# 8 Indefinite Integrals Case Study 8.1 Concepts of Indefinite Integrals

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8 Indefinite Integrals Case Study 8.1 Concepts of Indefinite Integrals
8.2 Indefinite Integration of Functions 8.3 Integration by Substitution 8.4 Integration by Parts 8.5 Applications of Indefinite Integrals Chapter Summary

Case Study Can you estimate the number of radioactive particles in the sample from your readings? By measuring the level of radioactivity with a counter, it is estimated that the number of radioactive particles, y, in the sample is decreasing at a rate of 1000e–0.046t per hour, where t is expressed in hours. I have already recorded the readings for the level of radioactivity. According to what we learnt in Section 7.5 (Rates of Change), we have In order to express y in terms of t, we need to find a function y of t such that its derivative is equal to –1000e–0.046t . The process of finding a function from its derivative is called integration and will be discussed in this chapter.

8.1 Concepts of Indefinite Integrals
A. Definition of Indefinite Integrals In previous chapters, we learnt how to find the derivative of a given function. Suppose we are given a function x2, by differentiation, we have As 2x is the derivative of x2, we call x2 the primitive function (or antiderivative) of 2x. Generally, for any differentiable function F(x), we have the following definition: Definition 8.1 If , then F(x) is called a primitive function of f(x). Although x2 is a primitive function of 2x, it is not the unique primitive function. If we add an arbitrary constant C to x2 and differentiate it, we have Thus, x2 + C is also a primitive function of 2x for an arbitrary constant C.

8.1 Concepts of Indefinite Integrals
A. Definition of Indefinite Integrals In order to represent all the primitive functions of a function f (x), we introduce the concept of indefinite integral as below: Definition 8.2 If , then the indefinite integral of f(x), which is denoted by , is given by , where C is an arbitrary constant. Note: In the notation of , f (x) is called the integrand, and ‘ ’ is called the integral sign. The process of finding the primitive function is called integration. 2. C is called the constant of integration (or integration constant).

8.1 Concepts of Indefinite Integrals
B. Basic Formulas of Indefinite Integrals As integration is the reverse process of differentiation, the basic formulas for integrations can be derived from the differentiation formulas. For example: Since , 8.1 , for all real numbers n  1. Note: 1. Formula 8.1 is also called the Power Rule for integration. When n  0, the left hand side of the formula becomes For convenience we usually express as

8.1 Concepts of Indefinite Integrals
B. Basic Formulas of Indefinite Integrals In addition to the Power Rule, we can use the similar way to derive the following integration formulas: , where k is a constant.

8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals Theorem 8.1 If k is a non-constant, then Proof: By definition, C ¢, where C ¢ is an arbitrary constant. On the other hand, Since C ¢ and kC are arbitrary constants, the expressions kg(x)  C ¢ and kg(x)  kC represent the same family of primitive functions.

8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals Theorem 8.2 Proof: Let and , where C1 and C2 are arbitrary constants. By definition, On the other hand, Since C1  C2 is an arbitrary constant, the expressions F(x)  G(x)  C and F(x)  G(x)  C1  C2 represent the same family of primitive functions.

Example 8.1T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Example 8.1T Find Solution: Use C to express the sum of all constants

Example 8.2T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Example 8.2T Find Solution: a3  b3  (a  b)(a2  ab  b2) Cancel the common factor

Example 8.3T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Example 8.3T Find Solution:

Example 8.4T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Example 8.4T Find Solution:

Example 8.5T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Example 8.5T Find Solution:

Example 8.6T 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Example 8.6T Let y  ln x – ln (x  1). Find Hence find Solution: (a) (b) By (a),

8.2 Indefinite Integration of Functions
A. Integration of Functions Involving the Expression (ax + b) The integration formulas mentioned in Section 8.1 enable us to find the indefinite integrals of simple functions such as ex, sin x and cos x. But how about e2x, sin 4x and cos (7x + 5)? , where n  –1 and a  0. Using the same method, we can obtain the following integration formulas: Suppose a and b are real numbers with a  0. 8.12 8.13 8.14 8.15

Example 8.7T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Example 8.7T Find Solution:

