

sec

tan �

x

x +


10.2

�

2 2x x+ +11. 2

�� x x+ +

12. 2

�

7 �� x �x

13.�

� 2x – x –14. 2

�

8 �x – x+15.

�

x – a x – b

16. 2

� �

2 �

x

x x –

+

+17. 2

2

�

x

x –

+18. 2

� 2� 2 �

xx x-

+ +

19.� 7

� �

x

x – x –

+20.

2

2

�

x

x – x

+21.

2

2

2 �

x

x x

+

+ +

22. 2

�2 �

xx – x

+-

23. 2

� �

� ��

x

x x

+

+ +.

Choose the correct answer in Exercises 2� and 2�.

24. 2 equals2 2dx

x x+ +ò(A) x tan�� (x � �) � C (�) tan�� (x � �) � C(C) (x � �) tan��x � C (D) tan��x � C

25. 2equals

� �

dx

x x-ò

(A) �� 8sin C

� 8x -æ ö +ç ÷

è ø(�) �� 8 �

sin C2 �

x -æ ö +ç ÷è ø

(C) �� 8sin C

� 8x -æ ö +ç ÷

è ø(D)

�� 8sin C

2 �x -æ ö +ç ÷

è ø

7.5 Integration by Partial FractionsRecall that a rational function is defined as the ratio of two polynomials in the form

�( )�( )

xx

, where � (x) and �(x) are polynomials in x and �(x) ¹ �. If the degree of �(x)

is less than the degree of �(x), then the rational function is called proper, otherwise, itis called improper. The improper rational functions can be reduced to the proper rational

INTEGRALS ��7

functions by long division process. Thus, if �( )�( )

xx

is improper, then ��( )�( )T( )

�( ) �( )xx

xx x

= + ,

where T(x) is a polynomial in x and ��( )�( )

xx

is a proper rational function. As we know

how to integrate polynomials, the integration of any rational function is reduced to theintegration of a proper rational function. The rational functions which we shall considerhere for integration purposes will be those whose denominators can be factorised into

linear and quadratic factors. Assume that we want to evaluate �( )�( )

xdx

xò , where �( )�( )

xx

is proper rational function. It is always possible to write the integrand as a sum ofsimpler rational functions by a method called partial fraction decomposition. After this,the integration can be carried out easily using the already known methods. The followingTable 7.2 indicates the types of simpler partial fractions that are to be associated withvarious kind of rational functions.

Table 7.2

S.No. Form of the rational function Form of the partial fraction

1.( � ) ( � )

px qx a x b

+ , a ¹ bA �

x – a x – b+

2. 2( � )px qx a

+2

A �x – a x – a

+

3.2

( � ) ( ) ( )px qx r

x a x – b x – c+ + A � C

x – a x – b x – c+ +

4.2

2( � ) ( )px qx r

x a x – b+ +

2

A � C( )x – a x – bx – a

+ +

5.2

2( � ) ( )px qx r

x a x bx c+ +

+ + 2

A � �Cxx – a x bx c

++ +

,

where x2 � bx � c cannot be factorised further

In the above table, A, � and C are real numbers to be determined suitably.


Example 11 Find ( �) ( 2)dx

x x+ +ò

Solution The integrand is a proper rational function. Therefore, by using the form ofpartial fraction �Table 7.2 (i)�, we write

�( �) ( 2)x x+ +

�A �

� 2x x+

+ +... (�)

where, real numbers A and � are to be determined suitably. This gives� �A (x � 2) � � (x � �).

Equating the coefficients of x and the constant term, we getA � � ��

and 2A � � ��Solving these equations, we get A �� and � � � �.Thus, the integrand is given by

�( �) ( 2)x x+ +

��

� 2x x+

+ +

Therefore, ( �) ( 2)dx

x x+ +ò �� 2

dx dx–

x x+ +ò ò

� log � log 2 Cx x+ - + +

��

log C2

xx

++

+

Remark The equation (�) above is an identity, i.e. a statement true for all (permissible)values of x. Some authors use the symbol �º� to indicate that the statement is anidentity and use the symbol �� to indicate that the statement is an equation, i.e., toindicate that the statement is true only for certain values of x.

Example 12 Find 2

2

��

xdx

x x+

- +ò

Solution �ere the integrand 2

2�

� �x

x – x+

+ is not proper rational function, so we divide

x2 � � by x2 � �x � � and find that

INTEGRALS ��

2

2�

� �x

x – x+

+ � 2

� � � ��

( 2) ( �)� �x – x –

x – x –x – x+ = +

+

Let� �

( 2) ( �)x –

x – x – �

A �2 �x – x –

+

So that 5x � � �A (x � �) � � (x � 2)Equating the coefficients of x and constant terms on both sides, we get A � � � �

and �A � 2� � �. Solving these equations, we get A � � � and � � ��

Thus,2

2�

� �x

x – x+

+ �

� ��

2 �x – x –- +

Therefore,2

2�

� �x

dxx – x

++ò �

��

2 �dx

dx dxx – x –

- +ò ò ò�x � � log �x � 2 � � �� log �x � �� C.

Example 13 Find 2

� 2( �) ( �)

xdx

x x-

+ +ò

Solution The integrand is of the type as given in Table 7.2 (�). �e write

2

� 2( �) ( �)

x –x x+ +

� 2

A � C� �( �)x xx

+ ++ ++

So that �x � 2 �A (x � �) (x � �) � � (x � �) � C (x � �)2

�A (x2 � �x � �) � � (x � �) � C (x2 � 2x � � )Comparing coefficient of x2, x and constant term on both sides, we get

A � C � �, �A � � � 2C � � and �A � �� C � � 2. Solving these equations, we get��

A � and C� 2 �

– –,= = = . Thus the integrand is given by

2

� 2( �) ( �)

xx x

-+ + � 2

�� ( �) �( �)2 ( �)

– –x xx+ ++

Therefore, 2

� 2( �) ( �)

xx x

-+ +ò � 2

�� 2 � �( �)

dx dx dx–

x xx-

+ ++ò ò ò

��

log �� log � C� 2 ( ��) �

x xx

+ - + +

��

log �C� �� 2 ( ��)

xx x

+

�2� MAT�EMATICS

Example 14 Find 2

2 2( �) ( �)x

dxx x+ +ò

Solution Consider 2

2 2( �) ( �)x

x x+ + and put x2 � y.

Then2

2 2( �) ( �)x

x x+ + �

( �) ( �)y

y y+ +

�rite( �) ( �)

yy y+ +

�A �

� �y y+

+ +

So that y � A (y � �) � � (y � �)Comparing coefficients of y and constant terms on both sides, we get A � � � �

and �A � � � �, which give

A ��

and ��

- =

Thus,2

2 2( �) ( �)x

x x+ + � 2 2

� ��( �) �( �)

–x x

++ +

Therefore,2

2 2( �) ( �)x dx

x x+ +ò � 2 2

� ��

dx dx–

x x+

+ +ò ò

� � �� tan tan C

� � 2 2– – x

– x + ´ +

� � �� 2tan tan C

� � 2– – x

– x + +

In the above example, the substitution was made only for the partial fraction partand not for the integration part. Now, we consider an example, where the integrationinvolves a combination of the substitution method and the partial fraction method.

