Chap 8 Techniques of Integration

Chap 8 Techniques of Integration - FIRST YEAR CALCULUS W W...

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Unformatted text preview: FIRST YEAR CALCULUS W W L CHEN c W W L Chen, 1994, 2005. This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 8 TECHNIQUES OF INTEGRATION 8.1. Integration by Substitution In this section, we discuss how we can use the Chain rule in differentiation to help solve problems in integration. This technique is usually called integration by substitution. As we shall not prove any result here, our discussion will be only heuristic. We emphasize that the technique does not always work. First of all, we have little or no knowledge of the antiderivatives of many functions. Secondly, there is no simple routine that we can describe to help us find a suitable substitution even in the cases where the technique works. On the other hand, when the technique does work, there may well be more than one suitable substitution! Occasionally, the possibility of substitution may not be immediately obvious, and a certain amount of trial and error does occur. The fact that one substitution does not appear to work does not mean that the method fails. It may very well be the case that we have used a bad substitution. INTEGRATION BY SUBSTITUTION – VERSION 1. If we make a substitution x = g (u), then dx = g (u) du, and f (x) dx = Example 8.1.1. Consider the indefinite integral √ 1 dx. 1 − x2 du = u + C = sin−1 x + C. page 1 of 26 f (g (u))g (u) du. If we make a substitution x = sin u, then dx = cos u du, and √ 1 dx = 1 − x2 cos u 1 − sin u 2 du = Chapter 8 : Techniques of Integration First Year Calculus c W W L Chen, 1994, 2005 On the other hand, if we make a substitution x = cos v , then dx = − sin v dv , and √ 1 dx = − 1 − x2 √ sin v dv = − 1 − cos2 v dv = −v + C = − cos−1 x + C. Example 8.1.2. Consider the indefinite integral 1 dx. 1 + x2 If we make a substitution x = tan u, then dx = sec2 u du, and 1 dx = 1 + x2 sec2 u du = 1 + tan2 u du = u + C = tan−1 x + C. On the other hand, if we make a substitution x = cot v , then dx = − csc2 v dv , and 1 dx = − 1 + x2 csc2 v dv = − 1 + cot2 v dv = −v + C = − cot−1 x + C. Example 8.1.3. Consider the indefinite integral √ x x + 1 dx. If we make a substitution x = u2 − 1, then dx = 2u du, and √ x x + 1 dx = = 2(u2 − 1)u2 du = 2 u4 du − 2 u2 du 2 2 25 23 u − u + C = (x + 1)5/2 − (x + 1)3/2 + C. 5 3 5 3 On the other hand, if we make a substitution x = v − 1, then dx = dv , and √ x x + 1 dx = = (v − 1)v 1/2 dv = v 3/2 dv − v 1/2 dv 2 5 /2 2 3 /2 2 2 −v + C = (x + 1)5/2 − (x + 1)3/2 + C. v 5 3 5 3 We can confirm that the indefinite integral is correct by checking that d dx 2 2 (x + 1)5/2 − (x + 1)3/2 + C 5 3 √ = x x + 1. INTEGRATION BY SUBSTITUTION – VERSION 2. Suppose that a function f (x) can be written in the form f (x) = g (h(x))h (x). If we make a substitution u = h(x), then du = h (x) dx, and f (x) dx = g (h(x))h (x) dx = g (u) du. Remark. Note that in Version 1, the variable x is initially written as a function of the new variable u, whereas in Version 2, the new variable u is written as a function of x. The difference, however, is minimal, as the substitution x = g (u) in Version 1 has to be invertible to enable us to return from the new variable u to the original variable x at the end of the process. Chapter 8 : Techniques of Integration page 2 of 26 First Year Calculus c W W L Chen, 1994, 2005 Example 8.1.4. Consider the indefinite integral x2 ex dx. Note first of all that the derivative of the function x3 is equal to 3x2 , so it is convenient to make the substitution u = x3 . Then du = 3x2 dx, and x2 ex dx = 3 3 1 3 3x2 ex dx = 3 1 3 eu du = 1u 13 e + C = ex + C. 3 3 3 A somewhat more complicated alternative is to note that the derivative of the function ex is equal to 3 3 3 3x2 ex , so it is convenient to make the substitution v = ex . Then dv = 3x2 ex dx, and x2 ex dx = 3 1 3 3x2 ex dx = 3 1 3 dv = 1 13 v + C = ex + C. 3 3 Example 8.1.5. Consider the indefinite integral x(x2 + 3)4 dx. Note first of all that the derivative of the function x2 + 3 is equal to 2x, so it is convenient to make the