UNIT – IV
INTEGRATION - II
4.1 INTEGRATION BY PARTS
Integrals of the form ∫x sin nx dx , ∫x cos nx dx, ∫xe dx,
n
∫x log xdx, ∫xlog x dx - Simple Problems
nx
4.2 BERNOULLI’S FORMULA
Evaluation of the integrals ∫x cos nx dx , ∫x sin nx dx,
m
nx
∫x e dx, when m ≤ 2 using Bernoulli’s formula. Simple
Problems.
m
m
4.3 DEFINITE INTEGRALS
Definition of definite integral, properties of definite integrals.
Simple problems.
4.1 INTEGRATION BY PARTS
Introduction:
W hen the integrand is a product of two functions and the method
of decomposition or substitution cannot be applied, then the method of
by parts is used.In differentiation , we have seen.
d
dv
du
(uv ) = u
+v
dx
dx
dx
(ie) d(uv) = udv+ vdu
Integrating both sides ;
∫d (uv)
= ∫udv + ∫vdu
uv
= ∫udv + ∫vdu
(ie) ∫ udv
= uv - ∫vdu
∴ ∫ udv = uv - ∫vdu is called Integration By Parts formula
103
The above formula is used by taking proper choice of ‘u’ and ‘dv’
‘u’ should be chosen based on the following order of preference
1. Inverse trigonometric functions
2. Logarithmic functions
3. Algebraic functions
4. Trigonometric function
5. Exponential Functions
Example:
1.
Evaluate ∫ x cosx dx
Solution :
∫ udv
= uv - ∫vdu
Choosing u =x
and
dv = cosx dx
∫dv= ∫cosxdx
du = dx
v = sin x
.
. .∫x cos x dx = xsinx -∫sinx dx
= xsinx + cosx + c
2.
Evaluate ∫ x sinx dx
2
Here we have to apply the integration by parts formula twice
Solution :
∫ udv
= uv - ∫vdu
Choosing u =x
2
and
dv = sinx dx
∫dv= ∫sinxdx
du = 2x dx
v = -cos x
.
2
2
. .∫x sin x dx = x (–cosx) -∫( –cosx) 2x dx
2
= -x cosx +2∫x cosx dx
2
= -x cosx+2I, W here I
Choosing u =x and
dv = cosx dx
104
=∫xcosx dx
∫dv= ∫cosxdx
du = dx
v = sin x
I=∫x cos x dx
= xsinx -∫sinx dx
= xsinx - (-cosx)
= xsinx + cosx)
∫x sin x dx = -x cosx + 2[xsinx+cosx] + c
2
2
4.1 WORKED EXAMPLES
PART – A
1.
.
x
Evaluate . .∫xe dx
Solution:
∫ udv = uv - ∫vdu
Choosing u = x
x
and
dv = e dx
∫dv= ∫ e dx
x
du = dx
v =e
.
x
. .∫x e dx
x
x
x
= x e -∫ e dx
x
x
=x e -e +c
2.
Evaluate ∫ x sinx dx
∫ udv = uv - ∫vdu
Choosing u =x
du =dx
and
dv = sinx dx
∫dv = ∫sinxdx
v = -cos x
.
. .∫xsin x dx
= x(–cosx) -∫( –cosx) dx
= -x cosx +∫ cosx dx
=-x cosx+sinx +c
105
PART - B
1.
Evaluate ∫ x log x dx
Solution :
∫ udv = uv - ∫vdu
Choosing u = logx and
du =
dv = xdx
1
dx
x
∫dv= ∫xdx
v=
∴∫x logx dx = log x
2.
x2
2
x2
x2 1
−
dx
2
2 x
∫
=
x2
1
log x −
2
2
=
x2
1 x2
log x −
+C
2
2 2
=
x2
x2
log x −
+C
2
4
∫ xdx
2 ax
Evaluate.∫x e dx
Solution:
∫ udv = uv - ∫vdu
Choosing u = x
2
and
ax
dv = e dx
∫dv = ∫ e dx
ax
du = 2x dx
v =
.
