c Dr Oksana Shatalov, Fall 2012
1
Fall 2012 Math 152
Week in Review 5
courtesy: Dr. Oksana Shatalov
(covering Sections 8.3,8.4&8.9)
8.3: Trigonometric Substitutions
integral with
substitution
identity
π
π
≤θ≤
2
2
1 − sin2 θ = cos2 θ
π
π
<θ<
2
2
1 + tan2 θ = sec2 θ
a2 − x2
x = a sin θ, −
a2 + x2
x = a tan θ,
−
x2 − a2
x = a sec θ,
0≤θ<
ax2 + bx + c
π
2
or
π≤θ<
3π
2
sec2 θ − 1 = tan2 θ
complete squares and then do the correct substitution
Examples
1. Evaluate the given integral:
Z p
(a)
1 − 3x2 dx
Z
1
√
(b)
dx
x2 25x2 − 9
Z 4/7
1
(c)
dx
(49x2 + 16)3/2
0
2. Use a trigonometric substitution to eliminate the root:
p
(a) (x + 1)2 − 64
p
(b) 4(x − 5)2 + 1
3. (a) Use a trigonometric substitution to eliminate the root:
Z
(x + 1)2
(b) Evaluate the integral
dx
(24 − x2 − 2x)3/2
√
24 − 2x − x2 .
8.4: Integration Of Rational Functions By Partial Fractions
Key Points
• Rational function: f (x) =
P (x)
, where P (x) and Q(x) are polynomials.
Q(x)
c Dr Oksana Shatalov, Fall 2012
2
• Partial Fraction Decomposition only works on proper rational functions ( degree of the numerator
is strictly less than the degree of denominator.) (Otherwise (for improper rational functions), you
must first do long division.)
• You have to factor a denominator as much as possible and “break down” the fraction using individual
denominators.
• For each factor in the denominator use the following table:
Factor in denominator
prime quadratic factor
ax2 + bx + c, where b2 − 4ac < 0
Term in partial fraction decomposition
A
ax + b
A1
A2
+
ax + b (ax + b)2
A1
A2
Ak
+
+ ··· +
ax + b (ax + b)2
(ax + b)k
Ax + B
ax2 + bx + c
repeated prime quadratic factor
(ax2 + bx + c)2 , where b2 − 4ac < 0
A1 x + B 1
A2 x + B2
+
ax2 + bx + c (ax2 + bx + c)2
repeated prime quadratic factor
(ax2 + bx + c)k , where b2 − 4ac < 0
A1 x + B 1
A2 x + B2
Ak x + B k
+
+ ··· +
2
2
2
ax + bx + c (ax + bx + c)
(ax2 + bx + c)k
linear factor ax + b
repeated linear factor (ax + b)2
repeated linear factor (ax + b)k
• Solve for numerators using roots of the denominator and/or matching powers of x. THEN INTEGRATE.
Examples
4. Write out the form of the partial fraction decomposition of the following rational functions. (Do
not try to solve)
(a)
3x
(x − 1)(3x + 12)
5x2
(x − 1)2 (x2 − 1)
7
(c)
x(x3 + x2 + x)
x+5
(d)
(x − 3)(x2 + 25)2
(b)
5. Compute the following integrals.
Z
x2 + 4x
dx
(a) I =
(x − 1)(x − 2)(x + 3)
Z
x2 − 3x + 7
(b) I =
dx
(x − 1)(x2 + 1)
Z 4
x − x3 − 12x2 + 10
(c) I =
dx
x3 − 4x2
Z
1
(d) I =
dx
2
(x + 1)(x2 + x + 1)
c Dr Oksana Shatalov, Fall 2012
3
8.9: Improper Integrals
Key Points
• TYPE
Z ∞ I: Infinite Interval
Z t and Continuous Integrand:
f (x) dx = lim
f (x) dx (likewise with −∞)
t→∞ a
a
• TYPE II: Discontinuous Integrand and Finite Interval:
Z b
Z b
f (x) dx (likewise with b− )
f (x) dx = lim
f (x) is discontinuous at x = a:
a
Z
•
∞
FACT: If a > 0 then
a
t→a+
t
1
dx is convergent as p > 1 and divergent as p ≤ 1.
xp
• Comparison Theorem: Let f (x) and g(x) be continuous and f (x) ≥ g(x) ≥ 0 for x ≥ a. Then
Z ∞
Z ∞
f (x) dx is convergent then
g(x) dx is convergent.
– if
a
a
Z ∞
Z ∞
(Note if
f (x) dx is divergent, no conclusion can be drawn about
g(x) dx.)
a
a
Z ∞
Z ∞
– if
g(x) dx is divergent then
f (x) dx is divergent.
a
a
Z ∞
Z ∞
(Note if
g(x) dx is convergent, no conclusion can be drawn about
f (x) dx.)
a
a
Examples
6. Compute the following integrals or show they diverge:
Z ∞
1
dx
(a) I =
x(ln(x)5 )
e
Z 0
(b) I =
(1 + x)ex dx
−∞
∞
Z
(c) I =
−∞
9
Z
(d) I =
0
5x4
dx
+ 5)3
(x5
√
3
1
dx
x−4
7. Determine whether the given integrals converge or diverge using the Comparison Theorem.
Z ∞
1
(a) I =
dx
2012
x
+ e2012x
0
Z ∞
x2
(b) I =
dx
x5/2 − x
5
Z ∞
sin4 (7x)
(c) I =
dx
x7
10