substitution u = x2 + 3. Then du = 2x dx, and x(x2 + 3)4 dx = 1 2 2x(x2 + 3)4 dx = 1 2 u4 du = 15 12 u +C = (x + 3)5 + C. 10 10 Example 8.1.6. Consider the indefinite integral 1 dx. x log x Note first of all that the derivative of the function log x is equal to 1/x, so it is convenient to make the substitution u = log x. Then du = (1/x) dx, and 1 dx = x log x 1 du = log |u| + C = log | log x| + C. u Example 8.1.7. Consider the indefinite integral tan3 x sec2 x dx. Note first of all that the derivative of the function tan x is equal to sec2 x, so it is convenient to make the substitution u = tan x. Then du = sec2 x dx, and tan3 x sec2 x dx = u3 du = 14 1 u + C = tan4 x + C. 4 4 Chapter 8 : Techniques of Integration page 3 of 26 First Year Calculus c W W L Chen, 1994, 2005 Example 8.1.8. Consider the indefinite integral sin3 x cos3 x dx. Note first of all that the derivative of the function sin x is equal to cos x, so it is perhaps convenient to make the substitution u = sin x. Then du = cos x dx, and sin3 x cos3 x dx = u3 (1 − u2 ) du = (u3 − u5 ) du = u4 u6 sin4 x sin6 x − +C = − + C. 4 6 4 6 Alternatively, note that the derivative of the function cos x is equal to − sin x, so it is convenient to make the substitution v = cos x. Then dv = − sin x dx, and sin3 x cos3 x dx = It can be checked that sin4 x sin6 x cos6 x cos4 x 1 − = − +. 4 6 6 4 12 Example 8.1.9. Recall Example 8.1.1. Since √ we have 1 /2 0 −(1 − v 2 )v 3 dv = (v 5 − v 3 ) dv = v6 cos6 x cos4 x v4 − +C = − +C . 6 4 6 4 1 dx = sin−1 x + C, 1 − x2 √ 1 dx = sin−1 x 1 − x2 1 /2 0 = sin−1 1 π − sin−1 0 = . 2 6 Note that we have in fact used the substitution x = sin u to show that √ 1 dx = 1 − x2 du = u + C, followed by an inverse substitution u = sin−1 x. Here, we need to make the extra step of substituting the values x = 0 and x = 1/2 to the indefinite integral sin−1 x. Observe, however, that with the substitution x = sin u, the variable x increases from 0 to 1/2 as the variable u increases from 0 to π/6. But then π /6 π /6 du = u 0 0 = π = 6 1/2 0 √ 1 dx, 1 − x2 so it appears that we do not need the inverse substitution u = sin−1 x. Perhaps we can directly substitute u = 0 and u = π/6 to the indefinite integral u. DEFINITE INTEGRAL BY SUBSTITUTION – VERSION 1. Suppose that a substitution x = g (u) satisfies the following conditions: (a) There exist α, β ∈ R such that g (α) = A and g (β ) = B . (b) The derivative g (u) > 0 for every u satisfying α < u < β . Then dx = g (u) du, and B β f (x) dx = A Chapter 8 : Techniques of Integration α f (g (u))g (u) du. page 4 of 26 First Year Calculus c W W L Chen, 1994, 2005 Remark. If condition (b) above is replaced by the condition that the derivative g (u) < 0 for every u satisfying β < u < α, then the same conclusion holds if we adopt the convention that β α f (g (u))g (u) du = − α β f (g (u))g (u) du. Example 8.1.10. To calculate the definite integral 1 0 1 dx, 1 + x2 we can use the substitution x = tan u, so that dx = sec2 u du. Note that tan 0 = 0 and tan(π/4) = 1, and that sec2 u > 0 whenever 0 < u < π/4. It follows that 1 0 1 dx = 1 + x2 π /4 0 sec2 u du = 1 + tan2 u π /4 π /4 du = u 0 0 = π π −0= . 4 4 We can compare this to first observing Example 8.1.2, so that 1 0 1 dx = tan−1 x 1 + x2 1 0 = tan−1 1 − tan−1 0 = π π −0= . 4 4 Example 8.1.11. To calculate the definite integral 3 0 √ x x + 1 dx, we can use the substitution x = g (u) = u2 − 1, so that dx = 2u du. Note that g (1) = 0 and g (2) = 3, and that g (u) = 2u > 0 whenever 1 < u < 2. It follows that 3 0 √ x x + 1 dx = 1 2 2(u2 − 1)u2 du = 25 23 u− u 5 3 2 = 1 64 16 − 5 3 − 22 − 53 = 62 14 116 − = . 5 3 15 DEFINITE INTEGRAL BY SUBSTITUTION – VERSION 2. Suppose that a substitution u = h(x) satisfies the following conditions: (a) There exists a function g (u) such that f (x) = g (h(x))h (x) for every x ∈ [A, B ]. (b) The derivative h (x) > 0 for every x satisfying A < x < B . Then du = h (x) dx, and B B h(B ) f (x) dx = A A g (h(x))h (x) dx = h(A) g (u) du. Remark. If condition (b) above is replaced by the condition that the derivative h (x) < 0 for every x satisfying A < x < B , then the same conclusion holds if we adopt the convention that h(B ) h(A) g (u) du = − h(A) h(B ) g (u) du. Chapter 8 : Techniques of Integration page 5 of 26 First Year Calculus c W W L Chen, 1994, 2005 Example 8.1.12. To calculate the definite integral 1 x(x2 + 3)4 dx, 0 we can use the substitution u = h(x) = x2 + 3, so that du = 2x dx. Note that h(0) = 3 and h(1) = 4, and that h (x) = 2x > 0 whenever 0 < x < 1. It follows that 1 0 1 x(x + 3) dx = 2 2 4 4 3 1 u5 u dx = 25 4 4 = 3 1 2 1024 243 − 5 5 = 781 . 