2
ax
. .∫x e dx
=
eax
a
eax
x 2 eax
-∫
2x dx
a
a
106
Choosing u = x
=
x 2 eax
a
−
2
ax
∫ xe dx
a
=
x 2 eax
a
−
2
ax
I W here I=∫x e dx
a
ax
and
dv = e
∫dv = ∫ e dx
ax
du =dx
e ax
a
v=
I= ∫x e dx
ax
= x
∴∫ x e dx
2
3.
ax
e ax
e ax
-∫
dx
a
a
=
x eax
a
−
1 ax
∫ e dx
a
=
x eax
a
−
1 e ax
a a
=
x 2 eax
a
−
2
a
⎡ x e ax eax ⎤
− 2 ⎥
⎢
a ⎥⎦
⎢⎣ a
Evaluate ∫log x dx
Solution:
∫ udv = uv - ∫vdu
Choosing u = logx and
du =
1
dx
x
dv = dx
∫dv= ∫dx
v =x
.
.
.∫logx dx
= logx x -∫x
1
dx
x
= xlogx-∫dx
= xlog x –x+c
107
+C
4. Evaluate ∫(x+3) cos 5x dx
Solution:
∫ udv = uv - ∫vdu
Choosing u =x+3 and
dv = cos5x dx
du = dx
∫dv= ∫cos5x dx
sin 5x
v=
5
sin 5x
sin 5x
.
−
dx
. .∫(x+3) cos 5x dx = (x + 3)
5
5
(x + 3) sin 5 x 1
− ∫ sin 5x dx
=
5
5
(x + 3) sin 5x 1 ⎛ − cos 5x ⎞
=
− ⎜
⎟+c
5
5⎝
5
⎠
∫
(x + 3) sin 5x cos 5x
+
+c
5
25
=
5.
Evaluate ∫x logx dx
n
Solution:
Choosing u = logx
du =
and
1
dx
x
n
dv = x dx
∫dv= ∫ x dx
n
v=
∫ x logx dx
n
= log x
x n +1
n +1
x n +1
x n +1 1
−∫
dx
n +1
n +1 x
=
x n +1 log x
1
x n dx
−
n +1
n +1 ∫
=
x n +1 log x
1 x n +1
−
+c
n +1
n +1 n +1
=
x n +1 log x
x n +1
−
+c
n +1
(n + 1)2
108
4.2 BERNOULLI’S FORM OF INTEGRATION BY PARTS
FORMULA
If u and v are functions of x, then Bernoulli’s form of integration
by parts formula is
∫udv = uv – u′v1+u′′v2 -u′′′v3+…….., W here u′,u′′, u′′′…. are
successive differentiation of the function u and v 1,v 2,v 3…. the
successive integration of the function v.
Example:
Evaluate ∫x e dx
2
and
Choosing u =x
u′ = 2x
2 ax
u′′ = 2
ax
dv = e dx
ax
∫dv= ∫ e dx
ax
e
v =
a
e ax
v1 = 2
a
v2 =
e ax
a3
∫udv = uv –u′v 1+u′′v 20-u′′′v3+…… ..,
x 2 e ax 2x e ax 2e ax
2 ax
+ 3 +c
∫x e dx =
a
a2
a
Note: The function ‘u’ gets differentiated till its derivative becomes a
constant.
4.2 WORKED EXAMPLE
PART – B
1.
Evaluate ∫x cos xdx
2
Solution:
∫udv = uv –u′v 1+u′′v 2-u′′′v 3+…… ..,
Choosing u = x
2
u′ = 2x
and
dv = cos x dx
∫dv= ∫ cos x dx
109
v =sinx
v 1 = - cosx
u′′ = 2
v 2 = -sinx
∴∫x cos x dx
2
2.
2
=
x sinx –2x (-cos x) + 2 (-sin x) + C
=
x sinx + 2x cos x - 2 sin x + C
2
Evaluate ∫x e dx
3
2x
Solution:
∫udv = uv –u′v 1+u′′v 2-u′′′v 3+…… ..,
3
2x
Choosing u = x
and
du = e dx
2
2x
u′ =3x
∫dv= ∫e dx
e 2x
u′′= 6x
v=
2
u′′′ =6
e 2x
4
e 2x
v2 =
8
v1 =
e 2x
16
6 xe2 x 6e 2 x
+
−
+c
8
16
v3 =
∫x
e dx =
3 2x
x 3 e 2x 3x 2e 2x
−
2
4
4.3 DEFINITE INTEGRALS
Definition of Definite Integrals:
Let ∫f(x)dx= F(x) + C, W here C is the arbitrary constant of
integration The value of the integral
when x = b, is F (b) + C
…1
and when x = a, is F (a) + C
…2
Subtracting (2) from (1) we have
F(b) –F(a) = (The value of the integral when x=b)
- (The value of the integral when x=a)
110
b
∫
(ie)
b
f ( x )dx = [F (x ) + c]
a
a
= [F(b) + c] - [F(a) +c]
= [F(b) +c] –[F(a) +c]
= F(b) –F(a)
b
Thus
∫ f(x ) dx is called the definite
integral, here a and b are
a
called the lower limit and upper limit of integral respectively.
