Download Integration Key Facts

Integration Key Facts
Facts which are in the formula book
Standard Integrals you need to learn:
1
 x dx  n  1 x
 e dx  e  c
n
x
n1
c
x
1
 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
2
Further generalisations using the above integrals:
 (ax  b)
n
dx 
1
(ax  b) n1  c
a( n  1)
1 axb
e
c
a
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
e
ax  b
dx 
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 2dx
a
General patterns you should remember:
 f ' ( x ) f ( x ) dx  n  1  f ( x )
 f ' ( x )e dx  e  c
1
n
f ( x)

n1
c
f ( x)
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
(1  cos 2 x )dx (double angle formula used)
2
1
2
 cos xdx  2  (1  cos 2 x )dx (double angle formula used)
1
n
n1
 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
n1
 tan x sec xdx   n  1 sec x  c
 sin
2
xdx 
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
 f ( x )dx  2  y
h
a
h
ba
n
0
 y n  2( y1  y2  ....  yn1 )
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
x sin ax
x cos ax
xn ln x
x ex
x2ex
x2sinx
x2cosx
lnx
x(2x + 3)4
x sec2xtanx
x
x
ln x
x
x2
parts will need to be applied
twice – this would apply to
other similar examples.
ln x
x
x
Trigonometric Identities which could be required:
sin 2 x  12 (1  cos 2 x)
cos2 x  12 (1  cos 2 x)
sin 2 x  cos2 x  1
tan 2 x  1  sec2 x
1  cot 2 x  cosec2 x
dv
needs integrating to find v
dx
sin ax
cos ax
xn
ex
ex
sinx
cosx
1
(2x + 3)4
sec2xtanx
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;
5) using a trigonometric identity (such as sin 2 x  12 (1  cos 2 x) or cos 2 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.

5x  2
dx
x
If you are asked to integrate an algebraic fraction
of the form
ax  b
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:
4

 (2 x  1)( x  2) dx

 x(4  x)dx



 2xe dx
 x cos( x  2)dx
 (cos x)e dx
 (2 x  5) ln( x  5 x)dx
x2
2
x
2
Use integration by parts to integrate other
products, especially when one function is a
simple polynomial. Examples are:

3

2sin x

2
To integrate even powers of sin x or cos x , use
the formulae sin 2 x  12 (1  cos 2 x) or
cos 2 x  12 (1  cos 2 x) .
Examples are:
cos 4 xdx


x
dx put u  x 2  1
1
sin x
 
put u  cos x
cos x
2x 1
  2
dx put u  x 2  x  1
2
( x  x  1)

4 x
Recognise that the integral is in the form
f '( x)  gf ( x) . Examples are:

Other quotients (where the denominator cannot
be factorised into linear factors) can usually be
integrated using a substitution.
Examples are:

 sin
2
xdx
 (3x  1)e dx
 x cos xdx
 2 x ln xdx
2x
2
3
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:
cos5 xdx



 sin
3
xdx
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
Find
Integration by substitution
This column gives the calculations
for changing the dx to du:
Example:
2
5
 x( x  3) dx .
Make the substitution u  x2  3
We get:  xu 5 dx   12 u 5 du
This gives:
1
12
 sin 3 x cos 4 xdx
Solution: We use the
sin (4x +3x) = sin4xcos3x + sin3xcos4x
sin (4x-3x) = sin4xcos3x – sin3xcos4x
y
4
. The region trapped between the lines
2x 1
x = 1, x = 4 and y = 0 is shaded. This region is
rotated completely about the x-axis. Find the volume
generated.
y
8
u  x2  3
du
 2x
Subtracting 1 and 2
dx
du  2 xdx
sin(4x+3x) – sin(4x-3x) = 2sin3xcos4x
So, xdx  12 du
u 6  c  121 ( x 2  3)6
6
4
So, sin3xcos4x = ½ sin7x – ½ sinx
Example 2: Use the substitution
x 1
t  2 x  1 to find 
dx .
2x  1
 sin 3 x cos 4 xdx =
– 2
t  2x  1
dt
2
dx
dt  2dx
dx  12 dt
x 1
1 x 1
dx  
dt
Solution: 
2 t
t
Since t  2 x  1 ,
2x  t  1  x  12 (t  1)
So we get
1 (t  1)
 2 t dt
1
2
1
4

2 3/ 2
3
t
But t  2 x  1 , so
1
4

2
3
This expands to give:
(2 x  1)
 (2 x  1)
1
2
1/ 2
c
6
8
– 2
2
 4 
Volume =   y dx    
 dx
 2x 1 
1
4
1
dx
= 16 
1 2x 1
2
0
Use the substitution u  x 2 .
4
 2e du  2e
u
u
4

0
= 2e4  2e0  2e4  2

3/ 2
x2
u 0
4
4
Find  4 xe .
2
2
2
Definite integrals using a substitution
2
0
(2 x  1)3/ 2  2(2 x  1)1/ 2  c .
1
6
14
x
 4 xe 
 2t1/ 2  c

=  1 cos 7 x  1 cos x  c
2
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
This gives:
2
1
1
 2 sin 7 x  2 sin xdx
u  x2
du
 2x
dx
du  2 xdx
4 xdx  2du
x=0→u=0
x=2→u=4
4
1

= 16  ln(2 x  1) 
2
1
1
1

= 16  ln 7  ln1  8 ln 7
2
2

x
1
4
6
 3x  2   c
3 6
2
6
  3x  2   c
9
1
1
 5  2 x dx  2 ln(5  2 x)
5
 4(3x  2) dx 
0
0


Example 1: Find
 2 xe dx .
Common examination questions
Example 1: Find  x ln xdx .
x
This is a suitable candidate for integration by parts
with u  2 x and
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
x
Example 2: Find
x
x
 2e x  c
 x cos xdx .
Here we take u  x and
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
dv
 ex :
dx
dv
 cos x :
dx
du
1
dx
dv
 cos x  v  sin x
dx
ux 
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 ).
1
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  14 x 2  c
b) The graph looks like:
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
y
3
2.5
2
 (2  x)e
c) Area is
x
2
dx .
0
1.5
1
du
u  2 x 
 1
dx
dv
 e  x  v  e  x
dx
0.5
0.5
1
1.5
So  (2  x)e x dx  (2  x)(e x )   1 e x dx
= (2  x)(e x )   e x dx
= (2  x)(e x )  e x
Now that we’ve integrated, we substitute in our
limits:
2
 (2  x)e
0
x
dx   (2  x)e  x  e  x 

2
2
 
2
0
 0e  e  2e  e0
 0.135  1  1.135
0

2
x