Introduction and RevieΒ Β· LIPET scheme for choosing and is often useful: (L)ogarithm (I)nverse Trig...
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Chapter 6A Notes Page 1 of 12 Introduction and Review Derivatives = () = β² () Evaluate derivative at =: | = = β² () = lim ββ0 (+β)β() β Geometric Interpretation: see figure slope of the line tangent to at =. Differentiation formulas = β1 ln|| = 1 = sin() = cos() cos() = βsin() tan() = sec 2 () sec() = sec() tan() sin β1 () = 1 β1 β 2
Transcript of Introduction and RevieΒ Β· LIPET scheme for choosing and is often useful: (L)ogarithm (I)nverse Trig...
Chapter 6A Notes Page 1 of 12
Introduction and Review
Derivatives π¦ = π(π₯)
ππ¦
ππ₯= πβ²(π₯)
Evaluate derivative at π₯ = π:
ππ¦
ππ₯|
π₯=π= πβ²(π) = lim
ββ0
π(π+β)βπ(π)
β
Geometric Interpretation: see figure
slope of the line tangent to π at π₯ = π.
Differentiation formulas π
ππ₯π₯π = ππ₯πβ1
π
ππ₯ln|π₯| =
1
π₯
π
ππ₯ππ₯ = ππ₯
π
ππ₯sin(π₯) = cos(π₯)
π
ππ₯cos(π₯) = βsin(π₯)
π
ππ₯tan(π₯) = sec2(π₯)
π
ππ₯sec(π₯) = sec(π₯) tan(π₯)
π
ππ₯sinβ1(π₯) =
1
β1 β π₯2
Chapter 6A Notes Page 2 of 12
π
ππ₯tanβ1(π₯) =
1
1 + π₯2
Review the sum, difference, product and quotient rules!
The chain rule
Let πΉ(π₯) = π(π(π₯)), then πΉβ²(π₯) = πβ²(π(π₯))πβ²(π₯)
Or let π¦ = π(π’), π’ = π(π₯), then ππ¦
ππ₯=
ππ¦
ππ’
ππ’
ππ₯
Example:
π
ππ₯sin(π₯2) = cos(π₯2)2π₯
Indefinite Integrals Indefinite integrals are also known as antiderivatives.
If πΉβ² is the derivative of πΉ then we write
β« πΉβ²(π₯)ππ₯ = πΉ(π₯) + πΆ
Example:
β« cos(π₯2) 2π₯ ππ₯ = sin(π₯2) + πΆ
Indefinite integral formulas
β« π₯πππ₯ =1
π + 1π₯π+1 + πΆ
β«1
π₯ ππ₯ = ln|π₯| + πΆ
β« ππ₯ππ₯ = ππ₯ + πΆ
β« cos(π₯) ππ₯ = sin(π₯) + πΆ
β« sin(π₯) ππ₯ = β cos(π₯) + πΆ
β« sec2(π₯)ππ₯ = tan(π₯) + πΆ
β« sec(π₯) tan(π₯) ππ₯ = sec(π₯) + πΆ
Chapter 6A Notes Page 3 of 12
Definite Integrals
Example: β«1
π₯ ππ₯
π
1
Geometric Interpretation: see figure
Evaluation theorem
β« πΉβ²(π₯)ππ₯ = πΉ(π₯)|ππ = πΉ(π) β πΉ(π)
π
π
Examples:
β«1
π₯ ππ₯ = ln(x)|1
eπ
1
= ln(π) β ln(1) = 1 β 0 = 1
β« cos(π₯2) 2π₯ ππ₯ = sin(π₯2)|01 = sin(1) β sin (0)
1
0
Integration by substitution Suppose an integral has the form: πΌ = β« πβ²(π(π₯))πβ²(π₯)ππ₯,
let π’ = π(π₯), then ππ’ = πβ²(π₯)ππ₯ (this is differential notation)
Substitute to obtain
πΌ = β« πβ²(π’)ππ’ = π(π’) + πΆ = π(π(π₯)) + πΆ
This is equivalent to taking the chain rule of differentiation backwards!
Example. πΌ = β« cos(π₯2) π₯ ππ₯
Let π’ = π₯2, ππ’ = 2π₯ ππ₯
πΌ = β« cos(π’) 1
2 ππ’ =
1
2 β« cos(π’) ππ’ =
1
2sin(π’) + πΆ =
1
2sin(π₯2) + πΆ.
Example. π½ = β« cos(π₯2) π₯ ππ₯3
2
Chapter 6A Notes Page 4 of 12
Let π’ = π₯2. If π₯ = 2 then π’ = 4. If π₯ = 3 then π’ = 9.
