Mathematics 11 (Vol- 2)

11.7.1 Decomposition method

Sometimes it is very difficult to integrate the given function directly. But it can be integrated after
decomposing it into a sum or difference of number of functions whose integrals are already known.

( )For example 1− x3 2 , x2 −x + 1 , cos 5x sin 3x , cos3 x , e2x − 1 , do not have direct formulae
x3 e
x

to integrate. But these functions can be decomposed into a sum or difference of functions, whose

individual integrals are known. In most of the cases the given integrand will be any one of the

algebraic, trigonometric or exponential forms, and sometimes combinations of these functions.

Example 11.15
Integrate the following with respect to x:

( ) (i) 1− x3 2 (ii) x2 −x +1
x3
Solution

(i) ( )∫ ∫1− x3 2 dx = (1− 2x3 + x6 )dx

= ³ dx 2³ x3dx ³ x6dx

= x − x4 + x7 + c.
27

(ii) ³ ∫x2 x 1 dx = ( x2 − x + 1 ) dx
x3 x3 x3 x3

= ∫ 1 dx − ∫ 1 dx + ∫ 1 dx.
x x2 x3

∫ x2 −x + 1 dx = log x + 1 − 1 + c.
x3 x 2x2

Example 11.16
Integrate the following with respect to x:

(i) cos 5x sin 3x (ii) cos3 x.

Solution

(i) ∫ cos 5x sin 3x dx = 1 ∫ 2 cos 5x sin 3x dx
2

= 1 ∫ (sin 8x − sin 2x ) dx
2

∫ cos 5x sin 3x dx = 1  − cos 8x + cos 2x  + c.
2  8 2 

(ii) ∫ cos3 x dx = 1 ∫ ( 3 cos x + cos 3x ) dx
4

= 1  3sin x + sin 3x  + c
4  3 

197 Integral Calculus

Example 11.17
Integrate the following with respect to x:

(i) e2x − 1 ( )(ii) e3x e2x −1 .
Solution e
x

(i) ∫ ∫e2x − 1 dx =  e2x − 1  dx
 ex ex 
ex  

∫ ( ) = ex − e−x dx = ex + e−x + c.

(ii) ∫ ( ) ∫ ( )e3x e2x −1 dx = e5x − e3x dx = e5x − e3x + c.
53

Example 11.18

∫ Evaluate : sin 2 1 x dx.
x cos2

Solution 1 sin 2 x + cos2 x
x cos2 sin 2 x cos2 x
∫ ∫sin2 x dx = dx

= ∫ 1 x dx + ∫ 1 x dx
cos2 sin 2

= ∫ sec2 xdx + ∫ cosec2xdx

= tan x − cot x + c.

Example 11.19

Evaluate : ∫ 1 sin x x dx.
+ sin

Solution sin x  sin x   1 − sin x 
+ sin  + sin   1 − sin x 
∫ 1 x dx = ∫ 1 x dx

sin x − sin2 x sin x − sin 2 x sin x sin 2 x
1− sin2 x cos2 x cos2 x cos2 x
∫ ∫ ∫ ∫ = dx = dx = dx − dx

= ∫ tan x sec xdx − ∫ tan2 xdx

= ∫ tan x sec xdx − ∫ (sec2 x −1)dx

= sec x − tan x + x + c.

Example 11.20

Evaluate : ∫ 1+ cos 2x dx.

Solution

∫ 1+ cos 2x dx = ∫ 2 cos2 x dx = 2 ∫ cos xdx = 2 sin x + c

XI - Mathematics 198

Example 11.21

∫ (x −1)2 dx.
Evaluate : x3 + x

Solution ∫ ∫(x −1)2 x2 +1 − 2xdx
x(x2 + 1)
x3 + x dx =

∫ =  (x2 +1) − 2x dx
 x(x2 +1) x(x2 +1) 


= ∫ 1 dx − 2∫ 1 dx
x 1+ x2

= log | x | −2 tan−1 x + c.

Example 11.22

Evaluate : ∫ (tan x + cot x)2 dx.

Solution

∫ ∫(tan x + cot x)2 dx = [tan2 x + 2 tan x cot x + cot2 x]dx

∫ = [(sec2 x −1) + 2 + (cosec2x −1)]dx

∫ = (sec2 x + cosec2x)dx

= tan x + (− cot x) + c

= tan x − cot x + c.

Example 11.23

Evaluate : ∫ 1 − cos x dx.
1 + cos x

Solution

∫ 1− cos x dx = ∫ 2 sin2 x dx = ∫ tan 2 x dx
1+ cos x 2 cos2 2 2
x

2

= ∫  sec2 x −1dx = tan x − x + c
 2 1 2

= 2 tan x − x + c. 2
2

Example 11.24

Evaluate : ∫ 1+ sin 2x dx.

Solution ∫ ∫1+ sin 2x dx = (cos2 x + sin2 x) + (2sin x cos x) dx


199 Integral Calculus

= ∫ (cos x + sin x)2 dx = ∫ (cos x + sin x)dx

= sin x − cos x + c

Example 11.25

Evaluate : ∫ x3 +2 dx.
x −1

Solution x3 +2 x3 −1+ 3  x3 −1 3
x −1 x −1  x −1
∫ dx = ∫ dx = ∫  + x −1  dx


= ∫ (x −1)(x2 + x + 1) + 3 dx
 x −1 x −1

= ∫  x2 + x +1+ x 3 dx
 −1

= x3 + x2 + x + 3log | (x −1) | +c.
32

Example 11.26

∫ Evaluate : (i) axexdx ∫(ii) exlog2ex dx.

Solution ∫ ∫axexdx = (ae)x dx = (ae)x + c
(i)
log(ae)
(ii)
∫ ∫ ∫exlog2exdx = elog2x exdx = 2x exdx

³ = (2e)x dx (2e)x c.
log(2e)

Example 11.27

Evaluate : ∫ (x − 3) x + 2 dx.

Solution

∫ (x − 3) x + 2 dx = ∫ (x + 2 − 5) x + 2 dx

= ∫ (x + 2) x + 2 dx − 5∫ x + 2 dx

31

= ∫ (x + 2)2 dx − 5∫ (x + 2)2 dx

53

( x + 2) 2 ( x + 2) 2
5 3
= − 5 +c

22 3
5
= 2 (x + − 10 (x + 2) 2 + c.
2) 2
53

XI - Mathematics 200

Example 11.28

Evaluate : ³ 1 dx.
x
Solution x 1

³ 1 x dx = ³ 1 ª x 1 x º dx
x 1 x 1 « x 1 x »
x ¬ ¼

x +1− x 2 dx
x +12 − x
( ) ( )∫ =

x +1− x
x +1−
( ) = ∫ dx = ∫
x x +1 − x dx

11

= ∫ x +1dx − ∫ xdx = ∫ (x +1)2 dx − ∫ x2dx

33

= (x +1)2 − x2 +c
3 3

22

= 2 ( x 3 − 3  + c.
3  
+ 1) 2 x2 

11.7.2 Decomposition by Partial Fractions

One of the important methods to evaluate integration is partial fractions. If the
integrand is in the form of an algebraic fraction and the integral cannot be evaluated by simple
methods, then the fraction need to be expressed in partial fractions  before integration takes

place. We will assume that we have a rational function p(x) , q(x) z 0 in which degree of

q(x)
p (x) < degree of q (x). If this is not the case, we can always perform long division.

Example 11.29

Evaluate : (i) ∫ 3x + 7 2 dx (ii) ∫ ( x + x+3 + 1) dx.
x2 − 3x + 2)2 (x

Solution

(i) ∫ 3x + 7 2 dx = ∫ 13 dx − ∫ 10 dx ­Resolving into
x2 − 3x + x−2 x −1 ¯®partial fractions
­Resolving into
= 13log x − 2 −10 log x −1 + c ®¯partial fractions

(ii) ∫ (x + x+3 + 1) dx = ∫ −2 dx − ∫ (x 1 dx + ∫ 2 dx
2)2 (x x+2 + 2)2 x +1

= −2∫ x 1 2 dx − ∫ (x 1 dx + 2∫ x 1 dx
+ + 2)2 +1

∫ = −2 log x + 2 − (x + 2)−2 dx + 2 log x +1 + c

= −2 log x + 2 + x 1 2 + 2 log | x +1| +c.
+

201 Integral Calculus

EXERCISE 11.5
Integrate the following functions with respect to x :

x3 + 4x2 − 3x + 2 (2)  x+ 1 2 (3) (2x − 5)(36 + 4x)
(1) x2  x 

(4) cot2 x + tan2 x (5) cos 2x − cos 2α (6) cos 2x x
cos x − cosα sin2 x cos2

(7) 3 + 4 cos x sin2 x (9)
sin2 x
(8)

1+ cos x

(10) cos3x cos2x (11) sin2 5x (12) 1 cos 4x
cot x tan x

(13) exlog aex (14) (3x + 4) 3x + 7 (15) 81+x + 41−x
2x

(16) 1 (17) (x + x +1 + 3) 1
x+3− x−4 2)( x (18) (x −1)(x + 2)2

(19) 3x − 9 (20) x3
(x −1)(x + 2)(x2 +1) (x −1)(x − 2)

11.7.3 Method of substitution or change of variable

The method of substitution in integration is similar to finding the derivative of function of
function in differentiation. By using a suitable substitution, the variable of integration is changed to
new variable of integration which will be integrated in an easy manner.

We know that, if u is a function of x then du = u′.
dx

Hence we can write ∫ f (u)u′dx = ∫ f (u)du

Thus, ∫ f [g(x)]g′(x)dx = ∫ f (u)du, where u = g(x)

The success of the above method depends on the selection of suitable substitution
either x = φ(u) or u = g(x).

Note 11.2
The substitution for the variable of integration is in trigonometric function, use a rough

diagram to find the re -substitution value for it. Suppose the variable of integration x is substituted

as x = tanθ . After integration suppose the solution is secθ + cosecθ

For example, if x tanT , then from

the figure 1+ x2

cosecT § 1 x2 · θ x
¨¨© x ¹¸¸ , 1

secT § 1 x2 ·
¨¨© 1 ¸¹¸

Then secT cosecT § 1 x2 · .
1 x2 ©¨¨ 1 ¸¸¹
XI - Mathematics
202

Example 11.30
Evaluate the following integrals :

³ ∫ (i) 2x 1 x2 dx (ii) e−x2 x dx (iii) ∫ 1 sin x x dx
+ cos

(iv) ∫ 1 1 2 dx (v) ∫ x (a − x)8 dx
+x

Solution

(i) ³ 2x 1 x2 dx Putting 1 x2 u, then 2x dx du



³ 2x 1 x2 dx = ∫ u du

3

∫ = 1 = u2 + c = 2 3 + c = 2 (1 + 3 + c .
3
u 2 du u2 x2 )2
33

2

∫ (ii) e−x2 xdx
=Putting x2 u=then 2x dx du

e x2 x dx = e−u du
2
³ ∫ Therefore,

= 1 e−udu = 1 (−e−u ) + c = − 1 e−u + c = − 1 e−x2 + c.
2 2 22
∫ (iii) sin x dx
∫ 1+ cos x

Putting 1 cos x u, then sin x dx du

Therefore, ³ sin x x dx = ³ du log | u | c log |1 cos x | c.
1 cos u

(iv) ∫ 1+1x2 dx

=Putting x ta=n u, then dx sec2 u du

∫ ∫ ∫ ∫1dx = 1 sec2 u u du = sec2 u du = du = u + c
+ tan2 sec2 u
1+ x2

∫1 dx = tan −1 x + c.

