
12-16 Calculus – Differentiation and Integration
We solve
5
5
2 2
( )−
∫
t
dt using substitution method.
Let
t = 5 sin ,q then dt d= 5 cosq q
5
5
5
5 5
5
1
5
5
1
5
2 2 2 2
2 2
3
( ) ( sin )
cos
(cos )
cos
sec
−
∫
=
−
∫
=
∫
=
t
dt d
d
q
q q
q
q q
qq q
q q q q
d
c
x
x x
x
∫
= + + +
=
+
− −
+
+
1
2 5
1
2 5
5 1
4 2
5
2
(sec tan log sec tan )
( )
log
++
− −
+
1
4 2
2
x x
c
∴
−
− −
∫
=
− −
−
+
− −
+
+ +
−
x
x x
dx
x x
x
x x
x4
4 2
1
2 4 2
1
2 5
5 1
4 2
5 1
4 2
2 2 2 2
( ) ( )
( )
log
xx x
c
−
+
2
.
Problems of this kind can be solved using a standard technique which we shall discuss in the next
chapter.
12.3 INTEGRATION BY PARTS
The method of integration by parts relies on the product rule
d
dx
uv u
dv
dx
v
du
dx
( ) = +
whose differential ...