MAT137 Study guide

April 2025

1 Integration Rules
1.1 Sine & Cosine:
Sine Reduction Formula
Let n ≥ 2. Then,
n−1
Z Z
n 1 n−1
sin (x)dx = sin (x) cos(x) + sinn−2 (x)dx
n n

Cosine Reduction Formula

Let n ≥ 2. Then,
n−1
Z Z
n 1 n−1
cos (x)dx = cos (x) sin(x) + cosn−2 (x)dx
n n

Z
Another type of possible integral: sinm (x) cosn (x)dx

Case 1:
• If power of sin(x) is ODD, keep 1 factor of sin(x), and use the Pythagorean identity
to change sin2 (x) = 1 − cos2 (x).
• If power of cos(x) is ODD, keep 1 factor of cos(x), and use the Pythagorean identity
to change cos2 (x) = 1 − sin2 (x).
• If both are odd, pick the one that will give a power of 2 (if possible).
Case 2:
• If both powers are EVEN, use half-angle formulas:
1 − cos(2x) 1 + cos(2x)
sin2 (x) = and cos2 (x) =
2 2
Then use u-substitution.

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1.2 Tangent & Secant
Z
Format: tanm (x) secn (x)dx

• If power of tan(x) is ODD, break off 1 factor of tan(x) sec(x), and use tan2 (x) =
sec2 (x) − 1.

• If power of sec(x) is EVEN, break off 1 factor of sec2 (x), and use sec2 (x) = tan2 (x) + 1.
Then use u-substitution.

1.3 Cotangent & Cosecant

Z
Format: cotm (x) cscn (x)dx

• If power of cot(x) is ODD, break off 1 factor of cot(x) csc(x), and use cot2 (x) =
csc2 (x) − 1.

• If power of sec(x) is EVEN, break off 1 factor of cot2 (x), and use csc2 (x) = cot2 (x) + 1.
d
(Remember dx
csc(x) = − csc(x) cot(x)).
Then use u-substitution.

1.4 Sine & Cosine with Different Angles

NOTE: For integrals where sine and cosine have different angles use these identities:

Product-to-Sum Formulas
Let m, n ∈ R. Then,
1
sin(mx) sin(nx) = [cos(mx − nx) − cos(mx + nx)]
2
1
sin(mx) cos(nx) = [sin(mx − nx) + sin(mx + nx)]
2
1
cos(mx) cos(nx) = [cos(mx − nx) + cos(mx + nx)]
2

1.5 Trigonometric Substitution

√ √ √
This idea works for a2 + x2 , a2 − x2 , x2 − a2 . Then, use 1 + tan2 x = sec2 x or 1 −
sin2 x = cos2 x. In trigonometric substitution we only use tan θ, sin θ, and sec θ. Here are
the main formulas.

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 Trigonometric Substitution
√ x
a2 + x2 : tan θ = =⇒ x = a tan θ
a
√ x
a2 − x2 : sin θ = =⇒ x = a sin θ
a
√ x
x2 − a2 : sec θ = =⇒ x = a sec θ
a

1.6 Rational Functions

1. Every polynomial can be decomposed/factored into a product of linear and/or irreducible quadratic
factors, i.e., ax + b and/or ax2 + bx + c
R(x)
2. Every rational, Q(x) , where deg R(x) < deg Q(x) can be decomposed into partial frac-
tions. For deg R(x) ≥ deg Q(x) do long division first.

R(x) A1 A2 An
= + + ··· +
Q(x) a1 x + b 1 a2 x + b 2 an x + b n

Rule for repeated factors:

A1 A2 An
+ 2
+ ··· +
ax + b (ax + b) (ax + b)n

The same holds true for repeated irreducible quadratic factors.

1.7 Fundemental Theorem of Calculus Questions

These questions come in the form of
Z b(x)
d
f (x)dx
dx a(x)

The solution is
f (b(x)) · b′ (x) − f (a(x)) · a′ (x)
Notice that if a(x) or b(x) are constant functions, then their derivatives become 0.

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2 Improper Integrals and Infinite Series
2.1 p-Series Test
p-Test for Infinite Limit
Let a > 0. Then,
Z ∞
1
If p > 1, then dx converges.
a xp
Z ∞
1
If p ≤ 1, then dx diverges.
a xp

p-Test for Discontinuity

Let a > 0. Then,
Z a
1
If p < 1, then dx converges.
0 xp
Z a
1
If p ≥ 1, then dx diverges.
0 xp

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3 Power Series
∞
X
an (x − c)n = a0 + a1 (x − c) + a2 (x − c)2 + · · ·
n=0

This is a power series in x, centered at c, a constant.

3.1 Convergence
1. At x = c only, R = 0, 0 ≤ x − c ≤ 0.
2. For all x, R = ∞, is −∞ ≤ x − c ≤ ∞.
3. For |x−c| < R, where |x−c| > R diverges. This means that the interval of convergence
is (c − R, c + R). We must also check the endpoints with a convergence test. Note that
the domain of the series will always be centered upon c.