Example 15 Find 2

�sin 2 cos

� cos �sin

–d

– –

f ff

f fòSolution Let y � sinf

Then dy �cosf df

INTEGRALS �2�

Therefore, 2

�sin 2 cos

� cos �sin

–d

– –

f ff

f fò � 2

(� �2)� (� ) �

y dy– – y – yò

� 2

� 2� �

y –dy

y – y +ò

� 2� 2

I (say)2

y –

y –=ò

Now, we write 2

� 2

2

y –

y – � 2

A �2 ( 2)y y

+- -

�by Table 7.2 (2)�

Therefore, �y � 2 �A (y � 2) � �Comparing the coefficients of y and constant term, we get A � � and � � 2A � � 2,

which gives A � � and � � �.Therefore, the required integral is given by

I � 2

� ��

2 ( 2)dy

y – y –ò � 2� ��2 ( 2)

dy dyy – y –ò ò

��log 2 � C

2y –

yæ ö

- + +ç ÷-è ø

� ��log sin 2 C

2 sin–f - + +

f

� ��log (2 sin ) �C

2 sin- f +

- f (since, 2 � sin f is always positive)

Example 16 Find 2

2

�( 2) ( �)x x dxx x

+ ++ +ò

Solution The integrand is a proper rational function. Decompose the rational functioninto partial fraction �Table 2.2(�)�. � rite

2

2

�( �) ( 2)

x xx x

+ ++ +

� 2

A � �C2 ( �)

xx x

++ +

Therefore, x2 � x � � �A (x2 � �) � (�x � C) (x � 2)

�22 MAT�EMATICS

Equating the coefficients of x2, x and of constant term of both sides, we getA � � ��, 2� � C � � and A � 2C � �. Solving these equations, we get

� 2 �A , � and C

� � �= = =

Thus, the integrand is given by

2

2

�( �) ( 2)

x xx x

+ ++ +

� 2

2 ��

�( 2) �

x

x x

++

+ + � 2

� � 2 ��( 2) � �

xx x

+æ ö+ ç ÷+ +è ø

Therefore,2

2

�( ��) ( 2)

x xdx

x x+ +

+ò � 2 2� � 2 � �� 2 � ��

dx xdx dx

x x x+ +

+ + +ò ò ò

� 2 �� log 2 log � tan C

� � �–x x x+ + + + +

EXERCISE 7.5Integrate the rational functions in Exercises � to 2�.

1. ( �) ( 2)x

x x+ + 2. 2�

�x –3.

� �( �) ( 2) ( �)

x –x – x – x –

4. ( �) ( 2) ( �)x

x – x – x – 5. 2

2� 2x

x x+ +6.

2�(� 2 )

– xx – x

7. 2( �) ( ��)x

x x+8. 2( �) ( 2)

xx – x +

9. � 2

� ��

xx – x x

+- +

10. 2

2 �( �) (2 �)

xx – x

-+

11. 2

�( �) ( �)

xx x+ -

12.�

2

��

x xx+ +

-

13. 2

2(� ) (� )x x- + 14. 2

� �( 2)

x –x + 15. �

��x -

16.�

( �)nx x + ��int: multiply numerator and denominator by x n � � and put xn � t �

17.cos

(��sin ) (2 �sin )x

x x ��int : �ut sin x � t�

INTEGRALS �2�

18.2 2

2 2

( �) ( 2)( �) ( �)x xx x

+ ++ +

19. 2 2

2( �) ( �)

xx x+ +

20. �

�( �)x x –

21.�

( �)xe – ��int : �ut ex � t�

Choose the correct answer in each of the Exercises 22 and 2�.

22.( �) ( 2)

x dxx x- -ò equals

(A)2( �)

log C2

xx

-+

-(�)

2( 2)log C

�xx-

+-

(C)2�

log C2

xx

-æ ö +ç ÷-è ø(D) log ( �) ( 2) Cx x- - +

23.2( �)

dxx x +ò equals

(A) 2�log log ( ��) �C

2x x- (�) 2�

log log ( ��) �C2

x x+

(C) 2�log log ( ��) �C

2x x- + (D) 2�

log log ( ��) �C2

x x+

7.6 Integration by PartsIn this section, we describe one more method of integration, that is found quite useful inintegrating products of functions.

If u and v are any two differentiable functions of a single variable x (say). Then, bythe product rule of differentiation, we have

( )d

uvdx

�dv du

u vdx dx

+

Integrating both sides, we get

uv �dv du

u dx v dxdx dx

+ò ò

ordv

u dxdxò �

duuv – v dx

dxò ... (�)

Let u �f (x) and dvdx

� g(x). Then

dudx

�f ¢(x) and v � ( )g x dxò


Therefore, expression (�) can be rewritten as

( ) ( )f x g x dxò � ( ) ( ) � ( ) � ( )f x g x dx – g x dx f x dx¢ò ò òi.e., ( ) ( )f x g x dxò � ( ) ( ) � ( ) ( ) �f x g x dx – f x g x dx dx¢ò ò ò

If we take f as the first function and g as the second function, then this formulamay be stated as follows:

“The integral of the product of two functions = (first function) × (integralof the second function) – Integral of [(differential coefficient of the first function)× (integral of the second function)]”

Example 17 Find cosx x dxòSolution �ut f (x) � x (first function) and g (x) � cos x (second function).Then, integration by parts gives

cosx x dxò � cos � ( ) cos �d

x x dx – x x dx dxdxò ò ò

� sin sinx x – x dxò � x sin x � cos x � C

Suppose, we take f (x) �cos x and g (x) � x. Then

cosx x dxò � cos � (cos ) �d

x x dx – x x dx dxdxò ò ò

�2 2

cos sin2 2x xx x dx+ ò

Thus, it shows that the integral cosx x dxò is reduced to the comparatively more

complicated integral having more power of x. Therefore, the proper choice of the firstfunction and the second function is significant.

Remarks(i) It is worth mentioning that integration by parts is not applicable to product of

functions in all cases. For instance, the method does not work for sinx x dxò .

The reason is that there does not exist any function whose derivative is

x sin x.

(ii) �bserve that while finding the integral of the second function, we did not addany constant of integration. If we write the integral of the second function cos x

INTEGRALS �2�

as sin x � k, where k is any constant, then

cosx x dxò � (sin ) (sin )x x k x k dx+ - +ò� (sin ) (sinx x k x dx k dx+ - -ò ò� (sin ) cos Cx x k x – kx+ - + � sin cos Cx x x+ +

This shows that adding a constant to the integral of the second function issuperfluous so far as the final result is concerned while applying the method ofintegration by parts.

(iii) �sually, if any function is a power of x or a polynomial in x, then we take it as thefirst function. �owever, in cases where other function is inverse trigonometricfunction or logarithmic function, then we take them as first function.

Example 18 Find log x dxòSolution To start with, we are unable to guess a function whose derivative is log x. �etake log x as the first function and the constant function � as the second function. Then,the integral of the second function is x.

�ence, (log .�)x dxò � log � � (log ) � �d

x dx x dx dxdx

-ò ò ò

��

(log ) � log Cx x x dx x x – xx

× = +ò .

Example 19 Find xx e dxòSolution Take first function as x and second function as ex. The integral of the secondfunction is ex.

Therefore, xx e dxò � �x xx e e dx- ×ò � xex � ex � C.

Example 20 Find �

2

sin

�

–x xdx

x-ò

Solution Let first function be sin � �x and second function be 2�

x

x-.

First we find the integral of the second function, i.e., 2�

x dx

x-ò .

�ut t �� x2. Then dt � � 2x dx


Therefore,2�

x dx

x-ò �

�2

dt–

tò � 2� �t x= - -

�ence,�

2

sin

�

–x xdx

x-ò � � 2 2

2

�(sin ) � ( � )

�– x – x – x dx

x- - -

-ò

� 2 �� sin C– x x x-- + + � 2 �� sin Cx – x x-- +

Alternatively, this integral can also be worked out by making substitution sin�� x � q andthen integrating by parts.

Example 21 Find sinxe x dxòSolution Take ex as the first function and sin x as second function. Then, integratingby parts, we have

I sin ( cos ) cosx x xe x dx e – x e x dx= = +ò ò�� ex cos x � I� (say) ... (�)

Taking ex and cos x as the first and second functions, respectively, in I�, we get

I� � sin sinx xe x – e x dxòSubstituting the value of I� in (�), we get

I �� ex cos x � ex sin x � I or 2I � ex (sin x � cos x)

�ence, I � sin (sin cos ) �C2

xx ee x dx x – x=ò

Alternatively, above integral can also be determined by taking sin x as the first functionand ex the second function.

7.6.1 Integral of the type � ( ) � ( )�xe f x f x dx¢ò�e have I � � ( ) � ( )�xe f x f x dx¢ò � ( ) � ( )x xe f x dx e f x dx¢ò ò

� � �I ( ) , where I � ( )x xe f x dx e f x dx¢+ ò ò ... (�)

Taking f (x) and ex as the first function and second function, respectively, in I� and

integrating it by parts, we have I� � f (x) ex – ( ) Cxf x e dx¢ +òSubstituting I� in (�), we get

I � ( ) ( ) ( ) Cx x xe f x f x e dx e f x dx¢ ¢- + +ò ò � ex f (x) � C

INTEGRALS �27

Thus, ¢ò [ ( ) ( )]xe f x + f x dx = ( ) Cxe f x +

Example 22 Find (i) �2

�(tan )

�x –e x

x+

+ò dx (ii) 2

2

( ��)( ��)

xx exò dx

Solution

(i) �e have I � �2

�(tan )

�x –e x dx

x+

+ò

Consider f (x) � tan� �x, then f ¢(x) � 2

�� x+

Thus, the given integrand is of the form ex �f (x) � f ¢(x)�.

