Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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 dy 1000e –0.046 t . dt 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. P. 2 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 d 2 ( x ) 2 x. dx 2 As 2x is the derivative of x , 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 d If [ F ( x)] f ( x), then F(x) is called a primitive function of f(x). dx 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 d 2 d 2 d ( x C ) ( x ) (C ) 2 x 0 2 x. have dx dx dx Thus, x2 + C is also a primitive function of 2x for an arbitrary constant C. P. 3 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 d If F ( x) f ( x) , then the indefinite integral of f(x), which is dx denoted by f ( x)dx , is given by f ( x)dx F ( x) C, where C is an arbitrary constant. Note: 1. In the notation of f ( x)dx , 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). P. 4 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. d For example: Since ( x n1 ) (n 1) x n, dx d 1 n1 i.e., x xn , dx n 1 8.1 n x dx 1 n1 x C , for all real numbers n 1. n 1 Note: 1. Formula 8.1 is also called the Power Rule for integration. 2. When n 0, the left hand side of the formula becomes x 0 dx 1dx . For convenience we usually express 1dx as dx . P. 5 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: 8.2 8.3 8.5 kdx kx C , where k is a constant. 1 x dx ln x C sin xdx cos x C 8.4 8.6 8.7 2 sec xdx tan x C 8.8 8.9 sec x tan xdx sec x C 8.10 x x e dx e C cos xdx sin x C 2 csc xdx cot x C csc x cot xdx csc x C P. 6 8.1 Concepts of Indefinite Integrals C. Basic Properties of Indefinite Integrals Theorem 8.1 If k is a non-constant, then kf ( x)dx k f ( x)dx . Proof: d [ g ( x)] f ( x). dx d d [kg ( x)] k [ g ( x)] kf ( x) dx dx Let f ( x)dx g ( x) C. Then By definition, kf ( x)dx kg ( x) C , where C is an arbitrary constant. On the other hand, k f ( x)dx k[ g ( x) C ] k g ( x) kC Since C and kC are arbitrary constants, the expressions kg(x) C and kg(x) kC represent the same family of primitive functions. kf ( x)dx k f ( x)dx P. 7 8.1 Concepts of Indefinite Integrals C. Basic Properties of Indefinite Integrals Theorem 8.2 [ f ( x) g ( x)]dx f ( x)dx g ( x)dx Proof: Let f ( x)dx F ( x) C1 and g ( x)dx G( x) C2 , where C1 and C2 are d d arbitrary constants. Then [ F ( x)] f ( x) and [G ( x)] g ( x). dx dx d [ F ( x) G ( x)] f ( x) g ( x) dx By definition, [ f ( x) g ( x)]dx F ( x) G( x) C. On the other hand, f ( x)dx g ( x)dx [ F ( x) C1 ] [G( x) C2 ] F ( x) G ( x) C1 C2 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. [ f ( x) g ( x)]dx f ( x)dx g ( x)dx P. 8 8.1 Concepts of Indefinite Integrals C. Basic Properties of Indefinite Integrals Example 8.1T Find (3 x 4 x 5 5)dx. Solution: 5 ( 3 x 4 x 5)dx 3 1 x 2 dx 4 5 x 2 dx 5dx 2 3 2 7 3 x 2 C1 4 x 2 C 2 5 x C3 3 7 8 7 Use C to express the sum 2 x3 x 5x C 7 of all constants P. 9 8.1 Concepts of Indefinite Integrals C. Basic Properties of Indefinite Integrals