Properties of Definite integrals
Certain properties of definite integral are useful in solving
problems. Some of the often used properties are given below . It is
assumed throughout that f′(x) = F(x)
a
1
∫ f(x )dx
a
b
2
∫
a
b
3
4
∫
a
b
5
a
∫
f (x )dx = − f (x )dx
b
∫ kdx
a
b
=0
= k(b − a ) where k is a cons tan t
b
∫
kf (x )dx = k f( x ) dx
a
∫ [f(x ) ± g(x )]dx
a
a
b
∫
∫
= f( x ) dx ± g( x ) dx
b
a
b
6
If f(x) ≥ 0 on [a, b] , then
∫ f(x ) dx ≥ 0
a
b
7
If f(x) ≤ 0 on [a, b] , then
∫ f(x ) dx ≤ 0
a
111
If f(x) ≥ g(x)on [a, b] , then
8
b
b
∫ f(x ) dx ≥ ∫ g(x ) dx
a
If a < c < b in [a, b] ,
9
10
b
b
a
a
a
b
c
b
a
a
c
∫ f(x ) dx = ∫ f(x ) dx + ∫ f(x ) dx
∫ f(x ) dx = ∫ f(t) dt
(i.e) value of the integral is independent of the variable of
integration.
a
11
a
∫
f (x )dx = f (a − x )dx
0
∫
o
b
12
b
∫
f (x ) dx = f(a + b − x ) dx
∫
a
a
−a
0
a
13
a
∫ f(x ) dx = 2∫ f(x ) dx, if f(x )is even
=0
2a
14
∫
i.e f (− x ) = f (x )
if f (x ) is odd
a
i.e f(− x ) = f ( x )
2a
∫
∫
f( x ) dx = f( x ) dx + f(2a − x ) dx
0
0
0
Examples:
9
1.
25
∫
Evaluate
x dx
if
25
∫
x dx =
9
196
3
Solution:
9
∫
25
∫
=−
x dx
25
b
x dx u sin g property
9
=−
a
Given that
∫
∫
b
196
3
7
2.
∫
a
f( x ) dx = − f ( x ) dx
7
f(x ) dx = 20
∫
4
∫
and f(x ) dx = 13, find f(x ) dx
2
4
112
2
Solution:
b
Using
c
b
a
7
c
4
7
2
2
4
∫ f(x ) dx = ∫ f(x ) dx + ∫ f(x ) dx if a < c < b,
a
∫ f(x ) dx = 20 = ∫ f(x ) dx + ∫ f(x ) dx
W e have
4
20 = ∫ f ( x ) dx + 13
2
4
∴ ∫ f (x ) dx = 20 − 13
=7
2
3
3.
Evaluate
∫
3
x dx
1
Solution:
3
∫
1
3
3
x dx =
∫
1
3
x dx
1
3
3
⎡ 4⎤
⎡ 1 +1 ⎤
⎢x3 ⎥
⎢x 3 ⎥
=⎢
⎥ =⎢ 4 ⎥
⎢ ⎥
⎢ 1 + 1⎥
⎢⎣ 3
⎥⎦1 ⎢⎣ 3 ⎥⎦ 1
3
4
4
⎤
3 ⎡ ⎤
3⎡
= ⎢ x 3 ⎥ = ⎢3 3 − 1⎥
4 ⎢ ⎥
4⎢
⎥
⎣ ⎦1
⎣
⎦
113
WORKED EXAMPLES
PART – A
2
1.
dx
x
1
∫
Evaluate
Solution:
2
Let I =
dx
x
1
∫
= [log x ]1
2
= log2 –log 1
(∵ log 1 = 0)
I = log 2
π
2
2.
Evaluate
∫ sin xdx
0
Solution:
π
2
Let I =
∫ sin xdx
0
π
= [− cos x ]0 2
= − cos
π
+ cos 0
2
= 0+1
I = 1
114
π
4
3.
Evaluate
∫ sec
2
xdx
0
Solution:
π
4
∫
Let I = sec 2 xdx
0
π
= [tan x ]04
= tan
π
4
-tan 0
= 1-0
I =1
1
4.
