Download Unit 7(Integrals)

Chapter
7
INTEGRALS
7.1 Overview
d
F (x) = f (x). Then, we write ∫ f ( x ) dx = F (x) + C. These integrals are
dx
called indefinite integrals or general integrals, C is called a constant of integration. All
these integrals differ by a constant.
7.1.1
Let
7.1.2
If two functions differ by a constant, they have the same derivative.
7.1.3
Geometrically, the statement
∫ f ( x ) dx = F (x) + C = y (say) represents a
family of curves. The different values of C correspond to different members of this
family and these members can be obtained by shifting any one of the curves parallel to
itself. Further, the tangents to the curves at the points of intersection of a line x = a with
the curves are parallel.
7.1.4
(i)
Some properties of indefinite integrals
The process of differentiation and integration are inverse of each other,
d
f ( x ) dx = f ( x ) and
dx ∫
arbitrary constant.
i.e.,
(ii)
∫ f ' ( x ) dx = f ( x ) + C ,
Two indefinite integrals with the same derivative lead to the same family of
curves and so they are equivalent. So if f and g are two functions such that
d
d
f ( x ) dx =
g ( x) dx , then
∫
dx
dx ∫
(iii)
where C is any
∫ f ( x ) dx
and
∫ g ( x ) dx are equivalent.
The integral of the sum of two functions equals the sum of the integrals of
the functions i.e.,
∫ ( f ( x ) + g ( x ) ) dx = ∫ f ( x ) dx + ∫ g ( x ) dx .
144
MATHEMATICS
(iv)
A constant factor may be written either before or after the integral sign, i.e.,
∫ a f ( x ) dx = a ∫ f ( x ) dx , where ‘a’ is a constant.
(v)
∫ (k
Properties (iii) and (iv) can be generalised to a finite number of functions
f1, f2, ..., fn and the real numbers, k1, k2, ..., kn giving
f ( x ) + k2 f 2 ( x ) + ...+, kn f n ( x ) ) dx = k1 ∫ f1 ( x ) dx + k2 ∫ f 2 ( x ) dx + ... + kn ∫ f n ( x ) dx
1 1
7.1.5
Methods of integration
There are some methods or techniques for finding the integral where we can not
directly select the antiderivative of function f by reducing them into standard forms.
Some of these methods are based on
1.
Integration by substitution
2.
Integration using partial fractions
3.
Integration by parts.
7.1.6
Definite integral
b
The definite integral is denoted by
∫ f ( x ) dx , where a is the lower limit of the integral
a
and b is the upper limit of the integral. The definite integral is evaluated in the following
two ways:
(i)
The definite integral as the limit of the sum
b
(ii)
∫ f ( x ) dx = F(b) – F(a), if F is an antiderivative of f (x).
a
7.1.7
The definite integral as the limit of the sum
b
The definite integral
∫ f ( x ) dx is the area bounded by the curve y = f (x), the ordia
nates x = a, x = b and the x-axis and given by
b
∫ f ( x ) dx = (b – a)
a
1
lim ⎡⎣ f (a) + f ( a + h ) + ... f ( a + ( n – 1) h ) ⎤⎦
n →∞ n
INTEGRALS
145
or
b
∫ f ( x ) dx = lim h ⎣⎡ f (a) + f ( a + h ) + ... + f ( a + ( n – 1) h )⎦⎤ ,
h →0
a
where h =
7.1.8
(i)
b–a
→ 0 as n → ∞ .
n
Fundamental Theorem of Calculus
Area function : The function A (x) denotes the area function and is given
x
by A (x) =
∫ f ( x ) dx .
a
(ii)
First Fundamental Theorem of integral Calculus
Let f be a continuous function on the closed interval [a, b] and let A (x) be
the area function . Then A′ (x) = f (x) for all x ∈ [a, b] .
(iii)
Second Fundamental Theorem of Integral Calculus
Let f be continuous function defined on the closed interval [a, b] and F be
an antiderivative of f.
b
∫ f ( x ) dx = [ F ( x )]
b
a
a
7.1.9
= F(b) – F(a).
