Thus
2
3
2x
x dx
−
−
∫
 =
2
4
−y dy
∫
=
2
1
1
4
4
sin
C
2
2
2
– y
y
−y
+
+
[using 7 6 2 (iii)]
=
2
1
1
1
(
1)
3
2
2 sin
C
2
2
–
x
x
x
x
+


+
−
−
+
+




EXERCISE 7 7
Integrate the functions in Exercises 1 to 9