Evaluate :
dx
∫
1 − x2
0
Solution:
1
Let I =
∫
0
dx
1 − x2
[
= sin−1 x
]
1
0
-1
-1
= sin 1 - sin 0
=
I =
π
−0
2
π
2
115
2
5.
Evaluate
∫ (x − x
2
)dx
1
Solution:
2
Let I = ∫ ( x − x 2 )dx
1
2
⎡ x2 x3 ⎤
−
⎥
⎢
3 ⎥⎦
⎣⎢ 2
1
⎡ 2 2 23 ⎤ ⎡ 1 1 ⎤
−
−
⎥ −
⎢
3 ⎥⎦ ⎢⎣ 2 3 ⎥⎦
⎢⎣ 2
8⎤ ⎡1 1⎤
⎡
⎢2 − 3 ⎥ − ⎢ 2 − 3 ⎥
⎣
⎦ ⎣
⎦
2 1 −4 − 1
− − =
3 6
6
5
=−
6
1
6.
Evaluate
∫x
2
dx
−1
Solution:
2
f(x) = x
2
2
Now f(-x) = (-x) = x = f(x)
∴f(x) is an even function
1
2
∫ x dx
−1
1
= 2∫ x 2dx Using property of even function
0
1
⎡ x3 ⎤
= 2⎢ ⎥
⎣⎢ 3 ⎦⎥ 0
=
⎡13 03 ⎤
= 2⎢ − ⎥
3 ⎥⎦
⎣⎢ 3
2
=
3
116
PART - B
π
2
1.
Evaluate:
∫ cos
2
x dx
2
x dx
0
Solution:
π
2
Let I =
∫ cos
0
π
2
=
I
1 + cos 2x
dx
2
0
1 + cos 2x ⎤
⎡
2
⎢∵ cos x =
⎥
2
⎣
⎦
∫
π
2
π
(1 + cos 2x )dx = 1
2
0
sin 2x ⎤ 2
⎡
⎢x + 2 ⎥
⎣
⎦0
=
1
2
=
π
⎡
⎤
sin 2
⎥
sin
2
(
0
)
1 ⎢π
⎫
⎧
2 − 0+
⎬⎥
⎨
⎢ +
2
2 ⎭⎥
2 ⎢2
⎩
⎣
⎦
=
1
2
⎡ π sin π
⎤
⎢ 2 + 2 − {0 + 0}⎥
⎣
⎦
=
1
2
⎡π
⎤
⎢ 2 + 0⎥
⎣
⎦
=
π
4
∫
117
π
2
2.
Evaluate :
∫ Sin x dx
3
0
Solution:
π
2
Let I =
∫ Sin x dx
3
0
=
1
4
sin 3x = 3 sin x − 4 sin3 x
1
∴ sin3 x = [3 sin x − sin 3x ]
4
π
2
∫ [3 sin x − sin 3x ]dx
0
π
=
1⎡
⎛ − cos 3x ⎞⎤ 2
⎟⎥
⎢− 3 cos x − ⎜
4⎣
3
⎝
⎠⎦ 0
=
1⎡
cos 3 x ⎤ 2
− 3 cos x +
⎢
4⎣
3 ⎥⎦ 0
π
=
=
⎡
⎤
⎛π⎞
cos 3⎜ ⎟
⎥
1 ⎢⎢
π
cos
0
⎫⎥
⎝ 2 ⎠ − ⎧− 3 cos 0 +
− 3 cos +
⎨
⎬
4⎢
2
3
3 ⎭⎥
⎩
⎢
⎥
⎣
⎦
1⎡
1 ⎫⎤
⎧
⎢0 + 0 − ⎨− 3(1) + ⎬⎥
4⎣
3 ⎭⎦
⎩
=
1⎡
1 ⎤ 1 ⎡ 9 − 1⎤
0+3− ⎥ = ⎢
⎢
4⎣
3 ⎦ 4 ⎣ 3 ⎥⎦
=
1⎛8⎞ 2
⎜ ⎟=
4⎝3⎠ 3
118
π
2
3.
Evaluation :
cos 2 x
∫ 1 + sin x dx
0
Solution:
π
2
Let
cos 2 x
∫ 1 + sin x dx
=
0
π
2
1 − sin2 x
∫ 1 + sin x dx
0
=
π
2
∫
=
(1 + sin x )(1 − sin×) dx
(1 + sin x )
0
π
2
∫ (1 − sin x )dx
=
0
π
[x − (− cos x )]02
=
π
2
0
[x + (cos x )]
=
π
π
+ cos − (0 + cos 0)
2
2
π
+ 0 − (0 + 1)
2
π
−1
2
=
=
I
=
π
2
4.