10 We can compare this to first observing Example 8.1.4, so that 1 x(x2 + 3)4 dx = 0 12 (x + 3)5 10 1 = 0 1024 243 781 − = . 10 10 10 Example 8.1.13. To calculate the definite integral 4 2 1 dx, x log x we can use the substitution u = h(x) = log x, so that du = h (x) dx, where h (x) = 1/x > 0 whenever 2 < x < 4. Note also that h(2) = log 2 and h(4) = log 4. It follows that 4 2 1 dx = x log x log 4 log 2 1 du = log |u| u log 4 = log log 4 − log log 2 = log log 2 log 4 log 2 = log 2. 8.2. Integration by Parts Recall the Product rule for differentiation, that (uv ) = uv + vu . Integrating with respect to x, we obtain (uv ) dx = uv dx + v u dx. Now the indefinite integral on the left hand side is of the form uv . Rewriting this equation, we have (1) uv dx = uv − v u dx. Equation (1) is called the formula for integration by parts for indefinite integrals. It is very useful if the indefinite integral v u dx is much easier to calculate than the indefinite integral uv dx. Example 8.2.1. Consider the indefinite integral xex dx. Writing u = x and v = ex , we have uv dx = Chapter 8 : Techniques of Integration xex dx. page 6 of 26 First Year Calculus c W W L Chen, 1994, 2005 Furthermore, v = ex and u = 1. It follows that uv − Hence xex dx = xex − ex + C. v u dx = xex − ex dx = xex − ex + C. Example 8.2.2. Consider the indefinite integral log x dx. Writing u = log x and v = 1, we have uv dx = Furthermore, v=x It follows that uv − Hence log x dx = x log x − x + C. v u dx = x log x − 1 x dx = x log x − x + C. x and u= 1 . x log x dx. Example 8.2.3. Consider the indefinite integral ex sin x dx. Writing u = ex and v = sin x, we have uv dx = Furthermore, v = − cos x and u = ex . It follows that uv − Hence (2) ex sin x dx = −ex cos x + ex cos x dx. v u dx = −ex cos x + ex cos x dx. ex sin x dx. We now need to study the indefinite integral ex cos x dx. Chapter 8 : Techniques of Integration page 7 of 26 First Year Calculus c W W L Chen, 1994, 2005 Writing u = ex and v = cos x, we have uv dx = Furthermore, v = sin x and u = ex . It follows that uv − Hence (3) ex cos x dx = ex sin x − ex sin x dx. v u dx = ex sin x − ex sin x dx. ex cos x dx. It looks like we are back to the same old problem. However, if we combine (2) and (3), then we obtain ex sin x dx = −ex cos x + ex sin x − so that 2 ex sin x dx = ex sin x − ex cos x = ex (sin x − cos x). ex sin x dx, Adding an arbitrary constant, which we may in view of Proposition 7C, we have ex sin x dx = 1x e (sin x − cos x) + C. 2 Example 8.2.4. Consider the indefinite integral x3 cos x dx. Writing u = x3 and v = cos x, we have uv dx = Furthermore, v = sin x and u = 3x2 . It follows that uv − Hence (4) x3 cos x dx = x3 sin x − 3 x2 sin x dx. v u dx = x3 sin x − 3 x2 sin x dx. x3 cos x dx. We now need to study the indefinite integral x2 sin x dx. Writing u = x2 and v = sin x, we have uv dx = Chapter 8 : Techniques of Integration x2 sin x dx. page 8 of 26 First Year Calculus c W W L Chen, 1994, 2005 Furthermore, v = − cos x and u = 2x. It follows that uv − Hence (5) Combining (4) and (5), we have (6) x3 cos x dx = x3 sin x + 3x2 cos x − 6 x cos x dx. x2 sin x dx = −x2 cos x + 2 x cos x dx. v u dx = −x2 cos x + 2 x cos x dx. We now need to study the indefinite integral x cos x dx. Writing u = x and v = cos x, we have uv dx = Furthermore, v = sin x and u = 1. It follows that uv − Hence (7) Combining (6) and (7), we have x3 cos x dx = x3 sin x + 3x2 cos x − 6x sin x + 6 sin x dx x cos x dx = x sin x − sin x dx. v u dx = x sin x − sin x dx. x cos x dx. = x3 sin x + 3x2 cos x − 6x sin x − 6 cos x + C. The technique is also valid for definite integrals, in view of the first Fundamental theorem of integral calculus. For definite integrals