Some properties of Definite Integrals
b
P0 :
∫
a
b
P1 :
P2 :
∫
a
f ( x ) dx =
b
∫ f ( t ) dt
a
a
f ( x ) dx = – ∫ f ( x ) dx , in particular,
b
b
c
b
a
a
c
∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx
a
∫ f ( x ) dx = 0
a
146
MATHEMATICS
b
∫
P3 :
f ( x ) dx =
a
∫
f ( x ) dx =
0
∫
f ( x ) dx =
0
∫ f ( x ) dx
0
a
P7 : (i)
a
(ii)
∫
∫ f ( a – x ) dx
a
∫
f ( x ) dx +
0
2a
P6 :
a
0
2a
P5 :
∫ f ( a + b – x ) dx
a
a
P4 :
b
∫
–a
a
∫ f ( 2a – x ) dx
0
⎧ a
= ⎪2∫ f ( x ) dx,if f (2a − x) = f ( x) ,
⎨ 0
⎪0, if f (2a − x) = − f ( x).
⎩
a
f ( x ) dx = 2 f ( x ) dx , if f is an even function i.e., f (–x) = f (x)
∫
0
f ( x ) dx = 0, if f is an odd function i.e., f (–x) = –f (x)
–a
7.2
Solved Examples
Short Answer (S.A.)
⎛ 2a b
⎞
– 2 + 3c 3 x 2 ⎟ w.r.t. x
Example 1 Integrate ⎜
⎝ x x
⎠
Solution
⎛ 2a b
⎞
– 2 + 3c 3 x 2 ⎟ dx
x
x
⎠
∫ ⎜⎝
2
–1
= ∫ 2a ( x ) 2 dx – ∫ bx –2 dx + ∫ 3c x 3 dx
5
3
= 4a x + b + 9 cx + C .
x
5
INTEGRALS
147
3ax
dx
b c2 x2
Example 2 Evaluate
2
Solution Let v = b2 + c2x2 , then dv = 2c2 xdx
Therefore,
3ax
∫ b2 + c2 x2 dx
3a dv
2c 2 v
=
=
3a
log b 2
2c 2
c2 x2
C.
Example 3 Verify the following using the concept of integration as an antiderivative.
x 3 dx
x 1
d
x2
x
–
Solution
dx
2
=1–
x2
2
x–
x3
– log x 1
3
C
C
3x 2
1
–
3
x 1
2x
2
=1–x+x –
1
2
Thus
x3
– log x 1
3
x
x3
=
.
x 1
1
⎛
⎞
x 2 x3
x3
+
– log x + 1 + C ⎟ = ∫
dx
⎜x–
2
3
x +1
⎝
⎠
Example 4 Evaluate
Solution Let I =
∫
1 x
dx , x ≠ 1.
1– x
1+ x
dx =
1– x
∫
1
1– x
2
dx +
x dx
1 – x2
= sin –1 x + I1 ,
148
MATHEMATICS
x dx
where I1 =
.
1 – x2
Put 1 – x2 = t2 ⇒ –2x dx = 2t dt. Therefore
I1 = – dt = – t + C = – 1 – x 2
Hence
I = sin–1x – 1 – x 2
C
C.
dx
Example 5 Evaluate
∫ ( x – α )( β – x ) , β > α
Solution Put x – α = t2. Then
– t2
–x =
=
and dx = 2tdt. Now
I =∫
2
2t dt
t2 (β – α – t2 )
dt
2
k – t2
Solution I =
∫ tan
8
∫
2 dt
(β – α – t )
, where k 2
–1 t
–1
= 2sin k + C = 2sin
Example 6 Evaluate
=
∫ tan
8
2
–
x–α
+ C.
β –α
x sec 4 x dx
x sec 4 x dx
=
8
2
2
∫ tan x ( sec x ) sec x dx
=
8
2
2
∫ tan x ( tan x + 1) sec x dx
– t2 –
= – t2 –
INTEGRALS
=
∫ tan
=
tan11 x tan 9 x
+
+C.
11
9
10
Example 7 Find
x sec 2 x dx + ∫ tan 8 x sec 2 x dx
x3
∫ x4 + 3x2 + 2 dx
Solution Put x2 = t. Then 2x dx = dt.
