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Chapter 6A Notes Page 1 of 12
Introduction and Review
Derivatives 𝑦 = 𝑓(𝑥)
𝑑𝑦
𝑑𝑥= 𝑓′(𝑥)
Evaluate derivative at 𝑥 = 𝑎:
𝑑𝑦
𝑑𝑥|
𝑥=𝑎= 𝑓′(𝑎) = lim
ℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Geometric Interpretation: see figure
slope of the line tangent to 𝑓 at 𝑥 = 𝑎.
Differentiation formulas 𝑑
𝑑𝑥𝑥𝑛 = 𝑛𝑥𝑛−1
𝑑
𝑑𝑥ln|𝑥| =
1
𝑥
𝑑
𝑑𝑥𝑒𝑥 = 𝑒𝑥
𝑑
𝑑𝑥sin(𝑥) = cos(𝑥)
𝑑
𝑑𝑥cos(𝑥) = −sin(𝑥)
𝑑
𝑑𝑥tan(𝑥) = sec2(𝑥)
𝑑
𝑑𝑥sec(𝑥) = sec(𝑥) tan(𝑥)
𝑑
𝑑𝑥sin−1(𝑥) =
1
√1 − 𝑥2
Chapter 6A Notes Page 2 of 12
𝑑
𝑑𝑥tan−1(𝑥) =
1
1 + 𝑥2
Review the sum, difference, product and quotient rules!
The chain rule
Let 𝐹(𝑥) = 𝑓(𝑔(𝑥)), then 𝐹′(𝑥) = 𝑓′(𝑔(𝑥))𝑔′(𝑥)
Or let 𝑦 = 𝑓(𝑢), 𝑢 = 𝑔(𝑥), then 𝑑𝑦
𝑑𝑥=
𝑑𝑦
𝑑𝑢
𝑑𝑢
𝑑𝑥
Example:
𝑑
𝑑𝑥sin(𝑥2) = cos(𝑥2)2𝑥
Indefinite Integrals Indefinite integrals are also known as antiderivatives.
If 𝐹′ is the derivative of 𝐹 then we write
∫ 𝐹′(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶
Example:
∫ cos(𝑥2) 2𝑥 𝑑𝑥 = sin(𝑥2) + 𝐶
Indefinite integral formulas
∫ 𝑥𝑛𝑑𝑥 =1
𝑛 + 1𝑥𝑛+1 + 𝐶
∫1
𝑥 𝑑𝑥 = ln|𝑥| + 𝐶
∫ 𝑒𝑥𝑑𝑥 = 𝑒𝑥 + 𝐶
∫ cos(𝑥) 𝑑𝑥 = sin(𝑥) + 𝐶
∫ sin(𝑥) 𝑑𝑥 = − cos(𝑥) + 𝐶
∫ sec2(𝑥)𝑑𝑥 = tan(𝑥) + 𝐶
∫ sec(𝑥) tan(𝑥) 𝑑𝑥 = sec(𝑥) + 𝐶
Chapter 6A Notes Page 3 of 12
Definite Integrals
Example: ∫1
𝑥 𝑑𝑥
𝑒
1
Geometric Interpretation: see figure
Evaluation theorem
∫ 𝐹′(𝑥)𝑑𝑥 = 𝐹(𝑥)|𝑎𝑏 = 𝐹(𝑏) − 𝐹(𝑎)
𝑏
𝑎
Examples:
∫1
𝑥 𝑑𝑥 = ln(x)|1
e𝑒
1
= ln(𝑒) − ln(1) = 1 − 0 = 1
∫ cos(𝑥2) 2𝑥 𝑑𝑥 = sin(𝑥2)|01 = sin(1) − sin (0)
1
0
Integration by substitution Suppose an integral has the form: 𝐼 = ∫ 𝑓′(𝑔(𝑥))𝑔′(𝑥)𝑑𝑥,
let 𝑢 = 𝑔(𝑥), then 𝑑𝑢 = 𝑔′(𝑥)𝑑𝑥 (this is differential notation)
Substitute to obtain
𝐼 = ∫ 𝑓′(𝑢)𝑑𝑢 = 𝑓(𝑢) + 𝐶 = 𝑓(𝑔(𝑥)) + 𝐶
This is equivalent to taking the chain rule of differentiation backwards!
Example. 𝐼 = ∫ cos(𝑥2) 𝑥 𝑑𝑥
Let 𝑢 = 𝑥2, 𝑑𝑢 = 2𝑥 𝑑𝑥
𝐼 = ∫ cos(𝑢) 1
2 𝑑𝑢 =
1
2 ∫ cos(𝑢) 𝑑𝑢 =
1
2sin(𝑢) + 𝐶 =
1
2sin(𝑥2) + 𝐶.