Therefore, �2

�I (tan )

�x –e x dx

x= +

+ò � ex tan� �x � C

(ii) �e have 2

2

( ��)I

( ��)

xx ex

= ò dx2

2

��)� �

( ��)x x –

e dxx

= ò2

2 2

� 2� �( ��) ( ��)

x x –e dx

x x= +ò 2

� 2� � �

�� ( ��)x x –

e dxx x

= ò

Consider �

( )�

xf x

x-

=+

, then 2

2( )

( �)f x

x¢ =

+Thus, the given integrand is of the form ex �f (x) � f ¢(x)�.

Therefore,2

2

� �C

�( �)x xx x

e dx exx

+ -= +

++ò

EXERCISE 7.6Integrate the functions in Exercises � to 22.

1. x sin x 2. x sin �x 3. x2 ex 4. x log x5. x log 2x 6. x2 log x 7. x sin� �x 8. x tan�� x

9. x cos�� x 10. (sin��x)2 11.�

2

cos

�

x x

x

-

-12. x sec2 x

13. tan��x 14. x (log x)2 15. (x2 � �) log x

�28 MAT�EMATICS

16. ex (sinx � cosx) 17. 2(� )

xx ex+ 18.

� sin� cos

x xe

xæ ö+ç ÷+è ø

19. 2

� ��xe

x xæ öç ÷è ø

20. �

( �)( �)

xx ex-- 21. e2x sin x

22.�

22

sin�

– xx

æ öç ÷+è ø

Choose the correct answer in Exercises 2� and 2�.

23.�2 xx e dxò equals

(A)��

C�

xe + (�)2�

C�

xe +

(C)��

C2

xe + (D)2�

C2

xe +

24. sec (� tan )xe x x dx+ò equals

(A) ex cos x � C (�) ex sec x � C(C) ex sin x � C (D) ex tan x � C

7.6.2 Integrals of some more types�ere, we discuss some special types of standard integrals based on the technique ofintegration by parts :

(i) 2 2x a dx-ò (ii) 2 2x a dx+ò (iii) 2 2a x dx-ò

(i) Let 2 2I x a dx= -òTaking constant function � as the second function and integrating by parts, we

have

I � 2 2

2 2

� 22

xx x a x dx

x a- -

-ò

�2

2 2

2 2

xx x a dx

x a- -

-ò �

2 2 22 2

2 2

x a ax x a dx

x a

- +- -

-ò

INTEGRALS �2�

� 2 2 2 2 2

2 2

dxx x a x a dx a

x a- - - -

-ò ò

� 2 2 2

2 2I

dxx x a a

x a- - -

-ò

or 2I � 2 2 2

2 2

dxx x a a

x a- -

-ò

or I = ò 2 2x – a dx =2

2 2 2 2– – log + – + C2 2x ax a x x a

Similarly, integrating other two integrals by parts, taking constant function � as thesecond function, we get

(ii) ò2

2 2 2 2 2 21+ = + + log + + + C2 2

ax a dx x x a x x a

(iii) ò2

2 2 2 2 –11– = – + sin + C

2 2a x

a x dx x a xa

Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometricsubstitution x � a secq in (i), x � a tanq in (ii) and x � a sinq in (iii) respectively.

Example 23 Find 2 2 �x x dx+ +òSolution Note that

2 2 �x x dx+ +ò � 2( �) �x dx+ +ò�ut x � � � y, so that dx � dy. Then

2 2 �x x dx+ +ò � 2 22y dy+ò

� 2 2� �� log � C

2 2y y y y+ + + + + �using 7.�.2 (ii)�

� 2 2�( �) 2 � 2 log � 2 � C

2x x x x x x+ + + + + + + + +

Example 24 Find 2� 2x x dx- -òSolution Note that 2 2� 2 � ( �)x x dx x dx- - = - +ò ò


�ut x � � � y so that dx � dy.

Thus 2� 2x x dx- -ò � 2� y dy-ò

� 2 �� sin C

2 2 2– y

y y- + + �using 7.�.2 (iii)�

�2 ��

( �) � 2 2 sin C2 2

– xx x x

+æ ö+ - - + +ç ÷è ø

EXERCISE 7.7Integrate the functions in Exercises � to �.

1. 2� x- 2. 2� �x- 3. 2 � �x x+ +

4. 2 � �x x+ + 5. 2� �x x- - 6. 2 � �x x+ -

7. 2� �x x+ - 8. 2 �x x+ 9.2

��x

+

Choose the correct answer in Exercises �� to ��.

10. 2� x dx+ò is equal to

(A) 2 2�� log � C

2 2x

x x x+ + + + +

(�)�

2 22(� ) C

�x+ + (C)

�2 22

(� ) C�

x x+ +

(D)2

2 2 2�� log � C

2 2x

x x x x+ + + + +

11. 2 8 7x x dx- +ò is equal to

(A) 2 2�( �) 8 7 �log � 8 7 C

2x x x x x x- - + + - + - + +

(�) 2 2�( �) 8 7 �log � 8 7 C

2x x x x x x+ - + + + + - + +

(C) 2 2�( �) 8 7 � 2 log � 8 7 C

2x x x x x x- - + - - + - + +

(D) 2 2� �( �) 8 7 log � 8 7 C

2 2x x x x x x- - + - - + - + +

INTEGRALS ��

7.7 Definite IntegralIn the previous sections, we have studied about the indefinite integrals and discussedfew methods of finding them including integrals of some special functions. In thissection, we shall study what is called definite integral of a function. The definite integral

has a unique value. A definite integral is denoted by ( )b

af x dxò , where a is called the

lower limit of the integral and b is called the upper limit of the integral. The definiteintegral is introduced either as the limit of a sum or if it has an anti derivative F in theinterval �a, b�, then its value is the difference between the values of F at the endpoints, i.e., F(b) � F(a). �ere, we shall consider these two cases separately as discussedbelow:

7.7.1 Definite integral as the limit of a sumLet f be a continuous function defined on close interval �a, b�. Assume that all thevalues taken by the function are non negative, so the graph of the function is a curveabove the x�axis.

The definite integral ( )b

af x dxò is the area bounded by the curve y � f (x), the

ordinates x � a, x � b and the x�axis. To evaluate this area, consider the region �RS��between this curve, x�axis and the ordinates x � a and x � b (Fig 7.2).

Divide the interval �a, b� into n equal subintervals denoted by �x�, x��, �x�, x2� ,...,�xr � �, xr�, ..., �xn � �, xn�, where x� � a, x� � a � h, x2 � a � 2h, ... , xr � a � rh and

xn � b � a � nh or .b an

h-

= �e note that as n ®¥, h ® �.

Fig 7.2


The region �RS�� under consideration is the sum of n subregions, where eachsubregion is defined on subintervals �xr � �, xr�, r � �, 2, �, � , n.

From Fig 7.2, we havearea of the rectangle (A�LC) � area of the region (A�DCA) � area of the rectangle

(A�DM) ... (�)!!!d!!!!!!!!!xr – xr�� ® �, i.e., h ® � all the three areas shown in (�) become

nearly equal to each other. Now we form the following sums.

sn �h �f(x�) � � � f (xn � �)� � �

�

( )n

rr

h f x-

=å ... (2)

and Sn � � 2�

�( ) ( ) ( )� ( )n

n rr

h f x f x f x h f x=

+ +¼+ = å ... (�)

�ere, sn and Sn denote the sum of areas of all lower rectangles and upper rectanglesraised over subintervals �xr��, xr� for r � �, 2, �, � , n, respectively.

In view of the inequality (�) for an arbitrary subinterval �xr��, xr�, we havesn �area of the region �RS�� Sn ... (�)

As n ® ¥ strips become narrower and narrower, it is assumed that the limitingvalues of (2) and (�) are the same in both cases and the common limiting value is therequired area under the curve.

