Download Calculus and Differential Equations I

The method of partial fractions
The purpose of the method of partial fractions is to find
antiderivatives of rational functions, i.e. functions of the form
P(x)
f (x) =
, where P and Q are polynomials.
Q(x)
Calculus and Differential Equations I
The method involves three steps:
MATH 250 A
1
If d o (P) ≥ d o (Q), first use long-division and re-write f as
Methods of integration II
f (x) = N(x) +
H(x)
,
Q(x)
d o (H) < d o (Q),
where N and H are polynomials. Then, apply the method to
the rational function H(x)/Q(x).
2
3
Methods of integration II
Calculus and Differential Equations I
The method of partial fractions (continued)
To do this, we need to be able to perform each of the steps
separately. They are:
1
Long-division of polynomials
2
Partial fraction decomposition of P(x)/Q(x) where
d o (P) < d o (Q)
3
Integration of terms that typically appear in a decomposition
into partial fractions. Such terms are of the form
A
(x − a)n
and
(x 2
Integrate each of the terms appearing in the partial fraction
decomposition of f to obtain an antiderivative of f .
Methods of integration II
Example for step 1: Divide x 3 by x 2 + 3x + 2.
Calculus and Differential Equations I
Calculus and Differential Equations I
Examples of application
1
We have already used partial fractions when solving the
logistic equation.
2
Solve the following differential equation
dy
(y + 1)(y 2 − 2y + 3)
=
.
dx
y2 + 5
3
Solve the differential equation
Bx +C
,
+ b x + c)n
where n ≥ 1 and x 2 + bx + c is irreducible.
Methods of integration II
If d o (P) < d o (Q), find the partial fraction decomposition of
P(x)/Q(x).
(y − 1)3
dy
=
dx
y4
with the following initial conditions
1
2
y (0) = 2
y (0) = 1
Methods of integration II
Calculus and Differential Equations I
Trigonometric substitutions
Trigonometric substitutions (continued)
√
For integrands that involve a2 − x 2 , a > 0, note that
|x| ≤ a, and try the substitution x = a sin(θ).
Since the integrand will involve cos2 (θ) and the dx will be
given by dx = a cos(θ) dθ, one can expect to be able to
simplify the integral after such a substitution.
x dx
√
Example: Show that
dx = arcsin
+ C.
a
a2 − x 2
√
Similarly, for integrands that involve x 2 − a2 , a > 0, one can
change variables so that x > 0 and then try x = a cosh(θ)
since x 2 ≥ a2 .
Examples: Show that
x 2 − a2 dx can be written as
a2 sinh2 (θ) dθ after a substitution.
Trigonometric substitutions take advantage of known algebraic
relationship between
Sines, cosines, and tangents
cos2 (θ) + sin2 (θ) = 1
d
d
cos(θ) = − sin(θ)
sin(θ) = cos(θ)
dθ
dθ
1
d
tan(θ) = 1 + tan2 (θ) =
dθ
cos2 (θ)
Hyperbolic sines, cosines and tangents
cosh2 (θ) − sinh2 (θ) = 1
d
d
cosh(θ) = sinh(θ)
sinh(θ) = cosh(θ)
dθ
dθ
1
d
tanh(θ) = 1 − tanh2 (θ) =
dθ
cosh2 (θ)
Methods of integration II
Methods of integration II
Calculus and Differential Equations I
Half-angle substitutions
Calculus and Differential Equations I
Partial fraction decomposition
P(x)
where d o (P) < d o (Q),
Q(x)
into partial fractions, proceed as follows.
To decompose the rational function
Half-angle substitutions are useful to find antiderivatives of
products and/or ratios of sines and cosines.
1
Indeed, let t = tan(θ/2). Then,
cos(θ) =
1 − t2
,
1 + t2
sin(θ) =
2t
,
1 + t2
1
dt = (1 + t 2 ) dθ.
2
2
Example: Show that
dθ
θ = ln tan
+ C.
sin(θ)
2 Methods of integration II
Calculus and Differential Equations I
For each factor of the form (x − a)n , the partial fraction
decomposition of P(x)/Q(x) will include terms of the form
Aj
An
A1
A2
, ··· ,
, ··· ,
.
,
2
j
x − a (x − a)
(x − a)
(x − a)n
A product or ratio of sines and cosines will thus be
transformed into a rational function of t, which we know how
to integrate (using partial fractions).
Factor the denominator Q(x) into terms of the form (x − a)n
and (x 2 + bx + c)n , where n ≥ 1 and x 2 + bx + c is irreducible.
3
4
To find An , multiply by (x − a)n and set x = a into the
resulting equation.
To find the Aj ’s, j =
n, multiply by (x − a)n , and substitute in
appropriate values of x.
Methods of integration II
Calculus and Differential Equations I
Partial fraction decomposition (continued)
5
Integration of a partial fraction decomposition
For each factor of the form (x 2 + bx + c)n , the partial fraction
decomposition of P(x)/Q(x) will include terms of the form
Typical terms in a partial fraction decomposition are of the form
A
(x − a)n
Bj x + Cj
Bn x + Cn
B1 x + C1
, ··· ,
.
, ··· ,
2
j
2
+bx +c
(x + b x + c)
(x + b x + c)n
and
x2
1
6
To find the Bj ’s and Cj ’s, multiply by (x 2 + b x + c)n , expand,
and equate the coefficients of the various powers of x in both
sides of the resulting equation.
Terms of the form
If n = 1, then
If n > 1, then
Example: Find the partial fraction decomposition of
x2 + 5
.
f (x) =
(x + 1)(x 2 − 2x + 3)
(x 2
Bx +C
.
+ b x + c)n
A
,n≥1
(x − a)n
A
dx = ln(|x − a|) + C .
x −a
A
−A
1
dx =
+ C.
(x − a)n
n − 1 (x − a)n−1
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Methods of integration II
Integration of a partial fraction decomposition (continued)
2
Bx +C
Terms of the form 2
,n≥1
(x + b x + c)n
1
Compare the numerator to the derivative of x 2 + b x + c.
Bx +C
dx
(x 2 + b x + c)n
=
=
+ b) − b2B + C
dx
(x 2 + b x + c)n
du
B
dx
+D
,
2
un
(x 2 + b x + c)n
bB
.
2
Thus, we can integrate provided we know how to find an
antiderivative of 1/(x 2 + b x + c)n .
3
Note that sincex 2 + bx+ c is irreducible, one can write
2
b
b2
+ d 2 , where d 2 = c − .
x 2 + bx + c = x +
2
4
Calculus and Differential Equations I
Integration of a partial fraction decomposition (continued)
3
To integrate x+
B
2 (2x
where u = x 2 + b x + c and D = C −
2
Methods of integration II
Calculus and Differential Equations I
x+
1
b 2
2
dx
b 2
2
+ d2
+ d2
x
b
+
. Then,
d
2d
n , let u =
n =
1
d 2n−1
(u 2
du
.
+ 1)n
If n = 1, then
du
dx
1
1
x
b
+C
=
=
arctan
+
2
d
(u 2 + 1)
d
d
2d
x + b2 + d 2
du
and
1 + u2
= cos2n−2 (θ) dθ.
If n > 1, let θ = arctan(u). Then, dθ =
du
n =
2
(u + 1)
dθ
n−1
(1 + tan2 (θ))
Alternatively, integrate by parts and find a recursive formula.
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Methods of integration II
Calculus and Differential Equations I
Methods of integration II
Calculus and Differential Equations I