Example. 𝐽 = ∫ cos(𝑥2) 𝑥 𝑑𝑥3
2
Chapter 6A Notes Page 4 of 12
Let 𝑢 = 𝑥2. If 𝑥 = 2 then 𝑢 = 4. If 𝑥 = 3 then 𝑢 = 9.
𝐽 = ∫ cos(𝑢)1
2𝑑𝑢
9
4
=1
2sin (𝑢)|4
9
6.1 Integration by parts
Chain rule for differentiation: 𝑑
𝑑𝑥𝑠𝑖𝑛(𝑥2) = (cos(𝑥2))2𝑥
Integration by substitution: let = 𝑥2 , 𝑑𝑢 = 2𝑥 𝑑𝑥
∫ (cos 𝑥2)2𝑥 𝑑𝑥 = ∫ cos(𝑢) 𝑑𝑢 = sin(𝑢) + 𝐶 = sin(𝑥2) + 𝐶
Product rule for differentiation
𝑑
𝑑𝑥[𝑓(𝑥)𝑔(𝑥)] = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)
Solve for𝑓(𝑥)𝑔′(𝑥):
𝑓(𝑥)𝑔′(𝑥) =𝑑
𝑑𝑥[𝑓(𝑥)𝑔(𝑥)] − 𝑓′(𝑥)𝑔(𝑥)
Take antiderivatives
(∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥 = 𝑓(𝑥)𝑔(𝑥) − ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥 (A)
This is the rule for integration of indefinite integrals by parts.
Shorthand formula: recall differential notation
Box: 𝑢 = 𝑓(𝑥) 𝑑𝑣 = 𝑔′(𝑥) 𝑑𝑥
𝑑𝑢 = 𝑓′(𝑥)𝑑𝑥 𝑣 = 𝑔(𝑥)
Substitute into (A) above to get
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 (B)
An easy-to-remember form of the integration by parts formula!
Chapter 6A Notes Page 5 of 12
Example 1. Evaluate 𝐼 = ∫ 𝑥 cos(𝑥) 𝑑𝑥.
Box: 𝑢 = 𝑥 𝑑𝑣 = cos(𝑥) 𝑑𝑥
𝑑𝑢 = 𝑑𝑥 𝑣 = sin (𝑥)
𝐼 = ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
= 𝑥 sin(𝑥) − ∫ sin (𝑥)𝑑𝑥
= 𝑥 sin(𝑥) + cos(𝑥) + 𝐶
Check:
𝑑
𝑑𝑥(𝑥 sin(𝑥) + cos(𝑥) + 𝐶) = sin(𝑥) + 𝑥 cos(𝑥) − sin(𝑥) = 𝑥 cos (𝑥) √√
In principle, pick 𝑢 so that 𝑑𝑢 is simpler and pick 𝑑𝑣 so that it is possible to integrate. The following
LIPET scheme for choosing 𝑢 and 𝑣 is often useful:
(L)ogarithm
(I)nverse Trig
(P)olynomials or powers of 𝑥
(E)xponential functions
(T)rig functions
Whichever piece of the integrand is higher on the list is 𝑢 and the other piece is 𝑑𝑣.
Example 2. Integration by parts twice
Evaluate 𝐼 = ∫ 𝑥2 sin(𝑥) 𝑑𝑥.
Box: 𝑢 = 𝑥2 𝑑𝑣 = sin(𝑥) 𝑑𝑥
𝑑𝑢 = 2𝑥 𝑑𝑥 𝑣 = −cos (𝑥)
𝐼 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
= −𝑥2 cos(𝑥) + ∫ cos(𝑥) 2𝑥 𝑑𝑥
= −𝑥2 cos(𝑥) + 2(𝑥 sin(𝑥) + cos(𝑥)) + 𝐶
Chapter 6A Notes Page 6 of 12
We used the result of example 1.