π½ = β« cos(π’)1
2ππ’
9
4
=1
2sin (π’)|4
9
6.1 Integration by parts
Chain rule for differentiation: π
ππ₯π ππ(π₯2) = (cos(π₯2))2π₯
Integration by substitution: let = π₯2 , ππ’ = 2π₯ ππ₯
β« (cos π₯2)2π₯ ππ₯ = β« cos(π’) ππ’ = sin(π’) + πΆ = sin(π₯2) + πΆ
Product rule for differentiation
π
ππ₯[π(π₯)π(π₯)] = πβ²(π₯)π(π₯) + π(π₯)πβ²(π₯)
Solve forπ(π₯)πβ²(π₯):
π(π₯)πβ²(π₯) =π
ππ₯[π(π₯)π(π₯)] β πβ²(π₯)π(π₯)
Take antiderivatives
(β« π(π₯)πβ²(π₯)ππ₯ = π(π₯)π(π₯) β β« πβ²(π₯)π(π₯)ππ₯ (A)
This is the rule for integration of indefinite integrals by parts.
Shorthand formula: recall differential notation
Box: π’ = π(π₯) ππ£ = πβ²(π₯) ππ₯
ππ’ = πβ²(π₯)ππ₯ π£ = π(π₯)
Substitute into (A) above to get
β« π’ ππ£ = π’π£ β β« π£ ππ’ (B)
An easy-to-remember form of the integration by parts formula!
Chapter 6A Notes Page 5 of 12
Example 1. Evaluate πΌ = β« π₯ cos(π₯) ππ₯.
Box: π’ = π₯ ππ£ = cos(π₯) ππ₯
ππ’ = ππ₯ π£ = sin (π₯)
πΌ = β« π’ ππ£ = π’π£ β β« π£ ππ’
= π₯ sin(π₯) β β« sin (π₯)ππ₯
= π₯ sin(π₯) + cos(π₯) + πΆ
Check:
π
ππ₯(π₯ sin(π₯) + cos(π₯) + πΆ) = sin(π₯) + π₯ cos(π₯) β sin(π₯) = π₯ cos (π₯) ββ
In principle, pick π’ so that ππ’ is simpler and pick ππ£ so that it is possible to integrate. The following
LIPET scheme for choosing π’ and π£ is often useful:
(L)ogarithm
(I)nverse Trig
(P)olynomials or powers of π₯
(E)xponential functions
(T)rig functions
Whichever piece of the integrand is higher on the list is π’ and the other piece is ππ£.
Example 2. Integration by parts twice
Evaluate πΌ = β« π₯2 sin(π₯) ππ₯.
Box: π’ = π₯2 ππ£ = sin(π₯) ππ₯
ππ’ = 2π₯ ππ₯ π£ = βcos (π₯)
πΌ = π’π£ β β« π£ ππ’
= βπ₯2 cos(π₯) + β« cos(π₯) 2π₯ ππ₯
= βπ₯2 cos(π₯) + 2(π₯ sin(π₯) + cos(π₯)) + πΆ
Chapter 6A Notes Page 6 of 12
We used the result of example 1.
Example 3. Recurrence of the original integral
Evaluate πΌ = β« πβπ₯ cos(π₯) ππ₯
Box: π’ = πβπ₯ ππ£ = cos(π₯) ππ₯
ππ’ = βπβπ₯ ππ₯ π£ = sin (π₯)
πΌ = π’π£ β β« π£ ππ’ = πβπ₯ sin(π₯) + β« πβπ₯ sin(π₯) ππ₯ (C)
Evaluate π½ = β« πβπ₯ sin(π₯) ππ₯
Box: π’ = πβπ₯ ππ£ = sin(π₯) ππ₯
ππ’ = βπβπ₯ππ₯ π£ = βcos (π₯)
π½ = π’π£ β β« π£ ππ’ = βπβπ₯ cos(π₯) β β« πβπ₯ cos(π₯) ππ₯ = βπβπ₯ cos(π₯) β πΌ (D)
Combine (C) and (D)
πΌ = πβπ₯ sin(π₯) + (βπβπ₯ cos(π₯) β πΌ)
2πΌ = πβπ₯(sin(π₯) β cos(π₯)) + 2πΆ
πΌ =1
2πβπ₯(sin(π₯) β cos(π₯)) + πΆ
We added an explicit constant of integration when we put πΌ on only one side of the equation. Since πΌ
represents an indefinite integral, it has an implicit constant of integration.
