1
Quick Review
1
Quick Review
1.1
Trig Integration
Z
I Integrals of the form
(sin x)m (cos x)n dx.
• (Method of substitution) When m = 1, 3, 5, 7, · · · or n = 1, 3, 5, 7, · · · ,
Z
Z
Z
n
n
2k+1
2
k
(sin x)
(cos x) dx =
(1 − cos x) (cos x) (sin x dx) = (1 − u2 )k un (−du) = · · ·
|
{z
}
u=cos x
Z
m
(sin x)
2k+1
(cos x)
Z
dx =
m
(sin x)
|
Z
• (Method of reduction) Let Sn =
Z
Sn =
2
k
(1 − sin x)
{z
u=sin x
Z
n
(sin x) dx and Cn =
Z
(cos x dx) =
}
um (1 − u2 )k du = · · ·
(cos x)n dx for convenience. When n ≥ 2,
integrate
differentiate
Z z
}|
{ z }| {
n−2
2
n−2
(sin x)
(1 − cos x) dx = Sn−2 − (sin x)
cos x · cos x dx
= Sn−2 −
IbP
(sin x)n−1 cos x
1
−
Sn .
n−1
n−1
Solving this for Sn , we have
Sn = −
(sin x)n−1 cos x n − 1
+
Sn−2 .
n
n
Similar trick gives
(cos x)n−1 sin x n − 1
+
Cn−2 .
n
n
(Formulas themselves are hard to memorize. Instead, remember the trick.) Thus if one of m and n is even, then
Z
Z
(sin x)2k (cos x)n dx =
(1 − cos2 x)k (cos x)n dx = Cn − kCn+2 + · · · + (−1)k Cn+2k
Cn =
Z
(sin x)
m
2k
(cos x)
Z
dx =
(sin x)m (1 − sin2 x)k dx = Sm − kSm+2 + · · · + (−1)k Sm+2k
and we can apply the reduction formulas.
• (Reducing the degree by half) We can avoid reduction formulas when both m and n are even. Use the half-angle
formulas
1 − cos 2x
1 + cos 2x
sin2 x =
and cos2 x =
2
2
to reduce the total degree by half. For example,
2
2
1 − cos 2x
1 + cos 2x
1 − cos 2x
1 + cos 2x
sin2 x cos2 x =
, sin4 x =
, cos4 x =
.
2
2
2
2
Applying the half-angle formulas again, this further simplifies to
1 cos 4x
3 cos 2x cos 4x
3 cos 2x cos 4x
sin2 x cos2 x = −
, sin4 x = −
+
, cos4 x = +
+
.
8
8
8
2
8
8
2
8
Then integrate the resulting formula.
1
1
Quick Review
1.2
Z
I Integrals of the form
Trig Substitution
(tan x)m (sec x)n dx.
• (Method of substitution I) When m = 1, 3, 5, 7, · · · is odd and n ≥ 1,
Z
Z
Z
(tan x)2k+1 (sec x)n dx =
(sec2 x − 1)k (sec x)n−1 (sec x tan x dx) = (u2 − 1)k un−1 du = · · · .
|
{z
}
u=sec x
• (Method of substitution II) When n = 2, 4, 6, 8, · · · ,
Z
Z
Z
(tan x)m (sec x)2k dx = (tan x)m (tan2 x + 1)k−1 (sec2 x dx) = um (u2 + 1)k−1 du = · · · .
|
{z
}
u=tan x
Z
• (Method of reduction I) Let Tn = (tan x)n dx for convenience. When n ≥ 2,
Z
Z
Tn = (tan x)n−2 (sec2 x − 1) dx = (tan x)n−2 (sec2 x dx) − Tn−2 .
Evaluating this integral using the substitution u = tan x, we get
Tn =
(tan x)n−1
− Tn−2
n−1
with T0 = x and T1 = ln | sec x|.
Z
• (Method of reduction II) Let En =
Z
En =
(sec x)n dx for convenience. When n ≥ 3,
integrate
differentiate
Z z
}|
{ z }| {
(sec x)n−2 (1 + tan2 x) dx = En−2 + (sec x)n−2 tan x · tan x dx
= En−2 +
IbP
1
(sec x)n−2 tan x
−
En .
n−2
n−2
Solving this for En , we get
En =
(sec x)n−2 tan x n − 2
+
En−2
n−1
n−1
with E1 = ln | sec x + tan x|. In fact, this formula holds for any n ≥ 2. Thus when m is even,
Z
Z
(tan x)2k (sec x)n dx =
(sec2 x − 1)k (sec x)n dx = En+2k − kEn+2k−2 + · · · + (−1)k En
and we can apply the reduction formula for En .
1.2
Trig Substitution
√
form in integrand
√
x2 + a2
√
a2 − x2
√
x2 − a2
substitution
√
x = a tan θ
√
x = a sin θ
√
x = a sec θ
2
x2 + a2 = a sec θ
dx = a sec2 θ dθ
a2 − x2 = a cos θ
dx = a cos θ dθ
x2
−
a2
= a tan θ
dx = a sec θ tan θ dθ
1
Quick Review
1.3
1.3
Partial Fractions
Partial Fractions
Let P (x) and Q(x) be polynomials, and denote the degree by deg P (x) and deg Q(x), respectively.