Symbolically, we write

lim Snn®¥ � lim nn

s®¥ � area of the region �RS�� ( )

b

af x dxò ... (�)

It follows that this area is also the limiting value of any area which is between thatof the rectangles below the curve and that of the rectangles above the curve. Forthe sake of convenience, we shall take rectangles with height equal to that of thecurve at the left hand edge of each subinterval. Thus, we rewrite (�) as

( )b

af x dxò �

�lim � ( ) ( ) ... ( ( ��) �h

h f a f a h f a n h®

+ + + + +

or ( )b

af x dxò �

�( � ) lim � ( ) ( ) ... ( ( ��) �

nb a f a f a h f a n h

n®¥+ + + + + ... (�)

where h ��

�b a

as nn

® ® ¥

The above expression (�) is known as the definition of definite integral as the limitof sum.

Remark The value of the definite integral of a function over any particular intervaldepends on the function and the interval, but not on the variable of integration that we

INTEGRALS ��

choose to represent the independent variable. If the independent variable is denoted by

t or u instead of x, we simply write the integral as ( )b

af t dtò or ( )

b

af u duò instead of

( )b

af x dxò . �ence, the variable of integration is called a dummy variable.

Example 25 Find 2 2

�( �)x dx+ò as the limit of a sum.

Solution �y definition

( )b

af x dxò �

�( � ) lim � ( ) ( ) ... ( ( ��) �,

nb a f a f a h f a n h

n®¥+ + + + +

where, h ��b an

In this example, a � �, b � 2, f (x) � x2 � �, 2 �� 2

hn n

= =

Therefore,

2 2

�( �)x dx+ò �

� 2 � 2 ( ��)2 lim � (�) ( ) ( ) ... ( )�

n

nf f f f

n n n n®¥+ + + +

�2 2 2

2 2 2� 2 � (2 �2)

2 lim �� ( �) ( �) ... � �n

nn n n n®¥

æ ö+ + + + + + +ç ÷

è ø

� 2 2 2

�

� �2 lim �(� � ... �) (2 � ... (2 �2) �2®¥

+ + + + + + +1442443n

n terms

nn n

�2

2 2 2� 22 lim � (� 2 ... ( ��) �

nn n

n n2®¥+ + + +

� � � ( �) (2 ��)2 lim � �

�n

n n nn

n n2®¥

-+

�� 2 ( �) (2 ��)

2 lim � ��n

n nn

n n®¥

-+

�2 � �

2 lim �� (� ) (2 � )��n n n®¥

+ - � �

2 ��

+ � ��


Example 26 Evaluate 2

�

xe dxò as the limit of a sum.

Solution �y definition

2

�

xe dxò �2 � 2 �2

��(2 ��) lim ...

nn n n

ne e e e

n®¥

é ù+ + + +ê ú

ê úë û

�sing the sum to n terms of a G.�., where a � �, 2nr e= , we have

2

�

xe dxò �

2

2

� ��2 lim � �

�

nn

nn

en

e®¥

- �

2

2

� ��2 lim

��n

n

en

e®¥

é ùê úê úë û

�2

2

2 ( ��)

��lim 22

n

n

e

e

n®¥

é ùê ú

×ê úê úë û

� e2 � � �using �

( �)lim �h

h

eh®

-= �

EXERCISE 7.8Evaluate the following definite integrals as limit of sums.

1.b

ax dxò 2.

�

�( �)x dx+ò 3.

� 2

2x dxò

4.� 2

�( )x x dx-ò 5.

�

�xe dx

-ò 6.� 2

�( )xx e dx+ò

7.8 Fundamental Theorem of Calculus7.8.1 Area function

�e have defined ( )b

af x dxò as the area of

the region bounded by the curve y � f (x),the ordinates x � a and x � b and x�axis. Let x

be a given point in �a, b�. Then ( )x

af x dxò

represents the area of the light shaded region Fig 7.3

INTEGRALS ��

in Fig 7.� ��ere it is assumed that f (x) � � for x Î �a, b�, the assertion made below isequally true for other functions as well�. The area of this shaded region depends uponthe value of x.

In other words, the area of this shaded region is a function of x. �e denote thisfunction of x by A(x). �e call the function A(x) as Area function and is given by

A (x) = ò ( )x

af x dx ... (�)

�ased on this definition, the two basic fundamental theorems have been given.�owever, we only state them as their proofs are beyond the scope of this text book.

7.8.2 First fundamental theorem of integral calculusTheorem 1 Let f be a continuous function on the closed interval �a, b� and let A (x) bethe area function. Then A¢(x) = f (x), for all x Î [a, b].

7.8.3 Second fundamental theorem of integral calculus�e state below an important theorem which enables us to evaluate definite integralsby making use of anti derivative.Theorem 2 Let f be continuous function defined on the closed interval �a, b� and F be

an anti derivative of f. Then ò ( )b

af x dx � [F( )] =b

ax F (b) – F(a).

Remarks

(i) In words, the Theorem 2 tells us that ( )b

af x dxò � (value of the anti derivative F

of f at the upper limit b � value of the same anti derivative at the lower limit a).(ii) This theorem is very useful, because it gives us a method of calculating the

definite integral more easily, without calculating the limit of a sum.(iii) The crucial operation in evaluating a definite integral is that of finding a function

whose derivative is equal to the integrand. This strengthens the relationshipbetween differentiation and integration.

(iv) In ( )b

af x dxò , the function f needs to be well defined and continuous in �a, b�.

For instance, the consideration of definite integral �

� 2 22

( ��)x x dx-ò is erroneous

since the function f expressed by f (x) � �

2 2( ��)x x is not defined in a portion� � � x � � of the closed interval �� 2, ��.


Steps for calculating ( )b

af x dxò .

(i) Find the indefinite integral ( )f x dxò . Let this be F(x). There is no need to keepintegration constant C because if we consider F(x) � C instead of F(x), we get

( ) �F ( ) C� �F( ) C��F( ) C� F( ) �F( )b b

aaf x dx x b a b a= + = + + =ò .

Thus, the arbitrary constant disappears in evaluating the value of the definiteintegral.

(ii) Evaluate F(b) � F(a) � �F ( )�bax , which is the value of ( )

b

af x dxò .

�e now consider some examples

Example 27 Evaluate the following integrals:

(i)� 2

2x dxò (ii)

�

��22(�� )

xdx

xò

(iii)2

� ( �) ( 2)x dx

x x+ +ò (iv) ��

sin 2 cos2t t dtp

òSolution

(i) Let � 2

2I x dx= ò . Since

�2 F ( )

�x

x dx x= =ò ,

Therefore, by the second fundamental theorem, we get

I � 27 8 ��

F (�) �F (2) ��

= =

(ii) Let �

��22

I

(�� )

xdx

x

= ò . �e first find the anti derivative of the integrand.

�ut �2 �

�� . Then �2

x t x dx dt= = or 2

��

x dx dt=

Thus, � 222

2�

�(�� )

x dtdx

tx

=ò ò � 2 �� t

é ùê úë û

� �2

2 � F ( )�

(�� )

x

x

é ùê ú =ê úê úë û

INTEGRALS ��7

Therefore, by the second fundamental theorem of calculus, we have

I �

�

�2

�

2 �F(�) �F(�)

�(�� )x

é ùê ú= ê úê úë û

�2 � �� (��27) ��8

é ù-ê ú

ë û �

2 � � �� 22 ��

é ù- =ê úë û

(iii) Let 2

�I

( �) ( 2)x dx

x x=

+ +ò

�sing partial fraction, we get �� 2

( �) ( 2) � 2x

x x x x= +

+ + + +

So( �) ( 2)

x dxx x+ +ò � �log � 2log 2 F( )x x x+ + + =

Therefore, by the second fundamental theorem of calculus, we haveI �F(2) � F(�) � �� log � � 2 log �� log 2 � 2 log ��

�� log � � log 2 � 2 log � � �2

log27

æ öç ÷è ø

(iv) Let ��

I sin 2 cos 2t t dtp

= ò . Consider �sin 2 cos2t t dtò

�ut sin 2t � u so that 2 cos 2t dt � du or cos 2t dt � �2

du

So �sin 2 cos2t t dtò � ��2

u duò

� � �� sin 2 F ( ) say

8 8u t t= =

Therefore, by the second fundamental theorem of integral calculus

I � � �� F ( ) �F (�) �sin �sin ��

� 8 2 8p p

= =


EXERCISE 7.9

Evaluate the definite integrals in Exercises � to 2�.