Now
x3 dx
1
t dt
= ∫ 2
I= ∫ 4
2
x + 3x + 2 2 t + 3t + 2
Consider
t
A
B
=
+
t + 3t + 2 t + 1 t + 2
2
Comparing coefficient, we get A = –1, B = 2.
Then
I=
=
1⎡
dt
dt ⎤
2∫
–∫
⎢
2⎣ t+2
t + 1 ⎥⎦
1
⎡ 2log t + 2 − log t +1 ⎦⎤
2⎣
= log
Example 8 Find
x2 + 2
+C
x2 + 1
dx
∫ 2sin 2 x + 5cos2 x
Solution Dividing numerator and denominator by cos2x, we have
I=
sec 2 x dx
2tan 2 x 5
149
150
MATHEMATICS
Put tanx = t so that sec2x dx = dt. Then
I=
=
1
dt
∫ 2t 2 + 5 = 2 ∫
dt
⎛ 5⎞
t +⎜
⎟
⎝ 2⎠
2
2
⎛ 2t ⎞
1 2
tan –1 ⎜⎜
⎟⎟ + C
2 5
⎝ 5⎠
=
⎛ 2 tan x ⎞
tan –1 ⎜⎜
⎟ + C.
10
5 ⎟⎠
⎝
1
2
7 x – 5 dx as a limit of sums.
Example 9 Evaluate
–1
Solution Here a = –1 , b = 2, and h =
2 +1
, i.e, nh = 3 and f (x) = 7x – 5.
n
Now, we have
2
∫ ( 7 x – 5) dx = lim h ⎡⎣ f ( –1) + f (–1 + h) + f ( –1 + 2h ) + ... + f ( –1 + ( n – 1) h )⎤⎦
h→0
–1
Note that
f (–1) = –7 – 5 = –12
f (–1 + h) = –7 + 7h – 5 = –12 + 7h
f (–1 + (n –1) h) = 7 (n – 1) h – 12.
Therefore,
2
∫ ( 7x –5) dx = lim h ⎡⎣( –12) + (7h – 12) + (14h –12) + ... + (7 ( n –1 ) h –12)⎤⎦ .
–1
h→0
h ⎡ 7 h ⎡⎣1 + 2 + ... + ( n – 1) ⎤⎦ – 12n ⎤⎦
= lim
h→0 ⎣
INTEGRALS
151
⎡ ( n – 1) n
⎤
⎡7
⎤
h ⎢7h
– .12n ⎥ = lim ⎢ ( nh )( nh – h ) – 12nh ⎥
= lim
h→0
h→0 2
2
⎣
⎦
⎣
⎦
=
7
3 3 – 0 – 12
2
3 =
7×9
–9
– 36 =
.
2
2
π
2
Example 10 Evaluate
tan 7 x
∫ cot 7 x + tan 7 x dx
0
Solution We have
π
2
I=
tan 7 x
∫ cot 7 x + tan 7 x dx
0
...(1)
⎛π
⎞
tan 7 ⎜ – x ⎟
2
⎝
⎠
dx
= ∫
π
⎛
⎞
⎛π
⎞
7
7
0 cot
⎜ – x ⎟ + tan ⎜ – x ⎟
⎝2
⎠
⎝2
⎠
π
2
=
π
2
∫ cot
0
cot 7 ( x ) dx
7
...(2)
x dx + tan 7 x
Adding (1) and (2), we get
π
2
⎛ tan 7 x + cot 7 x ⎞
2I = ∫ ⎜
⎟ dx
tan 7 x + cot 7 x ⎠
0⎝
π
2
= ∫ dx which gives I
0
by (P4)
π
.
4
152
MATHEMATICS
8
Example 11 Find
∫
2
10 – x
x + 10 – x
dx
Solution We have
8
10 – x
∫
I=
x + 10 – x
2
8
=
...(1)
10 – (10 – x)
10 – x
2
8
⇒
dx
10 – 10 – x
x
I=∫
10 – x + x
2
dx
dx
(2)
Adding (1) and (2), we get
8
2I
1dx
8–2 6
2
Hence
I=3
π
4
Example 12 Find
∫
1+ sin 2x dx
0
Solution We have
π
4
I=
∫
1+ sin 2 x dx =
0
π
4
=
∫ ( sin x + cos x ) dx
0
by (P3)
π
4
2
∫ ( sin x + cos x ) dx
0
INTEGRALS
=
153
π
( − cos x + sin x )04
I = 1.
x 2 tan –1 x dx .