over the interval [A, B ], we have B x=B B (8) A uv dx = uv x=A − A vu dx. Equation (8) is called the formula for integration by parts for definite integrals. Example 8.2.5. Consider the definite integral π /2 x3 cos x dx. 0 Writing u = x3 and v = cos x, we have π /2 π /2 uv dx = 0 Chapter 8 : Techniques of Integration 0 x3 cos x dx. page 9 of 26 First Year Calculus c W W L Chen, 1994, 2005 Furthermore, v = sin x and u = 3x2 . It follows that x=π/2 π /2 π /2 π /2 uv x=0 − 0 vu dx = x3 sin x 0 −3 0 x2 sin x dx. Hence π /2 π /2 π /2 (9) 0 x3 cos x dx = x3 sin x 0 −3 0 x2 sin x dx = π3 −3 8 π /2 x2 sin x dx. 0 We now need to study the definite integral π /2 x2 sin x dx. 0 Writing u = x2 and v = sin x, we have π /2 π /2 uv dx = 0 0 x2 sin x dx. Furthermore, v = − cos x and u = 2x. It follows that x=π/2 π /2 π /2 π /2 uv x=0 − 0 vu dx = −x2 cos x 0 +2 0 x cos x dx. Hence π /2 π /2 π /2 π /2 (10) 0 x2 sin x dx = −x2 cos x 0 +2 0 x cos x dx = 2 0 x cos x dx. Combining (9) and (10), we have π /2 (11) 0 x3 cos x dx = π3 −6 8 π /2 x cos x dx. 0 We now need to study the definite integral π /2 x cos x dx. 0 Writing u = x and v = cos x, we have π /2 π /2 uv dx = 0 0 x cos x dx. Furthermore, v = sin x and u = 1. It follows that x=π/2 π /2 π /2 π /2 uv x=0 − 0 vu dx = x sin x 0 − 0 sin x dx. Hence π /2 π /2 π /2 (12) 0 x cos x dx = x sin x 0 − 0 sin x dx = π − 2 π /2 sin x dx. 0 page 10 of 26 Chapter 8 : Techniques of Integration First Year Calculus c W W L Chen, 1994, 2005 Combining (11) and (12), we have π /2 x3 cos x dx = 0 π3 − 3π + 6 8 π /2 sin x dx = 0 π3 − 3π + 6 − cos x 8 π /2 = 0 π3 − 3π + 6. 8 8.3. Trigonometric Integrals In this section, we consider integrals involving the six trigonometric functions sin x, cos x, tan x, cot x, sec x and csc x. If we consider differentiation formulas involving these functions, then we can divide these into three groups: (a) sin x and cos x; (b) tan x and sec x; and (c) cot x and csc x. Note that the derivative of any of these functions can be expressed in terms of the two functions in the group to which it belongs. This division is also substantiated by integral formulas. It follows that given any indefinite integral f (x) dx, where the integrand f (x) involves trigonometric functions, it may be beneficial to try first to express f (x) in terms of trigonometric functions from only one of these three groups. Example 8.3.1. Consider the indefinite integral tan x + sec3 x cot x tan x sec3 x cot x dx = + dx cos2 x cos2 x cos2 x sin x sec5 x sin x sec5 x = + dx = dx + dx. cos3 x tan x cos3 x tan x Note that we can also write sin x dx = cos3 x However, the indefinite integral sec5 x dx tan x does not appear to be so simple. Let us consider first integrals involving sin x and cos x. Consider an integral of the form sinm x cosn x dx. When m = 1, the integral is simple to evaluate. Clearly sin x cosn x dx = − and sin x cos−1 x dx = − log | cos x| + C. Chapter 8 : Techniques of Integration page 11 of 26 tan x sec2 x dx = 1 tan2 x + C. 2 1 cosn+1 x + C n+1 if n = −1, First Year Calculus c W W L Chen, 1994, 2005 When n = 1, the integral is also simple to evaluate. Clearly sinm x cos x dx = and sin−1 x cos x dx = log | sin x| + C. In the general case, we may use standard trigonometric formulas like (13) (14) (15) sin2 x + cos2 x = 1, sin 2x = 2 sin x cos x, cos 2x = cos2 x − sin2 x. 1 sinm+1 x + C m+1 if m = −1, Note also that combining (13) and (15), we have (16) cos 2x = 2 cos2 x − 1 = 1 − 2 sin2 x. Example 8.3.2. Consider the indefinite integral sin5 x dx. Using (13), we can write sin5 x = sin4 x sin x = (1 − cos2 x)2 sin x = (1 − 2 cos2 x + cos4 x) sin x, so that sin5 x dx = = (1 − 2 cos2 x + cos4 x) sin x dx sin x dx − 2 cos2 x sin x dx + cos4 x sin x dx = − cos x + 2 1 cos3 x − cos5 x + C. 3 5 Example 8.3.3. Consider the indefinite integral sin3 x cos3 x dx. Using (13), we can write sin3 x cos3 x = cos2 x sin3 x cos x = (1 − sin2 x) sin3 x cos x = sin3 x cos x − sin5 x cos x, so that sin3 x cos3 x dx = = = Chapter 8 : Techniques of Integration (sin3 x cos x − sin5 x cos x) dx sin3 x cos x dx − sin5 x cos x dx 1 1 sin4 x − sin6 x + C. 4 6 page 12 of 26 First Year Calculus