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Trigonometric Substitution PartialFraction
Trigonometric Substitution and Integration ofRational Functions by Partial Fractions
Mathematics 54Elementary Analysis 2
Institute of Mathematics
University of the Philippines-Diliman
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
Trigonometric Substitution
We use trigonometric substitution for integrands that contain
expression of the form:a2u2
a2+u2u2a2
Example
1
49x2 dxx
2
dx(x3)225
5
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
Trigonometric Substitution
Case 1. Integrand contains the form
a2u2
Let u= asin, where
2,
2
. Note du= acosd.
a2
u2
=a2a2 sin2=a2(1 sin2)= a
cos2
= a|cos|
= acos
a2u2
ua
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
Case 1.
a2u2. Let u= asin.
Example 1. Find 49x2 dx
x.
Let x
=7sin. Then dx
=7cos d.49x2
xdx=
7cos7sin
7cos d
49x2
x7
=
7cos2
sind =
7(1sin2)
sind
= 7(cscsin) d= 7 ln|csccot|+7cos+C
= 7 ln
7
x
49x2x
+
49x2+C.
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
Trigonometric SubstitutionIntegrals involving
a2u2,
a2+u2 or
u2a2
Case 2. Integrand contains the form
a2+u2
Let u
=atan, where
2,
2 . Note du=asec2d.
a2+u2 =a2+a2 tan2
=
a2(1+ tan2)= a
sec2
= asec
a
u
a2+u2
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T i i S b i i P i lF i C C C E i
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
Case 2.
a2+u2. Let u= atan.Example 2. Find
dx(x26x+25)3
.
We have: x26x+25= (x3)2+42, Letx
3
=4tan
dx
=4sec2 d
4
x
3(x3)2+42
dx
(x26x+25)3= dx
(x3)2+423
=
4sec2
(4sec)3d
= 116
cos d
= 116
sin+C
=1
16
x
3
x26x+25 +C6/28
T i t i S b tit ti P ti lF ti C 1 C 2 C 3 E i
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
Trigonometric SubstitutionIntegrals involving
a2u2,
a2+u2 or
u2a2
Case 3. Integrand contains the form
u2a2
Let u
=asec, where
0,
2 , 3
2 . Note du=asec tand.
u2a2 =a2 sec2a2
=
a2(sec21)= a
tan2
= atan
a
u2a2
u
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
Case 3.
u2a2. Let u= asec.Example 3. Find
e4
e2
ln2 x4xln x
dx.
Let ln x= 2sec. Then 1x
dx= 2sec tan d.
2
ln2 x4ln x
e
4
e2
ln
2
x4xln x
dx=
3
0
2tan2sec
2sec tan d
=
3
0
2tan
2sec2sec tan d
=
3
02tan2 d
= 2
3
0(sec21) d= 2(tan)
3
0
= 23 3 8/28
Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
Example on Area
Example 4. Evaluate 1
0
1x2 dx
From the definition of the definite integral, the expression
1
01x2 dxis the area of the region
.
By trigonometric substitution, let x= sin. Then dx= cosd10
1x2 dx =
2
0 cos cosd=
2
0
1
2(1+cos2)d
=1
2 (+1
4 sin2) 2
0 =1
2
2 +1
4 sin01
4 sin0 = 4
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
Summary
a2
u2 Let u
=asin
a2+u2 Let u= atan
u2a2 Let u= asec
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Exercises
Exercises
Evaluate the following integrals.
1
dx
x2+2x152
16
e2x
ex dx
3
42
x24
xdx
4 x3 dx
1x25
dx
25+x23
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Case4 Exercises
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g
Introduction
Recall.
A function his a rational functionifh(x)= f(x)g(x)
, where fand g are
polynomial functions.
Consider
1x2+5x+6 dx.
Solvable using math 53:
1
x2
+5x+6dx
=1
x+
522 14
dx
=
1
2 1
2
ln x+3
x+2 +C
Note:1
x2+5x+6 dx=
1
x+3 1
x+2
dx= ln
x+3x+2
+C
This integral will be easier to evaluate.12/28
Trigonometric Substitution PartialFraction Case1 Case2 Case3 Case4 Exercises
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g
Goal.
To decompose a rational expression as a sum of two or more
simpler quotients, called partial fractions.
Remark.
1 We will consider rational functions h(x)= f(x)g(x)
with
deg(f(x))< deg(g(x)) (fand g are polynomial functions). Ifdeg(f(x)) deg(g(x)), then we divide first the numerator by thedenominator such that
h(x)
=q(x)
+
r(x)
g(x)
,
where deg(r(x))< deg(g(x)).2 Any polynomial with real number coefficients can be
expressed as a product of linear and quadratic polynomials.
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Case 1. Denominator of the rational function only has
distinct linear factors.
Ifg(x)= (a1x+b1)(a2x+b2) (anx+bn), where all factors aredistinct, then
p(x)
q(x)= A1
a1x+b1+ A2
a2x+b2+ + An
anx+bn.
Example 1. Find x dx
x25x+6 .
x
x25x+6 =x
(x3)(x2) =A
x3 +B
x2 =A(x2)+B(x3)
(x3)(x2) .
We have:
x=A(x2)+B(x3).
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Case 1. Factors of the denominator are linear.
Example 2. Find 2x34x215x+5
x2
2x
8
dx.
Since the degree of the numerator is greater than the denominator,
we divide first the numerator by the denominator. We have
2x34x215x+5
x22x8 =
2x
+
x+5
(x4)(x+2) =2x
+
A
x4 +
B
x+2.