Example 13 Find
2
–1
Solution I = x tan x dx
= tan –1 x ∫ x 2 dx –
1
x3
.
∫ 1 + x 2 3 dx
x3
x ⎞
1 ⎛
–1
= 3 tan x – 3 ∫ ⎜ x − 1 + x 2 ⎟ dx
⎝
⎠
x3
x2 1
–1
tan
x
–
+ log 1 + x 2 + C .
=
3
6 6
Example 14 Find
∫
10 – 4 x + 4 x 2 dx
Solution We have
10 – 4 x 4 x 2 dx
I=
=
2x – 1
2
3
2
dx
Put t = 2x – 1, then dt = 2dx.
Therefore,
I=
1
2
t 2 + ( 3) dt
∫
2
t2 9
2
=
1
t
2
=
1
( 2 x – 1)
4
9
log t
4
( 2 x – 1)
2
t2
+9 +
9
C
9
log ( 2 x – 1) +
4
( 2 x – 1) 2 + 9
+ C.
154
MATHEMATICS
Long Answer (L.A.)
x 2 dx
.
Example 15 Evaluate ∫ 4 2
x + x −2
Solution Let x2 = t. Then
x2
t
t
A
B
= 2
=
=
+
4
2
(
t
+
2)
(
t
−
1)
t
+
2
t
−1
x + x −2 t +t −2
So
t = A (t – 1) + B (t + 2)
Comparing coefficients, we get A =
2
1
, B= .
3
3
x2
2 1
1 1
=
+
4
2
2
3
x + x −2
x + 2 3 x 2 −1
So
Therefore,
x2
2
∫ x 4 + x 2 − 2 dx = 3
1
1
dx
∫ x 2 + 2 dx + 3 ∫ x 2 −1
2 1
x
1
x −1
tan –1
+ log
+C
= 3
x +1
2
2 6
Example16 Evaluate
x3 x
dx
x4 – 9
Solution We have
I=
x3 x
dx =
x4 – 9
x3
dx
x4 – 9
x dx
= I1+ I2 .
x4 – 9
INTEGRALS
x3
x –9
Now
I1 = ∫
Put
t = x4 – 9 so that 4x3 dx = dt. Therefore
4
I1 =
1 dt
1
= log t
4 t
4
C1 =
1
log x 4 – 9 + C1
4
x dx
x4 – 9 .
Again,
I2 =
Put
x2 = u so that 2x dx = du. Then
1
du
I2 = 2 u 2 – 3
=
Thus
2
=
1
2
6
log
u –3
u 3
C2
1
x2 – 3
log 2
+ C2 .
12
x +3
I = I1 + I2
=
1
1
x2 – 3
log x 4 – 9 + log 2
+ C.
4
12
x +3
π
sin 2 x
1
∫0 sin x + cos x = 2 log ( 2 + 1)
2
Example 17 Show that
Solution We have
π
sin 2 x
∫0 sin x + cos x dx
2
I=
155
156
MATHEMATICS
π
sin 2 ⎛⎜ – x ⎞⎟
⎝2
⎠
dx
= ∫0
π
π
sin ⎛⎜ – x ⎞⎟ + cos ⎛⎜ – x ⎞⎟
⎝2
⎠
⎝2
⎠
π
2
⇒
Thus, we get
(by P4)
π
2
cos 2 x
∫0 sin x + cos x dx
I=
2I =
1
π
2
dx
π⎞
⎛
0 cos x –
⎜
⎟
⎝
4⎠
∫
2
π
π
2
1 ⎡ ⎛
π⎞
π ⎞ ⎞⎤ 2
1
⎛
⎛
π⎞
⎛
log
sec
+
tan
x
–
x
–
sec
dx
=
=
x
–
⎜
⎟
⎜
⎟ ⎟⎥
⎢
⎜
⎜
⎟
∫ ⎝ 4⎠
⎝
4⎠
⎝
4 ⎠ ⎠⎦0
2⎣ ⎝
20
=
1 ⎡ ⎛
π
π
π
π ⎤
log ⎜ sec + tan ⎟⎞ – log sec ⎜⎛ – ⎟⎞ + tan ⎜⎛ − ⎟⎞ ⎥
⎢
4
4 ⎠
⎝ 4⎠
⎝ 4 ⎠⎦
2⎣ ⎝
=
1
⎡ log ( 2 + 1) – log ( 2 −1) ⎤
⎦
2⎣
=
1
log
2
⎛ ( 2 + 1)2 ⎞
1
2
log
log ( 2 + 1)
⎜
⎟ =
=
2
2
⎝
1
⎠
Hence
I=
1
2
1
Example 18
Find
∫ x ( tan
0
–1
log ( 2 + 1) .