The
P∞ ratio test is useful for determining the convergence of a power series. For a given series
n=1 an , let
an+1
L = lim .
n→∞ an
• If L < 1, the series is absolutely convergent.
• If L > 1, the series is divergent.
• If L = 1, the series may be divergent, conditionally convergent, or absolutely convergent
– no conclusion can be made.
After carrying out the ratio (or root) test, and we end up with something like

lim (n + 1)|x|,
n→∞

then we get two values:

1. limn→∞ (n + 1)|x| = ∞ if x ̸= 0, then by the ratio test, the power series is divergent.
2. limn→∞ (n + 1)|x| = 0 if x = 0, then by the ratio test, the power series is convergent.
First off, note that while this would be 0 · ∞, it would not be considered an indeterminate
form since we are not taking the limit of x – it is treated as a constant. Then, we can say
R = 0 (radius of convergence) for the given power series is 0.

• If we get L = limn→∞ an+1

an
= 0, then it is convergent for all x. The interval of
convergence would be (−∞, ∞).

• If we get L = limn→∞ an+1an

= k|x| where k ∈ R, then we take k|x| < 1 to obtain an
open interval of convergence by the ratio test. Then, we check the endpoints of the
open interval for convergence, to determine which endpoints should be included in the
interval of convergence.

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3.2 Power Series Representation for Functions
The idea is to find a function such that taking a derivative or integral of it would yield the
function that we want to make a power series out of. An example such as ln(1 − x) would
a
give the solution in the format of a solution to a geometric series, as in 1−r . Let the solution
equal its representation and integrate or take the derivative of both sides. If integrating, put
a +C on the side of the geometric series representation, and solve for it at the end. Here is
the complete example of finding the power series of ln(1 − x) valid on (−1, 1).

1. Find a function such that taking a derivative or integral, we get ln(x − 1):
Z
1
dx = − ln |x − 1| + C = − ln |1 − x| + C (this is only a negative sign off)
1−x

2. Notice geometric series:

1
1−x
is a sum of a geometric series with a = 1,r = x
∞ ∞
1 X X
= 1 · xn−1 = xn−1 = 1 + x + x2 + x3 + . . .
1 − x n=1 n=1

3. Take integral on both sides:

Let u = 1 − x, and so du = −dx.
Z Z
1
dx = 1 + x + x2 + x3 + · · · dx
1−x

x x3
=⇒ − ln(1 − x) = x + + + ··· + C
2 3
x x3
=⇒ ln(1 − x) = −x − − − ··· − C
2 3
4. Find +C:
First, let x = 0. Then, ln 1 = −C = 0
2 3 P∞ xn
Therefore, we can conclude that ln(x − 1) = −(x + x2 + x3 + · · · ) = − n=1 n on the
interval (−1, 1).

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4 Taylor and MacLaurin Series
Taylor Series:
∞
X f (n) (c) f ′′ (c)
f (x) = (x − c)n = f (c) + f ′ (c)(x − c) + (x − c)2
n=0
n! 2!

MacLaurin Series: This is a Taylor series with c = 0.

∞
X f (n) (0) n ′ f ′′ (0) 2
f (x) = x = f (0) + f (0)x + x
n=0
n! 2!

A function with a power series representation will have a Taylor series represen-
tation. However, the converse is not necessarily true. The nth derivative of a
Taylor series:
f (n) (c) = n!an

4.1 Finding a Taylor Series of a Function

1. Find many derivatives of f and look for a pattern.

2. Plug in x = c (center point of the Taylor series) into the derivative (useful for helping
finding pattern or writing explicit form).

3. Set up the series:

∞
X f (n) (c)
(x − c)n
n=0
n!

4. Find the interval of convergence. Here, the ratio test is useful.

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4.2 Common Maclaurin Series

Interval of
Function Maclaurin Series
Convergence
∞
1 X
1−x xn (−1, 1)
n=0
∞
1 X
1+x (−1)n xn (−1, 1)
n=0
∞
e x
X xn (−∞, ∞)
n=0
n!
∞
sin x
X x2n+1 (−∞, ∞)
(−1)n
n=0
(2n + 1)!
∞
cos x
X x2n
n (−∞, ∞)
(−1)
n=0
(2n)!
∞ n
n+1 x
X
ln(1 + x) (−1) (−1, 1]
n=1
n
∞
ln(1 − x)
X xn (−1, 1)
−
n=1
n
∞ 2n+1
n x
X
arctan x (−1) [−1, 1]
n=0
2n + 1
∞
arcsin x
X (2n)! [−1, 1]
n 2
x2n+1
n=0
4 (n!) (2n + 1)
∞
sinh x = ex −e−x X x2n+1 (−∞, ∞)
2
n=0
(2n + 1)!
∞
cosh x = ex +e−x X x2n (−∞, ∞)
2
n=0
(2n)!