1+ x2

(v) ∫ x (a − x)8 dx

Putting u = a − x, then du = −dx

∫ x (a − x)8 dx = ∫ x (a − x)8 dx

³ a u u 8 du

( ) = ∫ −a (u)8 + u9 du

203 Integral Calculus

= ∫ u9du − a∫ u8du

= u10 − a u9 + c
10 9

∫ x (a − x)8 dx (a x)10 a(a x)9 c.

10 9

11.7.4 Important Results

(1) ∫ f ′(x) dx = log | f (x) | +c
f (x)
(2)
Proof ∫ f ′(x)[ f (x)]n dx = [ f (x)]n+1 + c, n ≠ −1
(1)
n +1

Let I = ∫ f ′(x) dx
f (x)

Putting f (x) u then f c(x)dx du

Thus, I = ∫ du = log | u | +c
u

³Therefore, f c(x) dx = log | f (x) | +c.
f (x)

(2) Let I = ∫ f ′(x)[ f (x)]n dx

P utting f (x) u then f c(x)dx du

T hus, I ³ undu un 1 c

Therefore, ³ f c(x)[ f (x)]n dx n 1
[ f (x)]n 1 c.

n 1

Example 11.31
Integrate the following with respect to x.

(i) ∫ tan x dx (ii) ∫ cot x dx (iii) ∫ cosec x dx (iv) ∫ sec x dx

Solution Let I ³ tan x dx ³ sin x dx
(i) cos x

Putting cos x u then, sin x dx du

Thus, I 1 log | cos x | c log | sec x | c.
(ii)
³ u du log | u | c
Let I
³ cot x dx ³ cos x dx
sin x

= Putting sin x u=then, cos x dx du

1
Thus, I ³ u du log | u | c log | sin x | c.


XI - Mathematics 204

(iii) Let I = ∫ cosec xdx = ∫ cosec x(cosec x − cot x) dx
cosec x − cot x

= ∫ cosec2x − cosecx cot x dx
cosecx − cot x

Putting cosec x cot x u, then (cosec2x cosec x cot x) dx du

Thus, I = ∫ 1 du = log | u | +c = log | cosecx − cot x | +c.
u

(iv) Let I = ∫ sec xdx = ∫ sec x(sec x + tan x) dx = ∫ sec2 x + sec x tan x dx
sec x + tan x sec x + tan x

Putting sec x tan x u, then (sec2 x sec x tan x) dx du

Thus, I = ∫ 1 du = log u + c = log sec x + tan x +c
u

Therefore, ³ sec xdx log sec x tan x c.
Thus the following are the important standard results.

(1) ∫ tan x dx = log sec x + c


(2) ∫ cot x dx = log sin x + c

(3) ∫ cosecx dx = log cosec x − cot x + c

(4) ∫ sec x dx = log sec x + tan x + c

Example 11.32
Integrate the following with respect to x.

(i) ∫ x 2 2x + 4 6 dx ∫(ii) e ex 1 dx (iii) ∫ 1 dx
+ 4x + x− x log x

(iv) ∫ ssiinn x + cos x dx (v) ∫ (sin cos 2x x)2 dx
x − cos x x+ cos

Solution

(i) Let I = ∫ 2x + 4 6 dx
x2 + 4x +

P utting x2 4x 6 u, then (2x 4) dx du

Thus, I ³ du log u c log x2 4x 6 c
u

³Therefore, 2x 4 6 dx = log x2 + 4x + 6 + c.
x2 4x

205 Integral Calculus

(ii) ∫Let I = e ex 1 dx.
x−

Putting ex 1 u, then exdx du

(iii) Thus, I = ∫ du = log u +c = log ex −1 +c
u

³Therefore, ex dx = log(ex −1) + c.
ex 1

Let I = ∫ x 1 x dx.
log

1
P =utting log x u=, then x dx du

³Thus, I du log u c log log x c
u
Therefore, ò 1 dx = log | log x | +c.
x log x

(iv) Let I = ∫ sin x + cos x dx.
sin x − cos x

Putting sin x cos x u, then (cos x sin x) dx du

ccToohssuxxsd, xI == l∫odguu| s=inloxg− u+ c = log sin x − cos x +c
Therefore, cos x | +c
³ sin x
sin x

(v) Let I = ∫ (sin cos 2x x)2 dx = ∫ cos 2x dx.
x+ cos 1+ sin 2x

Putting 1+ sin 2x = u, then 2 cos 2x dx = du

Thus, I = ∫ du = 1 log u +c = 1 log 1+ sin 2x + c.
2u 2 2

EXERCISE 11.6

Integrate the following with respect to x

(1) x (2) x2 (3) ex − e−x (4) 10x9 +10x loge 10
1+ x2 1+ x6 ex + e−x 10x + x10

(5) sin x (6) cot x (7) cosecx (8) sin 2x
x log(sin x) a2 + b2 sin2 x
log  tan x 
(9) sin−1 x  2 
1− x2
(10) x (11) 1 (12) αβ xα −1e−β xα
(13) tan x sec x 1+ x x log x log(log x)

(14) x(1− x)17 (15) sin5x cos3 x (16) cos x
cos(x − a)

XI - Mathematics 206

11.7.5 Integration by parts

Integration by parts method is generally used to find the integral when the integrand is a product
of two different types of functions or a single logarithmic function or a single inverse trigonometric
function or a function which is not integrable directly. From the formula for derivative of product of
two functions we obtain this useful method of integration.

If u and v are two differentiable functions then we have

d (uv) vdu udv
udv d (uv) vdu

Integrating

³ udv ³ d(uv) ³ vdu
³ udv uv ³ vdu

∫ udv in terms of another integral ∫ vdu and does not give a final expression for the integral

∫ udv . It only partially solves the problem of integrating the product u dv . Hence the term ‘Partial

Integration’ has been used in many European countries. The term “Integration by Parts” is used in
many other countries as well as in our own.
The success of this method depends on the proper choice of u
(i) If integrand contains any non integrable functions directly from the formula, like logx,

tan−1 x etc., we have to take these non integrable functions as u and other as dv.
(ii) If the integrand contains both the integrable function, and one of these is xn (where n is a

positive integer) then take u = xn.

(iii) For other cases the choice of u is ours.

Example 11.33
Evaluate the following integrals

(i) ∫ xexdx (ii) ∫ x cos x dx (iii) ∫ log x dx ∫(iv) sin−1 x dx

Solution

(i) Let I = ∫ xexdx.

Since x is an algebraic function and ex is an exponential function,

so take u = x then du = dx

dv = exdx ⇒ v = ex

Applying Integration by parts, we get

∫ udv = uv − ∫ vdu

∫ ∫ ⇒ xexdx = xex − exdx

∫ That is, xexdx = xex − ex + c.

(ii) Let I = ∫ x cos x dx

Since x is an algebraic function and cos x is a trigonometric function,

207 Integral Calculus

so take u = x then du = dx

dv = cos xdx ⇒ v = sin x

Applying Integration by parts, we get

∫ udv = uv − ∫ vdu

⇒ ∫ x cos xdx = x sin x − ∫ sin x dx

⇒ ∫ x cos xdx = x sin x + cos x + c

(iii) Let I = ∫ log x dx

Take u = log x then du = 1 dx
dv = dx ⇒ v = x x

Applying Integration by parts, we get

∫ udv = uv − ∫ vdu

⇒ ∫ log x dx = x log x − ∫ x 1 dx
x

⇒ ∫ log x dx = x log x − x + c

(iv) ∫Let I = sin−1 x dx

u = sin−1(x), dv = dx

Then du = 1 , v = x
1− x2

∫ ∫sin−1 xdx = x sin−1 x − x dx
1− x2

x sin−1 1 dt , where t = 1− x2
∫ ∫sin−1 xdx = x + 2 t

= x sin−1 x + t + c

= x sin−1 x + 1− x2 + c

Example 11.34

∫ Evaluate : tan −1  2x  dx
 1− x2 
Solution

∫Let I = tan −1  2 x sedcx2
tan θ  − x
Putting x = 1 2

⇒ dx = θ dθ

Therefore, ∫I = tan −1  2 tanθ  sec2 θ dθ
 1− tan2 θ 

XI - Mathematics 208

∫ = tan−1(tan 2θ ) sec2 θ dθ 1+ x2 x

= ³ 2T sec2 T dT θ
= 2³ (T )(sec2 T dT ) 1

Applying integration by parts tanθ = x
secθ = 1+ x2
I = 2 θ tanθ − ∫ tanθ dθ 

= 2(θ tanθ − log secθ ) + c

³ tan 1 § 2x · dx = 2x tan 1 x 2 log | 1 x2 | c
¨© 1 x2 ¸¹

11.7.6 Bernoulli’s formula for Integration by Parts

If u and v are functions of x, then the Bernoulli’s rule is

∫ udv = uv − u′v1 + u′′v2 − 

where u′,u′′, u′′′,... are successive derivatives of u and

v, v1, v2, v3, are successive integrals of dv
Bernoulli’s formula is advantageously applied when u = xn ( n is a positive integer)
For the following problems we have to apply the integration by parts two or more times to find the
solution. In this case Bernoulli’s formula helps to find the solution easily.

Example 11.35 dv e5xdx
Integrate the following with respect to x.

(i) x2e5x (ii) x3 cos x (iii) x3e−x u x2 v e5x
5
Solution
e5x
∫ (i) x2e5xdx. uc 2x v1 52

Applying Bernoulli’s formula ucc 2 v2 e5x
53
∫ udv = uv − u′v1 + u′′v2 −
uccc 0 v3 e5x
54
§ e5x · § e5x · § e5x · § e5x ·
³ x2e5x dx x2 ¨ 5 ¸ (2x) ¨ 52 ¸ (2) ¨ 53 ¸ (0) ¨ 54 ¸ 0  0 c
© ¹ © ¹ © ¹ © ¹

= x2e5x − 2xe5x + 2e5x + c. dv = cos x dx
5 25 125
u = x3, v = sin x
(ii) ∫ x3 cos x dx.
u′ = 3x,2 v1 = − cos x
Applying Bernoulli’s formula u′′ = 6x, v2 = − sin x
u′′′ = 6, v3 = cos x
∫ udv = uv − u′v1 + u′′v2 −

∫ x3 cos x dx = ( x3 )(sin x) − (3x2 )(− cos x)

+ (6x)(−sin x) − (6)(cos x) + c

= x3 sin x + 3x2 cos x − 6x sin x − 6 cos x + c.