c W W L Chen, 1994, 2005 Example 8.3.4. Consider the indefinite integral sin4 4x dx. Using (16), we can write sin4 4x = 1 1 (1 − cos 8x)2 = (1 − 2 cos 8x + cos2 8x) 4 4 1 1 31 1 = 1 − 2 cos 8x + (1 + cos 16x) = − cos 8x + cos 16x, 4 2 82 8 so that sin4 4x dx = 31 1 − cos 8x + cos 16x dx 82 8 3 1 1 = dx − cos 8x dx + cos 16x dx 8 2 8 3 1 1 = x− sin 8x + sin 16x + C. 8 16 128 Example 8.3.5. Consider the indefinite integral sin2 x cos4 x dx. Using (14) and (16), we can write sin2 x cos4 x = cos2 x(sin x cos x)2 = 1 1 1 (1 + cos 2x) sin2 2x = sin2 2x + cos 2x sin2 2x 8 8 8 1 1 1 1 1 = (1 − cos 4x) + cos 2x sin2 2x = − cos 4x + cos 2x sin2 2x, 16 8 16 16 8 so that sin2 x cos4 x dx = 1 1 1 − cos 4x + cos 2x sin2 2x dx 16 16 8 1 1 1 = dx − cos 4x dx + cos 2x sin2 2x dx 16 16 8 1 1 1 = x− sin 4x + sin3 2x + C. 16 64 48 Let us consider next integrals involving tan x and sec x. Consider an integral of the form tanm x secn x dx. When m = 1, the integral is simple to evaluate. Clearly tan x secn x dx = and tan x dx = − log | cos x| + C. Chapter 8 : Techniques of Integration page 13 of 26 1 secn x + C n if n = 0, First Year Calculus c W W L Chen, 1994, 2005 When n = 2, the integral is also simple to evaluate. Clearly tanm x sec2 x dx = and tan−1 x sec2 x dx = log | tan x| + C. In the general case, we may use standard trigonometric formulas like (17) 1 + tan2 x = sec2 x. 1 tanm+1 x + C m+1 if m = −1, Example 8.3.6. Consider the indefinite integral tan3 x dx. Using (17), we can write tan3 x = tan2 x tan x = (sec2 x − 1) tan x = sec2 x tan x − tan x, so that tan3 x dx = = = (sec2 x tan x − tan x) dx sec2 x tan x dx − tan x dx 1 tan2 x + log | cos x| + C. 2 Example 8.3.7. Consider the indefinite integral tan4 x dx. Using (17), we can write tan4 x = tan2 x tan2 x = (sec2 x − 1) tan2 x = sec2 x tan2 x − tan2 x = sec2 x tan2 x − sec2 x + 1, so that tan4 x dx = = = (sec2 x tan2 x − sec2 x + 1) dx sec2 x tan2 x dx − sec2 x dx + dx 1 tan3 x − tan x + x + C. 3 Chapter 8 : Techniques of Integration page 14 of 26 First Year Calculus c W W L Chen, 1994, 2005 Example 8.3.8. Consider the indefinite integral sec3 x dx. Writing u = sec x and v = sec2 x, we have uv dx = sec3 x dx. Furthermore, v = tan x and u = tan x sec x. It follows that uv − Hence (18) sec3 x dx = sec x tan x − tan2 x sec x dx. v u dx = sec x tan x − tan2 x sec x dx. We now need to study the indefinite integral tan2 x sec x dx. Using (17), we can write tan2 x sec x = (sec2 x − 1) sec x = sec3 x − sec x, so that (19) tan2 x sec x dx = (sec3 x − sec x) dx = sec3 x dx − sec x dx. Combining (18) and (19), we have sec3 x dx = sec x tan x − so that sec3 x dx = 1 1 sec x tan x + 2 2 sec x dx = 1 1 sec x tan x + log | sec x + tan x| + C. 2 2 sec3 x dx + sec x dx, Example 8.3.9. Consider the indefinite integral tan2 x sec3 x dx. Writing u = tan2 sec x and v = sec2 x, we have uv dx = tan2 x sec3 x dx. Furthermore, v = tan x and u = 2 tan x sec3 x + tan3 x sec x. It follows that uv − v u dx = tan3 x sec x − (2 tan2 x sec3 x + tan4 x sec x) dx. page 15 of 26 Chapter 8 : Techniques of Integration First Year Calculus c W W L Chen, 1994, 2005 Hence (20) tan2 x sec3 x dx = tan3 x sec x − = tan3 x sec x − 2 We now need to study the indefinite integral tan4 x sec x dx. Using (17), we can write (the reader should check this) tan4 x sec x = tan2 x sec3 x − sec3 x + sec x, so that (21) tan4 x sec x dx = tan2 x sec3 x dx − sec3 x dx + sec x dx. (2 tan2 x sec3 x + tan4 x sec x) dx tan2 x sec3 x dx − tan4 x sec x dx. Combining (20) and (21), we have tan2 x sec3 x dx = 1 tan3 x sec x + 4 1 = tan3 x sec x + 4 1 1 sec3 x dx − sec x dx 4 4 1 1 tan x sec x − log | tan x + sec x| + C. 8 8 Occasionally, it may be necessary to convert an expression involving tan x and sec x to one involving sin x and cos x instead. Example 8.3.10. Consider the indefinite integral tan4 7x dx. sec5 7x Here the identity (17) does not help very much. However, we have tan4 7x = sin4 7x cos 7x, sec5 7x so that tan4 7x dx = sec5 7x sin4 7x cos 7x dx = 1 sin5 7x + C. 35 Let us consider finally integrals involving cot x and csc x. Consider an integral of the form cotm x cscn x dx. When m = 1, the integral is