Aviation Mathematics, AERO 1010
Integration Practice
The examples below relate to integration using either the power rule or the chain rule.
Break the function up into its component parts u(x) and u'(x) ready for integration.
1
n +1
( u ( x) ) + c
n +1
Power Rule:
∫ ( u ( x) )
Chain Rule:
∫ f (u( x)).u '( x)dx = ∫ f (u )du (Also known as integration by
n
u '( x)dx =
chain rule derivative)
1.
∫(x
2
+ 2 ) .2 xdx
2.
∫(x
3
+ 3 x 2 − 1) (3 x 2 + 6 x)dx
3.
∫x
4.
∫ (x
5.
∫ sin( x) ( cos( x) + 1)
3
−3
x 2 + 4dx
2
+ 1) sin( x 3 + 3 x ) dx
4
dx
Answers
1.
u ( x) = x 2 + 2; u '( x) = 2 x
∫(x
2.
3
4
1
1
+ 2 ) .2 xdx = ∫ u 3 .u '( x)dx = ∫ u 3 du = u 4 + c = ( x 2 + 2 ) + c
4
4
u ( x) = x3 + 3x 2 − 1; u '( x) = 3 x 2 + 6 x
∫(x
3.
2
3
+ 3x 2 − 1) (3 x 2 + 6 x)dx = ∫ u −3 .u '( x)dx = ∫ u −3 du =
−3
u ( x) = x 2 + 4; u '( x) = 2 x
2
∫ x x + 4dx =
4.
−2
1 −2
−1
u + c = ( x3 + 3 x 2 − 1) + c
2
−2
3
1
1 12
1 12
1 2 32
1 2
2
2 +c
x
+
x
dx
=
u
u
x
dx
=
u
du
=
u
+
c
=
x
+
4(2
)
.
'(
)
*
4
(
)
2∫
2∫
2∫
2 3
3
u ( x) = x3 + 3x; u '( x) = 3x 2 + 3
1
1
1
2
3
3
2
∫ ( x + 1) sin( x + 3x)dx = 3 ∫ sin(x + 3).(3x + 3)dx = 3 ∫ sin(u).u '( x)dx = 3 ∫ sin(u)du
−1
1
= (− cos(u ) + c = cos( x3 + 3x) + c
3
3
G R Lockwood, University of South Australia, 2003
Aviation Mathematics, AERO 1010
5.
u ( x) = cos( x) + 1; u '( x) = − sin( x)
∫ sin( x) ( cos( x) + 1)
=−
4
dx = − ∫ ( cos( x) + 1) (− sin( x)dx = − ∫ ( u ( x) ) .u '( x)dx = − ∫ ( u ( x) ) du
1 5
1
5
( u ) + c = − ( cos( x) + 1) + c
5
5
G R Lockwood, University of South Australia, 2003
4
4
4