1.�

�( �)x dx

-+ò 2.

�

2

�dx

xò 3.2 � 2

�(� �� )x x x dx+ +ò

4. �

�sin 2x dx

p

ò 5. 2

�cos 2x dx

p

ò 6.�

�

xe dxò 7. ��

tan x dxp

ò

8. �

�

cosec x dxp

pò 9.�

� 2��

dx

xò 10.

�

2��dx

x+ò 11.�

22 �dx

x -ò

12. 22�

cos x dxp

ò 13.�

22 �x dx

x +ò 14.�

2�

2 ��

xdx

x++ò 15.

2�

�

xx e dxò

16.22

2�

�� x

x x+ +ò 17. 2 ��

(2sec 2)x x dxp

+ +ò 18. 2 2�

(sin �cos )2 2x x

dxp

ò

19.2

2�

� ��

xdx

x++ò 20.

�

�( sin )

�x x

x e dxp

+òChoose the correct answer in Exercises 2� and 22.

21.�

2� �dx

x+ò equals

(A)�p

(�)2�p

(C)�p

(D)�2p

22.2�

2� � �dx

x+ò equals

(A)�p

(�)�2p

(C)2�p

(D)�p

7.9 Evaluation of Definite Integrals by SubstitutionIn the previous sections, we have discussed several methods for finding the indefiniteintegral. �ne of the important methods for finding the indefinite integral is the methodof substitution.

INTEGRALS ��

To evaluate ( )b

af x dxò , by substitution, the steps could be as follows:

�. Consider the integral without limits and substitute, y � f (x) or x � g(y) to reducethe given integral to a known form.

2. Integrate the new integrand with respect to the new variable without mentioningthe constant of integration.

�. Resubstitute for the new variable and write the answer in terms of the originalvariable.

�. Find the values of answers obtained in (�) at the given limits of integral and findthe difference of the values at the upper and lower limits.

ANote In order to quicken this method, we can proceed as follows: Afterperforming steps �, and 2, there is no need of step �. �ere, the integral will be keptin the new variable itself, and the limits of the integral will accordingly be changed,so that we can perform the last step.

Let us illustrate this by examples.

Example 28 Evaluate � � �

�� x x dx

-+ò .

Solution �ut t � x� � �, then dt � �x� dx.

Therefore, � �� x x dx+ò � t dtò � �22

�t �

�� 22

( �)�

x +

�ence,� � �

�� x x dx

-+ò �

�� 2

��

2( �)

�x

é ù+ê ú

ê úë û

��

� �2 22(� �) � (��) �

�

é ù+ +ê ú

ê úë û

�� 2 22

2 ��

é ù-ê ú

ê úë û �

2 � 2(2 2)

� �=

Alternatively, first we transform the integral and then evaluate the transformed integralwith new limits.


Let t �x� � �. Then dt � � x� dx.Note that, when x �� , t � � and when x � �, t � 2Thus, as x varies from � � to �, t varies from � to 2

Therefore� � �

�� x x dx

-+ò �

2

�t dtò

�

2� � �2 2 2

�

2 22 ��

� �t

é ù é ù=ê ú ê ú

ê ú ê úë û ë û �

2 � 2(2 2)

� �=

Example 29 Evaluate ��

2�

tan�

xdx

x+ò

Solution Let t � tan � �x, then 2�

�dt dx

x=

+. The new limits are, when x � �, t � � and

when x � �, �

tp

= . Thus, as x varies from � to �, t varies from � to �p

.

Therefore��

2�

tan�

xdx

x+ò �2 �

��

�2t

t dt

pp é ù

ê úë û

ò � 2 2�

��2 �� 2

é ùp p=ê ú

ë û

EXERCISE 7.10Evaluate the integrals in Exercises � to 8 using substitution.

1.�

2� �x

dxx +ò 2. �2

�sin cos d

p

f f fò 3.� ��

2�

2sin

�x

dxx

æ öç ÷+è øò

4.2

�2x x +ò (�ut x � 2 � t2) 5. 2

2�

sin� cos

xdx

x

p

+ò

6.2

2� ��dx

x x+ò 7.�

2� 2 �dx

x x- + +ò 8.2 2

2�

� ��

2xe dx

x xæ öç ÷è øò

Choose the correct answer in Exercises � and ��.

9. The value of the integral

��

� ��

( )x xdx

x-

ò is

(A) � (�) � (C) � (D) �

10. If f (x) � �

sinxt t dtò , then f ¢(x) is

(A) cosx � x sin x (�) x sinx(C) x cosx (D) sinx � x cosx

INTEGRALS ��

7.10 Some Properties of Definite Integrals�e list below some important properties of definite integrals. These will be useful inevaluating the definite integrals more easily.

P0 : ( ) ( )b b

a af x dx f t dt=ò ò

P1 : ( ) � ( )b a

a bf x dx f x dx=ò ò . In particular, ( ) �

a

af x dx =ò

P2 : ( ) ( ) ( )b c b

a a cf x dx f x dx f x dx= +ò ò ò

P3 : ( ) ( )b b

a af x dx f a b x dx= + -ò ò

P4 : � �( ) ( )

a af x dx f a x dx= -ò ò

(Note that �� is a particular case of ��)

P5 :2

� � �( ) ( ) (2 )

a a af x dx f x dx f a x dx= + -ò ò ò

P6 :2

� �( ) 2 ( ) , if (2 ) ( )

a af x dx f x dx f a x f x= - =ò ò and

� if f (2a � x) � � f (x)

P7 : (i) �

( ) 2 ( )a a

af x dx f x dx

-=ò ò , if f is an even function, i.e., if f (� x) � f (x).

(ii) ( ) �a

af x dx

-=ò , if f is an odd function, i.e., if f (� x) � � f (x).

�e give the proofs of these properties one by one.Proof of P0 It follows directly by making the substitution x � t.Proof of P1 Let F be anti derivative of f. Then, by the second fundamental theorem of

calculus, we have ( ) F ( ) �F ( ) ��F ( ) F ( )� ( )b a

a bf x dx b a a b f x dx= = - = -ò ò

�ere, we observe that, if a � b, then ( ) �a

af x dx =ò .

Proof of P2 Let F be anti derivative of f. Then

( )b

af x dxò �F(b) � F(a) ... (�)

( )c

af x dxò �F(c) � F(a) ... (2)

and ( )b

cf x dxò �F(b) � F(c) ... (�)


Adding (2) and (�), we get ( ) ( ) F( ) �F( ) ( )c b b

a c af x dx f x dx b a f x dx+ = =ò ò ò

This proves the property �2.Proof of P3 Let t � a � b � x. Then dt � � dx. � hen x � a, t � b and when x � b, t � a.Therefore

( )b

af x dxò � ( � )

a

bf a b t dt- +ò

� ( � )b

af a b t dt+ò (by ��)

� ( � )b

af a b x+ò dx by ��

Proof of P4 �ut t � a � x. Then dt � � dx. � hen x � �, t � a and when x � a, t � �. Nowproceed as in ��.

Proof of P5 �sing �2, we have 2 2

� �( ) ( ) ( )

a a a

af x dx f x dx f x dx= +ò ò ò .

Let t = 2a – x in the second integral on the right hand side. Thendt �� dx. � hen x � a, t � a and when x � 2a, t � �. Also x � 2a � t.

Therefore, the second integral becomes2

( )a

af x dxò �

�� (2 � )

af a t dtò �

�(2 � )

af a t dtò �

�(2 � )

af a x dxò

�ence2

�( )

af x dxò �

� �( ) (2 )

a af x dx f a x dx+ -ò ò

Proof of P6 �sing ��, we have 2

� � �( ) ( ) (2 )

a a af x dx f x dx f a x dx= + -ò ò ò ... (�)

Now, if f (2a � x) �f (x), then (�) becomes2

�( )

af x dxò �

� � �( ) ( ) 2 ( ) ,

a a af x dx f x dx f x dx+ =ò ò ò

and if f (2a � x) �� f (x), then (�) becomes2

�( )

af x dxò �

� �( ) ( ) �

a af x dx f x dx- =ò ò

Proof of P7 �sing �2, we have

( )a

af x dx

-ò ��

�( ) ( )

a

af x dx f x dx

-+ò ò . Then

Let t �� x in the first integral on the right hand side.dt �� dx. � hen x � � a, t � a and whenx ��, t � �. Also x � � t.