209 Integral Calculus

∫ (iii) x3e−xdx. dv = e−xdx
u = x3, v = −e−x
Applying Bernoulli’s formula u′ = 3x2 , v1 = +e−x
u′′ = 6x, v2 = −e−x
∫ udv = uv − u′v1 + u′′v2 − u′′′ = 6 , v3 = e−x

( )( ) ( )( )∫ x3e−x dx = x3 −e−x − 3x2 e−x
+ (6x)(−e−x ) − (6)(e−x ) + c

= − x3e−x − 3x2e−x − 6xe−x − 6e−x + c.

EXERCISE 11.7

Integrate the following with respect to x:

(1) (i) 9xe3x (ii) x sin 3x (iii) 25xe−5x (iv) x sec x tan x
(iv) x3 sin x
(2) (i) x log x (ii) 27 x2e3x (iii) x2 cosx

(3) (i) x sin−1 x (ii) x5ex2 (iii) tan −1  8x  (iv) sin −1  2x 
1− x2  1−16x2   1+ x2 

11.7.8 Integrals of the form (i) ∫eax sinbxdx (ii) ∫eax cosbx dx

The following examples illustrate that there are some integrals whose integration continues

forever. Whenever we integrate function of the form eax cos bx or eax sin bx, we have to apply the

Integration by Parts rule twice to get the similar integral on both sides to solve.

Result 11.1

∫ eax sin bx dx eax [a bx b bx] c
(i) = a2 + b2 sin − cos +

∫ (ii) eax cos bx dx = a eax [a cos bx + b sin bx] + c
2 + b2

Proof : (i) ∫Let I = eax sin bx dx



Take u = sin bx; du = b cos bx dx

dv = eax ; v = eax ,
a
Applying Integration by parts, we get

= ∫I eax sin bx − eax b cos bx dx
a a

∫I = eax sin bx − b eax cos bx dx
a a

Take u = cos bx; du = −b sin bx dx,

dv = eax ; v = eax ,
a
Again applying integration by parts, we get

∫I eax − b  eax + eax b dx
= a sin bx a  a cos bx a sin bx

XI - Mathematics 210

∫I eax bx b eax b2 eax sin bx dx
= a sin − a a cos bx − a2

I = eax sin bx − b eax cos bx − b2 I
a a2 a2

1 + b2  I = aeax sin bx − beax cos bx
 a2  a2


 a2 + b2  I = eax[a sin bx − b cos bx]
 a2  a2
 

Therefore, I = a eax [a sin bx − b cos bx] + c
2 + b2

∫ Therefore, eax sin bx dx = eax [a sin bx − b cos bx] + c
a2 + b2

∫ eax cos bx dx = eax [a cos bx b sin bx] c
Similarly, a2 + b2 + +

∫ eax sin bx dx = eax [a sin bx − b cos bx] + c
a2 + b2

∫ eax cos bx dx = a eax [a cos bx + b sin bx] + c
2 + b2

Caution

In applying integration by parts to specific integrals, the pair of choice for u and dv once initially
assumed should be maintained for the successive integrals on the right hand side. (See the above two
examples). The pair of choice should not be interchanged.

Examples 11.36
Evaluate the following integrals

∫ (i) e3x cos 2x dx ∫(ii) e−5x sin 3x dx

∫ (i) e3x cos 2x dx

Using the formula

∫ eax cos bx dx = a eax 2 [a cos bx + b sin bx] + c
2 +b

For a = 3 and b = 2 , we get

∫ e3x cos 2x dx =  e3x  ( 3 cos 2x + 2 sin 2x) + c
 32 + 22 


=  e3x  (3 cos 2x + 2 sin 2x) + c.
 13 


211 Integral Calculus

∫(ii) e−5x sin 3x dx

Using the formula

∫ eax sin bx dx = eax [a sin bx − b cos bx] + c
a2 + b2

for a = −5, b = 3 , we get

∫ e−5x sin 3x dx =  e −5 x 32  ( −5 sin 3x − 3cos 3x) + c
 
(−5)2 +

∫ e−5x sin 3x dx = −  e −5 x  ( 5 sin 3x + 3cos 3x) + c.
 34 
 

EXERCISE 11.8

Integrate the following with respect to x

(1) (i) eax cos bx (ii) e2x sin x (iii) e−x cos 2x

(2) (i) e−3x sin 2x (ii) e−4x sin 2x (iii) e−3x cos x

Result 11.2 ∫ ex[ f (x) + f ′(x)]dx = ex f (x) + c

Proof

Let I = ∫ ex[ f (x) + f ′(x)]dx

= ∫ ex f (x)dx + ∫ ex f ′(x)dx
Take u = f (x); du f c(x) dx, in the first integral

dv = ex ; v = ex ,

∫ ∫That is, I = ex f (x) − ex f ′(x)dx + ex f ′(x)dx + c

Therefore, I = ex f (x) + c.

Examples 11.37
Evaluate the following integrals

∫ (i) ex  1 1  ex (iii) ex  11+−xx2 2
 x x2  dx ∫ ∫(ii) (sin x  dx
− + cos x)dx

Solution

(i) ∫Let I = ex  1 − 1  dx
 x x2 

Take f(x) = 1, then f ′( x) = −1
x x2

XI - Mathematics 212

This is of the form ∫ ex[ f (x) + f ′(x)]dx

∫ ex  1 − 1  dx = ex 1 + c.
 x x2  x
(ii)
Let I = ∫ ex (sin x + cos x)dx

Take f(x) = sin x, then f ′(x) = cos x

This is of the form ∫ ex[ f (x) + f ′(x)]dx

∫ ex (sin x + cos x)dx = ex sin x + c.

ex  1− x 2
 1+ x2 
∫ ∫ ( )(iii) dx (1− x)2

Let I = ex 1+ x2 2 dx

( )∫ = ex 1+ x2 − 2x dx
( )1+ x2
2

 1 2x 
x  1+ x2 1+ x2 
( ) ( )∫ = e  −  dx
2

 

If f ( x ) = (1 1 ) , then f ′ ( x ) = − (1 2x )2
+ x2 + x2

Using ∫ ex ( f ( x) + f ′( x)) dx = ex f ( x) + c

 1− x 2 § 1 2x · ex 1 c.
 1+ x2  ¨ 1 x2 1 x2 ¸ 1 x2
∫ ³ex dx = e x ¨ ¸ dx

2

© ¹

EXERCISE 11.9

Integrate the following with respect to x:

(1) ex (tan x + log sec x) (2) ex  x −1 (3) ex sec x(1+ tan x)
 2x2 

(4) ex  2 + sin 2x  (5) etan−1 x 1+ x + x2  (6) log x
 1+ cos 2x    (1+ log x)2
 1+ x2 

11.7.9 Integration of Rational Algebraic Functions

In this section we are going to discuss how to integrate the rational algebraic functions whose

numerator and denominator contains some positive integral powers of x with constant coefficients.

213 Integral Calculus

Type I

dx dx dx , dx
a2 ± x2 x2 − a2 a2 ± x2 x2 − a2
∫ ∫ ∫ ∫ Integrals of the form, ,

(i) ∫ dx = 1 a x c
log a x
a2 − x2
2a

(ii) ∫ dx = 1 x a c
log x a
x2 − a2
2a

(iii) ∫ dx = 1 tan −1  x  + c
a  a 
a2 + x2

(iv) ∫ dx = sin −1  x  + c
 a 
(v) a2 − x2

(vi) ∫ dx = log x + x2 − a2 + c
Proof
x2 − a2

∫ dx = log x + x2 + a2 + c

x2 + a2

∫Let dx .
(i) I = a2 − x2

= ∫ (a − dx + x)
x)(a

= 1 ∫ 1 x + a 1 x  dx Resolving into
2a  a + −  partial fractions

= 1 log a + x − log a − x  + c
2a

= 1 log a+ x +c
2a a− x

³ dx = 1 log a+x +c
(ii) 2a a−x
a2 x2

∫Let dx .
I = x 2 −a 2

= ∫ (x − dx + a)
a)( x

= 1 ∫  1 a − 1 a  dx Resolving into
2a  x− x+  partial fractions

XI - Mathematics 214

= 1 [log x − a − log x + a ]+c
2a

= 1 log x−a +c
2a x+a

∫Therefore  x2 dx 2 = 1 log x−a +c
(iii) −a 2a x+a

∫Let I = a 2 dx 2 .
+x

Putting x = a tanT Ÿ T tan 1 x
a

dx = a sec2 θ

a sec2 θ dθ a sec2 θ dθ a sec2 θ dθ 1 dθ
+ a2 tan2 θ a2 (1+ tan2 θ ) a2 sec2 θ a
∫ ∫ ∫ ∫I = a2 = = =

= 1θ + c = 1 tan −1  x + c
a a  a 

³Therefore, dx = 1 tan −1  x  + c
a2 x2 a  a 

(iv) ∫Let I = dx .
a2 − x2

Putting x = a sinθ ⇒θ = sin −1  x 
 a 

dx = a cosθ

I = ∫ ∫a cosθ dθ = a cosθ θ) dθ = ∫ a cosθ dθ = ∫ dθ
a2 (1− sin2 a cosθ
a2 − a2 sin2 θ

= θ + c = sin −1  x + c
 a 

³Therefore, dx = sin −1  x  + c
a2 x2  a 

∫ (v) Let I = dx
x2 − a2

Putting x = a secθ ⇒θ = sec−1  x 
 a 

dx = a secθ tanθ dθ

I = ∫ ∫a secθ tanθ dθ = a secθ tanθ dθ = ∫ a secθ tanθ dθ = ∫ secθ dθ
a2 (sec2 θ −1) a tanθ
a2 sec2 θ − a2

= log | secT tanT | c

215 Integral Calculus

= log ax x2a a2 c x
= log x + x2 − a2 − log a + c
x2 − a2

θ
a

= log x + x2 − a2 + c1 where c1 = c − log a secθ = x
a

³ Therefore, dx = log x+ x2 − a2 + c1 tanθ = x2 − a2
x2 a2 a

(vi) ∫Let I = dx
a2 + x2

Putting x = a tan θ ⇒θ = tan −1  x 
 a 
a2 + x2

dx = a sec2 θ dθ x

I = a sec2 θ dθ = a sec2 θ dθ θ a
∫ ∫a2 tan2 θ + a2 tanθ = x
a2 (tan2 θ +1)
a

= ∫ aasseecc2θθ dθ = ∫ secθ dθ secθ = a2 + x2
a
= log | secT tanT | c

= log ax + ax22 +1 + c

= log x + x2 + a2 − log a + c

= log x + x2 + a2 + c1 where c1 = c − log a

³ dx = log x + x2 + a2 + c1
a2 x2

Remark: Remember the following useful substitution of the given integral as a functions of
a2 x2 , a2 x2 and x2 a2.