Example 3. Recurrence of the original integral
Evaluate 𝐼 = ∫ 𝑒−𝑥 cos(𝑥) 𝑑𝑥
Box: 𝑢 = 𝑒−𝑥 𝑑𝑣 = cos(𝑥) 𝑑𝑥
𝑑𝑢 = −𝑒−𝑥 𝑑𝑥 𝑣 = sin (𝑥)
𝐼 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 = 𝑒−𝑥 sin(𝑥) + ∫ 𝑒−𝑥 sin(𝑥) 𝑑𝑥 (C)
Evaluate 𝐽 = ∫ 𝑒−𝑥 sin(𝑥) 𝑑𝑥
Box: 𝑢 = 𝑒−𝑥 𝑑𝑣 = sin(𝑥) 𝑑𝑥
𝑑𝑢 = −𝑒−𝑥𝑑𝑥 𝑣 = −cos (𝑥)
𝐽 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 = −𝑒−𝑥 cos(𝑥) − ∫ 𝑒−𝑥 cos(𝑥) 𝑑𝑥 = −𝑒−𝑥 cos(𝑥) − 𝐼 (D)
Combine (C) and (D)
𝐼 = 𝑒−𝑥 sin(𝑥) + (−𝑒−𝑥 cos(𝑥) − 𝐼)
2𝐼 = 𝑒−𝑥(sin(𝑥) − cos(𝑥)) + 2𝐶
𝐼 =1
2𝑒−𝑥(sin(𝑥) − cos(𝑥)) + 𝐶
We added an explicit constant of integration when we put 𝐼 on only one side of the equation. Since 𝐼
represents an indefinite integral, it has an implicit constant of integration.
Example 4: combine parts and substitution
Evaluate 𝐼 = ∫ sin−1(𝑥) 𝑑𝑥
𝑑
𝑑𝑥sin−1(𝑥) =
1
√1−𝑥2
Box: 𝑢 = sin−1(𝑥) 𝑑𝑣 = 𝑑𝑥
𝑑𝑢 =1
√1−𝑥2𝑑𝑥 𝑣 = 𝑥
𝐼 = ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
= 𝑥 sin−1(𝑥) − ∫𝑥
√1−𝑥2𝑑𝑥
Chapter 6A Notes Page 7 of 12
Evaluate 𝐽 = ∫𝑥
√1−𝑥2𝑑𝑥
Let 𝑡 = 1 − 𝑥2, 𝑑𝑡 = −2𝑥 𝑑𝑥, −1
2𝑑𝑡 = 𝑥 𝑑𝑥
Then 𝐽 = −1
2 ∫ 𝑡−
1
2 𝑑𝑡 = −1
2 (2 𝑡
1
2) + 𝐶 = −𝑡1
2 + 𝐶
Combine to get
𝐼 = 𝑥 sin−1(𝑥) + (1 − 𝑥2)1
2 + 𝐶
Definite Integrals Evaluation Theorem
∫ 𝐹′(𝑥)𝑑𝑥 = 𝐹(𝑥)|𝑎𝑏 = 𝐹(𝑏) − 𝐹(𝑎)
𝑏
𝑎
Product Rule
𝑑
𝑑𝑥𝑓(𝑥)𝑔(𝑥) = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)
𝑓(𝑥)𝑔′(𝑥) =𝑑
𝑑𝑥[𝑓(𝑥)𝑔(𝑥)] − 𝑓′(𝑥)𝑔(𝑥)
integrate from 𝑥 = 𝑎 to 𝑥 = 𝑏:
∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥 = ∫𝑑
𝑑𝑥[𝑓(𝑥)𝑔(𝑥)]𝑑𝑥 = ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥
𝑏
𝑎
𝑏
𝑎
𝑏
𝑎
∫ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥𝑏
𝑎= 𝑓(𝑥)𝑔(𝑥)|𝑎
𝑏 − ∫ 𝑓′(𝑥)𝑔(𝑥)𝑑𝑥𝑏
𝑎 (E)
Shorthand using differential notation
Box: 𝑢 = 𝑓(𝑥) 𝑑𝑣 = 𝑔′(𝑥) 𝑑𝑥
𝑑𝑢 = 𝑓′(𝑥)𝑑𝑥 𝑣 = 𝑔(𝑥)
Write as for an indefinite integral:
∫ 𝑢 𝑑𝑣 = 𝑢 𝑣 − ∫ 𝑣 𝑑𝑢
Replace 𝑢, 𝑑𝑢, 𝑣 and 𝑑𝑣 with appropriate expressions in 𝑥 and add limits of integration to get (E)
Chapter 6A Notes Page 8 of 12
Example 1: 𝐼 = ∫ ln(𝑥) 𝑑𝑥𝑒
1
Box: 𝑢 = ln (𝑥) 𝑑𝑣 = 𝑑𝑥
𝑑𝑢 =1
𝑥𝑑𝑥 𝑣 = 𝑥
∫ 𝑢 𝑑𝑣 = 𝑢 𝑣 − ∫ 𝑣 𝑑𝑢
∫ ln(𝑥) 𝑑𝑥 = 𝑥 ln(x)|1e − ∫ dx
e
1
𝑒
1
= (𝑒 ln(𝑒) − 1 ln(1)) − (𝑒 − 1)
= (𝑒 − 0) − (𝑒 − 1)
= 1
Example 2: Combine substitution with parts
𝐼 = ∫ 𝑒√𝑥4
1𝑑𝑥
Let 𝑡 = √𝑥 = 𝑥1
2
𝑑𝑡 =1
2𝑥−
1
2 𝑑𝑥
2 𝑥1
2 𝑑𝑡 = 𝑑𝑥
2𝑡 𝑑𝑡 = 𝑑𝑥
If 𝑥 = 1 then 𝑡 = 1. If 𝑥 = 4 then 𝑡 = 2.
