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Math 106: Review for Exam II - SOLUTIONS
INTEGRATION TIPS
- Substitution: usually let u = a function that’s “inside” another function, especially if du (possibly off
by a multiplying constant) is also present in the integrand.
u dv = uv −
v du or
uv
′ dx = uv −
u
′ v dx
How to choose which part is u? Let u be the part that is higher up in the LIATE mnemonic below.
(The mnemonics ILATE and LIPET will work equally well if you have learned one of those instead;
in the latter A is replaced by P, which stands for “polynomial.”)
Logarithms (such as ln x)
Inverse trig (such as arctan x, arcsin x)
Algebraic (such as x, x
2 , x
3
Trig (such as sin x, cos 2x)
Exponentials (such as e
x , e
3 x )
- Rational Functions (one polynomial divided by another): if the degree of the numerator is greater than
or equal to the degree of the denominator, do long division then integrate the result.
Partial Fractions: here’s an illustrative example of the setup.
3 x
2
(x + 1)(x − 3) 2 (x 2
A
x + 1
B
x − 3
C
(x − 3) 2
Dx + E
x 2
Each linear term in the denominator on the left gets a constant above it on the right; the squared
linear factor (x − 3) on the left appears twice on the right, once to the second power. Each irreducible
quadratic term on the left gets a linear term (Dx + E here) above it on the right.
- Trigonometric Substitutions: some suggested substitutions and useful formulae follow.
Radical Form
a 2 − x 2
a 2
x 2 − a 2
Substitution x = a sin t x = a tan t x = a sec t
sin
2 x + cos
2 x = 1 tan
2 x + 1 = sec
2 x
sin
2 x =
cos(2x)
cos
2 x =
cos(2x)
sin(2x) = 2 sin x cos x
- Powers of Trigonometric Functions: here are some strategies for dealing with these.
sin
m x cos
n x dx Possible Strategy Identity to Use
m odd Break off one factor of sin x and substitute u = cos x. sin
2 x = 1 − cos
2 x
n odd Break off one factor of cos x and substitute u = sin x. cos
2 x = 1 − sin
2 x
m, n even Use sin
2 x + cos
2 x = 1 to reduce to only powers of sin x sin
2 x =
cos(2x)
or only powers of cos x, then use integration by parts cos
2 x =
cos(2x)
or identities shown to right of this box.
tan
m x sec
n x dx Possible Strategy Identity to Use
m odd Break off one factor of sec x tan x and substitute u = sec x. tan
2 x = sec
2 x − 1
n even Break off one factor of sec
2 x and substitute u = tan x. sec
2 x = tan
2 x + 1
m even, n odd Use identity at right to reduce to powers of sec x alone. tan
2 x = sec
2 x − 1
Then use integration by parts or reduction formula (if allowed).
Useful Trigonometric Derivatives and Antiderivatives
d
dx
tan x = sec
2 x
d
dx
sec x = sec x tan x
sec x dx = ln | sec x + tan x| + C
- Improper integrals: look for ∞ as one of the limits of integration; look for functions that have a vertical
asymptote in the interval of integration. It may be useful to know the following limits.
lim x→∞
e
x = ∞
lim x→∞
e
−x = 0 Note: this is the same as lim x→−∞
e
x .
lim x→∞
1 /x = 0 Note: the answer is the same for lim x→∞
1 /x
2 and similar functions.
lim x→ 0
1 /x = ∞ Note: the answer is the same for lim x→ 0
1 /x
2 and similar functions.
lim x→∞
ln x = ∞
lim x→ 0 +
ln x = −∞
lim x→∞
arctan x = π/ 2
- Evaluate the following.
(a) Let u = sin x, so du = cos x dx.
sin
6 x cos
3 x dx =
sin
6 x(1 − sin
2 x) cos x dx Use cos
2 x = 1 − sin
2 x.
u
6 (1 − u
2 ) du
(u
6 − u
8 ) du
u 7
u 9
+ C
sin
7 x
sin
9 x
+ C
(b) Let x = 10 tan t, so dx = 10 sec
2 t dt.
x
y
t
x
2
2 = y
2 ⇒ y =
x 2
sec t =
hyp
adj
x^2 + 100
tan t =
opp
adj
x
3 x
2
- 2x − 13 = (Ax + B)(x − 3) + C(x
2
Let x = 3. Then 20 = C(10), so C = 2.