x ) dx
2
2 +1
2 –1
INTEGRALS
1
I=
Solution
∫ x ( tan
–1
x ) dx .
2
0
Integrating by parts, we have
1
1 2 tan –1 x
x2 ⎡
2 1
–1
⎤
(
)
–
tan x ⎦ 0 2 ∫ x .2 1 + x 2 dx
2 ⎣
0
I=
1
π2
x2
–
.tan –1 x dx
=
32 ∫0 1 + x 2
π2
=
– I1 , where I1 =
32
1
x2
–1
∫ 1+ x 2 tan xdx
0
1
Now
I1 =
x2 + 1 – 1
∫0 1 + x 2 tan–1x dx
1
1
1
tan –1 x dx
2
+
x
1
0
–1
= ∫ tan x dx – ∫
0
= I2 –
1
1
(
( tan –1 x )2 )0
2
= I2 –
1
Here
I2 =
=
Thus
I1 =
π2
32
1
x
dx
1 + x2
0
–1
∫ tan x dx = ( x tan –1 x )0 – ∫
1
0
(
π 1
– log 1 + x 2
4 2
π 1
π2
– log 2 −
4 2
32
)
1
0
=
π 1
– log 2 .
4 2
157
158
MATHEMATICS
π2 π
–
I =
32 4
Therefore,
=
1
log 2
2
π2
π2 π 1
– + log 2
=
32
16 4 2
π 2 – 4π
+ log 2 .
16
2
Example 19 Evaluate
∫ f ( x) dx , where f (x) = |x + 1| + |x| + |x – 1|.
–1
⎧ 2 – x, if
⎪
as f ( x ) = ⎨ x + 2, if
⎪ 3x , if
⎩
Solution We can redefine f
2
∫
Therefore,
–1
0
1
2
–1
0
1
–1 < x ≤ 0
0 < x ≤1
1< x ≤ 2
f ( x ) dx = ∫ ( 2 – x ) dx + ∫ ( x + 2 ) dx + ∫ 3 x dx
0
(by P2)
1
2
=
5 5 9 19
+ + =
.
2 2 2 2
⎛
⎛ x2
⎞ ⎛ 3x 2 ⎞
x2 ⎞
+
2
x
–
+
2
x
= ⎜
⎟
⎜
⎟ +⎜
⎟
⎝
2 ⎠ –1 ⎝ 2
⎠0 ⎝ 2 ⎠1
1⎞ ⎛1
⎛
⎞
⎛4 1⎞
= 0 – ⎜ –2 – ⎟ + ⎜ + 2 ⎟ + 3 ⎜ – ⎟
⎝
2⎠ ⎝2
⎠
⎝2 2⎠
Objective Type Questions
Choose the correct answer from the given four options in each of the Examples from
20 to 30.
Example 20
∫ e ( cos x – sin x ) dx is equal to
x
(A) e x cos x + C
(B) e x sin x + C
(C) – e x cos x + C
(D) – e x sin x + C
INTEGRALS
159
x
x
Solution (A) is the correct answer since ∫ e ⎡⎣ f ( x ) + f '( x ) ⎤⎦ dx = e f ( x ) + C . Here
f (x) = cosx, f′ (x) = – sin x.