Given Substitution
a2 − x2 x = a sinq
a2 + x2 x = a tanq
x2 − a2 x = a secq

XI - Mathematics 216

Examples 11.38
Evaluate the following integrals

1 ∫ x2 dx (iii) 1 dx 1 dx
2)2 x2 + 5 1+ 4x2 4x2 − 25
(i) ∫ ( x − dx ∫(ii) (iv) ∫
+1

Solution

(i) ∫ ∫Let I = − 1 + dx = − 1 + 12 dx
(x 2)2 1 (x 2)2

Putting x - 2 = t ⇒ dx = dt

∫Thus, I = t2 1 dt = tan −1 (t ) + c = tan −1 ( x − 2) + c
(ii) + 12

Let ∫I = x2 dx
x2 + 5

= ∫ x2 + 5 − 5dx = ∫ 1 − 5 5  dx = ∫ dx − ∫ 5 5 dx
x2 + 5 x2 +  x2 +

1
x2 +
( )∫ = x − 5 5 2 dx

= x − 5 1 tan −1  x  + c
5  5 

I = x − 5 tan −1  x  + c
 5 

(iii) Let I = ∫ 1 dx = ∫ 1 dx
1+ 4x2 1+ (2x)2

Putting 2x = t ⇒ 2 dx = dt ⇒ dx = 1 dt
2

∫Thus, 1 1 dt
(iv) I = 2 12 + t 2

I = 1 log t+ t2 +1 + c = 1 log 2x + (2x)2 +1 + c
2 2

I = 1 log 2x + 4x2 +1 + c
2

Let I = ∫ 1 25 dx = ∫ 1 dx
4x2 − (2x)2 − 25

Putting 2x = t ⇒ 2 dx = dt ⇒ dx = 1 dt
2

217 Integral Calculus

∫Thus, 1 1 dt
I = 2 t2 − 52

= 12 log t + t2 − 52 + c

I = 1 log 2x + 4x2 − 25 + c
2

Type II

Integrals of the form ∫ ax2 dx and ∫ dx
+ bx + c ax2 + bx + c

First we express ax2 + bx + c as sum or difference of two square terms that is, any one of the
forms to Type I. The following rule is used to express the expression ax2 + bx + c as a sum or
difference of two square terms.

(1) Make the coefficient of x2 as unity.
(2) Completing the square by adding and subtracting the square of half of the coefficient of x.

That is, ax2 + bx + c = a  x2 + b x + c 
 a a 

= a  x + b 2 + 4ac − b2 
2a  4a2 


Examples 11.39
Evaluate the following integrals

³ 1 ∫ 1 dx (iii) ∫ 1 dx
(i) dx (ii) x2 +12x +11 12 + 4x − x2
x2 2x 5

Solution

(i) ∫ x2 − 1 + 5 dx = ∫ x2 − 1 (1)2 + 4 dx
2x 2(1)x +

∫ = (x 1 + 22 dx
−1)2

∫ x2 1 dx = 1 tan −1  x −1  + c
2x 2  2 
− + 5
(ii)
∫ x2 1 dx = ∫ 1 dx
+12x +11 (x + 6)2 − 25

1 dx
= ∫ (x + 6)2 − 52

= log | x + 6 + (x + 6)2 − 52 | +c

XI - Mathematics 218

∫Therefore, 1 dx = log | x + 6 + x2 +12x +11 | +c
x2 +12x +11

(iii) ∫ 1 x2 dx = ∫ 1 dx
12 + 4x − 12 − (x2 − 4x)

= ∫ 1 dx

12 −{(x − 2)2 − 4}

1 dx
= ∫ 42 − (x − 2)2

= sin −1  ( x − 2)  + c
 4 

EXERCISE 11.10

Find the integrals of the following :

(1) (i) 4 1 2 (ii) 1 (iii) 1
−x 25 − 4x2 9x2 − 4

(2) (i) 1 (ii) ( x 1 − 25 (iii) 1
6x − 7 − x2 + 1)2 x2 + 4x + 2

(3) (i) 1 (ii) 1 (iii) 1
(2 + x)2 −1 x2 − 4x + 5 9 +8x − x2

Type III

Integrals of the form ∫ px + q dx and ∫ px + q dx
ax2 + bx + c ax2 + bx + c

To evaluate the above integrals, first we write

px + q = A d (ax2 + bx + c) + B
dx

px + q = A(2ax + b) + B

Calculate the values of A and B, by equating the coefficients of like powers of x on both sides

(i) The given first integral can be written as

∫ px +q c dx = ∫ A(2ax + b) + B dx
ax2 + bx + ax2 + bx + c

= A∫ 2ax + b c dx + B∫ ax2 1 + c dx
ax2 + bx + + bx

(The first integral is of the form ∫ f ′(x) dx)
f (x)

∫ = Alog | ax2 + bx + c | +B ax2 1 + c dx
+ bx

The second term on the right hand side can be evaluated using the previous types.

219 Integral Calculus

(ii) The given second integral can be written as

∫ ∫px + q dx = A(2ax + b) + B dx
ax2 + bx + c ax2 + bx + c

= A∫ 2ax +b c dx + B∫ 1 dx
ax2 + bx + ax2 + bx + c

(The first integral is of the form ∫ f ′(x)[ f (x)]n dx)

( ) = A 2 ax2 + bx + c + B 1 dx
∫ ax2 + bx + c

The second term on the right hand side can be evaluated using the previous types.

Examples 11.40
Evaluate the following integrals

(i) ∫ x2 3x + 5 7 dx (ii) ∫ x2 x +1 dx (iii) ∫ 2x + 3 dx (iv) ∫ 5x − 7 dx
+ 4x + − 3x +1 x2 + x +1 3x − x2 − 2

Solution

(i) Let I = ∫ x2 3x + 5 7 dx
+ 4x +

3x + 5 = A d (x2 + 4x + 7) + B
dx

3x + 5 = A(2x + 4) + B

Comparing the coefficients of like terms, we get

2A = 3 ⇒ A = 3 ; 4A + B = 5 ⇒ B = −1
2

3 (2x + 4) −1
2
I = ∫ x2 + 4x + 7 dx

I = 3 ³ 2x 4 dx ³ x2 1 dx
2 x2 4x 7 4x 7

1
(x + 2)2 +
( )∫ = 32 log x2 + 4x + 7 − 2 dx
3

= 32 log x2 + 4x + 7 − 1 tan −1  x +2  + c
3  3 

(ii) Let I = ∫ x2 x +1 dx
− 3x +1

x + 1 = Ad (x2 − 3x +1) + B
dx

XI - Mathematics 220

x + 1 = A(2x − 3) + B

Comparing the coefficients of like terms, we get

2A = 1 ⇒ A = 1 ; −3A + B = 1 ⇒ B = 5
22

1 (2x − 3) + 5
2 2
I = ∫ x2 +1 dx
− 3x

I = 1 ∫ 2x −3 1 dx + 5 ∫ x2 1 dx
2 x2 − 3x + 2 − 3x +1

∫ = 12 log x2 − 3x +1 + 5 1 5 2 dx
2
 x − 3 2 − 
 2   2 
 

= 12 log x2 − 3x +1 + 5 1 log x−3− 5 +c
2  2 2
5  5
2  2  x − 3 + 2
 2


= 12 log x2 − 3x +1 + 5 log 2x − 3 − 5 +c
2 2x −3+ 5

(iii) Let I = ∫ 2x + 3 dx
x2 + x +1

2x + 3 = Ad (x2 + x +1) + B
dx

2x + 3 = A(2x + 1) + B

Comparing the coefficients of like terms, we get

2A = 2 ⇒ A = 1; A + B = 3 ⇒ B = 2

I = ∫ (2x +1) + 2 dx
x2 + x +1

I = ∫ 2 x +1 1 dx + 2∫ 1 dx
x2 +x+ x2 + x +1

= 2 x2 + x +1 + 2∫ 1 dx
3 2
 x + 1 2 +  
 2   2 


221 Integral Calculus

= 2 x2 + x +1 + 2 log x + 1 +  x + 1 2 +  3 2 +c
2  2   2 

Therefore, I = 2 x2 + x +1 + 2 log x + 1 + x2 + x +1 + c
2

(iv) ∫ 5x − 7 dx
3x − x2 − 2
d
5x − 7 = A dx (3x − x2 − 2) + B

5x - 7 = A(3 − 2x) + B

Comparing the coefficients of like terms, we get

−2 A = 5 ⇒ A = − 5 ;3A + B = −7 ⇒ B = 1
2 2

−5 (3 − 2x) + 1
2 2dx
I = ∫ 3x − x2 + 2

I = − 5 ∫ 3− 2x 2 dx + 1 ∫ 1 dx
2 3x − x2 + 2 3x − x2 + 2

=  − 52  2 3x − x2 + 2 + 1 1 dx
∫ 2 2
 17   3 2
 2  −  x − 2 


x2 1 −1  x − 3 
 2 
= −5 3x − + 2 + sin   + c
2  17 
2

Thus, I = −5 3x − x2 + 2 + 1 sin −1  2x −3  + c
2  17 

EXERCISE 11.11

Integrate the following with respect to x :

(1) (i) x2 2x 3 (ii) 5x 2 (iii) 3x 1
4x 12 2 2x x2 2x2 2x 3

(2) (i) 2x 1 (ii) x 2 (iii) 2x + 3
9 4x x2 x2 1 x2 + 4x +1

XI - Mathematics 222

Type IV

∫ ∫ Integrals of the form a2 ± x2 dx, x2 − a2 dx

Result 11.3

∫ (1) a2 − x2 dx = x a2 − x2 + a2 sin −1  x  + c
2  a 
2

∫ (2) x2 − a2 dx = x x2 − a2 − a2 log x + x2 − a2 + c

22

∫ (3) x2 + a2 dx = x x2 + a2 + a2 log x + x2 + a2 + c
Proof : 22

(1) ∫Let I = a2 − x2 dx

Take u = a2 − x2 then du = −2x dx
2 a2 − x2
dv = dx ⇒ v = x

Applying Integration by parts, we get

∫ udv = uv − ∫ vdu −x2 dx
∫⇒ I = x a2 − x2 − a2 − x2

= x a2 − x2 − a2 − x2 − a2 dx

∫ a2 − x2

∫ = x a2 − x2 −  a2 − x2 + (−a2 )  dx
 a2 − x2 a2 − x2 
 

∫ ∫ = x a2 − x2 − a2 dx
a2 − x2 dx + a2 − x2

∫ = x a2 − x2 − I + a2 1 dx
a2 − x2

2I = x a2 − x2 + a2 sin −1  x 
 a 

Therefore, I = x a2 − x2 + a2 sin −1  x  + c
2 2  a 

Similarly we can prove other two results.