INTEGRALS ��

Therefore ( )a

af x dx

-ò ��

�� (�) ( )

a

af t dt f x dx+ò ò

��

(� ) ( )a a

f x dx f x dx+ò ò (by ��) ... (�)

(i) Now, if f is an even function, then f (�x) � f (x) and so (�) becomes

� � �( ) ( ) ( ) 2 ( )

a a a a

af x dx f x dx f x dx f x dx

-= + =ò ò ò ò

(ii) If f is an odd function, then f (�x) � � f (x) and so (�) becomes

� �( ) ( ) ( ) �

a a a

af x dx f x dx f x dx

-= - + =ò ò ò

Example 30 Evaluate 2 �

��x x dx

-ò

Solution �e note that x� � x ³ � on �� , �� and x� � x £ � on ��, �� and thatx� � x ³ � on ��, 2�. So by �2 we write

2 ��

�x x dx-ò �

� � 2� � �

� � �( � ) �( � ) ( � )x x dx x x dx x x dx

-+ +ò ò ò

�� 2� � �

� � �( � ) ( � ) ( � )x x dx x x dx x x dx

-+ +ò ò ò

�

� � 2� 2 2 � � 2

��

� � �� 2 2 � � 2x x x x x xé ù é ù é ù

+ +ê ú ê ú ê úë û ë û ë û

��

� � � ��2 � �� 2 2 � � 2

æ ö æ ö æ ö+ +ç ÷ ç ÷ ç ÷è ø è ø è ø

��

� 2� 2 2 � � 2

+ + - + - + � � � ��

22 � �

- + =

Example 31 Evaluate 2��

sin x dxp

pòSolution �e observe that sin2 x is an even function. Therefore, by �7 (i), we get

2��

sin x dxp

pò � 2��

2 sin x dxp

ò


� ��

(� cos 2 )2

2x

dxp -

ò � ��

(� cos 2 )x dxp

-ò

��

�

�� sin 22

x x

p

é ùê úë û

� � �

� sin �� 2 2 � 2p p pæ ö =ç ÷

è ø

Example 32 Evaluate 2�

sin� cos

x xdx

x

p

+ò

Solution Let I � 2�

sin� cos

x xdx

x

p

+ò . Then, by ��, we have

I � 2�

( ) sin ( )� cos ( )

x x dxx

p p - p -+ p -ò

� 2�

( ) sin� cos

x x dxx

p p -+ò � 2�

sinI

� cosx dx

x

pp -

+ò

or 2 I � 2�

sin� cos

x dxx

pp

+ò

or I � 2�

sin2 � cos

x dxx

pp+ò

�ut cos x � t so that � sin x dx � dt. � hen x � �, t � � and when x � p, t � � �.Therefore, (by ��) we get

I ��

2�

�2 �

dtt

-p+ò �

�

2�2 �dt

t-

p+ò

��

2� �dt

tp

+ò (by �7, 2

�since

� t+ is even function)

�2��

�tan tan ��tan � ��

� �t - p pé ùé ù é ùp = p = p =ê úë û ë û ë û

Example 33 Evaluate � � ��

sin cosx x dx-ò

Solution Let I � � � �

�sin cosx x dx

-ò . Let f(x) � sin� x cos� x. Then

f (� x) � sin� (� x) cos� (� x) � � sin� x cos� x � � f (x), i.e., f is an odd function.Therefore, by �7 (ii), I � �

INTEGRALS ��

Example 34 Evaluate �

2� ��

sinsin cos

xdx

x x

p

+ò

Solution Let I � �

2� ��

sinsin cos

xdx

x x

p

+ò ... (�)

Then, by ��

I �

�

2� � �

sin ( )2

sin ( ) cos ( )2 2

xdx

x x

pp

-

p p- + -ò �

�2

� ��

coscos sin

xdx

x x

p

+ò ... (2)

Adding (�) and (2), we get

2I ��

22 2� ��

sin cos��

2sin cosx x

dx dx xx x

pp p+ p

= = =+ò ò

�ence I ��p

Example 35 Evaluate �

�� tan

dxx

p

p +ò

Solution Let I � � �

� �

cos

� tan cos sin

x dxdxx x x

p p

p p=

+ +ò ò ... (�)

Then, by �� I � �

�

cos� �

cos sin� � � �

x dx

x x

p

p

p pæ ö+ -ç ÷è ø

p p p pæ ö æ ö+ - + + -ç ÷ ç ÷è ø è ø

ò

� �

�

sinsin cos

xdx

x x

p

p +ò ... (2)

Adding (�) and (2), we get

2I � � �

� ��

dx xp p

p p

p p p= = - =ò . �ence I

�2p

=


Example 36 Evaluate 2�

log sin x dxp

ò

Solution Let I � 2�

log sin x dxp

òThen, by ��

I � 2 2� �

log sin log cos2

x dx x dxp p

pæ ö- =ç ÷è øò ò

Adding the two values of I, we get

2I � 2�

log sin logcosx x dxp

+ò

� 2�

log sin cos log 2 log 2x x dxp

+ -ò (by adding and subtracting log2)

� 2 2� �

log sin 2 log 2x dx dxp p

-ò ò (� hy?)

�ut 2x � t in the first integral. Then 2 dx � dt, when x � �, t � � and when 2

xp

= ,

t � p.

Therefore 2I ��

�log sin log 2

2 2t dt

p p-ò

� 2�

2log sin log 2

2 2t dt

pp

-ò �by �� as sin (p � t) � sin t)

� 2�

log sin log 22

x dxp

p-ò (by changing variable t to x)

� I log 22p

-

�ence 2�

log sin x dxp

ò ��

log 22p

.

INTEGRALS ��7

EXERCISE 7.11

�y using the properties of definite integrals, evaluate the integrals in Exercises � to ��.

1. 22�

cos x dxp

ò 2. 2�

sinsin cos

xdx

x x

p

+ò 3.

�2

2� ��2 2

sin

sin cos

x dx

x x

p

+ò

4.�

2� ��

cossin cos

x dxx x

p

+ò 5.�

�� 2 �x dx

-+ò 6.

8

2�x dx-ò

7.�

�(� )nx x dx-ò 8. �

�log (� tan )x dx

p

+ò 9.2

�2x x dx-ò

10. 2�

(2log sin log sin 2 )x x dxp

-ò 11. 22�2

sin x dxp

pò

12.�� sin

x dxx

p

+ò 13. 72�2

sin x dxp

pò 14.2 �

�cos x dx

p

ò

15. 2�

sin cos� sin cos

x xdx

x x

p-

+ò 16.�

log (� cos )x dxp

+ò 17. �

a x dxx a x+ -ò

18.�

��x dx-ò

19. Show that � �

( ) ( ) 2 ( )a a

f x g x dx f x dx=ò ò , if f and g are defined as f (x) � f(a � x)

and g(x) � g(a � x) � �Choose the correct answer in Exercises 2� and 2�.

