Integration techniques

751 views

Published on

Published in: Education
0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
751
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
15
Comments
0
Likes
1
Embeds 0
No embeds

No notes for slide

Integration techniques

  1. 1. INTEGRATION TECHNIQUES 1. Review of formulas and techniques Summary of known integral formulas. ˆ xr+1 x dx = + c, for r = −1 (power rule) r+1 ˆ r ˆ ˆ sin x dx = − cos x + c cos x dx = sin x + c ˆ ˆ 2 sec x dx = tan x + c sec x tan x dx = sec x + c ˆ ˆ 2 csc x dx = − cot x + c csc x cot x dx = − csc x + c ˆ ˆ x e−x dx = −e−x + c x e dx = e + c ˆ ˆ √ tan x dx = − ln | cos x| + c ˆ 1 dx = ln |x| + c for x = 0 x ˆ 1 dx = tan−1 x + c 2 1+x 1 dx = sin−1 x + c 2 1−x 1 √ dx = sec−1 x + c 2−1 |x| x Example 1.1. (A Simple Substitution) ˆ Evaluate sin(ax)dx, for a = 0. Example 1.2. (Generalizing a Basic Integration Rule) ˆ 1 Evaluate dx, for a = 0. 2 + x2 a Example 1.3. (An Integrand That Must Be Expanded) ˆ Evaluate (x2 − 5)2 dx. Example 1.4. (An Integral Where We Must Complete the Square) ˆ 1 √ dx. Evaluate −5 + 6x − x2 Example 1.5. (An Integral Requiring Some Imagination) ˆ 4x + 1 Evaluate dx. 2 + 4x + 10 2x 1
  2. 2. 2 INTEGRATION TECHNIQUES 2. Integration by Parts Remember that if u and v are differentiable functions of x, then u · v is also differentiable and d d d (u · v) = u · v + v · u. dx dx dx Using differential forms this may be written as d(uv) = udv + vdu. Taking the integral of the both sides of (1), we obtain ˆ ˆ ˆ d(uv) = udv + vdu. Note that ´ (1) (2) d(uv) = uv. Therefore (2) can be written as ˆ ˆ uv = udv + vdu, or, equivalently, ˆ ˆ udv = uv − vdu. Remark. This technique requires the separation of the integral into two parts as u and dv (which includes the differential of integral, say dx) such that • u gets simpler when differentiated. • dv is easy to integrate. Example 2.1. Evaluate the following integrals. ˆ (1) x sin x dx ˆ (2) x2 ex dx ˆ (3) ex cos x dx ˆ (4) x ln x dx ˆ (5) ln x dx ˆ (6) sin−1 x dx ˆ 2 (7) x3 ex dx ˆ (8) x sec2 x dx ˆ (9) x3x dx ˆ (10) cos(ln x) dx ˆ (11) x4 ex dx
  3. 3. INTEGRATION TECHNIQUES ˆ (12) (13) (14) (15) (16) (17) (18) (19) ˆ ˆ ˆ ˆ ˆ ˆ ˆ 3 √ ln x dx √ e x dx x2 sin x dx xe2x dx xe−x dx e2x cos 3x dx tan−1 x dx 2 ˆ x5 e−x dx −1 esin x dx ˆ √ (21) x 1 − x dx ˆ 2 (22) x3 ln x dx (20) 1 3. Trigonometric Integrals 3.1. Products of Powers of Sines and Cosines. If we are given an integral of the form ˆ sinm x cosn x dx, where m, n ≥ 0, There are three possible cases: Case 1. If m is odd, we write m as 2k + 1 and use the identity sin2 x = 1 − cos2 x to obtain sinm x = sin2k+1 x = (sin2 x)k sin x = (1 − cos2 x)k sin x, and, then ˆ ˆ m n sin x cos x dx = (1 − cos2 x)k cosn x sin x dx. Using the substitution u = cos x, we get ˆ ˆ ˆ m n 2 k n sin x cos x dx = (1 − cos x) cos x sin x dx = − (1 − u2 )k un du. Now, the integral is easy to evaluate. Case 2. If m is even and n is odd, we write n as 2k + 1 and use the identity cos2 x = 1 − sin2 x and repeat what we have done in Case 1 changing the places of sine and cosine. Case 3. If both m and n are even, we substitute 1 − cos 2x 1 + cos 2x , cos2 x = 2 2 to reduce the integrand to one in lower powers of cos 2x. sin2 x =