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4.3 Integration with Taylor/MacLaurin Series
We will find the following integral: Z
2
e−x dx

We can’t use any of our known integration techniques, so we’ll write the expression as a
MacLaurin series by using the already known MacLaurin series of ex .
∞ ∞ ∞
x
X xn −x2
X (−x2 )n X (−1)n x2n
e = =⇒ e = =
n=0
n! n=0
n! n=0
n!

Now, we can integrate the new expression, where we use the power rule for integration.
Remember, we must only integrate for the variable x as that is specified as the variable of
integration. The n act as coefficients.

(−1)n x2n
Z Z
−x2
e dx = dx
n!
∞
X (−1)n x2n+1
= +C
n=0
n!(2n + 1)

4.4 Computing Limits with Taylor Polynomials

We will compute the following limit:

(sin x − x + 16 x3 )(ex − 1)
lim
x→0 x6
First, we will write the Taylor polynomials for sin x and ex centered around 0 to obtain
reasonable approximations.

x2 x3 x4 x5
sin x ≈ sin(0) + x · sin′ (0) + · sin′′ (0) + · sin(3) (0) + · sin(4) (0) + · sin(5) (0)
2! 3! 4! 5!
x3 x5
=x− +
3! 5!
3
x x5
=x− +
6 120

d x x2 d2 x x3 d3 x
ex ≈ e0 + x ·
e + · e + · e
dx x=0 2! dx2 x=0 3! dx3 x=0
x2 x3
=1+x+ +
2! 3!
2
x x3
=1+x+ +
2 6

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We can now compute the limit:

(sin x − x + 16 x3 )(ex − 1) x3 x5 x3 x2 x3
     
1
lim ≈ lim x− + −x+ 1+x+ + −1
x→0 x6 x→0 6 120 6 2 6 x6
x5 x2 x3 1
 
= lim x+ +
x→0 120 2 6 x6
x2 x3
 
1
= lim x+ +
x→0 120x 2 6
2 3
x x x
= lim + +
x→0 120x 2 6
2 3
1 x x
= lim + +
x→0 120 2 6
1
=
120

4.5 Finding nth Derivative of a Taylor Series

For this, we can make use of the aforementioned derivative formula:

f (n) (c) = n0 ! · an

Here, an refers to the series without the x term at the new n value after setting n to the
term x is raised to, and n0 is the ith derivative to take.

4.5.1 Example 1
Define a function B, such that
∞
X x2n
B(x) = 2
.
n=0
(n!)

We will compute B (2019) (0).

Notice that x is raised to an even power in each term of the series, so the function B(x) is
even.
First, consider B (2019) (0). Since 2019 is odd, and the Taylor expansion of an even function
about 0 contains only even powers, we conclude:

B (2019) (0) = 0. □

Now, we will compute B (42) (0). Since 42 = 2n, we let n = 21. Then, using the formula
for the nth derivative at 0 in terms of the coefficient of x2n , we get:
1 42!
B (42) (0) = (2n)! · an = 42! · 2
= . □
(21!) (21!)2

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4.5.2 Example 2
Define a function F , such that

F (x) = 1 + x4 sin 2x2 .

We will compute F (42) (0).

First, recall that

∞
X (−1)n x( 2n + 1)
sin x = .
n=0
(2n + 1)!
Then,
∞ 2n+1 ∞
2

X (−1)n (2x2 ) X (−1)n · 22n+1 · x4n+2
sin 2x = =
n=0
(2n + 1)! n=0
(2n + 1)!
So, we arrive at
∞ ∞
4

2

X (−1)n · 22n+1 · x4n+2 4
X (−1)n · 22n+1 · x4n+2
1+x sin 2x =1· +x ·
n=0
(2n + 1)! n=0
(2n + 1)!
∞ ∞
X (−1)n · 22n+1 · x4n+2 X (−1)n · 22n+1 · x4n+6
= +
n=0
(2n + 1)! n=0
(2n + 1)!

For the summation on the left side we have 4n + 2 = 42 =⇒ n = 10 and for the summation
on the right, we have 4n + 6 = 9 =⇒ n = 9 Now we will compute F (42) (0).

F (42) (0) = n0 ! · an1 + n0 ! · an2

= n! (an1 + an2 )
(−1)10 · 22(10)+1 (−1)9 · 22(9)+1
 
= 42! +
(2(10) + 1)! (2(9) + 1)!
 21 19

2 2
= 42! −
21! 19!

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5 Value Theorems
Extreme Value Theorem
A continuous function on a compact set attains its supremum and infi-
mum.

Intermediate Value Theorem

If f is continuous on [a, b] and α is any number between f (a) and f (b),
then there is some c ∈ (a, b) for which f (c) = α.

Rolle’s Theorem
Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). If
f (a) = f (b), then there exists c ∈ (a, b) where f ′ (c) = 0.

Mean Value Theorem

Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). If
f (a) = f (b), then there exists c ∈ (a, b) where

f (b) − f (a)
f ′ (c) = .
b−a

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