Example 8.8T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Example 8.8T Find Solution: Rationalize the denominator

Example 8.9T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Example 8.9T Find Solution: Cancel the common factor

Example 8.10T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Example 8.10T Find Solution:

Example 8.11T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Example 8.11T Find Solution:

Example 8.12T 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Example 8.12T Find Solution: If y  ln 10, then ey  10 by definition, i.e., eln 10  10.

cos 2A  2 cos2 A  1 or cos 2A  1  2 sin2 A
8.2 Indefinite Integration of Functions B. Integration of Trigonometric Functions To find integrals where the integrand is the product or power of trigonometric functions, we can first use double angle formulas and product-to-sum formulas to express the integrand in the sum of trigonometric functions. cos 2A  2 cos2 A  1 or cos 2A  1  2 sin2 A Product-to-sum Formulas sin A cos B  [sin (A  B)  sin (A  B)] cos A sin B  [sin (A  B)  sin (A  B)] cos A cos B  [cos (A  B)  cos (A  B)] sin A sin B   [cos (A  B)  cos (A  B)]

Example 8.13T 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions Example 8.13T Find Solution: sin 2A  2 sin A cos A

Example 8.14T 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions Example 8.14T Find Solution: sin 2A  2 sin A cos A cos 2A  1  2 sin2 A

Example 8.15T 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions Example 8.15T Find Solution: Product-to-sum formula

Example 8.16T 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions Example 8.16T Find Solution: cos 2A  1  2 sin2 A cot A 

8.3 Integration by Substitution
A. Change of Variables In Section 8.2, we learnt some basic formulas to find the indefinite integrals of functions. However, not all functions can be integrated directly using these formulas. In this case, we have to use the method of integration by substitution. The following shows the basic principle of this method. Let and u  g (x).  f (u) g¢(x)  f [g(x)] g¢(x) By the definition of integration,

8.3 Integration by Substitution
A. Change of Variables For an integral , we can transform it into a simpler integral , by the following steps. Step 1: Separate the integrand into two parts: f [g(x)] and g¢(x)dx. Step 2: Replace every occurrence of g(x) in the integrand by u. Step 3: Replace the expression ‘g¢(x)dx’ by ‘du’. Let us use this method to find the integral together. Note that , so we let u  x2 + 1, such that  2x.

Example 8.17T 8.3 Integration by Substitution Solution:
A. Change of Variables Example 8.17T Find Solution: Let u  1 – x2. Then Express the answer in terms of x

Example 8.18T 8.3 Integration by Substitution Solution:
A. Change of Variables Example 8.18T Find Solution: Let u  x2 – 7. Then Rewrite x2 as (u  7)

8.3 Integration by Substitution
A. Change of Variables With the method of integration by substitution, we can find the integrals of trigonometric functions other than sine and cosine, as shown in the following example.

Example 8.19T 8.3 Integration by Substitution Solution:
A. Change of Variables Example 8.19T Find Solution: Alternative Solution: Let u  csc x – cot x. Let u  csc x – cot x.

8.3 Integration by Substitution
A. Change of Variables It is tedious to write u and du every time when finding the integrals by substitution, as shown in the previous examples. After becoming familiar with the method of integration by substitution, the working steps can be simplified by omitting the use of the variable u. Let us study the following example.

Example 8.20T 8.3 Integration by Substitution Solution:
A. Change of Variables Example 8.20T Find Solution:

8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric Functions Sometimes we need to handle indefinite integrals that involve the products of powers of trigonometric functions, such as or , where m and n are integers. In the following discussion, we will see how to apply different strategies according to different values of m and n. Strategies for finding integrals in the form Case 1: m is an odd number. Use sin x dx  –d(cos x) and express all the other sine terms as cosine terms. Case 2: n is an odd number. Use cos x dx  d(sin x) and express all the other cosine terms as sine terms. Case 3: both m and n are even numbers. Use the double-angle formula to reduce the powers of the functions.