Note 11.3
The above problems can also be solved by substituting x a sinT

Examples 11.41
Evaluate the following :

(i) ³ 4 x2 dx (ii) ³ 25x2 9 dx (iii) ∫ x2 + x +1dx (iv) ∫ (x − 3)(5 − x) dx

223 Integral Calculus

Solution Let I = 4 − x2 dx
(i)

∫ = 22 − x2 dx

= 2x 22 − x2 + 22 sin −1  x  + c
2  2 

Therefore, I = x 4 − x2 + 2 sin −1  x  + c
(ii) 2  2 

Let I = ∫ 25x2 − 9dx

∫ = (5x)2 − 32 dx

= 15  5x (5x)2 − 32 − 32 log 5x + (5x)2 − 32  + c
 2 2 


Therefore, I = 1 5x 25x2 − 9 − 9 log 5x + 25x2 − 9  + c
5  2 2 

(iii) Let I = ∫ x2 + x +1 dx

∫ =  x + 1 2 +  3 2 dx
 2   2 

1 2

§ 3·
¨ ¸
x § 1 2 § 3 2 © 2 ¹ ª 1 § 1 2 § 3 2 º
2 ©¨ 2 ¨©¨ 2 « 2 ©¨ 2 ¨©¨ 2 »
= 2 x · · log « x x · · » c
¸¹ ¸¸¹ ¹¸ ¹¸¸ ¼»
2 «¬

Therefore, I = 2x +1 x2 + x +1 + 3 log x + 1 + x2 + x +1 + c
4 82

(iv) Let I = ∫ (x − 3)(5 − x) dx

= ∫ 8x − x2 −15 dx

∫ = 12 − (x − 4)2 dx

= x − 4 12 − (x − 4)2 + 1 sin −1  x − 4 + c
2 2  1 

Therefore, I = x−4 8x − x2 −15 + 1 sin−1(x − 4) + c
2 2

XI - Mathematics 224

EXERCISE 11.12

Integrate the following functions with respect to x :

(1) (i) x2 + 2x +10 (ii) x2 − 2x − 3 (iii) (6 − x)( x − 4)

(2) (i) 9 − (2x + 5)2 (ii) 81+ (2x +1)2 (iii) (x +1)2 − 4

EXERCISE 11.13

Choose the correct or the most suitable answer from given four alternatives.

(1) If ∫ f (x)dx = g(x) + c, then ∫ f (x)g′(x)dx (4) ∫(g(x))2 dx

(1) ∫( f (x))2 dx (2) ∫ f (x)g(x)dx (3) ∫ f ′(x)g(x)dx

1 1

3x k (3x ) c, then the value of k is

(2) If ³ x2 dx (2) − log 3 (3) − 1 (4) 1
log 3 log 3
(1) log 3

∫ (3) If f ′(x)ex2 dx = (x −1)ex2 + c , then f(x) is

(1) 2x3 − x2 + x + c (2) x3 + 3x2 + 4x + c (3) x3 + 4x2 + 6x + c (4) 2x3 x2 x c
22 3

(4) The gradient (slope) of a curve at any point (x, y) is x2 − 4 . If the curve passes through the
point (2, 7), then the equation of the curve is x2

(1) y = x + 4 + 3 (2) y = x + 4 + 4 (3) y = x2 + 3x + 4 (4) y = x2 − 3x + 6
x x (4) cos(xex ) + c

∫ (5) ex (1+ x) dx is (2) sec(xex ) + c (3) tan(xex ) + c
cos2 (xex )

(1) cot(xex ) + c

(6) ò tan x dx is
sin 2x

(1) tan x + c (2) 2 tan x + c (3) 1 tan x + c (4) 1 tan x + c
2 4
(7) ∫ sin3 xdx is

(1) −3 cos x − cos 3x + c (2) 3 cos x + cos 3x + c
4 12 4 12

(3) −3 cos x + cos 3x + c (4) −3 sin x − sin 3x + c
4 12 4 12

∫ (8) e6log x − e5log x dx is
e4log x − e3log x
(2) x3 + c 3 1
(1) x + c 3 (3) x3 + c (4) x2 +c

225 Integral Calculus

(9) ∫ sec x dx is
cos 2x

(1) tan−1(sin x) + c (2) 2sin−1(tan x) + c (3) tan−1(cos x) + c (4) sin−1(tan x) + c

∫ (10) tan−1 1− cos 2x dx is

1+ cos 2x

(1) x2 + c (2) 2x2 + c (3) x2 + c (4) − x2 + c
2 2

∫ (11) 23x+5 dx is

(1) 3(23x+5 ) 23x+5 (3) 23x+5 23x+5
+ c (2) +c 2 log 3 + c (4) +c

log 2 2 log(3x + 5) 3log 2

∫ (12) sin8 x − cos8 x dx is
1− 2 sin2 x cos2 x

(1) 1 sin 2x + c (2) 1 1 cos 2x + c (4) − 1 cos 2x + c
2 sin 2x c (3) 2
22

∫ (13) ex (x2 tan−1 x + tan−1 x +1) dx is
x2 +1

(1) ex tan−1(x +1) + c (2) tan−1(ex ) + c (3) ex (tan−1 x)2 + c (4) ex tan−1 x + c
2

∫ (14) x2 + cos2 x cosec2 xdx is
x2 +1

(1) cot x + sin−1 x + c (2) −cot x + tan−1 x + c

(3) − tan x + cot−1 x + c (4) − cot x − tan−1 x + c

(15) ∫ x2 cos x dx is

(1) x2 sin x + 2x cos x − 2sin x + c (2) x2 sin x − 2x cos x − 2sin x + c

(3) −x2 sin x + 2x cos x + 2sin x + c (4) −x2 sin x − 2x cos x + 2sin x + c

(16) ∫ 1− x dx is
1+ x

(1) 1− x2 + sin−1 x + c (2) sin−1 x − 1− x2 + c

(3) log | x + 1− x2 | − 1− x2 + c (4) 1− x2 + log | x + 1− x2 | +c

(17) ³ dx is
ex 1

(1) log | ex | log | ex 1| c (2) log | ex | log | ex 1| c

(3) log | ex 1| log | ex | c (4) log | ex 1| log | ex | c

XI - Mathematics 226

∫ (18) e−4x cos x dx is

(1) e−4x [4 cos x − sin x] + c (2) e−4x [−4 cos x + sin x] + c
17 17

(3) e−4x [4 cos x + sin x] + c (4) e−4x [−4 cos x − sin x] + c
17 17

³ (19) sec2 x 1 dx
tan2 x

(1) 2 log 1 tan x c (2) log 1 tan x c
1 tan x 1 tan x

(3) 1 tan x 1 c (4) 1 tan x 1 c
log tan x 1 log tan x 1

2 2

³ (20) e 7x sin 5x dx is

(1) e−7x [−7 sin 5x − 5cos 5x] + c (2) e−7x [7 sin 5x + 5cos 5x] + c
74 74

(3) e−7x [7 sin 5x − 5cos 5x] + c (4) e−7x [−7 sin 5x + 5cos 5x] + c
74 74
x

∫ (21) x2e2dx is
x xx x xx

(1) x2e2 − 4xe2 − 8e2 + c (2) 2x2e2 − 8xe2 −16e2 + c

x xx

(3) x x x + c (4) x2 e2 − xe 2 + e2 +c

2x2e2 − 8xe2 + 16e 2 248

(22) ∫ x + 2 dx is
x2 −1

(1) x2 −1 − 2 log | x + x2 −1 | +c (2) sin−1 x − 2 log | x + x2 −1 | +c

(3) 2 log | x + x2 −1 | − sin−1 x + c (4) x2 −1 + 2 log | x + x2 −1 | +c

(23) ∫ x 1 dx is
(log x)2 − 5

(1) log | x + x2 − 5 | +c (2) log | log x + log x − 5 | +c

(3) log | log x + (log x)2 − 5 | +c (4) log | log x − (log x)2 − 5 | +c

(24) ∫ sin xdx is

( ) ( ) (1) 2 − x cos x + sin x + c (2) 2 − x cos x − sin x + c
( ) ( ) (3) 2 − x sin x − cos x + c (4) 2 − x sin x + cos x + c

(25) ò e x dx is

(1) 2 x (1− e x ) + c (2) 2 x (e x −1) + c

(3) 2e x (1− x ) + c (4) 2e x ( x −1) + c

227 Integral Calculus

SUMMARY

Derivatives Antiderivatives

d (c) = 0, where c is a constant ∫ 0 dx = c , where c is a constant
dx

d (kx) = k, where k is a constant ∫ k dx = kx + c where c is a constant
dx

d  x n+1  = xn ∫ xndx = xn+1 + c, n ≠ −1 (Power rule)
dx  n +1 
  n +1

d log x =  1  ∫ 1 dx = log x +c
dx  x  x

d (− cos x ) = sin x ∫ sin x dx = − cos x + c
dx

d (sin x ) = cos x ∫ cos x dx = sin x + c
dx

d ( tan x) = sec2x ∫ sec2 x dx = tan x + c

dx

d (− cot x) = cosec2x ∫ cosec2x dx = − cot x + c

dx

d (sec x ) = sec x tan x ∫ sec x tan x dx = sec x + c
dx

d (−cosec x) = cosec x cot x ∫ cosec x cot x dx = −cosec x + c

dx

( )d ex = ex ∫ exdx = ex + c
∫ axdx = ax + c
dx
log a
d  ax  = a x
dx  log  ∫ 1 dx = sin−1 x + c
 a 
1− x2
( )d sin−1 x = 1

dx 1− x2

( )d 1 ∫1 tan −1
+x 1+ x2
dx
tan−1 x = 1 2 dx = x + c

XI - Mathematics 228

(1) If k is any constant, then ∫ kf (x)dx = k ∫ f (x) dx
∫ ∫ ∫(2) ( f1(x) ± f2 (x) ) dx = f1(x)dx ± f2 (x)dx

If ³ f (x) dx g(x) c, then ³ f (ax b) dx 1
g(ax b) c
a

(1) ∫ tan x dx = log sec x + c

(2) ∫ cot x dx = log sin x + c

(3) ∫ cosecx dx = log cosec x − cot x + c

(4) ∫ sec x dx = log sec x + tan x + c

Bernoulli’s formula for integration by Parts:
If u and v are functions of x, then the Bernoulli’s rule is

∫ udv = uv − u′v1 + u′′v2 + ...

where u′, u′′, u′′′,... are successive derivatives of u and
v, v1, v2, v2,..., are successive integrals of dv

∫ eax sin bxdx = a eax [a sin bx − b cos bx] + c
2 + b2

∫ eax cos bxdx = eax 2 [a cos bx + b sin bx] + c
2 +b
a

³ dx 1 § a x · ³ dx = sin 1 § x · c
a2 x2 2a ¨© a x ¸¹ a2 x2 ©¨ a ¸¹
= log c

³ dx = 1 log § x a · c ³ dx = 1 tan 1 § x · c
x2 a2 2a ¨© x a ¸¹ a2 x2 a ©¨ a ¹¸

³ dx = log x x2 a2 c ³ dx = log x+ x2 + a2 + c
x2 a2 x2 a2

³ a2 x2 dx x a2 x2 a2 sin 1 § x · c
2 2 ¨© a ¹¸

³ x2 a2 dx x x2 a2 a2 log x x2 a2 c

22

³ x2 a2 dx x x2 a2 a2 log x x2 a2 c

22

229 Integral Calculus

ICT CORNER 11(a)

Integral Calculus

Expected Outcome

Step 1
Open the Browser type the URL Link given below (or) Scan the QR Code.
GeoGebra Workbook called “XI standard Integration” will appear. In that
there are several worksheets related to your lesson.