Evaluate
∫ Sin
3
x cos x dx
3
x cos x dx
0
Solution:
Let I =
∫ Sin
Put u = sin x
du = cosx dx
119
∫
= u3 du
=
u4
4
=
sin 4 x
4
π
2
π
⎡ sin4 x ⎤ 2
Now Sin x cos x dx = ⎢
⎥
⎣ 4 ⎦0
0
4
⎤
1 ⎡⎛
π⎞
4
⎢
=
⎜ sin ⎟ − (sin 0) ⎥
4 ⎢⎝
2⎠
⎥⎦
⎣
1
1 4
= (1) − 0
=
4
4
∫
3
[
]
π
2
5.
Evaluate :
∫ cos 5x sin 3x dx
0
Solution:
Let
=
1
2
∫ cos 5x sin 3xdx
cos A sin B
∫ [sin(5x + 3x )− sin(5x − 3x )]dx
=
1
[sin 8x − sin 2x ] dx
2
1 ⎡ − cos 8x (− cos 2x ) ⎤
= ⎢
−
⎥
2⎣
8
2
⎦
=
∫
1 ⎡ cos 8x cos 2x ⎤
= ⎢−
+
2⎣
8
2 ⎥⎦
π
2
Now
∫ cos 5x sin 3x dx
0
π
1 ⎡ cos 8 x cos 2x ⎤ 2
+
= ⎢−
2⎣
8
2 ⎥⎦ 0
120
1
[sin(A + B) − sin(A − B)]
2
⎡
⎤
⎛π⎞
⎛π⎞
cos 8⎜ ⎟ cos 2⎜ ⎟
⎥
cos
0
cos
0
1 ⎢⎢
2
2
⎧
⎫
⎝ ⎠+
⎝ ⎠− −
+
= −
⎨
⎬⎥
2
8
2 ⎭⎥
8
2⎢
⎩
⎢
⎥
⎣
⎦
1 ⎡ cos 4π cos π ⎧ 1 1 ⎫⎤
= ⎢−
+
− ⎨− + ⎬⎥
2⎣
8
2
⎩ 8 2 ⎭⎦
1⎡ 1 1 1
= ⎢− − + −
2⎣ 8 2 8
=
1
[− 1] = − 1
2
2
9
6.
Evaluate
x
∫
x + 9−x
0
1⎤
2 ⎥⎦
dx
Solution:
Using
a
a
0
0
∫ f(x )dx = ∫ f(a − x ) dx W e get
9
=
I
∫
0
x
x+
9
=
∫
0
∫
0
dx
9−x
…(1)
9 − x 9 − (9 − x )
9
I
9−x
9−x
9−x + x
dx
dx
…(2)
Adding (1) and (2)
9
2I
=
∫
0
9
=
∫
0
x
x + 9−x
x + 9−x
x + 9−x
9
dx + ∫
0
dx
121
9−x
9−x + x
dx
9
=
∫ 1dx = [x ]0
9
0
= 9-0
2I
= 9
I
=
9
2
π
2
7.
Evaluate
sin x
∫ sin x + cos x dx
0
Solution:
a
Using the result
∫ f(a)dx
0
I
=
π
2
a
= ∫ f(a − x ) dx W e get
0
sin x
∫ sin x + cos x dx
…(1)
0
⎛π
⎞
sin⎜ − x ⎟
2
⎝
⎠
dx
= ∫
π
⎞
⎛π
⎞
0 sin⎛
⎜ − x ⎟ + cos⎜ − x ⎟
⎝2
⎠
⎝2
⎠
π
2
π
2
=
cos x
∫ cos x + sin x dx
…(2)
0
Adding (1) and (2)
π
2
2I
π
2
sin x
cos x
dx + ∫
dx
= ∫
sin x + cos x
cos x + sin x
0
0
π
2
=
sin x + cos x
∫ sin x + cos x dx
0
122
π
2
=
π
∫ 1dx = [x ]20
0
π
−0
2
π
=
2
π
=
4
=
2I
I
π
2
8.