Example 8.21T 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric Functions Example 8.21T Find Solution:

Example 8.22T 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric Functions Example 8.22T Find Solution:

Example 8.23T 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric Functions Example 8.23T Find Solution:

8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric Functions Similarly, integrals in the form may be found by using the method of integration by substitution. Strategies for finding integrals in the form Case 1: m is an odd number. Use tan x sec x dx  d(sec x) and then express all other tangent terms as secant terms. Case 2: n is an even number. Use sec2x dx  d(tan x) and then express all other secant terms as tangent terms.

Example 8.24T 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric Functions Example 8.24T Find Solution:

Example 8.25T 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric Functions Example 8.25T Find Solution:

8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric Functions In the above examples, the case that m is even while n is odd is not considered. This is because there is no standard technique and the method varies from case to case. For example, to find (m  0 and n  1), we may follow the method in Example 8.19. In some other cases, such as (m  2 and n  1), we may need to use the technique ‘integration by parts’, which will be discussed later in this chapter.

cos2q  1  sin2q , sec2q  1  tan2q , tan2q  sec2q  1
8.3 Integration by Substitution C. Integration by Trigonometric Substitution If an indefinite integral involves radicals in the form , or , we can use the method of integration by substitution to eliminate the radicals. The following three trigonometric identities are very useful for the elimination: cos2q  1  sin2q , sec2q  1  tan2q , tan2q  sec2q  1 For example, if we substitute x  a sin q into the expression , we have Then we can express the integrand in terms of q. After finding the indefinite integral in terms of q (say, 3q + C), the final answer should be expressed in terms of the original variable, say, x.

8.3 Integration by Substitution
C. Integration by Trigonometric Substitution In order to express q in terms of x, let us first introduce the following notations: Inverse of Trigonometric Functions Let x be a real number. 1. sin–1x is defined as the angle q such that sin q  x (where –1  x  1) and 2. cos–1x is defined as the angle q such that cos q  x (where –1  x  1) and 0  q  p. 3. tan–1x is defined as the angle q such that tan q  x and

Example 8.26T 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution Example 8.26T Find Solution: Let x  sinq. Then dx  cosq dq. Since sin q  x,

Example 8.27T 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution Example 8.27T Find Solution: Let x  sinq. Then dx  cosq dq. Since sin q  x,

Example 8.28T 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution Example 8.28T Find Solution: Let x  3tanq. Then dx  3sec2q dq. Since

Example 8.29T 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution Example 8.29T Find Solution: q 1 x + 2 __________ (x  2)2  1 Let x + 2  secq. Then dx  secq tanq dq. Since secq  x + 2,

8.4 Integration by Parts Some indefinite integrals such as , and
cannot be found by using the techniques we have learnt so far. To evaluate them, we need to introduce another method called integration by parts. Theorem Integration by Parts If u(x) and v(x) are two differentiable functions, then In other words, Proof: Suppose u and v are two differentiable functions. Since (uv)  uv¢ + vu¢, by definition,

8.4 Integration by Parts From Theorem 8.3, we can see that the problem of finding can be transformed into the problem of finding instead. If the integral is much simpler than , then the original integral can be found easily. If we want to apply the technique of integration by parts to an integral, such as , we need to transform the integral into the form first, such as or

8.4 Integration by Parts Example 8.30T Find Solution:

8.4 Integration by Parts Example 8.31T Find Solution:

8.4 Integration by Parts In some cases, there may be more than one choice for u and v. For example, we can transform into or However, if we choose the former, then As a result, we get an integrand x2 cos x which is more complicated than the original one x sin x. Thus we should try sin x dx  d(–cos x) instead.

8.4 Integration by Parts Example 8.32T Find Solution:

Example 8.33T 8.4 Integration by Parts Solution: Show that
Hence find Solution: (b)

8.4 Integration by Parts Example 8.34T Find Solution:

8.4 Integration by Parts Example 8.35T Find Solution: Therefore,

8.5 Applications of Indefinite Integrals
A. Geometrical Applications In previous chapters, we learnt that of a curve y  F(x) is the slope function of the curve. Since integration is the reverse process of differentiation, if we let  f(x), then by the definition of integration, we have where C is an arbitrary constant. Thus we can see that the equation of a family of curves y  F(x) + C can be found by integration, providing that the slope function of the curve is known.