∫ sin
Example 21
2
dx
is equal to
x cos 2 x
(A) tanx + cotx + C
(B) (tanx + cotx)2 + C
(C) tanx – cotx + C
(D) (tanx – cotx)2 + C
Solution (C) is the correct answer, since
dx
I= ∫ 2
=
sin x cos 2 x
∫
( sin 2 x + cos2 x ) dx
sin 2 x cos 2 x
2
2
= ∫ sec x dx + ∫ cosec x dx = tanx – cotx + C
Example 22 If
3ex – 5 e– x
∫ 4 e x + 5 e – x dx = ax + b log |4ex + 5e–x| + C, then
(A) a =
–1
7
, b=
8
8
1
7
(B) a = , b =
8
8
(C) a =
–1
–7
, b=
8
8
1
–7
(D) a = , b =
8
8
Solution (C) is the correct answer, since differentiating both sides, we have
3ex – 5 e– x
( 4 e x – 5 e– x ) ,
=
a
+
b
4 ex + 5 e– x
4 ex + 5 e– x
giving 3ex – 5e–x = a (4ex + 5e–x) + b (4ex – 5e–x). Comparing coefficients on both
sides, we get 3 = 4a + 4b and –5 = 5a – 5b. This verifies a =
–1
7
, b= .
8
8
160
MATHEMATICS
b+c
Example 23
∫
f ( x ) dx is equal to
a+c
b
(A)
∫
b
f ( x – c ) dx
(B)
a
a
b–c
b
(C)
∫ f ( x + c ) dx
∫ f ( x ) dx
(D)
∫
f ( x ) dx
a –c
a
Solution (B) is the correct answer, since by putting x = t + c, we get
I=
b
b
a
a
∫ f ( c + t ) dt = ∫ f ( x + c ) dx .
Example 24 If f and g are continuous functions in [0, 1] satisfying f (x) = f (a – x)
a
and g (x) + g (a – x) = a, then
∫ f ( x ) . g ( x ) dx
is equal to
0
(A)
a
2
a
(C)
∫
(B)
a
a
2
∫ f ( x ) dx
0
a
(D) a ∫ f ( x ) dx
f ( x ) dx
0
0
a
Solution B is the correct answer. Since I =
∫ f ( x ) . g ( x ) dx
0
a
=
∫
f ( a – x ) g ( a – x ) dx =
0
a
∫ f ( x ) ( a – g ( x ) ) dx
0
a
a
a
0
0
0
= a ∫ f ( x ) dx – ∫ f ( x ) . g ( x ) dx = a ∫ f ( x ) dx – I
INTEGRALS
161
a
or
I
a
f ( x ) dx .
2 ∫0
=
y
Example 25 If x =
∫
0
(A) 3
dt
1 + 9t 2
and
d2y
= ay, then a is equal to
dx 2
(B) 6
(C) 9
y
∫
Solution (C) is the correct answer, since x =
0
dt
1 + 9t 2
(D) 1
⇒
dx
1
=
dy
1 + 9 y2
18 y
dy
d2y
=
= 9y.
2 .
2
2
1
+
9
y
dx
dx
which gives
1
x3 + x +1
∫ x 2 + 2 x +1 dx is equal to
–1
Example 26
(A) log 2
(B) 2 log 2
(C)
1
log 2
2
(D) 4 log 2
1
x3 + x +1
dx
Solution (B) is the correct answer, since I = ∫ 2
x + 2 x +1
–1
1
1
x3
x +1
+∫ 2
dx = 0 + 2
= ∫ 2
x + 2 x + 1 –1 x + 2 x + 1
–1
1
x +1
∫ ( x +1)
2
dx
0
[odd function + even function]
1
=2
∫ ( x +1)
0
1
x +1
2
dx = 2 ∫
0
1
dx
x +1
1
= 2 log x + 1 0 = 2 log 2.
162
MATHEMATICS
1
If
Example 27
(A) a – 1 +
et
∫0 1 + t dt = a, then
e
2
1
et
∫ (1 + t )
2
dt is equal to
0
e
2
(B) a + 1 –
(C) a – 1 –
e
2
(D) a + 1 +
1
et
dt
Solution (B) is the correct answer, since I = ∫
1+ t
0
1
1
t
e
1 t
dt = a (given)
e +∫
=
2
1 + t 0 0 (1 + t )
1
Therefore,
et
∫ (1 + t )
2
=a–
0
e
+ 1.