Example 8.2T Find 3 x 1 dx. x 1 Solution: (3 x ) 2 3 x 1 x 1 3 3 x 1 dx ( x 1) 3 x 1 dx 2 (x 3 2 x 3 dx 5 3 3 x 5 1 x3 1)dx 1 x 3 dx 4 3 3 x 4 a3 b3 (a b)(a2 ab b2) Cancel the common factor dx xC P. 10 8.1 Concepts of Indefinite Integrals C. Basic Properties of Indefinite Integrals Example 8.3T 2 Find 3x 2 dx. x Solution: 2 1 3 x 2 dx 3 xdx 2 dx 2 dx x x 2 x 3 2 x 2 ln x C 2 3 x 2 2 x 2 ln x C 2 P. 11 8.1 Concepts of Indefinite Integrals C. Basic Properties of Indefinite Integrals Example 8.4T 5 Find 3 x 2 2e x dx. x Solution: 1 2 5 x 2 x 3 x 2 e dx 3 x dx 5 dx 2 e dx x x 3 x 3 5 ln x 2e x C 3 x 3 5 ln x 2e x C P. 12 8.1 Concepts of Indefinite Integrals C. Basic Properties of Indefinite Integrals Example 8.5T Find (1 tan 2 x)dx. Solution: (1 tan 2 x)dx sec 2 xdx tan x C P. 13 8.1 Concepts of Indefinite Integrals C. Basic Properties of Indefinite Integrals Example 8.6T Let y ln x – ln (x 1). dy (a) Find . dx 1 dx . (b) Hence find x( x 1) Solution: dy 1 1 d ( x 1) dx x x 1 dx 1 1 x x 1 1 1 (b) By (a), dx ln x ln x 1 C x x 1 x 1 x x( x 1) dx ln x ln x 1 C 1 dx ln x ln x 1 C x( x 1) (a) P. 14 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)? d Consider (ax b) n1 a(n 1)(ax b) n . dx n 1 ( ax b ) 8.11 (ax b) n dx C, where n –1 and a 0. a(n 1) Using the same method, we can obtain the following integration formulas: Suppose a and b are real numbers with a 0. 1 1 8.12 dx ln ax b C ax b a 1 8.13 e axb dx e axb C a 1 8.14 sin( ax b)dx cos(ax b) C a 1 8.15 cos(ax b)dx sin( ax b) C a P. 15 8.2 Indefinite Integration of Functions A. Integration of Functions Involving the Expression (ax + b) Example 8.7T 9 dx. Find 3 2 x Solution: 3 1 9 dx 9 (2 x) 3 dx 2 x 1 1 9 (2 x) 3 C 1 1 3 2 27 (2 x) 3 C 2 P. 16 8.2 Indefinite Integration of Functions A. Integration of Functions Involving the Expression (ax + b) Example 8.8T 1 dx. Find 2x 3 2x 5 Solution: 1 1 2x 3 2x 5 dx dx 2x 3 2x 5 2x 3 2x 5 2x 3 2x 5 Rationalize the denominator 2x 3 2x 5 dx (2 x 3) (2 x 5) 1 ( 2 x 3 2 x 5 )dx 8 3 3 1 1 (2 x 3) 2 (2 x 5) 2 C 1 1 8 2 1 8 2 1 2 2 3 3 1 [(2 x 3) 2 (2 x 5) 2 ] C 24 P. 17 8.2 Indefinite Integration of Functions A. Integration of Functions Involving the Expression (ax + b) Example 8.9T 3x 4 dx. Find 2 6 x 5x 4 Solution: 3x 4 3x 4 dx (3x 4)(2 x 1) dx 6 x 2 5x 4 1 dx 2x 1 Cancel the common factor 1 ln 2 x 1 C 2 P. 18 8.2 Indefinite Integration of Functions A. Integration of Functions Involving the Expression (ax + b) Example 8.10T 4t 1 e 3e t dt. Find 2t e Solution: e 4t 1 e 4t 1 3e t et e 2t dt e2t 3 e2t dt (e 2t 1 3e t )dt e 2t 1 3e t C 2 1 1 e 2t 1 3e t C 2 P. 19 8.2 Indefinite Integration of Functions A. Integration of Functions Involving the Expression (ax + b) Example 8.11T Find [2t 2 sin(3t 2)]dt. Solution: [2t 2 sin(3t 2)]dt 2t 2 dt sin(3t 2)dt 2t 3 cos(3t 2) C 3 3 2 3 1 t cos(3t 2) C 3 3 P. 20 8.2 Indefinite Integration of Functions A. Integration of Functions Involving the Expression (ax + b) Example 8.12T Find 2 x 5 x dx. Solution: x x x 2 5 dx 10 