Z
I Integrals of the form
P (x)
dx when deg Q(x) = 2 or deg Q(x) = 3.
Q(x)
• If P (x)/Q(x) is not proper, apply long division to write P (x)/Q(x) = g(x) + R(x)/Q(x) such that g(x) is
polynomial and R(x)/Q(x) is proper.
• If P (x)/Q(x) is proper and deg Q(x) = 2 , exactly one of the following is possible:
P (x)
A
B
=
+
.
Q(x)
x−a x−b
P (x)
A1
A2
.
(b) Q(x) = k(x − a)2 . Then
=
+
Q(x)
x − a (x − a)2
(c) Q(x) has no real root. Then complete the square and apply the tangent trig substitution.
(a) Q(x) = k(x − a)(x − b) for a 6= b. Then
• If P (x)/Q(x) is proper and deg Q(x) = 3 , exactly one of the following is possible:
P (x)
A
B
C
=
+
+
.
Q(x)
x−a x−b x−c
P (x)
A
B1
B2
(b) Q(x) = k(x − a)(x − b)2 for distinct a, b. Then
=
+
+
.
Q(x)
x − a x − b (x − b)2
P (x)
A1
A2
A3
(c) Q(x) = k(x − a)3 . Then
=
+
+
.
Q(x)
x − a (x − a)2
(x − a)3
P (x)
A
Bx + C
(d) Q(x) = k(x − a)(x2 + bx + c) with x2 + bx + c having no real root. Then
=
+
.
Q(x)
x − a x2 + bx + c
(a) Q(x) = k(x − a)(x − b)(x − c) for distinct a, b, c. Then
1.4
Improper Integrals
• (Types of improper integral)
(a) (Simply) Improper. There are two types of improper integrals:
Z
Z
infinite interval
Z R
∞
f (x) dx = lim
f (x) dx
R→∞
a
b
a
b
f (x) dx = lim
R→−∞
R→b−
a
Z
−∞
infinite discontinuity at endpoint
Z b
Z R
f (x) dx = lim
f (x) dx
Z
f (x) dx
b
R→a+
a
(b) Doubly Improper. They are defined as a sum of two improper integrals:
Z ∞
Z 0
Z
f (x) dx =
lim
f (x) dx + lim
Z
S→−∞
∞
b
lim
S→a−
R→∞
S
Z
f (x) dx =
a
b
f (x) dx = lim
R
−∞
a
Z
!
f (x) dx
S
lim
R→∞
!
R
f (x) dx ,
0
Z
+
f (x) dx
R
R
!
f (x) dx
b
and similarly for two other possible combinations. (cf. p448, formula (2) and Exercise 8.6.75)
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1
Quick Review
1.4
Improper Integrals
(c) Remark. Unlike usual improper integrals, doubly improper integrals require ‘two-sided check’ for convergence.
This is because they are defined as sum of two improper integrals.
• (Evaluating improper integral) Just follow the definition. Of course, you may first identify the type of integral in
order to figure out an appropriate definition involved.
(a) (Step 1) Identify the type of integral. Is it doubly improper, or simply improper?
(b) (Step 2) Calculate the integral according to the corresponding definition.
• (Testing convergence) Using various convergence tests are easier and faster when only the convergence/divergence
is our concern (so that the exact value is irrelevant). The following test is very useful for this purpose:
Theorem (Comparison Test). Assume f (x) ≥ g(x) ≥ 0 for x ≥ a. Then we always have
Z b
Z ∞
g(x) dx ≥ 0.
f (x) dx ≥
a
a
In particular,
R∞
R∞
(a) a f (x) dx < ∞ implies a g(x) dx < ∞.
R∞
R∞
(b) a g(x) dx = ∞ implies a f (x) dx = ∞.
This result is valid for any type of improper integrals.
But it is not always easy to find a function to compare. So the following theorems are more practical in many
situations:
Theorem (Limit Comparison Test; Infinite Interval). Assume that both f (x) and g(x) are ≥ 0 and continuous
on the interval [a, ∞). If f (x) ∼ g(x) as x → ∞, then
Z ∞
Z ∞
f (x) dx and
g(x) dx
a
a
both converge or diverge simultaneously.
Its analogues for other types of improper integrals are also true. For example, we have
Theorem (Limit Comparison Test; Infinite discontinuity at x = a). Assume that both f (x) and g(x) are ≥ 0
and continuous on the interval (a, b]. If f (x) ∼ g(x) as x → a+, then
Z b
Z b
f (x) dx and
g(x) dx
a
a
both converge or diverge simultaneously.
Warning! These tests are only for simply improper integrals and NOT for doubly improper integrals. They require
‘two-sided check’.
To be completed· · ·
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