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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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