𝐼 = ∫ 𝑒𝑡2𝑡 𝑑𝑡 = 2 ∫ 𝑒𝑡2
1𝑡
2
1 𝑑𝑡
Now use integration by parts
Box: 𝑢 = 𝑡 𝑑𝑣 = 𝑒𝑡𝑑𝑡
𝑑𝑢 = 𝑑𝑡 𝑣 = 𝑒𝑡
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
∫ 𝑡𝑒𝑡𝑑𝑡 = 𝑡 𝑒𝑡|12 − ∫ 𝑒𝑡𝑑𝑡
2
1
2
1
= (2𝑒2 − 𝑒) − 𝑒𝑡|12
= (2𝑒2 − 𝑒) − (𝑒2 − 𝑒)
Chapter 6A Notes Page 9 of 12
= 𝑒2
Then 𝐼 = 2𝑒2 is the answer.
6.2A Trigonometric Integrals
Combine trig identities and substitution to evaluate some useful integrals.
Recall the identities:
sin2(𝑥) + cos2(𝑥) = 1 (1)
𝑑
𝑑𝑥sin(𝑥) = cos (𝑥)
𝑑
𝑑𝑥cos(𝑥) = −sin (𝑥)
We will consider two cases of integrals. The first is
∫ sin𝑚(𝑥) cos𝑛(𝑥)𝑑𝑥, (A)
where 𝑚 and 𝑛 are integers.
If 𝑚 or 𝑛 are odd, substitute 𝑢 = sin (𝑥) or 𝑢 = cos (𝑥).
Example. Evaluate 𝐼 = ∫ sin4(𝑥) cos3(𝑥)𝑑𝑥
Cosine is raised to an odd power. Factor out one power of cosine.
𝐼 = ∫ sin4(𝑥) cos2(𝑥) cos(𝑥) 𝑑𝑥
Use (1) to express the rest of the integrand as powers of sine.
𝐼 = ∫ sin4(𝑥)(1 − sin2(𝑥)) cos (𝑥) 𝑑𝑥
Substitute 𝑢 = sin (𝑥), 𝑑𝑢 = cos(𝑥) 𝑑𝑥
𝐼 = ∫ 𝑢4(1 − 𝑢2)𝑑𝑢 = ∫ 𝑢4 − 𝑢6 𝑑𝑢 =1
5𝑢5 −
1
7𝑢7 + 𝐶
𝐼 =1
5sin5(𝑥) −
1
7sin7(𝑥) + 𝐶 ■
Chapter 6A Notes Page 10 of 12
Example. Evaluate 𝐼 = ∫ tan(𝑥) 𝑑𝑥
Recognize 𝐼 = ∫sin (𝑥)
cos (𝑥) dx
Sine is raised to an odd power. Let 𝑢 = cos(𝑥), 𝑑𝑢 = − sin(𝑥) 𝑑𝑥
𝐼 = ∫ −𝑑𝑢
𝑢= − ln|𝑢| + 𝐶 = − ln |cos (𝑥)| + 𝐶 ■
If both 𝑚 and 𝑛 are even in case (A), use the following half angle identities
cos2(𝑥) =1
2(1 + cos(2𝑥)) (2)
sin2(𝑥) =1
2(1 − cos(2𝑥)) (3)
sin(𝑥) cos(𝑥) =1
2sin (2𝑥) (4)
Example. Evaluate 𝐼 = ∫ sin2(𝑥) cos2(𝑥)𝑑𝑥𝜋
20
By (4): sin2(𝑥) cos2(𝑥) =1
4sin2(2𝑥)
By (3): sin2(2𝑥) =1
2(1 − cos(4𝑥))
Thus sin2(𝑥) cos2(𝑥) =1
8(1 − cos(4𝑥))
𝐼 =1
8∫ (1 − cos(4x)) dx
𝜋
20
=1
8∫ 𝑑𝑥
𝜋
20
−1
8∫ cos(4𝑥) 𝑑𝑥
𝜋
20
Let 𝑢 = 4𝑥, 𝑑𝑢 = 4 𝑑𝑥
If 𝑥 = 0 then 𝑢 = 0,
If 𝑥 =𝜋
2 then 𝑢 = 2𝜋.