simple to evaluate. Clearly cot x cscn x dx = − Chapter 8 : Techniques of Integration 1 cscn x + C n if n = 0, page 16 of 26 First Year Calculus c W W L Chen, 1994, 2005 and cot x dx = log | sin x| + C. When n = 2, the integral is also simple to evaluate. Clearly cotm x csc2 x dx = − and cot−1 x csc2 x dx = − log | cot x| + C. The details are similar to the case of tan x and sec x. 1 cotm+1 x + C m+1 if m = −1, 8.4. Trigonometric Substitutions In this section, we shall consider techniques to handle integrals involving square roots of the form √ √ √ a2 − b2 x2 , a2 + b2 x2 or b2 x2 − a2 . Without loss of generality, assume that a, b > 0. Let us consider first the case √ a2 − b2 x2 . If we use the substitution x= then a2 − b2 x2 = while dx = a cos θ dθ. b a2 (1 − sin2 θ) = √ a2 cos2 θ = a| cos θ|, a sin θ, b Example 8.4.1. Consider the indefinite integral √ If we use the substitution x = 2 sin θ, then 4 − x2 = 2| cos θ| so that √ Suppose that cos θ > 0. Then √ Chapter 8 : Techniques of Integration 1 dx. 4 − x2 and dx = 2 cos θ dθ, 1 dx = 4 − x2 1 2 cos θ dθ. 2| cos θ| 1 dx = 4 − x2 dθ = θ + C = sin−1 x + C. 2 page 17 of 26 First Year Calculus c W W L Chen, 1994, 2005 Example 8.4.2. Consider the indefinite integral x3 If we use the substitution x = 3 2 9 − 4x2 dx. sin θ, then and dx = 3 cos θ dθ. 2 9 − 4x2 = 3| cos θ| Suppose that cos θ > 0. Then x3 9 − 4x2 dx = 243 sin3 θ cos2 θ dθ 16 243 = (1 − cos2 θ) sin θ cos2 θ dθ 16 243 243 = sin θ cos2 θ dθ − sin θ cos4 θ dθ 16 16 81 243 = − cos3 θ + cos5 θ + C. 16 80 Next, note that cos2 θ = 1 − sin2 θ = 1 − 4 x2 , so that 9 x3 9 − 4x2 dx = − √ 81 16 4 1 − x2 9 3 /2 + 243 80 4 1 − x2 9 5 /2 + C. Let us consider next the case a2 + b2 x2 . If we use the substitution x= a tan θ, b √ then a2 + b2 x2 = while dx = a sec2 θ dθ. b a2 (1 + tan2 θ) = a2 sec2 θ = a| sec θ|, Example 8.4.3. Consider the indefinite integral x2 If we use the substitution x = tan θ, then 1 + x2 = | sec θ| Suppose that sec θ > 0. Then x2 We have shown earlier that tan2 θ sec3 θ dθ = Chapter 8 : Techniques of Integration 1 + x2 dx. and dx = sec2 θ dθ. 1 + x2 dx = tan2 θ sec3 θ dθ. 1 1 1 tan3 θ sec θ + tan θ sec θ − log | tan θ + sec θ| + C. 4 8 8 page 18 of 26 First Year Calculus c W W L Chen, 1994, 2005 Next, note that sec2 θ = 1 + tan2 θ = 1 + x2 , so that x2 1 + x2 dx = 13 1 1 x (1 + x2 )1/2 + x(1 + x2 )1/2 − log |x + (1 + x2 )1/2 | + C. 4 8 8 √ b2 x2 − a2 . If we use the substitution x= then b2 x2 − a2 = while dx = a tan θ sec θ dθ. b a2 (sec2 θ − 1) = a2 tan2 θ = a| tan θ|, a sec θ, b Let us consider finally the case Example 8.4.4. Consider the indefinite integral √ x2 − 4 dx. x If we use the substitution x = 2 sec θ, then x2 − 4 = 2| tan θ| Suppose that tan θ > 0. Then √ x2 − 4 dx = 2 x tan2 θ dθ = 2 (sec2 θ − 1) dθ = 2 sec2 θ dθ − 2 dθ = 2 tan θ − 2θ + C. and dx = 2 tan θ sec θ dθ. Next, note that tan2 θ = sec2 θ − 1 = so that √ x2 − 4 dx = 2 x 12 x −1 4 1 /2 12 x −1 4 and θ = sec−1 x , 2 − 2 sec−1 x +C = 2 x2 − 4 − 2 sec−1 x + C. 2 8.5. Completing Squares In this section, we shall consider techniques to handle integrals involving square roots of the form αx2 + βx + γ , where α = 0. Our task is to show that such integrals can be reduced to integrals discussed in the previous section. Note that αx2 + βx + γ = α x2 + γ β x+ α α = α x2 + β2 β x+ 2 α 4α + γ− b2 4α =α x+ β 2α 2 + γ− β2 4α . Chapter 8 : Techniques of Integration page 19 of 26 First Year Calculus c W W L Chen, 1994, 2005 Suppose first of all that we use a substitution y =x+ Then dy = dx and αx2 + βx + γ = αy 2 + δ, where δ=γ− It now follows that αx2 + βx + γ is of the form a2 − b2 y 2 a2 + b2 y 2 b2 y 2 − a2 if α < 0 and δ > 0, if α > 0 and δ > 0, if α > 0 and δ < 0. β2 . 4α β . 