Example 4: combine parts and substitution
Evaluate πΌ = β« sinβ1(π₯) ππ₯
π
ππ₯sinβ1(π₯) =
1
β1βπ₯2
Box: π’ = sinβ1(π₯) ππ£ = ππ₯
ππ’ =1
β1βπ₯2ππ₯ π£ = π₯
πΌ = β« π’ ππ£ = π’π£ β β« π£ ππ’
= π₯ sinβ1(π₯) β β«π₯
β1βπ₯2ππ₯
Chapter 6A Notes Page 7 of 12
Evaluate π½ = β«π₯
β1βπ₯2ππ₯
Let π‘ = 1 β π₯2, ππ‘ = β2π₯ ππ₯, β1
2ππ‘ = π₯ ππ₯
Then π½ = β1
2 β« π‘β
1
2 ππ‘ = β1
2 (2 π‘
1
2) + πΆ = βπ‘1
2 + πΆ
Combine to get
πΌ = π₯ sinβ1(π₯) + (1 β π₯2)1
2 + πΆ
Definite Integrals Evaluation Theorem
β« πΉβ²(π₯)ππ₯ = πΉ(π₯)|ππ = πΉ(π) β πΉ(π)
π
π
Product Rule
π
ππ₯π(π₯)π(π₯) = πβ²(π₯)π(π₯) + π(π₯)πβ²(π₯)
π(π₯)πβ²(π₯) =π
ππ₯[π(π₯)π(π₯)] β πβ²(π₯)π(π₯)
integrate from π₯ = π to π₯ = π:
β« π(π₯)πβ²(π₯)ππ₯ = β«π
ππ₯[π(π₯)π(π₯)]ππ₯ = β« πβ²(π₯)π(π₯)ππ₯
π
π
π
π
π
π
β« π(π₯)πβ²(π₯)ππ₯π
π= π(π₯)π(π₯)|π
π β β« πβ²(π₯)π(π₯)ππ₯π
π (E)
Shorthand using differential notation
Box: π’ = π(π₯) ππ£ = πβ²(π₯) ππ₯
ππ’ = πβ²(π₯)ππ₯ π£ = π(π₯)
Write as for an indefinite integral:
β« π’ ππ£ = π’ π£ β β« π£ ππ’
Replace π’, ππ’, π£ and ππ£ with appropriate expressions in π₯ and add limits of integration to get (E)
Chapter 6A Notes Page 8 of 12
Example 1: πΌ = β« ln(π₯) ππ₯π
1
Box: π’ = ln (π₯) ππ£ = ππ₯
ππ’ =1
π₯ππ₯ π£ = π₯
β« π’ ππ£ = π’ π£ β β« π£ ππ’
β« ln(π₯) ππ₯ = π₯ ln(x)|1e β β« dx
e
1
π
1
= (π ln(π) β 1 ln(1)) β (π β 1)
= (π β 0) β (π β 1)
= 1
Example 2: Combine substitution with parts
πΌ = β« πβπ₯4
1ππ₯
Let π‘ = βπ₯ = π₯1
2
ππ‘ =1
2π₯β
1
2 ππ₯
2 π₯1
2 ππ‘ = ππ₯
2π‘ ππ‘ = ππ₯
If π₯ = 1 then π‘ = 1. If π₯ = 4 then π‘ = 2.
πΌ = β« ππ‘2π‘ ππ‘ = 2 β« ππ‘2
1π‘
2
1 ππ‘
Now use integration by parts
Box: π’ = π‘ ππ£ = ππ‘ππ‘
ππ’ = ππ‘ π£ = ππ‘
β« π’ ππ£ = π’π£ β β« π£ ππ’
β« π‘ππ‘ππ‘ = π‘ ππ‘|12 β β« ππ‘ππ‘
2
1
2
1
= (2π2 β π) β ππ‘|12
= (2π2 β π) β (π2 β π)
Chapter 6A Notes Page 9 of 12
= π2
Then πΌ = 2π2 is the answer.
6.2A Trigonometric Integrals
Combine trig identities and substitution to evaluate some useful integrals.
Recall the identities:
sin2(π₯) + cos2(π₯) = 1 (1)
π
ππ₯sin(π₯) = cos (π₯)
π
ππ₯cos(π₯) = βsin (π₯)
We will consider two cases of integrals. The first is
β« sinπ(π₯) cosπ(π₯)ππ₯, (A)
where π and π are integers.
If π or π are odd, substitute π’ = sin (π₯) or π’ = cos (π₯).
Example. Evaluate πΌ = β« sin4(π₯) cos3(π₯)ππ₯
Cosine is raised to an odd power. Factor out one power of cosine.
πΌ = β« sin4(π₯) cos2(π₯) cos(π₯) ππ₯
Use (1) to express the rest of the integrand as powers of sine.
