
11-26 Calculus – Differentiation and Integration
Example 11.21 Evaluate
dx
x
1 2
0
+
∫
cos
.
p
Solution:
I
dx
x
=
+
∫
1 2
0
cos
p
(1)
=
+
∫
=
+
∫
=
+
−
−
dx
dx
x
x
x
1 2
11
1 2
2
1 2
0
0
cos( )
cos
cos
co
( )
p
p
p
by property
ss
.
x
dx
0
p
∫
Thus,
I dx
x
x
=
+
∫
2
1 2
0
cos
cos
p
(2)
Adding Eqs. (1) and (2), we obtain
2
1
1 2
2
1 2
1
1 2
2
1 2
0 0
I dx dx
x
x
x
x
x
x
=
+
∫
+
+
∫
=
+
+
+
cos
cos
cos
cos
cos
cos
p p
∫
=
+
+
∫
=
∫
dx
dx dx
x
x
0
0 0
1 2
1 2
p
p p
(by property 5)
cos
cos
==
[ ]
=
x
0
p
p
(by FTC)
.
∴ =
+
∫
=I
dx
x
1 2
2
0
cos
.
p
p
Example 11.22 Evaluate
ln .
3
3
3
3
−
+
∫
−
x
x
dx
Solution: Let
g x
x
x
( ) ln=
−
+
3
3
Now consider
g x
x
x
x x
( ) ln
ln( ) ln( )
− =
+
−
= + − −
3
3
3 3