2α Example 8.5.1. Consider the indefinite integral √ We have 3 − 2x − x2 = −(x2 + 2x − 3) = −(x2 + 2x + 1) + 4 = −(x + 1)2 + 4 = −y 2 + 4, where we use the substitution y = x + 1. Note that α = −1 < 0 and δ = 4 > 0. Then √ We have shown earlier that 1 4 − y2 It follows that √ 1 dx = sin−1 3 − 2x − x2 x+1 2 + C. dy = sin−1 y + C. 2 1 dx = 3 − 2x − x2 1 4 − y2 dx. 1 dx. 3 − 2x − x2 Example 8.5.2. Consider the indefinite integral √ x2 − 4x dx. x−2 We have x2 − 4x = (x2 − 4x + 4) − 4 = (x − 2)2 − 4 = y 2 − 4, where we use the substitution y = x − 2. Note that α = 1 > 0 and δ = −4 < 0. Then √ x2 − 4x y2 − 4 dx = dy. x−2 y Chapter 8 : Techniques of Integration page 20 of 26 First Year Calculus c W W L Chen, 1994, 2005 We have shown earlier that y2 − 4 dy = y It follows that √ x2 − 4x dx = x−2 y 2 − 4 − 2 sec−1 y + C. 2 (x − 2)2 − 4 − 2 sec−1 x−2 2 +C = x2 − 4x − 2 sec−1 x−2 2 + C. 8.6. Partial Fractions In this section, we shall consider techniques to handle integrals of the form p(x) dx, q (x) where p(x) and q (x) are polynomials in x. If the degree of p(x) is not smaller than the degree of q (x), then we can always find polynomials a(x) and r(x) such that p(x) r(x) = a(x) + , q (x) q (x) where r(x) = 0 or has degree smaller than the degree of q (x). Example 8.6.1. Consider the indefinite integral x5 + 2x4 + 4x3 + x + 1 dx. x2 + x + 1 Note that x5 + 2x4 + 4x3 + x + 1 2x + 4 = (x3 + x2 + 2x − 3) + 2 , 2+x+1 x x +x+1 so that x5 + 2x4 + 4x3 + x + 1 dx = x2 + x + 1 (x3 + x2 + 2x − 3) dx + 2x + 4 dx. x2 + x + 1 It does not take a genius to work out the indefinite integral (x3 + x2 + 2x − 3) dx. We can therefore restrict our attention to the case when the polynomial p(x) is of lower degree than the polynomial q (x). The first step is to factorize the polynomial q (x) into a product of irreducible factors. It is a fundamental result in algebra that a real polynomial q (x) can be factorized into a product of irreducible linear factors and quadratic factors with real coefficients. Chapter 8 : Techniques of Integration page 21 of 26 First Year Calculus c W W L Chen, 1994, 2005 Example 8.6.2. Suppose that q (x) = x4 − 4x3 + 5x2 − 4x + 4. Then q (x) can be factorized into a product of irreducible linear factors in the form (x − 2)2 (x2 + 1). Suppose that a linear factor (ax + b) occurs n times in the factorization of q (x). Then we write down a decomposition A1 An A2 + ... + , + 2 ax + b (ax + b) (ax + b)n where the constants A1 , . . . , An will be determined later. Suppose that a quadratic factor (ax2 + bx + c) occurs n times in the factorization of q (x). Then we write down a decomposition A1 x + B1 An x + Bn A2 x + B2 + ... + , + 2 + bx + c 2 + bx + c)2 ax (ax (ax2 + bx + c)n where the constants A1 , . . . , An and B1 , . . . , Bn will be determined later. We proceed to add all the decompositions and equate their sum to p(x) , q (x) and then calculate all the constants by equating coefficients. Example 8.6.3. Suppose that p(x) 2x3 − 11x2 + 17x − 16 2x3 − 11x2 + 17x − 16 =4 = . q (x) x − 4x3 + 5x2 − 4x + 4 (x − 2)2 (x2 + 1) We now write 2x3 − 11x2 + 17x − 16 c1 c2 c3 x + c4 = + . +2 2 (x2 + 1) 2 (x − 2) x − 2 (x − 2) x +1 Now c1 c2 c1 (x − 2)(x2 + 1) + c2 (x2 + 1) + (c3 x + c4 )(x − 2)2 c3 x + c4 + = , +2 2 x − 2 (x − 2) x +1 (x − 2)2 (x2 + 1) so that c1 (x − 2)(x2 + 1) + c2 (x2 + 1) + (c3 x + c4 )(x − 2)2 = 2x3 − 11x2 + 17x − 16. Note now that c1 (x − 2)(x2 + 1) + c2 (x2 + 1) + (c3 x + c4 )(x − 2)2 = c1 (x3 − 2x2 + x − 2) + c2 (x2 + 1) + c3 (x3 − 4x2 + 4x) + c4 (x2 − 4x + 4) = (c1 + c3 )x3 + (−2c1 + c2 − 4c3 + c4 )x2 + (c1 + 4c3 − 4c4 )x + (−2c1 + c2 + 4c4 ). Equating coefficients, we have c1 + c3 = 2, −2c1 + c2 − 4c3 + c4 = −11, c1 + 4c3 − 4c4 = 17, −2c1 + c2 + 4c4 = −16. Chapter 8 : Techniques of Integration page 22 of 26 First Year Calculus c W W L Chen, 1994, 2005 This system has solution c1 = 1, c2 = −2, c3 = 1 and c4 = −3. Hence 2x3 − 11x2 + 17x − 16 x−3 1 2 +2 = − , 4 − 4x3 + 5x2 − 4x + 4 2 x x − 2 (x − 2) x +1 so that 2x3 − 11x2 + 17x − 16 dx = x4 − 4x3 + 5x2 − 4x + 4 1 dx − x−2 2 dx + (x − 2)2 x−3 dx. x2 + 1 We shall calculate the three indefinite integrals on the right hand side later. To calculate the indefinite integrals that arise, note that these indefinite integrals are of the form (22) or (23) Ax + B dx, + bx + c)k A dx, (ax + b)k (ax2 where A and B are constants and k is a positive integer. The integral (22) is simple. If k = 1, then we have A A dx = − + C. (ax + b)k (k − 1)a(ax + b)k−1 On the other hand, we have A A dx = log |ax + b| + C. ax + b a The integral (23) is a bit more complicated. Note that Ax + B A dx = (ax2 + bx + c)k 