  4. 4. 4 INTEGRATION TECHNIQUES 3.2. Integrals of Powers of tan x and sec x. We know how to integrate the tangent and secant and their squares. To integrate higher powers we use the identities tan2 x = sec2 x − 1 and sec2 x = tan2 x + 1, and integrate by parts when necessary to reduce the higher powers to lower powers. 3.3. Products of Sines and Cosines. To evaluate the integrals of the form ˆ ˆ ˆ sin mx sin nx dx, sin mx cos nx dx, and cos mx cos nx dx, we may apply the method of integration by parts, but two such integrations are required in each case. So, it is better to use the identities 1 sin mx sin nx = (cos(m − n)x − cos(m + n)x) 2 1 sin mx cos nx = (sin(m − n)x + cos(m + n)x) 2 1 cos mx cos nx = (cos(m − n)x + cos(m + n)x) 2 Example 3.1. Evaluate the following integrals. ˆ (1) cos4 x sin x dx ˆ (2) cos4 x sin3 x dx ˆ √ (3) sin x cos5 x dx ˆ (4) cos3 x dx ˆ (5) sin3 x cos7 x dx ˆ (6) sin2 x dx ˆ (7) cos4 x dx ˆ (8) sin2 x cos2 x dx ˆ (9) tan3 x sec3 x dx ˆ (10) tan2 x dx ˆ (11) tan2 x sec4 x dx ˆ (12) tan3 x dx ˆ (13) tan3 x sec5 x dx
  5. 5. INTEGRATION TECHNIQUES 5 ˆ (14) sec x dx ˆ √ tan x sec4 x dx (15) ˆ (16) cot4 x dx ˆ (17) csc x dx ˆ (18) cot3 x csc3 x dx ˆ sin5 x √ (19) dx cos x 3.4. Trigonometric Substitutions. Trigonometric substitutions can be effective in transforming complicated integrals involving a2 − x2 , a2 + x2 and x2 − a2 into integrals we can evaluate directly. The following table gives the suitable trigonometric substitution in each case and the trigonometric identity to be used. Expression Trigonometric Substitution Related Trigonometric Identity a2 + x 2 x = a tan θ 1 + tan2 θ = sec2 θ 2 2 a −x x = a sin θ 1 − sin2 θ = cos2 θ x 2 − a2 x = a sec θ sec2 θ − 1 = tan2 θ Remark. We want any substitution we use in an integration to be reversible so that we can change back to the original variable afterward. For example, if x = a tan θ, we want to be able x after the integration takes place. As we know, the functions in these to set θ = tan−1 a substitution have inverses only for selected values of θ. For reversibility, x π π x = a tan θ requires θ = tan−1 with − < θ < , a 2 2 π π ≤θ≤ , 2 2   0 ≤ θ < π if x ≥ 1,   2 a −1 x x = a sec θ requires θ = sec with  a  π < θ ≤ π if x ≤ −1,  2 a To simplify calculations with the substitution x = a sec θ, we will restrict its use to integrals in √ x π which ≥ 1. This will place θ in 0, and make tan θ ≥ 0. We will then have x2 − a2 = a 2 √ a2 tan2 θ = |a tan θ| = a tan θ, free of absolute values, provided that a > 0. x = a sin θ requires θ = sin−1 x a with − Example 3.2. (1) Evaluate the following integrals. ˆ dx √ (a) 2 2 ˆ x 4−x dx √ (b) 2 ˆ √9 2+ x x − 25 (c) dx, for x ≥ 5. x