πΌ = β« sin4(π₯)(1 β sin2(π₯)) cos (π₯) ππ₯
Substitute π’ = sin (π₯), ππ’ = cos(π₯) ππ₯
πΌ = β« π’4(1 β π’2)ππ’ = β« π’4 β π’6 ππ’ =1
5π’5 β
1
7π’7 + πΆ
πΌ =1
5sin5(π₯) β
1
7sin7(π₯) + πΆ β
Chapter 6A Notes Page 10 of 12
Example. Evaluate πΌ = β« tan(π₯) ππ₯
Recognize πΌ = β«sin (π₯)
cos (π₯) dx
Sine is raised to an odd power. Let π’ = cos(π₯), ππ’ = β sin(π₯) ππ₯
πΌ = β« βππ’
π’= β ln|π’| + πΆ = β ln |cos (π₯)| + πΆ β
If both π and π are even in case (A), use the following half angle identities
cos2(π₯) =1
2(1 + cos(2π₯)) (2)
sin2(π₯) =1
2(1 β cos(2π₯)) (3)
sin(π₯) cos(π₯) =1
2sin (2π₯) (4)
Example. Evaluate πΌ = β« sin2(π₯) cos2(π₯)ππ₯π
20
By (4): sin2(π₯) cos2(π₯) =1
4sin2(2π₯)
By (3): sin2(2π₯) =1
2(1 β cos(4π₯))
Thus sin2(π₯) cos2(π₯) =1
8(1 β cos(4π₯))
πΌ =1
8β« (1 β cos(4x)) dx
π
20
=1
8β« ππ₯
π
20
β1
8β« cos(4π₯) ππ₯
π
20
Let π’ = 4π₯, ππ’ = 4 ππ₯
If π₯ = 0 then π’ = 0,
If π₯ =π
2 then π’ = 2π.
πΌ =1
8(
π
2) β
1
8β« cos(π’)
2π
0
1
4ππ’
where β« cos(π’)ππ’ = sin(u)|02Ο = 0
2π
0
πΌ =π
16 β
Chapter 6A Notes Page 11 of 12
Recall the identities
tan2(π₯) + 1 = sec2(π₯) (5)
π
ππ₯tan(π₯) = sec2(π₯)
π
ππ₯sec(π₯) = tan(π₯) sec(π₯)
The second case of integrals we consider is
β« tanπ(π₯) secπ(π₯) ππ₯ (B)
If π is even use the substitution π’ = tan (π₯)
Example. Evaluate πΌ = β« tan4(π₯) sec4(π₯) ππ₯
By (5): πΌ = β« tan4(π₯)(tan2(π₯) + 1) sec2(π₯) ππ₯
Let π’ = tan (π₯), ππ’ = sec2(π₯) ππ₯
πΌ = β« π’4(π’2 + 1)ππ’ = β« π’6 + π’4 ππ’ =1
7π’7 +
1
5π’5 + πΆ =
1
7tan7(π₯) +
1
5tan5(π₯) + πΆ β
If π is odd use the substitution π’ = sec (π₯)
Example. Evaluate πΌ = β« tan3(π₯) sec3(π₯) ππ₯
πΌ = β« tan2(π₯) sec2(π₯) tan(π₯) sec(π₯) ππ₯ = β« (sec2(π₯) β 1) sec2(π₯) tan (π₯) sec(π₯) ππ₯
Let π’ = sec(π₯), ππ’ = tan(π₯) sec(π₯) ππ₯
πΌ = β« (π’2 β 1)π’2 ππ’ = β« π’4 β π’2 ππ’ =1
5π’5 β
1
3π’3 + πΆ =
1
5sec5(π₯) β
1
3sec3(π₯) + πΆ β
If both π is even and π is odd, express the integrand entirely in terms of sec (π₯). This is the hardest case.
Example. Evaluate πΌ = β« tan2(π₯) sec(π₯) ππ₯
From (5): tan2(π₯) = sec2(π₯) β 1
πΌ = β« sec3(π₯)ππ₯ β β« sec(π₯) ππ₯
Chapter 6A Notes Page 12 of 12
β« sec(π₯) ππ₯ = β« π ππ(π₯) sec(π₯)+tan (π₯)
sec(π₯)+tan (π₯) dx
= β«sec2(π₯)+sec(π₯)tan (π₯)
sec(π₯)+tan (π₯) dx
Let π’ = sec(π₯) + tan(π₯)
then ππ’
ππ₯= tan(π₯) sec(π₯) + sec2(π₯) and ππ’ = (tan(π₯) sec(π₯) + sec2(π₯)) ππ₯
Then β« sec(π₯) ππ₯ = β«ππ’
π’= ln|π’| + πΆ = ln | sec(π₯) + tan (π₯)| + πΆ
(6)
The other integral required to evaluate πΌ is treated by example 8 in section 6.2 of our text. The result is:
β« sec3(π₯) ππ₯ =1
2(sec(π₯) tan(π₯) + ln | sec(π₯) + tan(π₯) | ) + π· (7)
Results (6) and (7) can be combined to obtain an expression for πΌ. β