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Case4 Exercises
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2x34x215x+5x22x8 = 2x+ Ax4 + Bx+2
x+
5
(x4)(x+2) =A
x4 +B
x+2 x+5=A(x+2)+B(x4)
Substitute2 for x: 3=B(6)B= 12
Substitute 4 for x: 9=A(6)A=3
2
Hence,
2x34x215x+5
x22x8 dx =
2x+ x+5(x4)(x+2)
dx
= 2x22+
32(x4)
12(x+2)
dx
= x2+ 32
ln |x4| 12
ln |x+2|+C.
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Case 2. Factors of denominator are all linear but some are
repeated.
If (ax+b)n, n> 1 is a factor of the denominator, then the partialfractions corresponding to this factor are
A1
ax
+b+ A2
(ax
+b)2
+ + An(ax
+b)n
.
Example 3. Find
(3x2) dx
2x
3
x2
.
3x22x3x2 =
3x2x2(2x1) =
A
x+ B
x2+ C
2x1 .
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3x22x3x2 = Ax+ Bx2 + C2x1
Multiplying both sides byx2(2x1)
3x2=Ax(2x1)+B(2x1)+Cx2
let x=
0
B=
2
let x= 12
C=2let x= 1 A= 1
(3x2) dx
2x3x2 =
1
x+ 2
x2 2
2x1
dx
= ln |x|+ 2x1
1 ln |2x1|+C
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More Examples
Example 4. Find
(x5) dx
(x+1)2(x2) .
x5
(x+1)2
(x2) =
A
x+1 +
B
(x+1)2
+
C
x2.
We have x5=A(x+1)(x2)+B(x2)+C(x+1)2.Substitute1 for x: 6=B(3)B= 2
Substitute 2 for x: 3=C(3)2 C= 13
Substitute 0 for x: 5=A(1)(2)+2(2)+ 1
3 (1)
2
A=1
3x5
(x+1)2(x2) dx =
1
3(x+1) +2
(x+1)2 1
3(x2)
dx
=
1
3
ln
|x
+1
|+
2(x+1)1
1
1
3
ln
|x
2
|+C
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Case 3. Factors of the denominator are linear and
quadratic and none of the quadratic factors is repeated.
Ifax
2
+bx+ cwhere a= 0 is a factor of the denominator that is notrepeated , then the corresponding partial fraction to this factor isAx+B
ax2+bx+c.
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Case 3. Factors of the denominator are linear and
quadratic and none of the quadratic factors is repeated.
Example 5. Find 5x2+3x2
x31 dx.
5x2+3x2
x31 dx=
5x2+3x2(x1)(x2+x+1) dx=
A
x1 +Bx+C
x2+x+1
dx
Consider 5x2
+3x
2=
A(x2
+x+
1)+
(Bx+
C)(x
1).
Substitute x= 1: 6=A(3)A= 2Substitute x= 0: 2= 2(1)+C(1)C= 4
Substitute x=1: 0= 2(1)+ (B+4)(2)B= 3
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Case4 Exercises
5x2+3x 2 2 3x+4
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5x2+3x2x31 = 2x1 + 3x+4x2+x+1
5x2+3x2
x31 dx
= 2
x
1dx+
3x+4x2
+x+
1dx
=
2
x1 dx+ 3
2(2x+1)
x2+x+1 dx+ 5
2
x2+x+1 dx
=2 ln
|x
1
|+3
2
ln
|x2
+x
+1
|+5
2 1
3/2tan1
x+ 12
3/2+C
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Case4 Exercises
d i f f h d i
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Case 4. Some quadratic factors of the denominator are
repeated.
If (ax2+bx+c)n, n> 1 is a factor of the denominator , then the
corresponding partial fractions are
A1x+b1ax2+bx+c+
A2x+b2(ax2+bx+ c)2 + +
Anx+bn(ax2+bx+c)n
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Case4 Exercises
C 4 S d i f f h d i
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Case 4. Some quadratic factors of the denominator are
repeated.
Example 6. Find
x3+1
(x2
+4)2
dx.
x3+1
(x2+4)2 dx=
Ax+Bx2+4 dx+
Cx+D
(x2+4)2 dx
x3
+1=
(Ax+
B)(x2
+4)+
(Cx+
D)=
Ax3
+Bx2
+(4A
+C)x
+D
By comparing coefficients, we have A= 1, B= 0, D= 1.Also, ifx= 1 then 2=A+B+ (4A+C)+D.C=4
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C ti ti
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Continuation...
x3+1
(x2+4)2 dx =
x
x2+4 dx+
14x(x2+4)2 dx
=1
2 2x
x2+4 dx+ 14x(x2+4)2 dx
= 12
2x
x2+4 dx2
2x
(x2+4)2 dx+
dx
(x2+4)2
=1
2 ln |x2
+4|2(x2
+4)1
1 + dx(x2+4)2 (trig.sub.)
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Case4 Exercisesx3+1 d 1 l | 2+4|+ 2 +
dx
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x +1
(x2+4)2 dx= 12 ln |x2+4|+ 2x2+4 +
dx(x2+4)2
Let x= 2tan dx= 2sec2 d
2
x
x2+22
dx
(x2+4)2 =
2sec2 d
16sec4
= 18
cos2 d
=1
8 1+cos22 d
= 116+ 1
32sin2+C
=
1
16
+
1
32 2sincos
+C
Hence,x3+1
(x2+4)2 dx=1
2ln |x2+4|+ 2
x2+4+1
16tan1
x2
+ 1
16
xx2+4
2x2+4
+C.
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Trigonometric Substitution PartialFraction Case1 Case2 Case3 Case4 Exercises
E ercises
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Exercises
Decompose the following into partial fractions.
1(4x3)
x22x32 1
x3(x1)3
2x+1x2(x2+2)
4 x3
2x(x2+1)2
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