  6. 6. 6 INTEGRATION TECHNIQUES ˆ (d) ˆ (e) ˆ (f) ˆ (g) ˆ (h) (i) (j) ˆ ˆ ˆ dx (1 + x2 )2 √ 4 − x2 dx √ x2 − 1 dx x dx √ 2 x2 − 2 x √ 9x2 − 4 dx x dx 2 + 2x + 10 x √ ex 1 − e2x dx √ cos x dx 1 + sin2 x dx (l) 2 + 4x + 5)2 ˆ (4x 2x + 2 (m) dx 2 − 4x + 8 ˆ x 5x + 3 dx (n) 2 − 4x + 8 ˆ x x2 (o) 5 dx (x2 − 1) 2 (2) Find the volume of the solid generated by revolving about the x-axis the region in the 2 first quadrant enclosed by the coordinate axes, the curve y = , and the line x = 1. 1 + x2 (k) ˆ 4. Integration of Rational Functions Using Partial Fractions The method based on the fact that “every rational function can be written as a sum of simpler fractions that we can integrate with the techniques we already know”. General Description of the Method of Partial Fractions. To write a rational function P (x) as a sum of partial fractions, do the following: Q(x) (1) The degree of P (x) must be less than the degree of Q(x). (Otherwise, divide P (x) by Q(x) and work on the remainder.) (2) Factor out the polynomial Q(x). • For each linear factor, assign the sum of m partial fractions to this factor where m is the exponent of that linear factor as follows: A1 A2 Am + + ··· + 2 x − r (x − r) (x − r)m • For each irreducible quadratic factor (a polynomial of second degree that cannot be written as a product of linear factors), assign the sum of n partial fractions to
  7. 7. INTEGRATION TECHNIQUES this factor where n is the exponent of that quadratic factor as: B1 x + C1 B2 x + C2 Bn x + Cn + 2 + ··· + 2 2 + px + q 2 x (x + px + q) (x + px + q)n Several Examples. We assume that the degree of P (x) is less than the degree of Q(x). (1) Distinct Linear Factors: P (x) A B C P (x) = = + + . Q(x) (x − r1 )(x − r2 )(x − r3 ) x − r1 x − r2 x − r3 (2) Repeated Linear Factors: P (x) P (x) B A C D = + = + + . 4 2 3 Q(x) (x − r) x − r (x − r) (x − r) (x − r)4 (3) Some distinct, some repeated linear factors: P (x) A B C D E F P (x) + + = 2 = + 2+ + . 3 2 Q(x) x (x − r1 )(x − r2 ) x x x − r1 x − r2 (x − r2 ) (x − r3 )3 (4) Distinct irreducible quadratic factors: P (x) P (x) Ax + B Cx + D = 2 = 2 + 2 . 2+p x+q ) Q(x) (x + p1 x + q1 )(x x + p1 x + q 1 x + p2 x + q2 2 2 (5) Repeated irreducible quadratic factors: P (x) P (x) Ax + B Cx + D Ex + F = 2 = 2 + 2 + 2 . 3 2 Q(x) (x + p1 x + q1 ) x + p1 x + q1 (x + p1 x + q1 ) (x + p1 x + q1 )3 (6) General Case: P (x) P (x) = 2 (x2 + p x + q )2 (x2 + p x + q ) Q(x) x(x − r) 1 1 2 2 A B C Dx + E Fx + G Hx + I = + + + 2 + 2 + 2 . 2 2 x x − r (x − r) x + p1 x + q1 (x + p1 x + q1 ) x + p2 x + q 2 Example 4.1. Evaluate the following integrals. ˆ 1 (1) dx 2+x−2 ˆ x 2 3x − 7x − 2 (2) dx x3 − x ˆ 2x3 − 4x2 − 15x + 5 (3) dx x2 − 2x − 8 ˆ 5x2 + 20x + 6 (4) dx 3 2 ˆ x 2 + 2x + x 2x − 5x + 2 (5) dx x3 + x ˆ 5x2 + 6x + 2 dx (6) 2 ˆ (x + 2)(x + 2x + 5) 3x − 1 (7) dx (x + 2)(x − 3) 7
  8. 8. 8 INTEGRATION TECHNIQUES ˆ (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18) (19) (20) (21) (22) (23) (24) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ x2 + 1 dx x(x − 1)(x − 2)(x − 3) x dx (x − 3)2 x−1 dx (x + 1)3 x2 + 1 dx (x − 1)2 (x + 1) x+1 dx 2 + 1)(x − 2) (x 6x2 − 15x + 22 dx (x + 3)(x2 + 2)2 2x + 1 dx 3x + 1 x5 − x4 − 3x + 5 dx x4 − 2x3 + 2x2 − 2x + 1 x3 + 1 dx x2 − x sec x dx x3 dx x2 + 6x + 5 x4 dx x4 − 1 dx 3−1 x dx x − e2x e sin x dx cos x + cos 2x sin4 x dx cos3 x dx 3 2 2 ˆ x(1 − x ) 3 2x + 5x2 + 8x + 4 (25) dx (x2 + 2x + 2)2