dx (e ln 10 ) x dx If y ln 10, then ey 10 by definition, i.e., eln 10 10. e (ln 10 ) x C ln10 10 x C ln 10 P. 21 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 1 sin A cos B [sin (A B) sin (A B)] 2 1 cos A sin B [sin (A B) sin (A B)] 2 1 cos A cos B [cos (A B) cos (A B)] 2 1 sin A sin B [cos (A B) cos (A B)] 2 P. 22 8.2 Indefinite Integration of Functions B. Integration of Trigonometric Functions Example 8.13T 6 sin cos d . Find 2 2 Solution: 6 sin 2 cos 2 d 3 sin d 3 cos C sin 2A 2 sin A cos A P. 23 8.2 Indefinite Integration of Functions B. Integration of Trigonometric Functions Example 8.14T 2 x 2 x sin cos dx. Find 4 4 Solution: 2 x 1 2 x 2 x sin cos dx sin dx 4 4 2 2 1 2x sin dx 4 2 1 1 cos x dx 4 2 1 (1 cos x)dx 8 1 ( x sin x) C 8 sin 2A 2 sin A cos A cos 2A 1 2 sin2 A P. 24 8.2 Indefinite Integration of Functions B. Integration of Trigonometric Functions Example 8.15T Find sin 8t sin 7tdt. Solution: 1 sin 8 t sin 7 tdt 2 [cos(8 7)t cos(8 7)t ]dt 1 (cos t cos15t )dt 2 Product-to-sum formula 1 sin 15t sin t C 2 15 sin t sin 15t C 2 30 P. 25 8.2 Indefinite Integration of Functions B. Integration of Trigonometric Functions Example 8.16T 2 cos Find d . cos 2 1 Solution: 2 cos cos d cos 2 1 cos 2 1 d 2 cos d 2 sin ( cot csc )d csc C cos 2A 1 2 sin2 A cot A cos A sin A P. 26 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 f (u )du F (u ) C and u g (x). d d du Since [ F (u )] [ F (u )] dx du dx f (u) g(x) f [g(x)] g(x) By the definition of integration, f [ g ( x)]g' ( x)dx F (u) C. f [ g ( x)]g' ( x)dx f (u)du P. 27 8.3 Integration by Substitution A. Change of Variables For an integral f [ g ( x)]g' ( x)dx, we can transform it into a simpler integral f (u)du , 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 2 x x 2 1dx together. Note that 2 x x 2 1dx x 2 1 2 xdx , du so we let u x2 + 1, such that 2x. dx 2 x 1 2xdx udu 3 2 u2 C 3 3 2 2 ( x 1) 2 C 3 P. 28 8.3 Integration by Substitution A. Change of Variables Example 8.17T 3 Find 2 x 1 x 2 dx. Solution: Let u 1 – x2. Then du 2 x . dx 3 3 2 2 1 x (2 x)dx 2 x 1 x dx 3 udu 4 3 u3 C 4 4 3 (1 x 2 ) 3 C 4 Express the answer in terms of x P. 29 8.3 Integration by Substitution A. Change of Variables Example 8.18T Find x 3 ( x 2 7)10 dx. Solution: Let u x2 – 7. Then du 2x. dx 3 2 10 2 2 10 1 x ( x 7 ) dx x ( x 7 ) 2 xdx 2 1 Rewrite x2 as (u 7) (u 7)u10 du 2 1 (u11 7u10 )du 2 1 1 7 u12 u11 C 2 12 11 1 7 ( x 2 7)12 ( x 2 7)11 C 24 22 P. 30 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. P. 31 8.3 Integration by Substitution A. Change of Variables Example 8.19T Find csc xdx. Solution: csc x cot x dx csc x cot x csc 2 x csc x cot x dx csc x cot x csc xdx csc x Let u csc x – cot x. du Then csc 2 x csc x cot x. dx 1 csc xdx du u ln u C ln csc x cot x C Alternative Solution: d csc x csc x cot x dx d and cot x csc 2 x dx d (csc x cot x) dx csc x(csc x cot x) Let u csc x – cot x. du Then u csc x. dx 1 csc xdx du u ln csc x cot x C P. 32 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. P. 