Evaluate:
∫ x cos x dx
−
π
2
Solution:
f(x) = x cos x
f(-x) = (-x) cos (-x) = -x cos x
= -f(x)
∴ f(x) is an odd function
π
2
∫ x cos x dx = 0 Using property of odd function
−
π
2
EXERCISE
PART – A
1. Find the value of
∫ xe
2. Evaluate
∫ x cos x dx
3. Evaluate
∫ xe
−x
2x
dx
dx
4. Evaluate the following
1
(i)
dx
5 − 3x
0
∫
1
(ii)
∫ (5 − 3x )dx
0
123
∫ (x + 2x
2
(iii)
1
4
)dx
5−x
0
(iv)
(vi)
(ix)
∫
(viii)
∫ tan
0
1
dx
0 1+
dx
∫ sin xdx
0
π
4
π
(vii) ∫ cos xdx
0
1
∫ cos x
0
π
dx
∫
(v)
2
π
2
x2
(x)
∫
2
xdx
ex
0 1+ e
x
dx
PART – B
1.
2.
Integrate the following with respect t x :
2x
(i) x cos nx
(ii) (2x-1) e
2
2 3x
(iii) x sin x
(iv) x e
2
3
(v) x cos x
(vi) x log x
2 -x
2
(vii) x e
(viii) x sec x
Evaluate the following
π
2
(i)
∫ sin
2
π
2
xdx
(ii)
0
π
2
(iii)
cos 2 x
∫ 1 − sin x dx
0
π
2
(v)
0
π
2
(vii)
sin x
∫ cos
(iv)
5
x sin x dx
xdx
sin2 x
∫ 1 + cos x dx
0
π
2
dx
3
0
π
2
2
∫ 1 − cos x
∫ cos
(vi)
∫ (2 + sin x )
3
cos x dx
0
π
2
(viii)
0
∫
0
124
sin x cos x dx
π
6
(ix)
∫ sin 2x cos 3x dx
0
1
(xi)
π
4
∫ (2x + 3)
4
(x)
0
2
dx
(xii)
0
π
4
(xiii)
(xv)
∫x
cos x
∫ 1 + sin x
(xvii)
(xix)
∫
∫ sin
(xvi)
π
e x + e−x
dx
(xviii)
2
∫ (sin x + cos x ) dx
2
∫ (3x
1
1 + sin x dx
(i) Evaluate
(xx)
0
∫
1
x
x + 5−x
a
(ii) Evaluate
∫x
2
(a −
3
x )2
dx
dx
0
1
(iii) Evaluate
x cos xdx
0
5
∫ x(1 − x )
3
dx
0
ANSWERS
PART – A
1.
7
0
0
3.
dx
∫ 1 + e− x
0
π
2
dx
∫ e x − e −x
0
π
2
x dx
x 3 + 1 dx
2
1
(xiv)
0
1
2
0
∫ (cos 2x + sin 4x ) dx
0
π
2
∫ tan x sec
xe 2 x 1 2 x
− e +c
2
4
125
2
)
+ 1 (x − 1) dx
2.
xsinx +cos x +c
3
.-xe -e +c
4
.(i)
-x
-x
1 ⎛5⎞
log⎜ ⎟
3 ⎝2⎠
(ii)
7
2
(iv) 1
(v) log
(vii) 0
(viii) 1-
(iii)
5 −1
5 −4
π
4
37
6
(vi) 2
(ix)
π
4
(iii)
π
+1
2
⎛ 1+ e ⎞
(x) log⎜
⎟
⎝ 2 ⎠
PART – B
1
x
1
sin n x + 2 cos nx +c
n
n
2x
(ii) e (x-1) =c
2
(iii) –x cosx +2x sin x +2cos x +c
(i)
(iv)
e 3 x ⎛ 2 2x 2 ⎞
+ ⎟+c
⎜x −
3 ⎝
3 9⎠
(v)
x2
1
+ x sin 2x + cos 2x + c
4
4
(vi)
x 4 log x x 4
−
+c
4
16
(vii) − x 2e − x − 2xe− x 2e − x + c
(viii) x tanx - log sec x +c
2.
(i)
π
4
(ii)
2
3
126
(iv)
π
−1
2
(vii).
1
6
(x)
1
2
(xiii) 1
1
8
(xix) 2
(xvi)
3
3. (i)
2
(v)
π
+1
2
(vi)
65
4
2
3
(ix)
3 3 −4
10
(xii)
52
9
(viii)
(xi) 288.2
(xiv) log
1+ e
2
(xvii) log
x
e2 + 1
(xviii) + 1
2e
2
(xv) log 2
(xx) 4
a
16a 2
(ii)
3 /5
127
(iii)
1
20