Download Techniques of Integration

Techniques of Integration
By Danang Mursita
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Integration By Part
d
dx
f ( x ) g( x ) 
 f(x)
d g( x ) 
dx
 g( x )
df ( x ) 
dx
d
dg(x)
df(x)


f
(
x
)
g
(
x
)
dx

f
(
x
)
dx

g
(
x
)
dx
 dx


dx
dx
f(x)g(x)   f(x)
 f ( x )g ' ( x ) dx
dg(x)
df(x)
dx   g(x)
dx
dx
dx
 f ( x )g ( x ) 
 udv  uv 

v du
 g( x )f ' ( x )dx
b
b
a
a
b
 u dv  uv a 
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 v du
Trick and Tips
• If the integrand has two factors ie the
polynomial and the trigonometric function
or transcendental function then substitute u
= the polynomial
• If the integrand has a factor trigonometric
function and transcendental function then
substitute u = the trigonometric function or
u = the transcendental function
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Examples
Use the integration by part for calculating this
integral
(1).  x  2cos 2x dx
1


(2).  x 2  1 ex dx
0

x
(3).  e cos x dx
1
(4).  ln(x  2) dx
1
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Trigonometric Integral
We calculate integral of f(x) with integrand
f(x) is represented by
1. Sinnx, cosnx
2. sinmx cosnx
3. sin (ax) cos (bx),
4. sin (ax) sin(bx),
5. cos (ax) cos (bx)
6. tanmx secnx ,
7. cotmx cscnx
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n
n
 cos x dx &  sin x dx
1.Let n is odd. Then
– Represent cosn x = cos x cosn-1x and sinn x = sin
x sinn-1x
– Use sin2x + cos2x =1, d(sin x) = cos x dx and
d(cos x) = - sinx dx
2.Let n is even. Then
– Use cos2x = 2 cos2x – 1 = 1 – 2 sin2x or
– Represent cosnx = cos x cosn-1x and sinnx = sin x
sinn-1x then use integral by part for calculating
this integral
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Examples
Calculate this integral
(1).  cos 3 x dx

(2).
4

sin2 2x dx
0
(3).  cos 4 3x  dx

(4).
2

sin5 x dx
0
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
sin
m
x cos
n
x dx
1. n is odd, substitute u = sin x, identity :
cos2x = 1 – sin2x
2. m is odd, substitute u = cos x, identity
: sin2x = 1 – cos2x
3. m and n are even, use identity to
reduce the powers on sin and cos,
identity : sin2x = ½ ( 1 – cos 2x) and
cos2x = ½ ( 1 + cos 2x)
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Examples
Calculate this integral
(1).  cos 5 x sin x dx
(2).  cos 4 x sin3 x dx

(3).
3
4
3
sin 3x cos 3x dx

0

(4).
2

0
x
x
2
2
sin cos dx
2
2
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 sin ax cos bx dx ,  sin ax sin bx dx ,  cos ax cos bx dx
• Use this identity
sin  ax cos  bx  
1
2
sin  ax sin  bx   
cos  ax cos  bx  
sin  a  b  x
1
2
1
2
 sin  a  b  x 
cos  a  b  x
 cos  a  b  x 
cos  a  b  x
 cos  a  b  x 
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Examples
Calculate this integral
(1).  cos 3x sin 2 x dx
x
(2).  sin x sin dx
2

6
(3).  cos2x cos4x dx
0

(4).
2

0
x
sin cos xdx
2
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m
n
m
n
 tan x sec x dx &  cot x csc x dx
1. m = 1 and n = 0, substitute : d(cos x) = - sin
x dx and d(sin x) = cos x dx
2. m = 0 and n = 1, substitute : d(sec x ) = sec
x tan x dx and d(csc x ) = - csc x cot x dx
3. m > 1 and n = 0 or m = 0 and n > 1, use
identity : tan2x = sec2x – 1 and cot2x =
csc2x – 1
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Examples
Calculate this integral
(1).  tan x dx
2
(5).  tan x dx
(2).  cot x dx
3
(6).  cot x dx
(3).  sec x dx
3
(7).  sec x dx
(4).  csc x dx
4
(8).  csc x dx
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m
n
 tan x sec x dx
1. n is even, substitute u = tan x,
identity : sec2x = tan2x + 1
2. m is odd, substitute u = sec x,
identity : tan2x = sec2x – 1
3. m is even and n is odd, reduce to
powers of sec
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m
n
 cot x csc x dx
1. n is even, substitute u = cot x,
identity : csc2x = cot2x + 1
2. m is odd, substitute u = csc x,
identity : cot2x = csc2x – 1
3. m is even and n is odd, reduce to
powers of csc
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Examples
Calculate this integral
(1).  tan3 x sec 4 x dx
(2).  cot 3 x csc3 x dx