2
2
Example 28
∫ x cos πx dx is equal to
–2
(A)
8
π
(B)
4
π
(C)
2
π
(D)
2
Solution (A) is the correct answer, since I =
∫
2
x cos πx dx = 2
–2
3
⎧ 12
⎫
2
2
⎪
⎪
= 2 ⎨ ∫ x cos πx dx + ∫ x cos πx dx + ∫ x cos πx dx ⎬ =
1
3
⎪0
⎪
2
2
⎩
⎭
Fill in the blanks in each of the Examples 29 to 32.
Example 29
sin 6 x
∫ cos8 x dx = _______.
1
π
∫ x cos πx dx
0
8
.
π
e
2
INTEGRALS
tan 7 x
+C
7
Solution
a
Example 30
∫
f ( x ) dx = 0 if f is an _______ function.
–a
Solution Odd.
2a
Example 31
∫
0
a
f ( x ) dx = 2 ∫ f ( x ) dx , if f (2a – x) = _______.
0
Solution f (x).
π
2
Example 32
Solution
sin n x dx
∫0 sin n x + cosn x = _______.
π
.
4
7.3 EXERCISE
Short Answer (S.A.)
Verify the following :
2x – 1
1.
∫ 2 x + 3 dx = x – log |(2x + 3) | + C
2.
∫x
2
2x + 3
dx = log |x2 + 3x| + C
2
+ 3x
Evaluate the following:
3.
∫
( x 2 + 2 ) dx
x +1
4.
e6 log x – e5log x
∫ e4log x – e3log x dx
163
164
MATHEMATICS
(1 + cos x )
dx
x + sin x
5.
∫
7.
∫ tan
9.
∫
1 + sin xdx
10.
∫
x
dx
x +1
12.
∫
x2
14.
∫
16.
∫
18.
x
∫ x4 – 1 dx
20.
∫
2
x sec 4 x dx
(Hint : Put
x = z)
dx
6.
∫ 1 + cos x
8.
∫
11.
∫
a+x
a–x
13.
∫
1 + x2
dx
x4
15.
∫
17.
∫
19.
x2
∫ 1 – x 4 dx put x2 = t
sin x + cos x
1 + sin 2 x
dx
1
22.
∫
1+ x
3
4
dx
(Hint : Put x = z4)
dx
16 – 9 x
3x – 1
x2 + 9
2
dx
2
2ax – x dx
( cos5 x + cos 4 x ) dx
1 – 2cos3 x
21.
23.
∫
dt
3t – 2t 2
5 – 2x + x 2 dx
sin –1 x
3
(1 – x 2 ) 2
dx
sin 6 x + cos 6 x
∫ sin 2 x cos2 x dx
INTEGRALS
24.
∫
26.
∫x
x
3
a –x
3
dx
dx
25.
∫
165
cos x – cos 2 x
dx
1 – cos x
(Hint : Put x2 = sec θ)
x4 – 1
Evaluate the following as limit of sums:
2
27.
2
∫ ( x 2 + 3) dx
28.
0
∫e
x
dx
0
Evaluate the following:
π
2
1
29.
dx
∫0 e x + e – x
2
31.
30.
tan x dx
2
tan 2 x
∫1+ m
0
1
dx
∫ ( x – 1) (2 − x)
1
32.
xdx
∫
1+ x 2
0
1
2
π
33.
∫ x sin x cos
2
xdx
34.
0
∫ (1+ x
0
dx
2
) 1− x 2
(Hint: let x = sinθ)
Long Answer (L.A.)
35.
x 2 dx
∫ x 4 – x 2 – 12
π
37.
x
∫ 1 + sin x
0
36.
( x2
x 2 dx
a 2 )( x 2 b 2 )
2x – 1
38.
∫ ( x – 1)( x + 2 )( x – 3) dx
166
39.
MATHEMATICS
tan
∫e
–1
x
⎛ 1+ x + x 2 ⎞
⎜
⎟ dx
2
⎝ 1+ x ⎠
40.
∫ sin
–1
x
dx
a+x
(Hint: Put x = a tan2θ)
π
2
41.
1 + cos x
∫π
5
(1 − cos x) 2
3
43.
∫
∫e
−3 x
cos3 x dx
tan x dx (Hint: Put tanx = t2)
π
2
44.