  9. 9. INTEGRATION TECHNIQUES 9 Brief Summary of Integration Techniques. ˆ ˆ • Integration by Substitution: f (u(x))u (x) dx = f (u) du What to look for – Compositions of the form f (u(x)), where the integrand also contains u (x); for example, ˆ ˆ ˆ 2 2 2x cos(x ) dx = cos(x )2x dx = cos udu. – Compositions of the form f (ax + b); for example, ˆ ˆ x u−1 √ √ du. dx = u x+1 ´ ´ • Integration by Parts: u dv = uv − v du What to look for: products of different types of functions: xn , cos x, ex ; for example, ˆ ˆ 2x cos x dx = x sin x − sin x dx. • Trigonometric Substitutions: What to look √ for: – Terms like a2 − x2 : Let x = a sin θ, − π ≤ θ ≤ π , so that dx = a cos θ dθ and 2 2 √ a2 − x2 = a2 − a2 sin2 θ = a cos θ; for example, ˆ ˆ x2 √ dx = sin2 θ dθ. 2 1−x √ – Terms like x2 + a2 : Let x = a tan θ, − π < θ < π , so that dx = a sec2 θ dθ and 2 2 √ √ x2 + a2 = a2 tan2 θ + a2 = a sec θ; for example, ˆ ˆ x3 √ dx = 27 tan3 θ sec θ dθ. 2+9 x √ 2 − a2 : Let x = a sec θ, for θ ∈ [0, π ) ∪ ( π , π], so that dx = – Terms like x 2 2 √ √ a sec θ tan θ dθ and x2 − a2 = a2 sec2 θ − a2 = a tan θ; for example, ˆ ˆ √ 3 2 − 4 dx = 32 x x sec4 θ tan2 θ dθ. • Partial Fractions: What to look for: rational functions, for example, ˆ ˆ ˆ x+2 x+2 A B dx = dx = + 2 − 4x + 3 x (x − 1)(x − 3) x−1 x−3 dx.
  10. 10. 10 INTEGRATION TECHNIQUES 5. Improper Integrals ˆ 2 Example 5.1. Evaluate −1 1 dx. x2 When studying the definite integrals, we required two things. First, the domain of integration (from a to b) [a, b], be finite. Second, function is finite on the domain of integration. Question: What happens if the domain of integration is infinite? What if function becomes infinite in the domain of integration? Answer: Improper integration! The integrals of type described above are called improper integrals. If the limits exist, we evaluate them with the following definitions (1) If f is continuous on [a, ∞), then ˆ ∞ ˆ b f (x) dx = lim f (x) dx. b→∞ a a (2) If f is continuous on (−∞, b], then ˆ b ˆ b f (x) dx = lim f (x) dx. a→−∞ −∞ a (3) If f is continuous on [a, b) then ˆ b ˆ c f (x) dx = lim f (x) dx. − c→b a a (4) If f is continuous on (a, b] then ˆ b ˆ b f (x) dx = lim f (x) dx. + c→a a c In each case, if the limit exists and is finite we say that the improper integral converges and the limit is the value of the improper integral. Otherwise the improper integral diverges. Similarly, if f becomes infinite at an interior point d ∈ [a, b], then ˆ b ˆ d ˆ b f (x) dx = f (x) dx + f (x) dx. a a d This integral (on [a, b]) converges if both integrals (on [a, d] and on [d, b]) converges. Otherwise, the integral from a to b diverges. ˆ ˆ ∞ a Finally, if f is continuous on (−∞, ∞) and if ˆ ∞ f (x) dx converges and its value is −∞ ˆ ˆ ∞ −∞ f (x) dx + −∞ f (x) dx both converge, then a ˆ a f (x) dx = −∞ f (x) dx and ∞ f (x) dx. a If either one or both of the integrals on the right-hand side of this equation diverge, the integral diverges. Example 5.2. In each part, determine whether the improper integral converges or diverges, and find its value if it converges.