33 8.3 Integration by Substitution A. Change of Variables Example 8.20T sin(ln x) dx. Find x Solution: sin(ln x) 1 sin(ln x ) dx dx x x sin(ln x)d (ln x) cos(ln x) C P. 34 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 m n m n or sin x cos xdx tan x sec xdx , 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 sin m x cosn xdx 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. P. 35 8.3 Integration by Substitution B. Integrals Involving Powers of Trigonometric Functions Example 8.21T Find sin 3 x cos5 xdx. Solution: 3 5 sin x cos xdx sin 2 x cos5 ( sin x)dx (1 cos2 x) cos5 xd (cos x) (cos7 x cos5 x)d (cos x) cos8 x cos6 x C 8 6 P. 36 8.3 Integration by Substitution B. Integrals Involving Powers of Trigonometric Functions Example 8.22T Find 3 cos x sin 3 xdx. Solution: 3 cos x sin 3 xdx 3 cos x sin 2 x ( sin x)dx 3 cos x (1 cos2 x)d (cos x) 7 1 (cos x) 3 (cos x) 3 d (cos x) 10 4 3 3 (cos x) 3 (cos x) 3 C 10 4 P. 37 8.3 Integration by Substitution B. Integrals Involving Powers of Trigonometric Functions Example 8.23T 2 4 Find sin x cos xdx. Solution: 2 sin 2 x 2 cos xdx sin x cos xdx 2 1 sin 2 2 x cos 2 xdx 4 1 (1 cos 4 x)(1 cos 2 x)dx 16 1 (1 cos 2 x cos 4 x cos 2 x cos 4 x)dx 16 1 1 1 1 cos 2 x cos 4 x cos 6 x cos 2 x dx 16 2 2 1 1 1 1 cos 2 x cos 4 x cos 6 x dx 16 2 2 x 1 1 1 sin 2 x sin 4 x sin 6 x C 16 64 64 192 2 4 P. 38 8.3 Integration by Substitution B. Integrals Involving Powers of Trigonometric Functions Similarly, integrals in the form tan m x sec n xdx may be found by using the method of integration by substitution. Strategies for finding integrals in the form tan m x sec n xdx 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. P. 39 8.3 Integration by Substitution B. Integrals Involving Powers of Trigonometric Functions Example 8.24T 3 5 Find tan x sec xdx. Solution: 3 5 tan x sec xdx tan 2 x sec 4 x tan x sec xdx (sec 2 x 1) sec 4 xd (sec x) (sec 6 x sec 4 x)d (sec x) sec 7 x sec 5 x C 7 5 P. 40 8.3 Integration by Substitution B. Integrals Involving Powers of Trigonometric Functions Example 8.25T Find cot 2 x csc 4 xdx. Solution: 2 4 cot x csc xdx cot 2 x csc 2 x( csc 2 x)dx cot 2 x(cot 2 x 1)d (cot x) ( cot 4 x cot 2 x)d (cot x) cot 5 x cot 3 x C 5 3 P. 41 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 sec xdx (m 0 and n 1), we may follow the method in Example 8.19. In some other cases, such as tan 2 x sec xdx (m 2 and n 1), we may need to use the technique ‘integration by parts’, which will be discussed later in this chapter. P. 42 8.3 Integration by Substitution C. Integration by Trigonometric Substitution If an indefinite integral involves radicals in the form a 2 x 2 , a 2 x 2 or x 2 a 2 , we can use the method of integration by substitution to eliminate the radicals. The following three trigonometric identities are very useful for the elimination: cos2 1 sin2 , sec2 1 tan2 , tan2 sec2 1 For example, if we substitute x a sin into the expression a 2 x 2, we have a 2 x 2 a 2 a 2 sin 2 a 2 cos2 a cos Then we can express the integrand in terms of . After finding the indefinite integral in terms of (say, 3 + C), the final answer should be expressed in terms of the original variable, say, x. P. 43 8.3 Integration by Substitution C. Integration by Trigonometric Substitution In order to express 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 such that sin x (where –1 x 1) π π and . 