Step 2
Select the work sheet “Simple Integration”. You can enter any function in
the f(x)box. Graph of f(x) appear on left side and the Integrated function will
appear on right side. (Note: for x5 enter x^5) Move the slider “integration
constant” to change the constant value in integration.

Step1 Step2

Browse in the link:
XI standard Integration: https://ggbm.at/c63hdegc

XI - Mathematics 230

ICT CORNER 11(b)

Integral Calculus

Expected Outcome

Step 1
Open the Browser type the URL Link given below (or) Scan the QR Code.
GeoGebra Workbook called “XI standard Integration” will appear. In that
there are several worksheets related to your lesson.

Step 2
Select the work sheet “Algebraic type-1”. The graph of the function given on
left side and the Integration of the function appear on right side. Click on both
to see the graph. You can move the slider “a” to change the value. Algebraic
types are grouped as 4 types open other three Algebraic types and observe.

Step1 Step2

Browse in the link: Integral Calculus
XI standard Integration: https://ggbm.at/c63hdegc

231

Chapter 12 Introduction to
Probability Theory

The most important questions of life are, indeed, for the most part,
really only problems of probability
Pierre - Simon Laplace

12.1 Introduction

A gambler’s dispute in 1654 led to the creation of a mathematical
theory of probability by two famous French mathematicians, Blaise
Pascal and Pierre de Fermat. The fundamental principles of probability
theory were formulated by Pascal and Fermat for the first time. After an
extensive research, Laplace published his monumental work in 1812,
and laid the foundation to Probability theory. In statistics, the Bayesian
interpretation of probability was developed mainly by Laplace.

T he topic of probability is seen in many facets of the modern world.

Laplace From its origin as a method of studying games, probability has involved
1749-1827 in a powerful and widely applicable branch of mathematics. The uses of
probability range from the determination of life insurance premium, to the

prediction of election outcomes, the description of the behaviour of molecules in a gas. Its utility is

one good reason why the study of probability has found in the way into a school textbook.

The interpretation of the word ‘probability’ involves synonyms such as chance, possible, probably,

likely, odds, uncertainty, prevalence, risk, expectancy etc.

Our entire world is filled with uncertainty. We make decisions affected by uncertainty virtually
every day. In order to measure uncertainty, we turn to a branch of mathematics called theory of
probability. Probability is a measure of the likeliness that an event will occur.

Learning Objectives

On completion of this chapter, the students are expected to

• understand the classical theory of probability and axiomatic approach to probability.
• understand mutually exclusive, mutually inclusive and exhaustive events.
• understand the concepts of conditional probability and independent events.
• apply Bayes’ theorem.
• apply probability theory in day-to-day life.

12.2 Basic definitions

Before we study the theory of probability, let us recollect the definition of certain terms already
studied in earlier classes, which are frequently used.

XI - Mathematics 232

EXPERIMENT

Deterministic Experiment Random Experiment

(Genetic determination) (Hitting the target)

Definition 12.1
An experiment is defined as a process for which its result is well defined.

Definition 12.2
Deterministic experiment is an experiment whose outcomes can be predicted with certain,
under ideal conditions.

Definition 12.3
A random experiment (or non-deterministic) is an experiment
(i) whose all possible outcomes are known in advance,
(ii) whose each outcome is not possible to predict in advance, and
(iii) can be repeated under identical conditions.
A die is ‘rolled’, a fair coin is ‘tossed’ are examples for random experiments.

Definition 12.4
A simple event (or elementary event or sample point) is the most basic possible outcome of
a random experiment and it cannot be decomposed further.

Definition 12.5

A sample space is the set of all possible outcomes of a random experiment. Each point in
sample space is an elementary event.

Illustration 12.1

(1) (i) If a die is rolled, then the sample space S = {1, 2, 3, 4, 5, 6}

(ii) A coin is tossed, then the sample space S = {H , T}

(2) (i) Suppose we toss a coin until a head is obtained. One cannot say in advance how many
tosses will be required, and so the sample space.

S = {H , TH , TTH ,TTTH ,...} is an infinite set.

(ii) The sample space associated with the number of passengers waiting to buy train tickets

in counters is S =^0,1,2,...` .

(3) (i) If the experiment consists of choosing a number randomly between 0 and 1, then the
sample space is S = { x: 0< x <1}.

(ii) The sample space for the life length (t in hours) of a tube light is
S = { t: 0 < t<1000}.

233 Introduction to Probability Theory

From (2) and (3), one need to distinguish between two types of infinite sets, where one type is
significantly ‘larger’ than the other. In particular, S in (2) is called countably infinite, while the S in
(3) is called uncountably infinite. The fact that one can list the elements of a countably infinite set
means that the set can be put in one-to-one correspondence with natural numbers  . On the other
hand, you cannot list the elements in uncountable set.
From the above example, one can understand that the sample space may consist of countable or
uncountable number of elementary events.

Number of sample points
or elementary events in a sample space

Countable number of Uncountable number of sample points
sample points S = {x : 0 < x < 1}

Finite number of sample points Countably Infinite number of sample points
S ={1, 2,3, 4,5, 6} S = {H ,TH ,TTH ,TTTH ,...}

12.3 Finite sample space

In this section we restrict our sample spaces that have at most a finite number of points.

Types of events

Let us now define some of the important types of events, which are used frequently in this

chapter.

• Sure event or certain event • Impossible event

• Complementary event • Mutually exclusive events

• Mutually inclusive event • Exhaustive events

• Equally likely events • Independent events (defined after learning the concepts of

probability)

Definition 12.6
When the sample space is finite, any subset of the sample space is an  event. That is, all

elements of the power set (S )of the sample space are defined as events. An event is a collection
of sample points or elementary events.
The sample space S is called sure event or certain event. The null set ∅ in S is called an
impossible event.

Definition 12.7
For every event A, there corresponds another event A is called the complementary event to
A. It is also called the event ‘not A’.

Illustration 12.2
Suppose a sample space S is given by S = {1,2,3,4}.

Let the set of all possible subsets of S (the power set of S) be (S).

(S) = {∅, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4},
{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}

XI - Mathematics 234

(i) All the elements of (S ) are events.
(ii) ∅ is an impossible event.
(iii) {1},{2},{3},{4} are the simple events or elementary events.
(iv) {1, 2, 3, 4}is a sure event or certain event.

Definition 12.8
Two events cannot occur simultaneously are mutually exclusive events. are
mutually exclusive or disjoint events means that, Ai ∩ Aj = ∅ , for i ≠ j. A1, A2 , A3 ,..., Ak

Definition 12.9
Two events are mutually inclusive when they can both occur simultaneously.
A1, A2 , A3 ,..., Ak are mutually inclusive means that, Ai ∩ Aj ≠ ∅ , for i ≠ j

Illustration 12.3
When we roll a die, the sample space S = {1,2,3,4,5,6}.

(i) Since{1, 3}∩{2, 4, 5, 6}=∅, t he events {1,3}and{2, 4,5,6}are mutually exclusive events.

(ii) The events {1,6,},{2,3,5} are mutually exclusive.

(iii) The events {2,3,5},{5,6} are mutually inclusive, since {2, 3, 5}∩{5, 6}={5} ≠ ∅

Definition 12.10
A1, A2 , A3 ,..., Ak are called exhaustive events if, A1 ∪ A2 ∪ A3 ∪∪ Ak = S

Definition 12.11
A1, A2, A3 ,..., Ak are called mutually exclusive and exhaustive events if,
(i) Ai ∩ Aj ≠ ∅ , for i ≠ j (ii) A1 ∪ A2 ∪ A3 ∪ ∪ Ak = S

Illustration 12.4
When a die is rolled, sample space S = {1,2,3,4,5,6}.
Some of the events are {2,3},{1,3,5},{4,6},{6} and{1,5}.

(i) Since {2, 3}∪{1, 3, 5}∪{4, 6} = {1, 2, 3, 4, 5, 6} = S (sample space), the events

{2,3},{1,3,5},{4,6} are exhaustive events.
(ii) Similarly {2,3},{4,6}and{1,5} are also exhaustive events.
(iii) {1,3,5},{4,6},{6} and{1,5} are not exhaustive events.

(Since {1, 3, 5}∪{4, 6}∪{6}∪{1, 5} ≠ S )
(iv) {2,3},{4,6},and{1,5} are mutually exclusive and exhaustive events, since

{2,3}ˆ{4, 6} ‡, {2,3}ˆ{1,5} ‡,{4, 6}ˆ{1,5} ‡ and {2, 3}∪ {4, 6}∪ {1, 5} = S

Types of events associated with sample space are easy to visualize in terms of Venn diagrams,

as illustrated below. S SS

S

AB AB A A

BB

A and B are A and B are A and B are A and B are
Mutually exclusive Mutually inclusive Mutually exclusive Mutually inclusive

and exhaustive and exhaustive

Definition 12.12

The events having the same chance of occurrences are called equally likely events.

235 Introduction to Probability Theory

Example for equally likely events: Suppose a fair die is rolled.

Number on the face 1 2 3 4 5 6 4 5
Chance of occurrence 1 1 1 1 1 1 1 364
51
2

Example for not equally likely events: A colour die is shown in figure is rolled.

Colour on the face
Chance of occurrence 1 1 1 2 1

Similarly, suppose if we toss a coin, the events of getting a head or a tail are equally likely.

Methods to find sample space
Illustration 12.5
Two coins are tossed, the sample space is

(i) S = {H ,T}×{H ,T} = {(H , H ), (H ,T ), (T , H ), (T ,T )} or {HH , HT ,TH ,TT}
(ii) If a coin is tossed and a die is rolled simultaneously, then the sample space is

S = {H ,T}×{1, 2,3, 4,5, 6} = {H1, H 2, H 3, H 4, H 5, H 6,T1,T 2,T 3,T 4,T 5,T 6} or

S = {(H ,1), (H , 2), (H ,3), (H , 4), (H ,5), (H , 6), (T ,1), (T , 2), (T ,3), (T , 4), (T ,5), (T , 6)}.

Also one can interchange the order of outcomes of coin and die. The following table gives the
sample spaces for some random experiments.