𝐼 =1
8(
𝜋
2) −
1
8∫ cos(𝑢)
2𝜋
0
1
4𝑑𝑢
where ∫ cos(𝑢)𝑑𝑢 = sin(u)|02π = 0
2𝜋
0
𝐼 =𝜋
16 ■
Chapter 6A Notes Page 11 of 12
Recall the identities
tan2(𝑥) + 1 = sec2(𝑥) (5)
𝑑
𝑑𝑥tan(𝑥) = sec2(𝑥)
𝑑
𝑑𝑥sec(𝑥) = tan(𝑥) sec(𝑥)
The second case of integrals we consider is
∫ tan𝑚(𝑥) sec𝑛(𝑥) 𝑑𝑥 (B)
If 𝑛 is even use the substitution 𝑢 = tan (𝑥)
Example. Evaluate 𝐼 = ∫ tan4(𝑥) sec4(𝑥) 𝑑𝑥
By (5): 𝐼 = ∫ tan4(𝑥)(tan2(𝑥) + 1) sec2(𝑥) 𝑑𝑥
Let 𝑢 = tan (𝑥), 𝑑𝑢 = sec2(𝑥) 𝑑𝑥
𝐼 = ∫ 𝑢4(𝑢2 + 1)𝑑𝑢 = ∫ 𝑢6 + 𝑢4 𝑑𝑢 =1
7𝑢7 +
1
5𝑢5 + 𝐶 =
1
7tan7(𝑥) +
1
5tan5(𝑥) + 𝐶 ■
If 𝑚 is odd use the substitution 𝑢 = sec (𝑥)
Example. Evaluate 𝐼 = ∫ tan3(𝑥) sec3(𝑥) 𝑑𝑥
𝐼 = ∫ tan2(𝑥) sec2(𝑥) tan(𝑥) sec(𝑥) 𝑑𝑥 = ∫ (sec2(𝑥) − 1) sec2(𝑥) tan (𝑥) sec(𝑥) 𝑑𝑥
Let 𝑢 = sec(𝑥), 𝑑𝑢 = tan(𝑥) sec(𝑥) 𝑑𝑥
𝐼 = ∫ (𝑢2 − 1)𝑢2 𝑑𝑢 = ∫ 𝑢4 − 𝑢2 𝑑𝑢 =1
5𝑢5 −
1
3𝑢3 + 𝐶 =
1
5sec5(𝑥) −
1
3sec3(𝑥) + 𝐶 ■
If both 𝑚 is even and 𝑛 is odd, express the integrand entirely in terms of sec (𝑥). This is the hardest case.
Example. Evaluate 𝐼 = ∫ tan2(𝑥) sec(𝑥) 𝑑𝑥
From (5): tan2(𝑥) = sec2(𝑥) − 1
𝐼 = ∫ sec3(𝑥)𝑑𝑥 − ∫ sec(𝑥) 𝑑𝑥
Chapter 6A Notes Page 12 of 12
∫ sec(𝑥) 𝑑𝑥 = ∫ 𝑠𝑒𝑐(𝑥) sec(𝑥)+tan (𝑥)
sec(𝑥)+tan (𝑥) dx
= ∫sec2(𝑥)+sec(𝑥)tan (𝑥)
sec(𝑥)+tan (𝑥) dx
Let 𝑢 = sec(𝑥) + tan(𝑥)
then 𝑑𝑢
𝑑𝑥= tan(𝑥) sec(𝑥) + sec2(𝑥) and 𝑑𝑢 = (tan(𝑥) sec(𝑥) + sec2(𝑥)) 𝑑𝑥
Then ∫ sec(𝑥) 𝑑𝑥 = ∫𝑑𝑢
𝑢= ln|𝑢| + 𝐶 = ln | sec(𝑥) + tan (𝑥)| + 𝐶
(6)
The other integral required to evaluate 𝐼 is treated by example 8 in section 6.2 of our text. The result is:
∫ sec3(𝑥) 𝑑𝑥 =1
2(sec(𝑥) tan(𝑥) + ln | sec(𝑥) + tan(𝑥) | ) + 𝐷 (7)
Results (6) and (7) can be combined to obtain an expression for 𝐼. ■