2 2 2. cos–1x is defined as the angle such that cos x (where –1 x 1) and 0 p. π π 3. tan–1x is defined as the angle such that tan x and . 2 2 P. 44 8.3 Integration by Substitution C. Integration by Trigonometric Substitution Example 8.26T dx . Find 2 2 x 1 x Solution: Let x sin.Then dx cos d. dx cos d x 2 1 x 2 sin 2 1 sin 2 cos d 2 sin cos csc 2 d cot C Since sin x, cot csc2 1 1 1 2 sin 1 1 2 x 1 x 2 x dx 1 x2 C 2 2 x x 1 x P. 45 8.3 Integration by Substitution C. Integration by Trigonometric Substitution Example 8.27T Find x 2 1 x 2 dx. Solution: Since sin x, cos 1 x 2 , Let x sin.Then dx cosd. sin 4 2 cos 2 sin 2 2(1 2 sin 2 )(2 cos sin ) 2 2 x 1 x dx sin 2 1 sin 2 cos d sin cos d 2 sin 2 d 2 1 sin 2 2 d 4 1 1 cos 4 d 4 2 1 sin 4 C 8 4 2 4(1 2 x 2 )( x 1 x 2 ) 2 4( x 2 x 3 ) 1 x 2 2 2 x 1 x dx 1 1 4( x 2 x 3 ) 1 x 2 sin x C 8 4 1 [(2 x3 x) 1 x 2 sin 1 x] C 8 P. 46 8.3 Integration by Substitution C. Integration by Trigonometric Substitution Example 8.28T dx . Find 2 2 x 9 x Solution: x Since tan , 3 Let x 3tan.Then dx 3sec2 d. dx x2 csc x 1 cot 2 9 x 3 sec 2 d 2 9 tan 2 9 9 tan 2 3 sec 2 d 27 tan 2 sec 1 sec 2 d 9 tan 1 1 cos cot d 9 cos sin 1 tan 2 9 1 2 x 1 1 csc cot d 9 1 csc C 9 x2 x2 9 x 9 x2 C 2 9x 9 x dx P. 47 8.3 Integration by Substitution C. Integration by Trigonometric Substitution Example 8.29T dx . Find 2 x 4x 3 Solution: dx dx x+2 __________ (x 2)2 1 x 4x 3 ( x 2) 2 1 Let x + 2 sec. Then dx sectan d. 1 dx sec tan d sec tan sec d x 2 4 x 3 sec2 1 sec tan sec tan d d (sec tan ) tan sec tan ln sec tan C sec d 2 Since sec x + 2, tan sec 2 1 ( x 2) 2 1 x 2 4 x 3 dx ln x 2 x 2 4 x 3 C x2 4x 3 P. 48 8.4 Integration by Parts Some indefinite integrals such as ln xdx, xsin xdx and x xe dx 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 8.3 Integration by Parts If u(x) and v(x) are two differentiable functions, then uv'dx uv vu'dx. In other words, udv uv vdu . Proof: Suppose u and v are two differentiable functions. d Since (uv) uv + vu, by definition, (uv' vu' )dx uv C. dx uv'dx vu'dx uv C uv'dx uv vu'dx P. 49 8.4 Integration by Parts From Theorem 8.3, we can see that the problem of finding udv can be transformed into the problem of finding vdu instead. If the integral vdu is much simpler than udv, then the original integral can be found easily. If we want to apply the technique of integration by parts to an integral, such as e x sin xdx, we need to transform the integral into the form udv first, such as (sin x)d (e x ) or e x d ( cos x). P. 50 8.4 Integration by Parts Example 8.30T Find x ln xdx. Solution: x2 x