Random Total Sample space
Experiment Number of
Outcomes

Tossing a fair coin 21 = 2 {H , T}
Tossing two coins 22 = 4 {HH , HT , TH , TT}
Tossing three coins 23 = 8
Rolling fair die 61 = 6 {HHH , HHT , HTH , THH , HTT , THT , TTH , TTT}
{1, 2,3, 4, 5,6}

Rolling 62 = 36 {(1,1) ,(1,2), (1,3), (1,4), (1,5), (1,6),
Two dice
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
or (3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
single die two times. (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2),(6,3), (6,4), (6,5), (6,6)}

Drawing a card 521 = 52 Heart ♥ A 2 3 4 5 6 7 8 9 10 J Q K Red in colour
from a pack of 52 Diamond ♦ A 2 3 4 5 6 7 8 9 10 J Q K Red in colour
playing cards Spade ♠ A 2 3 4 5 6 7 8 9 10 J Q K Black in colour
Club ♣ A 2 3 4 5 6 7 8 9 10 J Q K Black in colour
XI - Mathematics
236

Notations Priori : Knowledge which
Let A and B be two events. precedes from theoretical
(i) A ∪ B stands for the occurrence of A or B or both. deduction or making
(ii) A ∩ B stands for the simultaneous occurrence of assumption. Not from
A and B. A ∩ B can also be written as AB experience or observation
(iii) A or A′ or Ac stands for non-occurrence of A
(iv) ( A ∩ B) stands for the occurrence of only A.

12.4 Probability

12.4.1 Classical definition (A priori) of probability Posteriori :
(Bernoulli’s principle of equally likely) Knowledge which

Earlier classes we have studied the frequency (A posteriori) precedes from
definition of probability and the problems were solved. Now experience or
let us learn the fundamentals of the axiomatic approach to observation.

probability theory.

The basic assumption of underlying the classical theory is that the outcomes of a random
experiment are equally likely. If there are n exhaustive, mutually exclusive and equally likely
outcomes of an experiment and m of them are favorable to an event A, then the mathematical

probability of A is defined as the ratio m . In other words, P( A) = m .
n n

Definition 12.13
Let S be the sample space associated with a random experiment and A be an event. Let n(S)
and n(A) be the number of elements of S and A respectively. Then the probability of the event A
is defined as

P( A) = n( A) = Number of cases favourable to A
n(S) Exhaustive number of cases in S

Every probabilistic model involves an underlying process is shown in the following figure.

Event A Probability P(A) P(C)
Event B P(B)

Random Event C
experiment

Sample space S Collection of subsets Event A Event B Event C

The classical definition of probability is limited in its application only to situations where there are
a finite number of possible outcomes. It mainly considered discrete events and its methods were mainly
combinatorial. This renders it inapplicable to some important random experiments, such as ‘tossing a coin
until a head appears’ which give rise to the possibility of infinite set of outcomes. Another limitation of the
classical definition was the condition that each possible outcome is ‘equally likely’.

237 Introduction to Probability Theory

These types of limitations in the classical definition of probability led to the evolution of the
modern definition of probability which is based on the concept of sets. It is known an axiomatic
approach.
The foundations of the Modern Probability theory were laid by Andrey
Nikolayevich Kolmogorov, a Russian mathematician who combined the notion
of sample space introduced by Richard von Mises, and measure theory and
presented his axiomatic system for probability theory in 1933. We introduce the
axiomatic approach proposed by A.N. Kolmogorov. Based on this, it is possible to
construct a logically perfect structure of the modern theory of probability theory.
The classical theory of probability is a particular case of axiomatic probability.
The axioms are a set of rules, which can be used to prove theorems of probability. A.N. Kolmogorov

12.4.2 Axiomatic approach to Probability

Axioms of probability
Let S be a finite sample space, let (S) be the class of events, and let P be a real valued

function defined on (S ) Then P( A) is called probability function of the event A , when the

following axioms are hold:

[P1] For any event A, P( A) ≥ 0 (Non-negativity axiom)
[P2 ] For any two mutually exclusive events
= P( A) + P(B) ((NAdodrmitiavliitzyataixoinomax)iom)
[P3] For the certain event P(A∪ B)
P(S) = 1

Note 12.1

(i) 0 ≤ P( A) ≤ 1

(ii) If A1, A2, A3,..., An are mutually exclusive events in a sample space S, then
P( A1 ∪ A2 ∪ A3 ∪∪ An ) = P( A1) + P( A2 ) + P( A3) + + P( An )

Theorems on finite probability spaces (without proof)
When the outcomes are equally likely Theorem 12.1 is applicable, else Theorem 12.2 is applicable.

Theorem 12.1

Let S be a sample space and, for any subset A of S, let P( A) = n( A) . Then P(A) satisfies axioms
n(S )
of probability [P1],[P2 ], and [P3] .

Theorem12.2

Let S be a finite sample space say S = {a1, a2, a3,..., an}. A finite probability space is obtained

by assigning to each point ai in S a real number pi , is called the probability of ai , satisfying the
following properties:
(i) Each pi ≥ 0 . ∑(ii) The sum of the pi is 1, that is, pi = p1 +p2 + p3 ++ p1 = 1 .

If the probability P( A), of an event [P1] is defined as the sum of the probabilities of the points in A ,

then the function P( A) satisfies the faixniiotemssamofpplerosbpaabcielityan[dPt1h],e[iPr2a]s,siagnnded[pPr3o] .babilities are given
Note: Sometimes the points in a
in the form of a table as follows:
Outcome a1 a2 a3 … an

Probability P1 P2 P3 … Pn

XI - Mathematics 238

Here is an illustration of how to construct a probability law starting from some common sense
assumptions about a model.

Illustration 12.6

(1) Let S = {1, 2,3}. Suppose (S)is the power set of S, and P( A) = n( A) .
n(S )
P ({1}) = 1, P ({2}) = 1, ({3}) 1
Then 3 3 and P = 3 ,

satisfies axioms of probability [P1],[P2 ], and [P3] . Here all the outcomes are equally likely.

(2) Let S = {1, 2,3}.Suppose (S)is the power set of S ,
If the probability P( A), of an event[P1] of S is defined as the sum of the probabilities of the
points in A ,

then P ({1}) = 1, P ({2}) = 1 , P ({3}) = 1 ,
2 4 4
satisfy the axioms of probability [P1],[P2 ], and [P3] .

(3) Let S = {1, 2,3} and (S) is the power set of S . If the probability P( A), of an event [P1] of

S is defined as the sum of the probabilities of the points in A ,

then P ({1}) = 0, P ({2}) = 1 , and P ({3}) = 1− 1 ,
22
and [P3 ] .
satisfy the above axioms [P1],[P2 ],


In (2) and (3), the outcomes are not equally likely.

Note 12.2
Irrational numbers also can act as probabilities.

Classroom Activity: Each student to flip a coin10 times,
Calculate: p = Number times heads occur

10
Find the cumulative ratio of heads to tosses. As number of tosses increases p → 1

2

Example 12.1
If an experiment has exactly the three possible mutually exclusive outcomes A, B, and C,
check in each case whether the assignment of probability is permissible.

(i) P( A) = 4 , P(B) = 1 , P(C) = 2 .
7 7 7

(ii) P( A) = 2 , P(B) = 1 , P(C ) = 3 .
5 5 5

(iii) P( A) = 0.3 , P(B) = 0.9 , P(C) = − 0.2.

11 P(C) 0.
(iv) P( A) , P(B) 1 , P(C) = 0.042.

33
(v) P( A) = 0.421, P(B) = 0.527

239 Introduction to Probability Theory

Solution

Since the experiment has exactly the three possible mutually exclusive outcomes A, B

and C, they must be exhaustive events. S
⇒ S = A ∪ B ∪ C

Therefore, by axioms of probability A

P( A) ≥ 0, P(B) ≥ 0, P(C) ≥ 0 and B
C
P ( A ∪ B ∪ C ) = P(A) + P(B) + P(C) P S 1

(i) Given that P( A) = 4 ≥ 0 , P(B) = 1 ≥ 0 , and P(C) = 2 ≥ 0
7 7 7

Also P (S ) = P( A) + P(B) + P(C) = 4 + 1 + 2 = 1
77 7

Therefore the assignment of probability is permissible.

(ii) Given that P( A) = 2 ≥ 0, P(B) = 1 ≥ 0 , and P(C) = 3 ≥ 0
5 5 5

But P (S ) = P( A) + P(B) + P(C) = 2 + 1 + 3 = 6 > 1
5 55 5

Therefore the assignment is not permissible.

(iii) Since P(C) = − 0.2 is negative, the assignment is not permissible.

(iv) The assignment is permissible because

P( A) 11
t 0 , P(B) 1 t 0, and P(C) 0 t 0
33

P (S ) = P(A) P(B) P(C) 11
1 0 1.
33

(v) Even though P( A) = 0.421 ≥ 0 , P(B) = 0.527 ≥ 0, and P(C) = 0.042 ≥ 0 ,
the sum of the probability

P (S ) = P( A) + P(B) + P(C) = 0.421+ 0.527 + 0.042 = 0.990 < 1 .

Therefore, the assignment is not permissible.

Example 12.2
An integer is chosen at random from the first ten positive integers. Find the probability
that it is (i) an even number (ii) multiple of three.
Solution
The sample space is

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} , n(S) = 10

Let A be the event of choosing an even number and
B be the event of choosing an integer multiple of three.

A = {2, 4, 6, 8, 10}, n( A) = 5,

B = {3, 6, 9}, n(B) = 3

XI - Mathematics 240

P (choosing an even integer) = P=( A) n=( A) 51
n(S ) =.

10 2

P (choosing an integer multiple of three) = P=(B) n=(B) 3
n(S ) .

10

Example 12.3
Three coins are tossed simultaneously, what is the probability of getting (i) exactly one head
(ii) at least one head (iii) at most one head?
Solution:

Notice that three coins are tossed simultaneously = one coin is tossed three times.

The sample space S = {H ,T}×{H ,T}×{H ,T}

S = {HHH , HHT , HTH ,THH , HTT ,THT ,TTH ,TTT}, n(S) = 8

Let A be the event of getting one head, B be the event of getting at least one head and C be

the event of getting at most one head.

A = {HTT ,THT ,TTH}; n( A) = 3

B = {HTT ,THT ,TTH , HHT , HTH ,THH , HHH}; n(B) = 7

C = {TTT , HTT ,THT ,TTH}; n(C) = 4 .

Therefore the required probabilities are

(i) P(A) = n( A) = 3
n( S) 8

(ii) P(B) = n( B) = 7
n( S) 8

(iii) P(C) = n(C) = 4 = 1 .
n(S) 8 2

Note 12.3

When the number of elements in sample space is considerably
small we can solve by finger-counting the elements in the events. But
when the number of elements is too large to count then combinatorics
helps us to solve the problems.

For the following problem, combinatorics is used to find the
number of elements in the sample space and the events.