20. The value of � �2

2

( cos tan �)x x x x dxp

-p+ + +ò is

(A) � (�) 2 (C) p (D) �

21. The value of 2�

� �sinlog

� �cosx

dxx

pæ ö+ç ÷+è ø

ò is

(A) 2 (�)��

(C) � (D) �2


Miscellaneous Examples

Example 37 Find cos � � sin �x x dx+òSolution �ut t � � � sin �x, so that dt � � cos �x dx

Therefore�2�

cos � � sin ��

x x dx t dt+ =ò ò

�� 2 2� 2 �

( ) C � (� sin � ) C� � �

t x´ + + +

Example 38 Find

��

�

( )x xdx

x-

ò

Solution �e have

��

� � �

� �

�(� )( )x x xdx dxx x

--=ò ò

�ut ��

� �� , so thatx t dx dt

x x- = = =

Therefore

��

�( ) �

�x x dx t dt

x-

=ò ò �

��

��

�

� � � �C � � C

� � ��t

xæ ö´ + - +ç ÷è ø

Example 39 Find �

2( �) ( �)x dx

x x- +ò

Solution �e have

�

2( �)( �)x

x x- + � � 2

�( �)

�x

x x x+ +

- + -

� 2

�( �)

( �) ( �)x

x x+ +

- +... (�)

Now express 2

�( �)( �)x x- +

� 2

A � C( �) ( �)

xx x

++

- +... (2)

INTEGRALS ��

So � �A (x2 � �) � (�x � C) (x � �)�(A � �) x2 � (C � �) x � A � C

Equating coefficients on both sides, we get A � � � �, C � � � � and A � C � �,

which give � �A , � C �

2 2= = = . Substituting values of A, � and C in (2), we get

2

�( �) ( �)x x- +

� 2 2

� � �2( �) 2 ( �) 2( �)

xx x x

- -- + +

... (�)

Again, substituting (�) in (�), we have�

2( �) ( �)x

x x x- + + � 2 2

� � �( �)

2( �) 2 ( �) 2( �)x

xx x x

+ + - -- + +

Therefore� 2

2 ��2

� � �log � � log ( �) � tan C

2 2 � 2( �) ( �)x x

dx x x x xx x x

= + + - + +- + +ò

Example 40 Find 2

�log (log )

(log )x dx

xé ù

+ê úë û

ò

Solution Let 2

�I log (log )

(log )x dx

xé ù

= +ê úë û

ò

� 2

�log (log )

(log )x dx dx

x+ò ò

In the first integral, let us take � as the second function. Then integrating it byparts, we get

I � 2

�log (log )

log (log )dx

x x x dxx x x

- +ò ò

� 2log (log )log (log )dx dx

x xx x

- +ò ò ... (�)

Again, consider logdx

xò , take � as the second function and integrate it by parts,

we have 2

� ��

log log (log )dx x

x dxx x xx

é ùì üæ ö= ê úí ýç ÷è øê úî þë û

ò ò ... (2)


�utting (2) in (�), we get

2 2I log (log )log (log ) (log )

x dx dxx x

x x x= - - +ò ò � log (log ) C

logx

x xx

- +

Example 41 Find cot tanx x dxé ù+ë ûòSolution �e have

I � cot tanx x dxé ù+ë ûò tan (� cot )x x dx= +ò�ut tan x � t2, so that sec2 x dx � 2t dt

or dx � �2�

t dtt+

Then I � 2 �� 2

�(� )

tt dt

t tæ ö+ç ÷ +è øò

�2 2 2

� 22

2

� �� ( �)2 �2 �2

�� 2

dt dtt t tdtt t tt t

æ ö æ ö+ +ç ÷ ç ÷+ è ø è øæ ö+ æ ö+ - +ç ÷ ç ÷è ø è ø

ò ò ò

�ut �

tt

- � y, so that 2

��

tæ ö+ç ÷è ø

dt � dy. Then

I � �� 22

�

2 2 tan C � 2 tan C2 22

tdy y t

y

æ ö-ç ÷è ø= + +

+ò

�2

�� tan �2 tan C � 2 tan C

2 2tant x

t x

æ ö- -æ ö+ +ç ÷ ç ÷ç ÷ è øè ø

Example 42 Find �

sin 2 cos 2

��cos (2 )

x x dx

xò

Solution Let �

sin 2 cos 2I

��cos 2

x xdx

x= ò

INTEGRALS ��

�ut cos2 (2x) � t so that � sin 2x cos 2x dx � � dt

Therefore �� 2

2

� � � �I � � sin C sin cos 2 C

� � � � ��

dt tx

t-æ ö é ù= = + = - +ç ÷ ê úè ø ë ûò

Example 43 Evaluate �2�

sin ( )x x dx-

pò

Solution �ere f (x) � �x sin px � � sin for � �

�sin for �

2

x x x

x x x

p - £ £ìïí

- p £ £ïî

Therefore�2�

� sin �x x dx-

pò ��

� 2� �

sin sinx x dx x x dx-

p + - pò ò

��

�2

� �sin sinx x dx x x dx

-p - pò ò

Integrating both integrals on righthand side, we get

�2�

� sin �x x dx-

pò �

��

2

2 2� �

� cos sin cos sinx x x x x x

-

p p - p pé ù é ù+ - +ê ú ê úp pp pë û ë û

� 2 2

2 � � � �é ù- - - = +ê úp p pp pë û

Example 44 Evaluate 2 2 2 2� cos sinx dx

a x b x

p

+ò

Solution Let I � 2 2 2 2 2 2 2 2� �

( )cos sin cos ( ) sin ( )

x dx x dxa x b x a x b x

p p p -=

+ p - + p -ò ò (using ��)

� 2 2 2 2 2 2 2 2� �cos sin cos sindx x dx

a x b x a x b x

p pp -

+ +ò ò

� 2 2 2 2�I

cos sindx

a x b x

pp -

+ò

Thus 2I � 2 2 2 2� cos sindx

a x b x

pp

+ò


or I � 22 2 2 2 2 2 2 2� �

22 2cos sin cos sin

dx dxa x b x a x b x

ppp p

= ×+ +ò ò (using ��)

�2�

2 2 2 2 2 2 2 2�

�cos sin cos sin

pp

p

é ùp +ê ú

+ +ê úë ûò ò

dx dxa x b x a x b x

� 2 2

2�2 2 2 2 2 2

��

sec cosectan cot

pp

p

é ùp +ê ú

+ +ê úë ûò ò

xdx x dxa b x a x b

� ��

2 2 2 2 2 2� �

tan t coté ùp - = =ê ú+ +ë ûò òdt du

put x and x ua b t a u b

� � �

��

� �

tan � tanp pé ù é ù

ê ú ê úë û ë û

bt auab a ab b �

�� tan tanp é ù+ê úë û

b aab a b �

2

2pab

Miscellaneous Exercise on Chapter 7Integrate the functions in Exercises � to 2�.

1. �

�x x-

2.�

x a x b+ + +3.

2

�

x ax x-��int: �ut x �

at

�

4. �2 � �

�

( �)x x +5. ��

�2

�

x x+ ��int: ��

�2 � �

� �

�x x x x

=æ ö

+ +ç ÷ç ÷è ø

, put x � t��

6. 2

�( �) ( �)

xx x+ +

7.sin

sin ( )x

x a- 8.�log �log

�log 2 log

x x

x xe ee e

--

9.2

cos

� sin

x

x-10.

8 8

2 2

sin cos� 2sin cos

xx x

--

11.�

cos ( ) cos ( )x a x b+ +

12.�

8�

x

x-13.

(� ) (2 )

x

x x

ee e+ +

14. 2 2

�( �) ( �)x x+ +

15. cos�x elog sinx 16. e� logx (x� � �)� � 17. f ¢ (ax � b) �f (ax � b)�n

18. �

�

sin sin ( )x x + a 19.� �

� �

sin cossin cos

x xx x

- -

- -

-+

, x Î ��, ��

INTEGRALS ��

20.��

xx

-+

21.2 sin 2� cos 2

xxe

x++

22.2

2

�( �) ( 2)

x xx x

+ ++ +

23. �� tan

�xx

-+

24.2 2

�

� log ( �) 2 logx x x

x

é ù+ + -ë û

Evaluate the definite integrals in Exercises 2� to ��.

25.2

� sin� cos

p

p-æ ö

ç ÷-è øò x xe dx

x26. �

� ��

sin coscos sin

x xdx

x x

p

+ò 27.2

22 2�

coscos �sin

x dxx x

p

+ò

28. �

�

sin cossin 2x x

dxx

p

p

+ò 29.

�

� �dxx x+ -ò 30. �

�

sin cos� ��sin 2

x xdx

x

p+

+ò

31. �2�

sin 2 tan (sin )x x dxp

-ò 32.�

tansec tan

x xdx

x xp

+ò

33.�

�� 2 � � ��x x x dx- + - + -ò

�rove the following (Exercises �� to ��)

34.�

2�

2 2log

� �( �)dx

x x= +

+ò 35.�

��xx e dx =ò

36.� �7 ��

cos �x x dx-

=ò 37. �2�

2sin�

x dxp

=ò

38. ��

2 tan � log 2x dxp

= -ò 39.� �

�sin �

2x dx- p

= -ò40. Evaluate

� 2 �

�

xe dx-ò as a limit of a sum.