Example 8.36T 8.5 Applications of Indefinite Integrals Solution:
A. Geometrical Applications Example 8.36T The equation of the slope of a curve at the point (x, y) is given by . If the curve passes through (2, 2), find the equation of the curve. Solution: where C is an arbitrary constant When x  2, y  2, we have ∴ The equation of the curve is

Example 8.37T 8.5 Applications of Indefinite Integrals Solution:
A. Geometrical Applications Example 8.37T It is given that at each point (x, y) on a certain curve, If the curve passes through and , find the equation of that curve. Solution: Since and lie on the curve, we have ∴ The equation of the curve is

8.5 Applications of Indefinite Integrals
B. Applications in Physics In Section 7.5 of Book 1, we learnt that for a particle moving along a straight line, its velocity v and acceleration a at time t are given by where s is the displacement of the particle at time t. Since integration is the reverse process of differentiation, we have

Example 8.38T 8.5 Applications of Indefinite Integrals Solution:
B. Applications in Physics Example 8.38T A particle moves along a straight line such that its velocity v at time t is given by Find the displacement s of the particle at time t, given that s  6 when t  2. Solution: When t  2, s  6, we have

Example 8.39T 8.5 Applications of Indefinite Integrals Solution:
B. Applications in Physics Example 8.39T A particle moves along a straight line so that its acceleration a at time t is given by a  for t > 0. When t  1, the velocity of the particle is 8 and its displacement is 24. Find the displacement of the particle at time t. Solution: When t  1, v  8, we have When t  1, s  24, we have

Chapter Summary 8.1 Concepts of Indefinite Integrals
1. If , then the indefinite integral of f(x) is defined by , where C is an arbitrary constant. 2. 3. 4. 5.

Chapter Summary 8.1 Concepts of Indefinite Integrals 6. 7. 8. 9. 10.
11. 12. 13.

Chapter Summary 8.2 Indefinite Integration of Functions
Let a and b be real numbers with a  0. 1. 2. 3. 4. 5.

Chapter Summary 8.3 Integration by Substitution
1. Let u  g(x) be a differentiable function. Then, 2. If the integrated involves terms like , and , we can simplify the integrand by substituting x  a sin q, x  a tan q or x  a sec q respectively.

Chapter Summary 8.4 Integration by Parts
If u(x) and v(x) are two differentiable functions, then

Chapter Summary 8.5 Applications of Indefinite Integrals
If the slope of a curve at point (x, y) is f(x), then the equation of the family of curves is given by where F ¢(x)  f(x). 2. Let s, v and a be the displacement, velocity and acceleration of a particle moving along a straight line respectively, then

Follow-up 8.1 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Follow-up 8.1 Find Solution:

Follow-up 8.2 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Follow-up 8.2 Find Solution: Cancel the common factor

Follow-up 8.3 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Follow-up 8.3 Find Solution: Express the answer with positive indices.

Follow-up 8.4 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Follow-up 8.4 Find Solution:

Follow-up 8.5 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Follow-up 8.5 Find Solution: cot A 

Follow-up 8.6 8.1 Concepts of Indefinite Integrals Solution:
C. Basic Properties of Indefinite Integrals Follow-up 8.6 Let y  x ln x. Find in terms of ln x. Hence find Solution: (a) Chain Rule (b) By (a),

Follow-up 8.7 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Follow-up 8.7 Find Solution:

Follow-up 8.8 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Follow-up 8.8 Find Solution:

Follow-up 8.9 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Follow-up 8.9 Find Solution: Cancel the common factor

Follow-up 8.10 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Follow-up 8.10 Find Solution:

Follow-up 8.11 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Follow-up 8.11 Find Solution:

Follow-up 8.12 8.2 Indefinite Integration of Functions Solution:
A. Integration of Functions Involving the Expression (ax + b) Follow-up 8.12 Find Solution: If y  ln 3, then ey  3 by definition, i.e., eln 3  3.

