Indefinite Integration
Advanced Level Pure Mathematics
Advanced Level Pure Mathematics
6
Calculus II
Indefinite Integration
2
Method of substitution
3
Integration by Parts
6
Special Integration
10
Integration of Trigonometric Function
14
Reduction Formula
15
1
 x dx  ln x  C
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1
Indefinite Integration
Advanced Level Pure Mathematics
INDEFINITE INTEGRATION
Definition
f ( x ) is said to be primitive function or anti-derivative of g( x) if f ' ( x )  g( x ) .
Example
d 2
( x )  2x
dx
Note
Primitive function is not UNIQUE.
Definition
For any function f ( x ) if F ( x ) is the primitive function of f ( x ) , i.e. F ' ( x)  f ( x) , then we

x 2 is the primitive function of 2x .
define the indefinite integral of f ( x ) w.r.t.x as
 f (x) dx F(x)  c , where
c is called the
constant of integration.
Theorem
Two function f ( x ) and h( x) differ by a constant if and only if they have the same primitive
function.
Standard Results
Theorem
1
1.
 x dx  ln x  c
2.
 e dx  e
3.
 cos xdx  sin x  c
4.
 sin xdx   cos x  c
5.
 sec
6.
 csc
7.
 secx tan xdx  secx  c
8.
 cscx cot xdx   csc x  c
9.
x
 a dx 
ax
c
ln a
10.

11*.

12*.
x
2
xdx  tan x  c
1
a2  x2
dx  sin 1
x
c
a
x
2
x
xdx   cot x  c
1
x2  a2
2
c
dx  ln
x  x2  a2
c
a
1
1
x
dx  tan 1  c
2
a
a
a
x2  a2  x
dx  ln
c
a
x2  a2
1
13.

(a)
 kf (x)dx  k  f (x)dx
(b)
 [f (x)  g(x)]dx   f (x)dx   g(x)dx .
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Indefinite Integration
Advanced Level Pure Mathematics
a
x
 a dx  ln a  c
x
Example
Prove
proof
Let y  a x .
ln y  x ln a
1 dy
 ln a
y dx


dy
 y ln a
dx
dy
=
 y ln a dx
y
=
ln a  y dx
x
 a dx
=
ax
c
ln a
 dx dx
METHOD OF SUBSTITUTION
Theorem
( CHANGE OF VARIABLE )
If x  g( t ) is a differentiable function,
Proof
 f (x)dx   f (g(t)) g '(t) dt .
Let F ( x ) is the primitive function of f ( x ) .
i.e.

dF ( x)
 f ( x)
dx
x  g( t )
d
F( x) =
dt
We have

dF ( x) dx

dx dt
=
dF ( x)
g ' (t )
dx
F( x )
=
 f (g(t))g '(t)dt
 f (x)dx
=
 f (g(t))g '(t)dt
dx  ln
x2  a2  x
c
a
1
Example
Prove
proof
sub x  a tan θ
1

dx
2
x  a2
x2  a2

=
dx  a sec2 θ dθ
1
2
 a secθ a sec θ dθ
=
 sec θ dθ
=
ln sec θ  tan θ  c
=
ln
(  secθ dθ  ln secθ  tan θ  c )
x2  a2  x
c
a
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Indefinite Integration
Advanced Level Pure Mathematics
Remark
By using substitution, the following two formulae can be derived easily.
f '(x)
(I)
 f (x) dx  ln f (x)  c ,
(II)
2
f ' ( x)
f ( x)
dx  f ( x)  c .
The following examples illustrate the use of the above results.
Example
 secθ dθ  ln secθ  tan θ  c
 cscθ dθ   ln cscθ  cot θ  c
and
proof
 tan θ dθ
Example
 cos θ dθ
=
1

d( cos θ )
cos θ
=
 ln cos θ  c
ln x
dx
x
 ln x d(ln x)
=
(ln x ) 2
c
2
=
cos x
 cot θ dθ
Example
Example
sin θ
=

 3  2 sin x dx
cos θ
=
 sin θ dθ
=
 sin
=
ln sinθ  c
(a)
e
1
dx
x
θ
d(sin θ )
(b)
x e
2 x3
dx
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Indefinite Integration
Advanced Level Pure Mathematics
Example

