8.1 Concepts of Indefinite Integrals

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

Can you estimate the number of radioactive particles in the sample from your readings? I have already recorded the readings for the level of radioactivity. Case Study 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. 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.

Definition 8.1 If , then F(x) is called a primitive function of f(x). 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: 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.

Definition 8.2 If , then the indefinite integral of f(x), which is denoted by , is given by , where C is an arbitrary constant. 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: • 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 , for all real numbers n  1. • 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 As integration is the reverse process of differentiation, the basic formulas for integrations can be derived from the differentiation formulas. For example: Since , Note: 1. Formula 8.1 is also called the Power Rule for integration.

8.2 , where k is a constant. 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 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:

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.

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

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

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

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

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

8.11 , where n –1 and a 0. Suppose a and b are real numbers with a 0. 8.12 8.13 8.14 8.15 8.2Indefinite 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)? Using the same method, we can obtain the following integration formulas:

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

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

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

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

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)] 8.2Indefinite 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 cos2A  1 or cos 2A  1  2 sin2A

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

sin 2A  2 sin A cos A cos 2A  1  2 sin2A 8.2Indefinite Integration of Functions B. Integration of Trigonometric Functions Example 8.14T Find Solution:

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

cos 2A  1  2 sin2A cot A  8.2Indefinite Integration of Functions B. Integration of Trigonometric Functions Example 8.16T Find Solution:

8.3Integration 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, .

8.3Integration 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.

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

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

8.3Integration 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.

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

8.3Integration 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.

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

8.3Integration 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 xdx –d(cos x) and express all the other sine terms as cosine terms. Case 2:n is an odd number. Use cos xdx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.

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

8.3Integration 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 sec2xdxd(tan x) and then express all other secant terms as tangent terms.

8.3Integration 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.

8.3Integration 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 sinq 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.

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

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

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

8.3Integration by Substitution C. Integration by Trigonometric Substitution Example 8.28T Find Solution: Letx 3tanq. Thendx 3sec2qdq. Since

x + 2 q 1 8.3Integration by Substitution C. Integration by Trigonometric Substitution Example 8.29T Find Solution: __________ (x  2)2  1 Letx + 2  secq. Thendx secq tanqdq. Since secqx + 2,

Theorem 8.3 Integration by Parts If u(x) and v(x) are two differentiable functions, then In other words, . 8.4Integration 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. Proof: Suppose u and v are two differentiable functions. Since (uv)  uv¢ + vu¢, by definition, .

8.4Integration by Parts From Theorem 8.3, we can see that the problem of finding can betransformed 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 .

8.1 Concepts of Indefinite Integrals