Integration Key Facts
Facts which are in the formula book
Standard Integrals you need to learn:
1
n
n1
x dx n 1 x c
dx e x c
x dx ln x c
cos xdx sin x c
sin xdx cos x c
sec xdx tan x c
sec x tan xdx sec x c
cos ec xdx cot x c
cos ecx cot xdx cos ecx c
2
Further generalisations using the above integrals:
1
n
n1
(ax b) dx a( n 1) (ax b) c
1 ax b
ax b
e dx a e c
1
1
ax b dx a ln( ax b) c
1
cos(ax b)dx a sin( ax b) c
1
sin(ax b)dx a cos(ax b) c
1
2
sec (ax b)dx a tan(ax b) c
1
sec(ax b) tan(ax b)dx a sec(ax b) c
1
2
cos ec (ax b)dx a cot(ax b) c
1
cos ec(ax b) cot axdx a cos ec(ax b) c
Area under a curve, where y = f(x) between x=a and x=b:
b
ydx
a
Volume of revolution formed by rotating y about the x axis, where y = f(x) between x=a and
x=b:
b
y 2 dx
a
General patterns you should remember:
1
n
n 1
f ' ( x ) f ( x ) dx n 1 f ( x ) c
f ' ( x )e
f ( x)
dx e f ( x ) c
f '( x)
dx ln f ( x ) c
f ( x)
f ' ( x ) cos f ( x )dx sin f ( x ) c
f ' ( x ) sin f ( x )dx cos f ( x ) c
f ' ( x ) sec f ( x )dx tan f ( x ) c
2
Integrating trigonometric powers
1
2
sin xdx 2 (1 cos 2 x )dx (double angle formula used)
1
2
cos xdx 2 (1 cos 2 x )dx (double angle formula used)
1
n
n1
sin x cos xdx n 1 sin x c
1
n
n 1
cos x sin xdx n 1 cos x c
1
n
n
sec x tan xdx n sec x c
1
n
2
n1
tan x sec xdx n 1 sec x c
Integrating using Addition Formula
To integrate sin(ax)sin(bx), cos(ax)cos(bx) cos (A B) = cosAcosB sinAsinB (C3 section) are
required.
To integrate sin(ax)cos(bx) sin (A B) = sinAcosB sinAcosB (C3 section) are required.
Other trigonometric integrals:
tan xdx ln(cos x ) c ln(sec x ) c
sec xdx ln(sec x tan x ) c
cot xdx ln(sin x ) c
cos ecxdx ln(cos ecx cot x ) c
Trapezium Rule (C2 section)
b
h
a f ( x )dx 2 y0 yn 2( y1 y2 .... yn1 )
ba
n
Separating variables:
dy
When
f ( x ) g( y )
dx
1
Then
dy f ( x )dx
g( y )
Integrating parametric equations:
When x = f(t), y = (t), the area under the curve:
dx
y dt
dt
Volume of revolution formed by rotating y about the x axis,
t2
y2
t1
dx
dt
dt
Integration By Parts
dv
du
u dx dx uv v dx c
How do you know which is u and
dv
?
dx
Integration by parts is used to integrate a product. One part of the product (u) should be easy to
dv
differentiate (and will usually simplify when differentiated). The other part should be easy to
dx
integrate and should not become too much harder when integrated.
Examples:
Function to be integrated
u needs differentiating to find
du
dx
dv
needs integrating to find v
dx
x sin ax
x cos ax
xn ln x
x ex
x2ex
2
x sinx
x2cosx
x
x
ln x
x
x2
parts will need to be applied
twice this would apply to
other similar examples.
ln x
x
x
sin ax
cos ax
xn
ex
ex
sinx
cosx
lnx
x(2x + 3)4
x sec2xtanx
1
(2x + 3)4
sec2xtanx
Trigonometric Identities which could be required:
sin 2 x 12 (1 cos 2 x)
cos 2 x 12 (1 cos 2 x)
sin 2 x cos 2 x 1
tan 2 x 1 sec 2 x
1 cot 2 x cos ec 2 x
Which method of integration do you use?
The methods of integrating an expression are:
1) directly writing it down (either by memory or by looking in the formula book);
2) writing the expression in partial fractions;
3) using the method of integration by parts;
4) using a substitution;
2
2
5) using a trigonometric identity (such as sin x 12 (1 cos 2 x) or cos x 12 (1 cos 2 x) )
First look to see if the question tells you (or gives you a hint) about which method of integration you
should be using!!
First look to see if the integral is easy to work out: Is it one that you should remember (see my sheet)
or is it one in the formula book? Also is the integral easier to work out than it looks e.g.
If you are asked to integrate an algebraic fraction
ax b
of the form
or
(cx d )(ex f )
ax 2 bx c
, you could try writing it in
(dx e) 2 ( fx g )
partial fractions.
(Note: you might be asked in the first part of the
question to express it in partial fractions)
Examples are:
(2 x 1)( x 2) dx
x(4 x)dx
2 xe dx
x cos( x 2)dx
(cos x)e dx
(2 x 5) ln( x 5 x)dx
x2
2sin x
To integrate even powers of sin x or cos x , use
2
the formulae sin x 12 (1 cos 2 x) or
cos 2 x 12 (1 cos 2 x) .
Examples are:
5x 2
dx
x
Other quotients (where the denominator cannot
be factorised into linear factors) can usually be
integrated using a substitution.
Examples are:
x
dx put u x 2 1
1
sin x
put u cos x
cos x
2x 1
dx put u x 2 x 1
2
( x x 1) 2
4 x
Recognise that the integral is in the form
f '( x ) gf ( x) . Examples are:
cos
sin
xdx
xdx
Use integration by parts to integrate other
products, especially when one function is a
simple polynomial. Examples are:
(3x 1)e dx
x cos xdx
2 x ln xdx
2x
To integrate odd powers of sin x or cos x , start
by using the formula sin 2 x cos 2 x 1 and
then use a substitution.