  11. 11. INTEGRATION TECHNIQUES ˆ (1) (2) (3) (4) (5) (6) (7) 1 dx √ . 1−x 0 ˆ 0 1 dx. 2 −1 x ˆ 1 1 √ dx. ˆ0 2 x 1 dx. 1 x−1 ˆ 2 1 dx. 2 ˆ−1 x ∞ 1 dx. 2 ˆ1 ∞ x 1 √ dx. x ˆ1 ∞ (8) (9) e−x dx. ˆ1 ∞ ˆ0 sin x dx. ∞ xe−x dx. (10) ˆ0 1 (11) ˆ−1 2 (12) (13) (14) (15) (16) (17) (18) 1 (20) dx. 1 √ dx. 2 ˆ0 3 4 − x dx 2 . 0 (x − 1) 3 ˆ 1 dx . x ˆ−1 1 dx . p 0 x ˆ −1 1 dx. −∞ x ˆ 0 1 dx. 2 ˆ−∞ (x − 1) ∞ 1 dx. 2 ˆ−∞ 1 + x ∞ (19) 1 x3 ˆ−∞ ∞ ˆ−∞ ∞ (21) 0 2 xe−x dx. e−x dx. 1 dx. (x − 1)2 11
  12. 12. 12 INTEGRATION TECHNIQUES ˆ (22) (23) ∞ ˆ1 ∞ ˆ−∞ ∞ 1 dx. xp e−|x| dx. cos x dx. (24) −∞ Exercises 5.3. In each part, determine whether the improper integral converges or diverges, and find its value if it converges. ˆ 1 dx (1) . 2 −1 x ˆ 16 dx √ . (2) 4 x ˆ0 1 dx √ (3) . 2 ˆ0 ∞ 1 − x ln x dx. (4) ˆ1 ∞ x dx (5) . 1.001 ˆ1 1 x dx (6) 1 . −8 x 3 ˆ 1 dx (7) . 0.999 ˆ0 ∞x x dx (8) 3 . 2 −∞ (x + 4) 2 ˆ ∞ 16 tan−1 x dx. (9) 2 ˆ0 ∞ 1 + x 2e−θ sin θ dθ. (10) ˆ0 1 (11) − ln x dx. ˆ0 ∞ dx (12) . 2 ˆ0 1 (x + 1)(x + 1) (13) −x ln |x| dx. ˆ−1 ∞ dx (14) . x + e−x ˆ−∞ e ∞ (15) ln(ln x). ee A comparison test. Theorem 5.1 (Direct Comparison Test). Let f and g be continuous on [a, ∞) and suppose that 0 ≤ f (x) ≤ g(x) for all x ≥ a. ˆ ∞ ˆ ∞ (i) If g(x) dx converges then f (x) dx converges. a a
  13. 13. INTEGRATION TECHNIQUES ˆ (ii) If ˆ ∞ f (x) dx diverges then a 13 ∞ g(x) dx diverges. a Example 5.4. In each part, determine whether the improper integral converges or diverges. ˆ ∞ 1 (1) dx. x ˆ0 ∞ x + e 2 e−x dx. (2) ˆ0 ∞ 2 + sin x √ (3) dx. x 1 ˆ ∞ sin2 x (4) dx. 2 ˆ1 ∞ x dx √ . (5) 4 ˆ1 ∞ x + 5 ln x √ dx. (6) ˆ1 ∞ x dx (7) . 5x + 2x 1 ˆ ∞ dx (8) . x ˆ0 ∞ 1 + e 1 + sin x dx. (9) x2 ˆπ ∞ dx (10) . ˆ2 1 ln x ln x √ dx. (11) x 0
  14. 14. 14 INTEGRATION TECHNIQUES Answers cos(ax) + c. a x 1 Answers 1.2. tan−1 + c. a a x5 10x3 − + 25x + c. Answers 1.3. 