Let x = 0. Then −13 = B(−3) + 2(1), so B = 5.
Let x = 1. Then −8 = (A(1) + 5)(−2) + 2(2), so A = 1.
3 x
2
(x − 3)(x 2
dx =
∫ [^
x + 5
x 2
x − 3
]
dx
∫ [^
x
x 2
x 2
x − 3
]
dx Let u = x
2
2 du
u
∫ [^
x 2
x − 3
]
dx
ln u
- 5 arctan x + 2 ln |x − 3 | + D
ln(x 2
(f) Since the degree of the numerator is greater than or equal to the degree of the denominator, we
do long division.
4 x
2 − 3 x + 2 +
x − 6
x − 6
4 x
3 − 27 x
2
4 x
3 − 24 x
2
− 3 x
2
− 3 x
2 +18x
2 x
2 x − 12
Now, we compute the integral. ∫ 4 x
3 − 27 x
2
x − 6
dx =
∫ [
4 x
2 − 3 x + 2 −
x − 6
]
dx =
4 x
3
3 x
2
(g) This integral is improper at x = 1 because the integrand has a vertical asymptote there, so we
split into two integrals.
5
− 1
(x − 1) 6
dx
− 1
(x − 1)^6
dx
5
− 1
(x − 1) 6
dx =
1
− 1
dx
(x − 1) 6
5
1
dx
(x − 1) 6
= lim a→ 1 −
∫ (^) a
− 1
dx
(x − 1) 6
b
dx
(x − 1) 6
= lim a→ 1 −
5(x − 1) 5
a
− 1
5(x − 1) 5
5
b
u
− 6 du =
u − 5
+ C
= lim a→ 1 −
[
5(a − 1) 5
5
]
[
5
5(b − 1) 5
]
Since lim a→ 1 −
5(a − 1) 5 = ∞ and lim b→ 1 +
5(b − 1) 5 = ∞, this integral diverges (to ∞).
- Find the second-degree Taylor polynomial for f(x) =
f(x) = (^) x
f(x) = x
x centered at xxx = 100= 100= 100.
f(x) = x
1 / 2 f(100) = 10
f
′ (x) =
x
− 1 / 2
x
f
′ (100) =
f
′′ (x) =
x
− 3 / 2
4 x 3 / 2
f
′′ (100) =
3 / 2
P 2 (x) = f(100) + f
′ (100)(x − 100) +
f ′′ (100)
(x − 100)
2
x − 100
(x − 100)
2
- What is the maximum possible error that can occur in your Taylor approximation from
the previous problem on the interval [100, 110]?
We know that |f(x) − Pn(x)| ≤
Kn+
(n + 1)!
|x − x 0 |
n+ .
In this case, n = 2, x 0 = 100, and x = 110 (the farthest from x 0 that we are considering).
K 3 = max of |f
′′′ (x)| on [100, 110] = max of |
8 x 5 / 2
| on [100, 110] =
5 / 2
Putting this all together, we have |f(x) − P 2 (x)| ≤
3 800 , 000
3!
3
- Use comparisons to show whether each of the following converges or diverges. If an
integral converges, also give a good upper bound for its value.
(a)
∞
1
6 + cos x
x^0.^99
dx
For all x ≥ 1, we have
6 + cos x
x
- 99
x
- 99
x
- 99
because the minimum value of cos x is −1.
Since
∞
1
5 dx
x
- 99
diverges (compute yourself or notice that p = 0. 99 < 1), we know that the integral
in question must diverge too.
(b)
∞
1
4 x
3 − 2 x
2
2 x 4
dx
For all x ≥ 1, we have
4 x
3 − 2 x
2
2 x^4 + x^5 + 1
4 x
3
x^5
x^2
. (We’ve made the denominator smaller and
the numerator larger, so the new fraction is larger.)
∞
1
dx
x 2
= 4 lim t→∞
t
1
dx
x 2
= 4 lim t→∞
x
t
1
= 4 lim t→∞
[
t
]
= 4[0 − (−1)]
Therefore, the original integral in question must converge to a value less than 4.