2a Ab 2ax + b dx + B − (ax2 + bx + c)k 2a 1 dx. (ax2 + bx + c)k The first integral on the right hand side is simple. If k = 1, then we have 2ax + b 1 dx = − + C. (ax2 + bx + c)k (k − 1)(ax2 + bx + c)k−1 On the other hand, we have 2ax + b dx = log |ax2 + bx + c| + C. ax2 + bx + c It remains to study the integral 1 dx. (ax2 + bx + c)k To do this, we may try the technique of completing squares as described in the previous section, and then use a trigonometric substitution to find the integral. Chapter 8 : Techniques of Integration page 23 of 26 First Year Calculus c W W L Chen, 1994, 2005 Example 8.6.4. Let us continue the discussion of Example 8.6.3. Note that 1 dx = log |x − 2| + C x−2 On the other hand, we have x−3 1 dx = x2 + 1 2 Clearly 2x dx = log |x2 + 1| + C. x2 + 1 Using the substitution x = tan θ, we have 1 dx = +1 sec2 θ dθ = 1 + tan2 θ dθ = θ + C = tan−1 x + C. 2x dx − 3 x2 + 1 1 dx. x2 + 1 and 1 1 dx = − + C. (x − 2)2 x−2 x2 It follows that 2x3 − 11x2 + 17x − 16 dx = x4 − 4x3 + 5x2 − 4x + 4 x−3 2 dx + dx (x − 2)2 x2 + 1 2 1 = log |x − 2| + + log |x2 + 1| − 3 tan−1 x + C. x−2 2 1 dx − x−2 Example 8.6.5. Consider the indefinite integral x2 + x − 3 dx. x3 − 2x2 − x + 2 Note first of all that x3 − 2x2 − x + 2 = (x − 2)(x + 1)(x − 1), so we consider partial fractions of the form x2 + x − 3 c1 c2 c3 = + + (x − 2)(x + 1)(x − 1) x−2 x+1 x−1 c1 (x + 1)(x − 1) + c2 (x − 2)(x − 1) + c3 (x − 2)(x + 1) = . (x − 2)(x + 1)(x − 1) It follows that (24) c1 (x + 1)(x − 1) + c2 (x − 2)(x − 1) + c3 (x − 2)(x + 1) = x2 + x − 3. We may equate coefficients and solve for c1 , c2 , c3 . Alternatively, substituting x = 2, −1, 1 into equation (24), we get respectively 3c1 = 3, 6c2 = −3 and −2c3 = −1, so that c1 = 1, c2 = −1/2 and c3 = 1/2. Hence x2 + x − 3 dx = x3 − 2x2 − x + 2 1 1 1 1 1 dx − dx + dx x−2 2 x+1 2 x−1 1 1 = log |x − 2| − log |x + 1| + log |x − 1| + C. 2 2 page 24 of 26 Chapter 8 : Techniques of Integration First Year Calculus c W W L Chen, 1994, 2005 Example 8.6.6. Consider the indefinite integral x6 − 2 dx. x4 + x2 Note that x6 − 2 x2 − 2 = x2 − 1 + 4 , 4 + x2 x x + x2 so that (25) x6 − 2 dx = x4 + x2 (x2 − 1) dx + x2 − 2 1 dx = x3 − x + x4 + x2 3 x2 − 2 dx. x4 + x2 Next, we study the integral x2 − 2 dx. x4 + x2 Note first of all that x4 + x2 = x2 (x2 + 1), so we consider partial fractions of the form x2 − 2 c3 x + c4 c1 c2 c1 x(x2 + 1) + c2 (x2 + 1) + (c3 x + c4 )x2 = + 2+ 2 = . x2 (x2 + 1) x x x +1 x2 (x2 + 1) It follows that c1 x(x2 + 1) + c2 (x2 + 1) + (c3 x + c4 )x2 = x2 − 2. Equating coefficients, we have c1 c2 c1 c2 + c3 = + c4 = = 0, 1, 0, = −2. This system has solution c1 = 0, c2 = −2, c3 = 0 and c4 = 3. Hence (26) x2 − 2 dx = −2 x4 + x2 1 dx + 3 x2 1 2 dx = + 3 tan−1 x + C. +1 x x2 Combining (25) and (26), we obtain x6 − 2 1 2 dx = x3 − x + + 3 tan−1 x + C. 4 + x2 x 3 x Chapter 8 : Techniques of Integration page 25 of 26 First Year Calculus c W W L Chen, 1994, 2005 Problems for Chapter 8 1. Evaluate each of the following indefinite integrals: a) d) g) j) m) p) s) v) y) bb) ee) sin x cos 7x dx cos 2x dx 1 − sin 2x 1 dx x2 + 4x − 4 x2 + 3x − 1 dx 4 + x3 + x2 + x x 1 dx x2 − 5x + 4 √ x4 + x x + 1 dx x e4x+2 dx (log x)5 dx x 1 √ dx a2 − x2 x5 ex dx x−4 dx (x2 + 4)(x + 1) b) e) h) k) n) q) t) w) z) cc) e2x cos 3x dx 1 dx 16 − 3x + x2 x2 dx 3 + 3x2 + 3x + 1 x √ log(x6 ) dx e2x cos x dx √ x2 x − 1 dx xex dx 2x + 3 dx x2 + 3x − 4 √ ( x + 1)10 √ dx x 1 dx 2 − 4x + 3 x 2 c) f) i) l) o) r) u) x) aa) dd) x2 log x dx x sec2 x dx √ cot x csc4 x dx sin2 3x dx (x3 + √ x) dx x x2 + 4 dx log x dx x sin−1 x √ dx 1 − x2 1 dx x2 + a2 xex dx 2. Evaluate each of the following definite integrals: √ 3 4 x+1 √ a) x(1 + 2x2 )4 dx b) x 2 1 π /4 2 dx c) 1 4 x2 + 1 dx (x + 1)4 ex √ dx x x cos 2x dx √ d) 0 π /4 cos x dx (1 + sin x)2 cos2 2x dx 1 e) 0 x π /2 x2 √ + 1 dx f) 1 π /4 g) 0 1 /2 h) 0 2 − 2 cos x dx i) 0 j) 0 √ x dx 1 − x2 Chapter 8 : Techniques of Integration page 26 of 26 ...
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This note was uploaded on 02/01/2009 for the course MATH 3412341 taught by Professor Staff during the Spring '06 term at UCSD.

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