6

(3).
sec 3 x tan x dx
0

(4).
2


csc3 x cot xdx
4
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Trigonometric Substitutions
• This method can be used for calculating
integration with integrand has a factor of
2
2
(1). a  x
(2). a2  x 2
2
2
(3). x  a
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Trigonometric Substitutions
(1). Substitute x  a sin t 
2
2
a  x  a cos t, dx  a cos t dt
(2). Substitute x  a tan t 
a2  x 2  a sec t, dx  a sec 2 t dt
(3). Substitute x  a sec t 
2
2
x  a  a tan t, dx  a sec t tan t dt
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Examples
2
9  x dx
(1). 
(3). 
(4). 
dx
2
1 x

dx
(1).  x 3 16  x 2 dx
0
2
x
(2). 
4
2

(2).
2 x 2 1
x
2
1

3
x2  4 2
1 x2
dx
x
dx
3
(3). 
dx


2
2
0 4  9x
1
(4).

2
1
1  2x 2 
3
2
dx
2
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Integrals involving ax2 + bx + c
Integrals that involve ax2 + bx + c with b  0
can be solved by
1. Substitution u = x + b/2a, ie :
ax2 + bx + c = a (x2 + b/a x) + c
= a (x2 + b/a x + b2/4a2) + c – b2/4a
= a ( x + b/2a)2 + c – b2/4a
2. We have a simple form au2 + d and then
trigonometric substitution can be applied
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Examples
(1). 
(2). 
dx
x 2  6x  13
x
(3).
2
(4). 
x 2  6x  10
x3
x 2  2x  2
dx
dx
dx
2
4
x

x
1
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Integral of Rational Function
• The function f(x) = P(x)/ Q(x) is Rational Function,
ie : P(x) and Q(x) are Polynomials.
• Integral of f(x) which is rational function with
degree of numerator, P(x) less than degree of
denominator, Q(x) can be solved by the possibility
of factor of the denominator, Q(x) :
– Q(x) have linear factor ( repeat and non repeat
factor)
– Q(x) have non linear factor ( repeat and non repeat
factor)
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The Linear Factor
1. Q(x) = ( x – q1) ( x – q2)…(x – qn)
A1
A2
An
f(x) 

 ... 
x  q1 x  q2
x  qn
2. Q(x) = ( x – q1)2 ( x – q2)3.
A
B
C
D
E
f(x) 




x  q1 x  q12 x  q2 x  q22 x  q23
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Examples
(1). 
(2). 
(3). 
(4). 

x 2  1dx
(5). 
dx
x 2  4x  5
x 1
x 2  5x  6
2x 2  3
xx  12
x 2  4x  5
dx
dx
2x 2  2x  1
x3  x 2
(6). 
(7). 
dx
(8). 
x3
x2  x  6
x2
x  13
dx
dx
2x 2  3x  3
x  13
dx
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The Non Linear Factor
1. Q(x) = ( x2 + q1) ( x2 + px + q2)
f(x) 
A  Bx
2
x  q1

C  Dx
2
x  px  q2
2. Q(x) = ( x2 + q1)2 ( x2 + q2).
f(x) 
A  Bx
x 2  q1


C  Dx
E  Fx
2  q2

x

2
2
x  q1

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Soal latihan
• Selesaikan Integral tak tentu berikut

x  2
2
x  4x  3
dx
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Examples
x 3  3x 2  x  9
x 2 1x 2  4
dx
(2). 
x x 2  x  1
(1). 
(3). 
(4). 
3x 2  12x  2


2
2
x 4
dx
dx
4x  2
x 4  2x 3  x 2
dx
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Integrals Involving Rational Exponents
• The method can be used for calculating integrals
with integrand has rational power of x
– x1/n, substitute u = x1/n.
– x1/n and x1/m, substitute u = x1/mn.
• Examples :
(1).  x x  2 dx
8
(2). 
4
(3). 
x4
dx
x
dx
x 3 x
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Integrals involving Rational Expressions in sin
x and cos x
• We will calculate the integral of f(x) in form of
rational expressions in sin x and cos x
• Substitute :
– u = tan ½ x and du = ½ sec2 ½ x dx, - π < x < π and
– sin x = sin 2( ½ x ) = 2 sin ½ x cos ½ x
– cos x = cos 2( ½ x) = 1- 2sin2 ½ x
sin x 
2u
1  u2
cos x 
1  u2
1  u2
1  u2
u
½x
1
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Examples
dx
(1). 
1  sin x

dx
(2). 
 1  cos x
2
dx
(3). 
sin x  tan x
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