Choose the correct answers in Exercises �� to ��.

41.x x

dxe e-+ò is equal to

(A) tan�� (ex) � C (�) tan�� (e–x) � C(C) log (ex � e–x) � C (D) log (ex � e–x) � C

42. 2

cos 2(sin cos )

xdx

x x+ò is equal to

(A)��

Csin cosx x

++

(�) log �sin cos � Cx x+ +

(C) log �sin cos � Cx x- + (D)2

�(sin cos )x x+


43. If f (a � b � x) � f (x), then ( )b

ax f x dxò is equal to

(A) ( )2

b

a

a bf b x dx

+-ò (�) ( )

2b

a

a bf b x dx

++ò

(C) ( )2

b

a

b af x dx

-ò (D) ( )

2b

a

a bf x dx

+ò

44. The value of � �

2�

2 �tan

�x

dxx x

- -æ öç ÷+ -è øò is

(A) � (�) � (C) �� (D)�p

Summary

® Integration is the inverse process of differentiation. In the differential calculus,we are given a function and we have to find the derivative or differential ofthis function, but in the integral calculus, we are to find a function whosedifferential is given. Thus, integration is a process which is the inverse ofdifferentiation.

Let F( ) ( )d

x f xdx

= . Then we write ( ) F ( ) Cf x dx x= +ò . These integrals

are called indefinite integrals or general integrals, C is called constant ofintegration. All these integrals differ by a constant.

® From the geometric point of view, an indefinite integral is collection of familyof curves, each of which is obtained by translating one of the curves parallelto itself upwards or downwards along the y�axis.

® Some properties of indefinite integrals are as follows:

�. � ( ) ( )� ( ) ( )f x g x dx f x dx g x dx+ = +ò ò ò2. For any real number k, ( ) ( )k f x dx k f x dx=ò òMore generally, if f�, f2, f�, ... , fn are functions and k�, k2, ... ,kn are realnumbers. Then

� � 2 2� ( ) ( ) ... ( )�n nk f x k f x k f x dx+ + +ò� � � 2 2( ) ( ) ... ( )n nk f x dx k f x dx k f x dx+ + +ò ò ò

INTEGRALS ��

® Some standard integrals

(i)�

C�

nn x

x dxn

+

= ++ò , n ¹ � �. �articularly, Cdx x= +ò

(ii) cos sin Cx dx x= +ò (iii) sin �cos Cx dx x= +ò(iv) 2sec tan Cx dx x= +ò (v) 2cosec �cot Cx dx x= +ò(vi) sec tan sec Cx x dx x= +ò

(vii) cosec cot �cosec Cx x dx x= +ò (viii)�

2sin C

�

dxx

x-= +

-ò

(ix)�

2cos C

�

dxx

x-= - +

-ò (x)

�2 tan C

�dx

xx

-= ++ò

(xi)�

2 cot C�

dxx

x-= - +

+ò (xii) Cx xe dx e= +ò

(xiii) Clog

xx a

a dxa

= +ò (xiv) �

2sec C

�

dx xx x

-= +-

ò

(xv) �

2cosec C

�

dxx

x x-= - +

-ò (xvi)

�log � � Cdx x

x= +ò

® Integration by partial fractions

Recall that a rational function is ratio of two polynomials of the form �( )�( )

xx

,

where �(x) and �(x) are polynomials in x and �(x) ¹ �. If degree of thepolynomial �(x) is greater than the degree of the polynomial �(x), then we

may divide �(x) by � (x) so that ��( )�( )T ( )

�( ) �( )xx

xx x

= + , where T(x) is a

polynomial in x and degree of ��(x) is less than the degree of �(x). T(x)

being polynomial can be easily integrated. ��( )�( )

xx

can be integrated by


expressing ��( )�( )

xx

as the sum of partial fractions of the following type:

�.( ) ( )

px qx a x b

+- -

�A �

x a x b+

- -, a ¹ b

2. 2( )px qx a

+- � 2

A �( )x a x a

+- -

�.2

( ) ( ) ( )px qx r

x a x b x c+ +

- - - �A � C

x a x b x c+ +

- - -

�.2

2( ) ( )px qx r

x a x b+ +

- - � 2

A � C( )x a x bx a

+ +- --

�.2

2( ) ( )px qx r

x a x bx c+ +

- + + � 2

A � �Cxx a x bx c

+- + +

where x2 � bx � c can not be factorised further.

® Integration by substitutionA change in the variable of integration often reduces an integral to one of thefundamental integrals. The method in which we change the variable to someother variable is called the method of substitution. � hen the integrand involvessome trigonometric functions, we use some well known identities to find theintegrals. �sing substitution technique, we obtain the following standardintegrals.

(i) tan log sec Cx dx x= +ò (ii) cot log sin Cx dx x= +ò(iii) sec log sec tan Cx dx x x= + +ò(iv) cosec log cosec cot Cx dx x x= - +ò

® Integrals of some special functions

(i) 2 2

�log C

2dx x a

a x ax a-

= ++-ò

(ii) 2 2

�log C

2dx a x

a a xa x+

= +--ò (iii)

�2 2

�tan C

dx xa ax a

-= ++ò

INTEGRALS ��7

(iv) 2 2

2 2log C

dxx x a

x a= + - +

-ò (v)

�

2 2sin C

dx xaa x

-= +-

ò

(vi)2 2

2 2log � � C

dxx x a

x a= + + +

+ò

® Integration by partsFor given functions f� and f2, we have

� 2 � 2 � 2( ) ( ) ( ) ( ) ( ) ( )d

f x f x dx f x f x dx f x f x dx dxdx

é ù× = - ×ê úë ûò ò ò ò , i.e., the

integral of the product of two functions � first function � integral of thesecond function � integral of �differential coefficient of the first function �integral of the second function�. Care must be taken in choosing the firstfunction and the second function. �bviously, we must take that function asthe second function whose integral is well known to us.

® � ( ) ( )� ( ) Cx xe f x f x dx e f x dx¢+ = +ò ò® Some special types of integrals

(i)2

2 2 2 2 2 2log C2 2x a

x a dx x a x x a- = - - + - +ò

(ii)2

2 2 2 2 2 2log C2 2x a

x a dx x a x x a+ = + + + + +ò

(iii)2

2 2 2 2 �sin C2 2x a xa x dx a x

a-- = - + +ò

(iv) Integrals of the types 2 2or

dx dxax bx c ax bx c+ + + +

ò ò can be

transformed into standard form by expressing

ax2 � bx � c � 2 2

222 �

b c b c ba x x a x

a a a a a

é ùæ öé ù æ ö+ + = + + -ê úç ÷ç ÷ê úë û è øê úè øë û

(v) Integrals of the types 2 2or

px q dx px q dxax bx c ax bx c

+ ++ + + +

ò ò can be


transformed into standard form by expressing

2A ( ) � A (2 ) �d

px q ax bx c ax bdx

+ = + + + = + + , where A and � are

determined by comparing coefficients on both sides.

® �e have defined ( )b

af x dxò as the area of the region bounded by the curve

y � f (x), a £ x £ b, the x�axis and the ordinates x � a and x � b. Let x be a

given point in �a, b�. Then ( )x

af x dxò represents the Area function A (x).

This concept of area function leads to the Fundamental Theorems of IntegralCalculus.

® First fundamental theorem of integral calculus

Let the area function be defined by A(x) � ( )x

af x dxò for all x ³ a, where

the function f is assumed to be continuous on �a, b�. Then A¢ (x) � f (x) for allx Î �a, b�.

® Second fundamental theorem of integral calculusLet f be a continuous function of x defined on the closed interval �a, b� and

let F be another function such that F( ) ( )d

x f xdx

= for all x in the domain of

f, then ( ) F( ) C F ( ) F ( )b b

aaf x dx x b a= + = -ò .

This is called the definite integral of f over the range �a, b�, where a and bare called the limits of integration, a being the lower limit and b theupper limit.

—v—

integrals - byju's...integrals. ˜ut these integrals are very similar geometrically. thus, y ˜...

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