Examples are:
cos xdx
sin xdx
5
To integrate sinaxcosbx, use:
sin (A +B) = sinAcosB + sinBcosA
sin (A -B) = sinAcosB - sinBcosA
To integrate sinaxsinbx or cosaxcosbx, use:
cos (A -B) = sinAsinB + cosAcosB
cos (A +B) = sinAsinB - cosAcosB
sin 3 x cos 4 xdx
sin 3 x sin 4 xdx
Integration involving trigonometric identities
Volumes of revolution: rotating about x axis
The formula is:
Example:
2
cos xdx 12 (1 cos 2 x)dx 12 ( x 12 sin 2 x) c
V y 2 dx
Example: The diagram shows the graph of
8 2
2
4
6
2 2
4
6
8
Integration by substitution
Find
This column gives the calculations
for changing the dx to du:
Example:
2
5
x( x 3) dx .
Make the substitution u x 2 3
5
5
We get: xu dx 12 u du
This gives:
1
12
u 6 c 121 ( x 2 3) 6
Example 2: Use the substitution
x 1
dx .
t 2 x 1 to find
2x 1
x 1
1 x 1
dx
dt
Solution:
2 t
t
Since t 2 x 1 ,
2 x t 1 x 12 (t 1)
1 12 (t 1)
dt
So we get
2
t
But t 2 x 1 , so
1
4
t 2x 1
dt
2
dx
dt 2dx
dx 12 dt
This expands to give:
y
8
6
4
2
4 xe
2(2 x 1)
1/ 2
x2
2e du 2e
u
4
0
u 0
= 2e 2e 2e 2
4
c.
(2 x 1)3/ 2 12 (2 x 1)1/ 2 c
u x2
du
2x
dx
du 2 xdx
4 xdx 2du
x=0u=0
x=2u=4
Volume = y dx
Find 4 xe .
x2
Use the substitution u x .
(2 x 1)
1
6
1
1
cos 7 x cos x c
14
2
Definite integrals using a substitution
=
3/ 2
x = 1, x = 4 and y = 0 is shaded. This region is
rotated completely about the x-axis. Find the volume
generated.
2
3
Solution: We use the
sin (4x +3x) = sin4xcos3x + sin3xcos4x
sin (4x-3x) = sin4xcos3x sin3xcos4x
4
. The region trapped between the lines
2x 1
u x2 3
du
Subtracting 1 and 2
dx 2 x
du 2 xdx
sin(4x+3x) sin(4x-3x) = 2sin3xcos4x
So, xdx 12 du
So, sin3xcos4x = sin7x sinx
1
1
sin 3 x cos 4 xdx = 2 sin 7 x 2 sin xdx
Take the multipliers outside the integral:
(t 1)
1
dt 14 (t 1)t 1/ 2 dt 14 t1/ 2 t 1/ 2 dt
4
t
3/2
1/ 2
c
This gives: 14 23 t 2t
sin 3 x cos 4 xdx
= 16
= 16
= 16
1
dx
2x 1
4
1
ln(2 x 1)
2
1
1
1
ln 7 ln1 8 ln 7
2
2
4
dx
2 x 1
4
6
3x 2 c
3 6
2
6
3x 2 c
9
5 2 x dx 2 ln(5 2 x)
5
4(3x 2) dx
2 xe dx .
Common examination questions
Example 1: Find x ln xdx .
This is a suitable candidate for integration by parts
with u 2 x and
dv
ex :
dx
du
2
dx
dv
ex v ex
dx
u 2x
Substitute these into the formula:
2 xe dx 2 xe 2e dx 2 xe
x
Example 2: Find
2e x c
x cos xdx .
Here we take u x and
dv
cos x :
dx
du
ux
1
dx
dv
cos x v sin x
dx
Substitute these into the formula:
x cos xdx x sin x sin xdx x sin x ( cos x) c
x sin x cos x c
Note: Sometimes it is necessary to use the
integration by parts formula twice (e.g. with
2
x sin xdx ).
Example 1: Find
This can be found using integration by parts if we
dv
x.
dx
du 1
u ln x
dx 2 x
dv
x
x v
dx
2
1
ln 3 ln 5
2
1
1
5
ln 5 ln 3 ln
2
2 3
Definite integrals (using by parts)
Example: a) Find the points where the graph of
y (2 x)e x cuts the x and y axes.
take u ln x and
b) Sketch the graph of y (2 x)e x .
c) Find the area of the region between the axes and
the graph of y (2 x)e x .
a) Graph cuts y-axis when x = 0, i.e. at y = 2
Graph cuts the x-axis when y = 0, i.e. when x = 2..
Substitute these into the formula:
x2
1 x2
2
x
ln
xdx
ln
x
x 2 dx 12 x ln x 12 xdx
2
12 x 2 ln x 41 x 2 c
b) The graph looks like:
2.5
2
x
c) Area is (2 x)e dx .
1.5
du
1
dx
v e x
u 2 x
Example 2: Find ln xdx .
This can be thought of as 1ln xdx and so can be
integrated by parts with u ln x and
dv
1
dx
du 1
u ln x
dx x
dv
1 v x
dx
1
ln xdx x ln x x x dx x ln x 1dx
= x ln x x c
dv
e x
dx
0.5
0.5
1.5
x
x
x
So (2 x)e dx (2 x)( e ) 1 e dx
x
x
= (2 x)(e ) e dx
= (2 x)(e x ) e x
Now that weve integrated, we substitute in our
limits:
2
(2 x)e
0
dx (2 x)e x e x
2
0
0e e 2e e 0
0.135 1 1.135
0