5 3 x−3 Answers 1.4. sin−1 + c. 2 Answers 1.1. − Answers 1.5. ln x2 + 2 x + 5 − 3 tan−1 4 x+1 2 Answers 2.1. (1) sin x − x cos x + c. (2) x2 ex − 2xex + 2ex + c. ex (cos x + sin x) (3) + c. 2 2 2 x ln x x (4) − + c. 2 4 (5) x ln x − x + c. √ (6) x sin−1 x + 1 − x2 + c. 2 ex (x2 − 1) + c. (7) 2 (8) x tan x + ln(cos x) + c. 3x x3x − 2 + c. (9) ln 3 ln 3 x(cos(ln x) + sin(ln x)) (10) + c. 2 4 x 3 x 2 x (11) x e − 4x e + 12x e − 24xex + 24ex + c. x ln x x − + c. (12) 2 √ √x 2 √x (13) 2 xe − 2e + c. (14) −x2 cos x + 2x sin x + 2 cos x + c. xe2x e2x (15) − + c. 2 −x 4 −x (16) −xe − e + c. 2e2x cos(3x) 3e2x sin(3x) (17) + + c. 13 13 1 (18) x tan−1 x − ln(1 + x2 ) + c. 2 2 x4 e−x 2 2 (19) − − x2 e−x − e−x + c. 2 √ −1 esin x ( 1 − x2 + x) (20) + c. 2 5 3 2 2 (21) (1 − x) 2 − (1 − x) 2 + c. 5 3 15 (22) 4 ln 2 − + c. 16 + c.
  15. 15. INTEGRATION TECHNIQUES Answers 3.1. (1) − cos5 x + c. 5 cos7 x cos5 x − + c. 7 5 3 3 11 4 2 2 (3) sin 2 x − sin 2 x + sin 2 x + c. 3 7 11 sin3 x (4) sin x − + c. 3 cos10 x cos8 (5) − + c. 10 8 x sin(2x) + c. (6) − 2 4 3x sin(2x) sin(4x) (7) + + c. 8 4 32 x sin(4x) + c. (8) − 8 32 sec5 x sec3 x (9) − + c. 5 3 (10) tan x − x + c. tan5 x tan3 x (11) + + c. 5 3 tan2 x + ln | cos x| + c. (12) 2 7 sec x sec5 x (13) − + c. 7 5 (14) ln | sec x + tan x| + c. 7 3 2 2 (15) tan 2 + tan 2 +c. 7 3 1 3 (16) − cot x + cot x + x + c. 3 (17) − ln | csc x + cot x| + c. csc3 x csc5 x − + c. (18) 3 5 √ 5 9 4 2 (19) − cos x + cos 2 x − cos 2 x + c. 5 9 √ 4 − x2 Answers 3.2. (1) (a) − + c. 4x √ x + 9 + x2 + c. (b) ln 3 √ x (c) x2 − 25 − 5 sec−1 + c. 5 1 x −1 (d) tan x + + c. 2 2(1 + x2 ) √ x x 4 − x2 (e) 2 sin−1 + + c. 2 2 √ (f) √x2 − 1 − sec−1 x + c. x2 − 2 (g) + c. 2x (2) 15
  16. 16. 16 INTEGRATION TECHNIQUES √ 3x 9x2 − 4 − 2 sec−1 + c. 2 1 x+1 (i) tan−1 + c. 3 3 √ ex 1 − e2x 1 −1 x + c. (j) sin (e ) + 2 2 (k) ln 1 + sin2 x + sin x + c. 1 2x + 1 2x + 1 (l) tan−1 + + c. 32 2 16(4x2 + 4x + 5) x−2 (m) ln x2 − 4 x + 8 + 3 tan−1 + c. 2 5 13 x−2 (n) ln x2 − 4 x + 8 + tan−1 + c. 2 2 2 x3 (o) − 3 + c. 3(x2 − 1) 2 π(π + 2) (2) . 2 1 1 Answers 4.1. (1) ln |x − 1| − ln |x + 2| + c. 3 3 (2) 2 ln |x| + 4 ln |x + 1| − 3 ln |x − 1| + c. 3 1 (3) x2 + ln |x − 4| − ln |x + 2| + c. 2 2 9 (4) − + 6 ln |x| − ln |x + 1| + c. x+1 (5) −5 tan−1 x + 2 ln |x| + c. 3 7 x+1 (6) ln x2 + 2 x + 5 − tan−1 + 2 ln |x + 2| + c. 2 2 2 7 8 (7) ln |x − 3| + ln |x + 2| + c. 5 5 5 1 5 (8) − ln |x − 2| − ln |x| + ln |x − 1| + ln |x − 3| + c. 2 6 3 3 (9) − + ln |x − 3| + c. x−3 1 1 (10) − + + c. x + 1 (x + 1)2 1 1 1 (11) ln |x + 1| + ln |x − 1| − + c. 2 2 x−1 3 1 3 (12) − ln x2 + 1 − tan−1 x + ln |x − 2| + c. 10 5 5 √ √ 5 1 3 2 x 2 2 −1 (13) ln |3 + x| + − ln x + 2 + tan 2(x2 + 2) 2 2 2 2 1 (14) x + ln |3 x + 1| + c. 3 9 x2 1 (15) x + + ln x2 + 1 + tan−1 x − 2 ln |x − 1| − + c. 2 x−1 x2 (16) x + − ln |x| + 2 ln |x − 1| + c. 2 (h) + c.
  17. 17. INTEGRATION TECHNIQUES (17) ln |sec x + tan x| + c. 1 125 x2 − 6 x − ln |x + 1| + ln |5 + x| + c. (18) 2 4 4 1 1 1 (19) x + ln |x − 1| − ln |x + 1| − tan−1 x + c. 4 4 √ 2 √ 1 3 (2 x + 1) 3 1 2 −1 (20) − ln x + x + 1 − tan + ln |x − 1| + c. 6 3 3 3 −x x (21) −e + x − ln |e − 1| + c. 1 1 (22) − ln |2 cos x − 1| + ln |1 + cos x| + c. 3 3 1 3 3 1 (23) sin x − − ln |sin x + 1| + ln |sin x − 1| − + c. 4(sin x + 1) 4 4 4(sin x − 1) √ √ 1 1 1 (24) √ − ln 1 − x2 + 1 + ln 1 − x2 − 1 + c. 2 1 − x2 2 1 (25) ln x2 + 2 x + 2 − tan−1 (x + 1) − 2 + c. x + 2x + 2 Answers 5.1. See, Answers 5.2, (5). Answers 5.2. (1) Convergent, 2. (2) Divergent. (3) Convergent, 2. (4) Divergent. (5) Divergent. (6) Convergent, 1. (7) Divergent. 1 (8) Convergent, . e (9) Divergent. (10) Convergent, 1. (11) Convergent, 0. π (12) Convergent, . 2 √ 3 (13) Convergent, 3 + 3 2. (14) Divergent. (15) Divergent if p ≥ 1. Convergent if p < 1, (16) (17) (18) (19) (20) (21) 1 . 1−p Divergent. Convergent, 1. Convergent, π. Convergent, 0. Divergent. Divergent. (22) Divergent if p ≤ 1. Convergent if p > 1, (23) Convergent, 2. (24) Divergent. Answers 5.3. (1) Divergent. 1 . p−1 17
  18. 18. 18 INTEGRATION TECHNIQUES 32 . 3 π (3) Convergent, . 2 (4) Divergent. (5) Convergent, 1000. 9 (6) Convergent, − . 2 (7) Convergent, 1000. (8) Convergent, 0. (9) Convergent, 2π 2 . (10) Convergent, 1. (11) Convergent, 1. π (12) Convergent, . 4 (13) Convergent, 0. π (14) Convergent, . 2 Answers 5.4. (1) Convergent. (2) Convergent. (3) Divergent. (4) Convergent. (5) Convergent. (6) Divergent. (7) Convergent. (8) Convergent. (9) Convergent. (10) Divergent. (11) Convergent. (12) Divergent. (2) Convergent,

×