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8.3 Trig Substitution
An additional identity to recall for this section:
sin 2θ = 2 sin θ cos θ
More difficult integrals that are sometimes used in trig integrals.
Z
sec θ dθ = ln | sec θ + tan θ| + C
Z
sec3 θ dθ =
1
(sec θ tan θ + ln | sec θ + tan θ|) + C
2
The general idea of trig substitution is to transform an algebraic integral that involves one of the general
forms in the table below into a trig integral that can be integrated using the techniques from 8.2.
•
Expression
Substitution
Identity Used to Simplify
√
a2 − x2
x = a sin θ
1 − sin2 θ = cos2 θ
√
x 2 + a2
x = a tan θ
tan2 θ + 1 = sec2 θ
√
x 2 − a2
x = a sec θ
sec2 −1 = tan2 θ
Z √
16 − x2
dx
x2
1
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•
Z
x3
√
1
dx
x2 − 25
2
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•
Z √
x2 + 9
dx
x6
3
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•
Z
0
1/8
√
x2
dx
1 − 16x2
4
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•
Z
0
2/3
1
dx
(4 + 9x2 )5/2
5
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dx
− 8x
•
Z
√
•
Z
dx
(5 − 4x − x2 )3/2
x2
6
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8.4 Partial Fraction Decomposition
Fact used often in this section:
Z
x
1
1
dx = arctan
+C
2
2
x +a
a
a
The method of partial fraction decomposition can be used when:
• The integrand is a rational function (consists of a polynomial over a polynomial).
• The degree of the numerator must be less than the degree of the denominator. If not, you must first use
long division to simplify the expression.
The general idea is to take a more complicated rational expression and to “decompose” it into easier
integrable expressions.
1
2
3x + 10
=
+
by combining the right hand side.
x2 + 7x + 12
x+3
x+4
For example, it can be shown that
Therefore,
Z
3x + 10
dx =
2
x + 7x + 12
However, if you were just given
Z
Z
x2
1
dx +
x+3
Z
2
dx = ln |x + 3| + 2 ln |x + 4| + C
x+4
3x + 10
dx, how do you obtain the decomposition?
+ 7x + 12
Setting up the partial fraction decomposition:
Linear (Degree One) Factors:
x+2
x(x + 3)(x − 7)
Repeated Linear Factors:
(x2
x+2
− 16)2 (x + 5)
Irreducible Quadratic Factors:
x+7
+ 3)
x3 (x2
Repeated Irreducible Quadratic Factors:
x+7
(3x − 1)2 (x2 + 9)2
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Examples: Integrate the following.
•
Z
2x3
1
dx
+ 5x2 + 3x
8
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•
Z
x4 − 11x2 + 14x + 8
dx
x3 − 4x2 + 4x
9
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•
Z
3x2 + 5x + 33
dx
x2 + 10
10
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•
Z
3x3 + 18
dx
x2 (x2 + 9)
11
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•
Z
3x3 + 6x + 5
dx
(x2 + 2)(x2 + 4)
12