42.
dx
∫ (a 2 cos2 x + b2 sin 2 x)2
0
(Hint: Divide Numerator and Denominator by cos4x)
π
1
45.
∫ x log (1+ 2 x) dx
0
46.
∫ x log sin x dx
0
π
4
47.
∫π log (sin x + cos x)dx
−
4
Objective Type Questions
Choose the correct option from given four options in each of the Exercises from 48 to 63.
48.
∫
cos2 x – cos 2θ is equal to
dx
cos x – cosθ
(A) 2(sinx + xcosθ) + C
(B) 2(sinx – xcosθ) + C
(C) 2(sinx + 2xcosθ) + C
(D) 2(sinx – 2x cosθ) + C
INTEGRALS
49.
dx
is equal to
sin x – a sin x – b
(A) sin (b – a) log
sin( x – b)
+C
sin( x – a)
(C) cosec (b – a) log
50.
∫ tan
–1
sin( x – b)
+C
sin( x – a)
x – x tan –1 x + C
(C)
(B) cosec (b – a) log
(D) sin (b – a) log
sin( x – a)
+C
sin( x – b)
sin( x – a)
+C
sin( x – b)
x dx is equal to
(A) (x + 1) tan –1 x – x + C
(B) x tan –1 x – x + C
(D)
x – ( x + 1) tan –1 x + C
2
51.
x ⎛ 1– x ⎞
∫ e ⎜⎝ 1 + x 2 ⎟⎠ dx is equal to
–ex
+C
(B)
1 + x2
ex
+C
(A)
1 + x2
(C)
52.
167
ex
(1 + x 2 )2
∫ ( 4x
x9
2
+ 1)
6
+C
(1 + x 2 )2
+C
dx is equal to
–5
1 ⎛
1 ⎞
(A)
⎜4+ 2 ⎟ + C
5x ⎝
x ⎠
(C)
(D)
–ex
1
(1 + 4 ) –5 + C
10x
–5
1⎛
1 ⎞
(B) ⎜ 4 + 2 ⎟ + C
5⎝
x ⎠
–5
(D)
1⎛ 1
⎞
⎜ 2 + 4⎟ + C
10 ⎝ x
⎠
168
53.
54.
55.
MATHEMATICS
If
dx
∫ ( x + 2) ( x
2
+ 1)
= a log |1 + x2| + b tan–1x +
(A) a =
–1
–2
,b=
10
5
(B) a =
1
2
,b=–
10
5
(C) a =
–1
2
,b=
10
5
(D) a =
1
2
,b=
10
5
x3
∫ x + 1 is equal to
(A) x +
x 2 x3
+ – log 1 – x + C
2
3
(B) x +
x 2 x3
–
– log 1 – x + C
2
3
(C) x –
x 2 x3
– – log 1 + x + C
2
3
(D) x –
x 2 x3
+ – log 1 + x + C
2
3
x + sin x
∫ 1 + cos x dx is equal to
(A) log 1 + cos x + C
(C) x – tan
56.
1
log |x + 2| + C, then
5
If
x
+C
2
(D) x .tan
x
+C
2
3
x3 dx
1 x
(B) log x + sin x + C
2
a (1 x 2 ) 2
b 1 x2
C, then
(A) a =
1
,
3
b=1
(B) a =
–1
,
3
b=1
(C) a =
–1
,
3
b = –1
(D) a =
1
,
3
b = –1
INTEGRALS
π
4
57.
dx
∫ 1 + cos2x
169
is equal to
–π
4
(A) 1
(B) 2
(C) 3
(D) 4
(C) 2
(D) 2 ( 2 –1)
π
2
58.
∫
1 – sin 2xdx is equal to
0
(B) 2
(A) 2 2
π
2
59.
∫ cos x e
sin x
(
2 +1)
dx is equal to _______.
0
x+3
60.
∫ ( x + 4)
2
e x dx = ________.
Fill in the blanks in each of the following Exercise 60 to 63.
a
61.
If
1
∫ 1 + 4x
2
dx =
0
sin x
62.
∫ 3 + 4cos2 x dx
π
, then a = ________.
8
= ________.
π
63.
The value of
∫
−π
sin3x cos2x dx is _______.