ln xdx ln xd 2 x2 x2 (ln x) d (ln x) 2 2 x 2 ln x x2 1 dx 2 2 x x 2 ln x 1 xdx 2 2 x 2 ln x x 2 C 2 4 P. 51 8.4 Integration by Parts Example 8.31T 2 Find x log5 xdx. Solution: 2 x log5 xdx x 2 ln x dx ln 5 x3 1 ln xd ln 5 3 x3 1 x3 (ln x) d (ln x) ln 5 3 3 1 x 3 ln x 1 3 1 x dx ln 5 3 3 x 1 x 3 ln x 1 2 x dx ln 5 3 3 1 x3 x3 ln x C ln 5 3 9 P. 52 8.4 Integration by Parts In some cases, there may be more than one choice for u and v. For example, we can transform xsin xdx into x2 sin xd 2 or xd (cos x). However, if we choose the former, then x2 x2 x2 sin xd 2 (sin x) 2 2 d (sin x) 2 x sin x 1 2 x cos xdx 2 2 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. P. 53 8.4 Integration by Parts Example 8.32T Find x csc 2 xdx. Solution: 2 x csc xdx xd ( cot x) x( cot x) cot xdx cos x x cot x dx sin x 1 x cot x d (sin x) sin x x cot x ln sin x C P. 54 8.4 Integration by Parts Example 8.33T d x 1 cot . dx 2 cos x 1 x (b) Hence find dx . cos x 1 (a) Show that Solution: (a) d x 2 x d x cot csc dx 2 2 dx 2 1 2 x csc 2 2 1 2 x 2 sin 2 1 1 cos x 2 2 1 cos x 1 x x dx xd cot (b) 2 cos x 1 x x x cot cot dx 2 2x cos x 2 d x x cot 2 x 2 2 sin 2 x 1 x x cot 2 d sin x 2 2 sin 2 x x x cot 2 ln sin C 2 2 P. 55 8.4 Integration by Parts Example 8.34T Find x 2 x e dx. Solution: 2 x x 2 d (e x ) x e dx [ x 2 e x e x d ( x 2 )] ( x 2 e x 2 xe x dx) x 2 e x 2 xd (e x ) x 2 e x [2 xe x e x d (2 x)] x 2 e x 2 xe x 2 e x dx x 2 e x 2 xe x 2e x C ( x 2 2 x 2)e x C P. 56 8.4 Integration by Parts Example 8.35T Find cos(ln x)dx. Solution: cos(ln x)dx [cos(ln x)]x xd[cos(ln x)] 1 x cos(ln x) x sin(ln x) dx x x cos(ln x) sin(ln x)dx x cos(ln x) [sin(ln x)]x xd[sin(ln x)] 1 x cos(ln x) x sin(ln x) x cos(ln x) dx x x cos(ln x) x sin(ln x) cos(ln x)dx Therefore, 2 cos(ln x)dx x cos(ln x) x sin(ln x) C x cos(ln x)dx [sin(ln x) cos(ln x)] C 2 P. 57 8.5 Applications of Indefinite Integrals A. Geometrical Applications dy In previous chapters, we learnt that of a curve y F(x) is the slope function of the curve. dx Since integration is the reverse process of differentiation, dy if we let f(x), then by the definition of integration, we have dx y f ( x)dx F ( x) C where C is an arbitrary constant. Thus we can see that the equation of a family of curves y F(x) + C dy can be found by integration, providing that the slope function of the dx curve is known. P. 58 8.5 Applications of Indefinite Integrals A. Geometrical Applications Example 8.36T The equation of the slope of a curve at the point (x, y) is given by dy (2 x 3) 6. If the curve passes through (2, 2), find the equation of dx the curve. Solution: dy (2 x 3) 6 dx (2 x 3) 7 6 C , where C is an arbitrary constant y (2 x 3) dx 27 (2 x 3) 7 C 14 7 ( 2 2 3 ) When x 2, y 2, we have 2 C 14 27 C 14 (2 x 3) 7 27 . ∴ The equation of the curve is y 14 14 P. 59 8.5 Applications of Indefinite Integrals A. Geometrical Applications Example 8.37T d2y 2 It is given that at each point (x, y) on a certain curve, 2 sin x . If 2 dx 5 5 the curve passes through 0, and p, , find the equation of that curve. 