x
e
Example
(a)
Example
x
x
dx
( Let y  x )
ex  1
 e x  1dx
2
e sin 2 x sin 2 x
(b) 
dx
e 2x
1  x dx
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Indefinite Integration
Advanced Level Pure Mathematics
Example
(a)
(b)
(c)
e
x
cot(e x )dx
a 2x  b 2x
 e x dx
x2
 px  q dx
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Indefinite Integration
Advanced Level Pure Mathematics
INTEGRATION BY PARTS
*Theorem
( INTEGRATION BY PARTS )
If u , v are two functions of x , then
d
uv
dx
proof
=
u
 udv  uv   vdu .
dv
du
v
dx
dx
dv
d
du
=
uv  v
dx
dx
dx
We integrate both sides with respect to x to obtain
u
 udv
=
Example
(a)
 ln x dx
Example
(a)
I   x 2 cos xdx
dv
 u dx dx
=
uv   vdu
(b)
x
2
ln x dx
(b) I   x 2 sin xdx
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Indefinite Integration
Advanced Level Pure Mathematics
x
Example
I
xe
dx
(1  x) 2
Example
 tan
1
Example
 (ln x) dx
x dx
2
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Indefinite Integration
Advanced Level Pure Mathematics
Example
(a) Show that
d
x
1
.
tan 
dx
2 1  cos x
(b) Using (a), or otherwise, find 
x  sin x
dx
1  cos x
[HKALE 1997]
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Indefinite Integration
Advanced Level Pure Mathematics
SPECIAL INTEGRATION
We resolve the rational function
P( x )
by simple partial fraction for P( x), Q( x) being poly. The
Q( x )
integration of rational function is easily done by terms by terms integration.
x
dx
 a2
Example
(a)
Example
x 3  2x 2  1
 (x  1)( x  2)( x  3)2 dx
2
(b)
x1
dx
2
1
x
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Indefinite Integration
Advanced Level Pure Mathematics

2 x  x  3x  3x
dx .
x3  1
4
3
2
Example
Evaluate
Solution
By decomposing into partial fractions,
2x 4  x 3  3x 2  3x
1
2x
.
 2x  1 
 2
3
x1 x  x1
x 1
Hence,
Integration of
Example
Solution
Px  Q
ax 2  bx  c
Evaluate

4x  1

4x  1
dx .
5  4x  x 2
Observing that the derivative of 5  4x  x 2 is  (4  2x ) , we have
5  4x  x
2
dx =

 2( 4  2x)  9
dx
5  4x  x 2
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Indefinite Integration
Advanced Level Pure Mathematics
Integration of
Example
1
x  ax 2  bx  c
 x
dx
x2  1
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Indefinite Integration
Advanced Level Pure Mathematics
Integration of
 R ( x,
n
ax  b
)dx
cx  d
In solving such problems, we use the substitution u  n
Example
I
ax  b
cx  d
x2
dx
x x1
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Indefinite Integration
Advanced Level Pure Mathematics
INTEGRATION OF TRIGONOMETRIC FUNCTION
Integration of
 R( cosθ , sinθ )dθ
(1) If  R(  cosθ , sinθ )  R( cosθ , sinθ ) ,
put u  sinθ .
(2) If  R( cosθ , sinθ )  R( cosθ , sinθ ) ,
put u  cos θ .
(3) If R(  cosθ , sinθ )  R( cosθ , sinθ ) ,
put u  tanθ .
(4) Otherwise, put t  tan
θ
.
2

tan θ 
2t
1 t2
1 t2
cos θ 
1 t2
sin θ 
Example
(a)
Example
 sin
 cos
8
3
2t
1 t2
θ sin 2 θ dθ
(b)
 cos
2
θ sin 3 θ dθ
x cos 7 x dx
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Indefinite Integration
Advanced Level Pure Mathematics
REDUCTION FORMULA
Certain integrals involving powers of the variable or powers of functions of the variable can be related to
integrals of the same form but containing reduced powers and such relations are called REDUCTION
FORMULAS (遞推公式或歸約公式). Successive use of such formulas will often allow a given integral to
be expressed in terms of a much simpler one.
Example
Let I n   sinn x dx for n is non-negative integer.
1
n 1
Show that I n   cos x sinn1 x 
I n 2
n
n
Hence, find I 6 .
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Indefinite Integration
Advanced Level Pure Mathematics
Example
Show that if I n   cos n θ dθ , where n is a non-negative integer, then
sinθ cos n1 θ n  1

I n 2 , for n  2 .
n
n
Hence evaluate I 5 and I 6 .
In 
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Indefinite Integration
Advanced Level Pure Mathematics
Example
If I n   tan n x dx , where n is a non-negative integer, find a reduction formula for I n .
( In 
1
tan n1 x  I n 2 )
n 1
This formula relates I n with I n 2 , and if n is a positive integer, successive use of it will
ultimately relate with either
Since
 tan x dx
or
 dx .
 tan xdx  ln secx  c,  dx  x  c , and positive integral power of
tan x can
therefore be integrated.
Example
For non-negative integer n, I n   (ln x)n dx .
Find a reduction formula for I n and hence evaluate I 3 .
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Indefinite Integration
Advanced Level Pure Mathematics
Example
Let n be a positive integer and a  0 .
dx
 (*)
In  
2
(ax  bx  c)n
(a) Prove that n(4ac  b 2 )I n 1  2( 2n  1)a I n 
(b) Evaluate
 (x
2
2ax  b
.
(ax  bx  c) n
2
dx
.
 2x  2) 2
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