Example 12.4
Suppose ten coins are tossed. Find the probability to get (i) exactly two heads (ii) at most
two heads (iii) at least two heads
Solution

Ten coins are tossed simultaneously one time = one coin is tossed 10 times

Let S the sample space, 10 times

That is S = {H,T}×{H,T}×{H,T}××{H,T}

241 Introduction to Probability Theory

Let A be the event of getting exactly two heads,
B be the event of getting at most two heads, and
C be the event of getting at least two heads.
When ten coins are tossed, the number of elements in sample space is 2n = 210 = 1024

n(S) = 1024

n( A) = 10C2 = 45

n(B) = 10C0 +10 C1 +10 C2 = 1+10 + 45 = 56

n(C) = 10C2 +10 C3 +10 C4 + +10 C10

= n(S ) − (10C0 +10 C1) = 1024 −11 = 1013

The required probabilities are

(i) P( A) = n( A) = 45
n(S) 1024

(ii) P(B) = n(B) = 56 = 7
n(S ) 1024 128

(iii) P(C) = n(C) = 1013 .
n(S) 1024

Example 12.5
Suppose a fair die is rolled. Find the probability of getting

(i) an even number ( ii) multiple of three.

Solution
Let S be the sample space,
A be the event of getting an even number,

B be the event of getting multiple of three.

Therefore,

S = {1, 2, 3, 4, 5, 6}. ⇒ n (S) = 6

A = {2, 4, 6} ⇒ n ( A) = 3

B = {3, 6} ⇒ n(B) = 2

The required probabilities are

(i) P (getting an even number) = P( A) = n( A) = 3 = 1
n(S) 6 2

(ii) P (getting multiple of three) = P(B) = n(B) = 2 = 1.
n(S ) 6 3

XI - Mathematics 242

Example 12.6
When a pair of fair dice is rolled, what are the probabilities of getting the sum

(i) 7 (ii) 7 or 9 (iii) 7 or 12?
Solution

The sample space S = {1, 2, 3, 4, 5, 6}×{1, 2, 3, 4, 5, 6}
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Number of possible outcomes = 62 = 36 = n(S)

Let A be the event of getting sum 7, B be the event of getting the sum 9 and C be the
event of getting sum 12. Then

A = {(1, 6), (2,5), (3, 4), (4,3), (5, 2), (6,1)} ⇒ n( A) = 6

B = {(3, 6), (4,5), (5, 4), (6,3)} ⇒ n(B) = 4 S

C = {(6, 6)} ⇒ n(C) = 1 A (3(B,56, )4,)(,4(,65, )3,)

(1, 6), (2, 5),
(3, 4), (4, 3),
(5, 2), (6, 1)

(i) P (getting sum 7) = P(A) (6, 6) C

= n( A) = 6 = 1
n(S) 36 6

(ii) P (getting sum 7 or 9) = P( A or B) = P( A ∪ B)

= P( A) + P(B)

(Since A and B are mutually exclusive that is, A ˆ B ‡ )

= n( A) + n(B) = 6 + 4 = 5
n(S) n(S) 36 36 18

(iii) P (getting sum 7 or 12) = P( A or C) = P( A ∪ C)

= P( A) + P(C) (since A and C are mutually exclusive)

= n( A) + n(C) = 6+ 1 = 7 .
n(S ) n(S ) 36 36 36

Example 12.7
Three candidates X, Y, and Z are going to play in a chess competition to win
FIDE (World Chess Federation) cup this year. X is thrice as likely to win as Y
and Y is twice as likely as to win Z. Find the respective probability of X, Y and
Z to win the cup.

Solution
Let A, B, C be the event of winning FIDE cup respectively by X, Y, and Z this year.

243 Introduction to Probability Theory

Given that X is thrice as likely to win as Y.

A : B : : 3 : 1. (1)

Y is twice as likely as to win Z

B : C : : 2 : 1 (2)

From (1) and (2)

A : B : C : : 6 : 2 : 1

A = 6k, B =2k, C = k, where k is proportional constant.

Probability to win the cup by X is P( A) = 6k = 2
9k 3

Probability to win the cup by Y is P(B) = 2k = 2 and
9k 9

Probability to win the cup by Z is P(C) = k = 1 .
9k 9

Example 12.8 1A gambler's dispute in 1654 led to the C
Three letters are written to three different persons and addresses on
three envelopes are also written. Without looking at the addresses, what creation of a mathematical theory of
is the probability that (i) exactly one letter goes to the right envelopes probability by two famous French
(ii) none of the letters go into the right envelopes? mathematicians, Blaise Pascal and
Pierre de Fermat. The fundamental
Solution
Let A, B, and C denote the envelopes and 1, 2, and 3 denote the 2principles ofThprreoebcaabnilditiydattehseXor,yY,wanerdeZ are going to
corresponding letters.
formulated byplPayasicnalaacnhdeFssercmomatpfeotrittihoen to win FIDE
first time. Af(tWeroralnd eCxhteensssiFveedreersaetaiorcnh) ,cup this year.
Laplace pubXlisishethdricheiass lmikoelnyumtoewntianl as Y and Y is
work in 181t2w, icaendaslaildikefloyunadsattioonwtion Z. Find the

3Probability rtehsepoercy.tivIen psCtraoutbpisatbihciilssi,tyyethaert.oX iws itnhrictehaes likely to win

Bayesian intceurppr.TetharteioencaoanfsdiYpdraoatbneasdbYiXli,tiysYt,waicnedaZs liakreely as to win Z.
was developgeodinmgationlpylabyFyiinnLdaapthclaehcerese.sspAceocmtivpeetiptrioonbatboility to win the
gambler's diwspiunteFIDinE1(6W5co4urplld.eTdChhrteeoessthcFeaenddeirdaatitoens ) Xcu,pY, and Z are
creation of athims yaethaer.mXatiiscgatohlirnitgcheetoarspyllaikoyefilnyatochweisns acsoYmpetition to win
probability abnyd YtwisotwfiacmeFoIaDussElik(eFWlryeonarcsldhtoCwheinssZF. eFdinedration) cup this
mathematiciatnhse, reBsplaeicsteivyeePaaprs.rcoXablaisbtiahlnirtdiycetaos lwikienlytthoewin as Y and Y
Pierre de Fceurpm.Tath.reTe hceainsfduitndwdaitacemes eaXnst,alilYk,elayndas Ztoawrein Z. Find the

Bprinciples ofgopinrogbatobipliltayyretihsnpeoeacryctihvewesesrecpormobpaebtiitliiotyn toto win the
win FIDE (Wcourpld.TChhreeess cFaenddeirdaatitoesn) Xcu,pY, and Z are
this year. X isgtohirnigcetoaspllaikyeilnyatochweisns acsoYmpetition to win
and Y is twiceFIaDs Elik(eWlyoarlsdtoCwheinssZF. eFdinedration) cup this
the respectivyeeapr.roXbaisbtihlirtiycetaos lwikienlytthoewin as Y and Y
Acup.this year.isXtiws itchericaes alsikleilkyelaystotowwinin Z. Find the
respective probability to win the cup.this
year. X is thrice as likely to win as Y and Y
is twic
Three candidates X, Y, and Z are going to
play in a chess competition to win FIDE

The different combination of letters put Envelope Outcomes c6
into the envelopes are shown in the table. c1 c2 c3 c4 c5 3
A1 1 22 3 2
Let ci denote the outcomes of the events. 1
Let X be the event of putting the letters into
the exactly only one right envelopes.

Let Y be the event of putting none of the B2 3 13 1
letters into the right envelope. C3 2 31 2

S={c1, c2 , c3, c4 , c5, c6}, n (S) = 6

=X {=c2 , c3, c6}, n( X ) 3

=Y {=c4 , c5} n(Y ) 2

31 21
P( X )= = P(Y )= = .
62 63

Example 12.9

Let the matrix M = x y  . If x, y and z are chosen at random from the set {1, 2,3},
 z 1 

and repetition is allowed (i.e., x = y = z ), what is the probability that the given matrix M is a

singular matrix?

XI - Mathematics 244

Solution

If the given matrix M is singular, then

x y = 0.
z1
That is, x − yz = 0.

Hence the possible ways of selecting ( x, y, z) are

{(1,1,1),(2,1,2),(2,2,1),(3,1,3),(3,3,1)} = A(say)

The number of favourable cases n( A) = 5

The total number of cases are n(S) = 33 = 27
The probability of the given matrix is a singular matrix is

p = n( A) = 5 .
n(S) 27

Example 12.10

For a sports meet, a winners’ stand comprising of three wooden 1
blocks is in the form as shown in figure. There are six different colours
available to choose from and three of the wooden blocks is to be painted 3 2

such that no two of them has the same colour. Find the probability that the
smallest block is to be painted in red, where red is one of the six colours.

Solution

Let S be the sample space and A be the event that the smallest block is to be painted in
red.

n(S) = 6P3 = 6× 5× 4 = 120 6 63
n(S) 6 54
n( A) = 5× 4 = 20 n(A) 5 4 Red
P( A) = n( A) = 20 = 1 .

n(S) 120 6

12.4.3 ODDS

The word odds is frequently used in probability and statistics. Odds relate the chances in favour
of an event A to the chances against it. Suppose a represents the number of ways that an event can
occur and b represents the number of ways that the event can fail to occur.
The odds of an event A are a : b in favour of an event and

P( A) = a .
a+b

245 Introduction to Probability Theory

Further, it may be noted that the odds are a : b in favour of an event is the same as to say that the
odds are b : a against the event.

If the probability of an event is p , then the odds in favour of its occurrence are p to (1− p) and
the odds against its occurrence are (1− p) to p .

Illustration 12.7
(i) Suppose a die is rolled.
Let S be the sample space and A be the event of getting 5.

n(S) = 6 , n( A) = 1 and n( A) = 5 .
It can also be interpreted as

1 odds against A is 5 :1or 5 ,
Odds in favour of A is or , 1
1:5 5

and P( A) = n( A) = 1 = 1 = n( A) .
n( A) + n( A) 5 +1 6 n(S)

(ii) Suppose B is an event such that odds in favour of B is 3:5, then P(B) = 3
8

(iii) Suppose C is an event such that odds against C is 4:11, then P(C) = 11.
15

Example 12.11
A man has 2 ten rupee notes, 4 hundred rupee notes and 6 five hundred rupee notes in his
pocket. If 2 notes are taken at random, what are the odds in favour of both notes being of
hundred rupee denomination and also its probability?

Solution
Let S be the sample space and A be the event of taking 2 hundred rupee note.

Therefore, n(S) = 12c2 = 66 , n( A) = 4c2 = 6 and n( A) = 66 − 6 = 60
Therefore, odds in favour of A is 6: 60

That is, odds in favour of A is 1: 10, and P( A) = 111.

EXERCISE 12.1

(1) An experiment has the four possible mutually exclusive and exhaustive outcomes A, B,
C, and D. Check whether the following assignments of probability are permissible.

(i) P(A) = 0.15, P(B) = 0.30, P(C) = 0.43, P(D)= 0.12

(ii) P(A) = 0.22, P(B) = 0.38, P(C) = 0. 16, P (D) = 0.34

(iii) P(A) = 2 , P(B) = 3 , P(C) = − 1 , P(D) = 1
55 55
(2) If two coins are tossed simultaneously, then find the probability of getting

(i) one head and one tail (ii) at most two tails

(3) Five mangoes and 4 apples are in a box. If two fruits are chosen at random, find the
probability that (i) one is a mango and the other is an apple (ii) both are of the same
variety.

XI - Mathematics 246