4 4 Solution: 5 5 0 , Since and p, lie on the curve, we have d2y 2 4 4 2 sin x 5 1 dx 2 0 C2 dy 2 4 4 2 sin xdx C2 1 dx (1 cos 2 x)dx 5 p2 1 pC1 1 4 2 4 sin 2 x 2 x C1 p 2 pC1 sin 2 x 2 y x C1 dx p 2 C1 x 2 cos 2 x ∴ The equation of the curve is C1 x C2 2 4 x 2 cos 2 x px y 1. 2 4 2 P. 60 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 ds dv d 2 s v and a 2, dt dt dt where s is the displacement of the particle at time t. Since integration is the reverse process of differentiation, we have s vdt and v adt. P. 61 8.5 Applications of Indefinite Integrals 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 v t t 2 5 . Find the displacement s of the particle at time t, given that s 6 when t 2. Solution: 1 2 2 t 5 d ( t 5) 2 3 3 1 1 2 2 (t 5) 2 C (t 2 5) 2 C 3 2 3 When t 2, s 6, we have s t t 5dt 2 3 1 6 (2 2 5) 2 C 3 C 3 3 1 s (t 2 5) 2 3 3 P. 62 8.5 Applications of Indefinite Integrals B. Applications in Physics Example 8.39T A particle moves along a straight line so that its acceleration a at time t is 1 given by a for t > 0. When t 1, the velocity of the particle is 8 and t its displacement is 24. Find the displacement of the particle at time t. Solution: 1 v dt ln t C1 t When t 1, v 8, we have 8 ln1 C1 C1 8 v ln t 8 s (ln t 8)dt ln tdt 8t t ln t td (ln t ) 8t 1 t ln t t dt 8t t t ln t dt 8t t ln t 7t C2 When t 1, s 24, we have 24 1 ln1 7(1) C 2 C 2 17 s t ln t 7t 17 P. 63 Chapter Summary 8.1 Concepts of Indefinite Integrals d [ F ( x)] f ( x), then the indefinite integral of f(x) dx is defined by f ( x)dx F ( x) C , where C is an arbitrary constant. 1. If 2. n x dx 3. kdx kx C, where k is a constant. 4. 1 x dx ln x C 5. x x e dx e C 1 n1 x C , where n 1. n 1 P. 64 Chapter Summary 8.1 Concepts of Indefinite Integrals 6. 7. 8. 9. 10. 11. 12. 13. sin xdx cos x C cos xdx sin x C 2 sec xdx tan x C 2 csc xdx cot x C sec x tan xdx sec x C csc x cot xdx csc x C kf ( x)dx k f ( x)dx, where k is a non - zero constant. f ( x) g ( x)dx f ( x)dx g ( x)dx P. 65 Chapter Summary 8.2 Indefinite Integration of Functions Let a and b be real numbers with a 0. (ax b) n1 n 1. (ax b) dx C , where n 1. a(n 1) 1 1 dx ln ax b C 2. ax b a 1 3. e axb dx e axb C a 1 4. sin( ax b)dx cos(ax b) C a 1 cos( ax b ) dx sin( ax b) C 5. a P. 66 Chapter Summary 8.3 Integration by Substitution 1. Let u g(x) be a differentiable function. Then, f [ g ( x)]g' ( x)dx f (u)du. 2. If the integrated involves terms like a 2 x 2 , a 2 x 2 and x 2 a 2 , we can simplify the integrand by substituting x a sin , x a tan or x a sec respectively. P. 67 Chapter Summary 8.4 Integration by Parts If u(x) and v(x) are two differentiable functions, then udv uv vdu. P. 68 Chapter Summary 8.5 Applications of Indefinite Integrals 1. If the slope of a curve at point (x, y) is f(x), then the equation of the family of curves is given by f ( x)dx F ( x) C 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 s vdt and v adt. P. 69