(3)
Substituting (2) and (3) in (1), we obtain
2
1
2
3
2
5 – 4 –
+ sin
C
3
5
4
–
x
x
–
x
x
– x – x
+
+
=
+
∫
, where 
1
2
C
C
C
2
–
=
EXERCISE 7 4
Integrate the functions in Exercises 1 to 23 1 2
6
3
1
x
x +
2