(ii) We have
2
2
3
3
(
1)
x
dx
x dx
dx
+
=
+
∫
∫
∫
=
2
1
3
C
2
1
3
x
x
+
+
+
+
 = 
335
C
5
x
+x
+
(iii) We have 
3
3
2
2
1
1
(
2
)
2
x
x
x
e –
dx
x
dx
e dx –
dx
x
x
+
=
+
∫
∫
∫
∫
=
3
1
2
2
– log
+ C
3
1
2
x
x
e
x
+
+
+
=
25
2
2
– log
+ C
5
x
x
e
x
+
Example 3 Find the following integrals:
(i)
(sin
cos )
x
x dx
+
∫
(ii) cosec
(cosec 
cot
)
x
x
x dx
+
∫
(iii)
2
1
–cossin
xx dx
∫
Solution
(i)
We have
(sin
cos )
sin
cos
x
x dx
x dx
x dx
+
=
+
∫
∫
∫
= – cos
sin
C
x
x
+
+
INTEGRALS         297
(ii)
We have
2
(cosec
(cosec
+ cot
)
cosec
cosec
cot
x
x
x dx
x dx
x
x dx
=
+
∫
∫
∫
= – cot
cosec
C
x –
x +
(iii)
We have
2
2
2
1 sin
1
sin
cos
cos
cos
–
x
x
dx
dx –
dx
x
x
x
=
∫
∫
∫
= 
sec2
tan
sec
x dx –
x
x dx
∫
∫
= tan
sec
C
x –
x +
Example 4 Find the anti derivative F of  f defined by f (x) = 4x3 – 6, where F (0) = 3
Solution One anti derivative of f (x) is x4 – 6x since
(4
6 )
d
x – x
dx
 = 4x3 – 6
Therefore, the anti derivative F is given by
F(x) = x4 – 6x + C, where C is constant Given that
F(0) = 3, which gives,
3 = 0 – 6 × 0 + C    or    C = 3
Hence, the required anti derivative is the unique function F defined by
F(x) = x4 – 6x + 3 Remarks
(i)
We see that if F is an anti derivative of f, then so is F + C, where C is any
constant Thus, if we know one anti derivative F of a function f, we can write
down an infinite number of anti derivatives of f by adding any constant to F
expressed by F(x)  + C, C ∈ R