Follow-up 8.13 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions Follow-up 8.13 Find Solution: cos 2A  1  2 sin2 A

Follow-up 8.14 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions Follow-up 8.14 Find Solution: cos 2A  2 cos2 A  1 cos 2A  2 cos2 A  1

Follow-up 8.15 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions Follow-up 8.15 Find Solution:

Follow-up 8.16 8.2 Indefinite Integration of Functions Solution:
B. Integration of Trigonometric Functions Follow-up 8.16 Find Solution:

Follow-up 8.17 8.3 Integration by Substitution Solution:
A. Change of Variables Follow-up 8.17 Find Solution: Let u  1 + x3. Then Express the answer in terms of x

Follow-up 8.18 8.3 Integration by Substitution Solution:
A. Change of Variables Follow-up 8.18 Find Solution: Let u  x – 4. Then Rewrite x as (u + 4)

Follow-up 8.19 8.3 Integration by Substitution Solution:
A. Change of Variables Follow-up 8.19 Find Solution: Note that Let u  sin x. Then

Follow-up 8.20 8.3 Integration by Substitution Solution:
A. Change of Variables Follow-up 8.20 Find the following integrals. (a) (b) Solution: (a) (b)

Follow-up 8.21 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric Functions Follow-up 8.21 Find Solution:

Follow-up 8.22 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric Functions Follow-up 8.22 Find Solution:

Follow-up 8.23 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric Functions Follow-up 8.23 Find Solution:

Follow-up 8.24 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric Functions Follow-up 8.24 Find Solution:

Follow-up 8.25 8.3 Integration by Substitution Solution:
B. Integrals Involving Powers of Trigonometric Functions Follow-up 8.25 Find Solution:

Follow-up 8.26 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution Follow-up 8.26 Find Solution: Let x  3 sinq. Then dx  3 cosq dq.

Follow-up 8.27 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution Follow-up 8.27 Find Solution: Let Then Since sin q  2x,

Follow-up 8.28 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution Follow-up 8.28 Find Solution: Let Then Since

Follow-up 8.29 8.3 Integration by Substitution Solution:
C. Integration by Trigonometric Substitution Follow-up 8.29 Find Solution: Let x  8 secq . Then dx  8 tanq secq dq. q 8 x _______  x2  64

8.4 Integration by Parts Follow-up 8.30 Find Solution:

8.4 Integration by Parts Follow-up 8.31 Find Solution:

8.4 Integration by Parts Follow-up 8.32 Find Solution:

Follow-up 8.33 8.4 Integration by Parts Solution: Find Hence find
(a) Let y  2x. (b) ln y  x ln 2 Differentiate both sides with respect to x,

Follow-up 8.34 Find Solution:

8.4 Integration by Parts Follow-up 8.35 Find Solution: Therefore,

Follow-up 8.36 8.5 Applications of Indefinite Integrals Solution:
A. Geometrical Applications Follow-up 8.36 The equation of the slope of a curve at the point (x, y) is given by  16 sin x cos x. If the curve passes through , find its equation. Solution: where C is an arbitrary constant. When y  7, we have ∴ The equation of the curve is y  –4 cos 2x – 5.

Follow-up 8.37 8.5 Applications of Indefinite Integrals Solution:
A. Geometrical Applications Follow-up 8.37 It is given that at each point (x, y) on a certain curve,  18e–3x. If  1 and y  1 when x  0, find the equation of that curve. Solution: When x  0, y  1, we have ∴ The equation of the curve is y  2e–3x + 7x – 1. When x = 0,  1, we have

Follow-up 8.38 8.5 Applications of Indefinite Integrals Solution:
B. Applications in Physics Follow-up 8.38 A particle moves along a straight line such that its acceleration a at time t is given by Find the velocity v of the particle at time t if v  7 when t  4. Solution: When t  4, v  7, we have

Follow-up 8.39 8.5 Applications of Indefinite Integrals Solution:
B. Applications in Physics Follow-up 8.39 A particle is moving along a straight line such that its acceleration a cm s–2 after t seconds is given by a  6(t – 1). The displacement of the particle is 13 cm after 2 seconds and 45 cm after 4 seconds. Find the displacement of the particle when its velocity is at its minimum. Solution: (2) – (1), Substituting C1  6 into (1), When t  2, s  13, we have When t  4, s  45, we have ∴ The velocity is minimum when t = 1. When t  1, ∴ The displacement is 9 cm.

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