MATH1231 CALCULUS. Session II Dr John Roberts (based on notes of A./Prof. Bruce Henry) Red Center Room 3065

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1 MATH1231 CALCULUS Session II Dr John Roberts (based on notes of A./Prof. Bruce Henry) Red Center Room 3065 MATH1231 CALCULUS p.1/66

2 Overview Systematic Integration Techniques Ordinary Differential Equations Convergence of Sequences Convergence of Series Power Series Applications of Integration MATH1231 CALCULUS p.2/66

3 Systematic Integration Techniques Integration is the undoing of differentiation. F (x) = df dx MATH1231 CALCULUS p.3/66

4 Systematic Integration Techniques Integration is the undoing of differentiation. F (x) dx = F (x) = df dx d f(x) dx = f(x) + C dx MATH1231 CALCULUS p.3/66

5 Systematic Integration Techniques Integration is the undoing of differentiation. F (x) dx = F (x) = df dx d f(x) dx = f(x) + C dx (0.0) Chain Rule d dx f(g(x)) = g (x)f (g(x)) MATH1231 CALCULUS p.3/66

6 Systematic Integration Techniques Integration is the undoing of differentiation. F (x) dx = F (x) = df dx d f(x) dx = f(x) + C dx (0.0) Chain Rule d dx f(g(x)) = g (x)f (g(x)) g (x)f (g(x)) dx = d f(g(x)) dx = f(g(x)) + C dx MATH1231 CALCULUS p.3/66

7 Examples 6x(1 + x 2 ) 2 dx = cosh x cosh(sinh x) = 2e 2x cos(e 2x ) dx = tan(x) dx = MATH1231 CALCULUS p.4/66

8 Examples 6x(1 + x 2 ) 2 dx = (1 + x 2 ) 3 + C cosh x cosh(sinh x) = sinh(sinh x) + C 2e 2x cos(e 2x ) dx = sin(e 2x ) + C tan(x) dx = log(cos x) + C MATH1231 CALCULUS p.4/66

9 (0.1) Substitution MATH1231 CALCULUS p.5/66

10 (0.1) Substitution I = f(x) dx Put x = g(u) then f(x) = f(g(u)) and dx = g (u) du. MATH1231 CALCULUS p.5/66

11 (0.1) Substitution I = f(x) dx Put x = g(u) then f(x) = f(g(u)) and dx = g (u) du. I = f(g(u))g (u) du = h(u) du MATH1231 CALCULUS p.5/66

12 (0.1) Substitution I = f(x) dx Put x = g(u) then f(x) = f(g(u)) and dx = g (u) du. I = f(g(u))g (u) du = h(u) du This is only useful if h(u) du is easier than f(x) dx. MATH1231 CALCULUS p.5/66

13 Example cos x 1 + sin x dx MATH1231 CALCULUS p.6/66

14 Example cos x 1 + sin x dx Put sin x = u cos x dx = du MATH1231 CALCULUS p.6/66

15 Example cos x 1 + sin x dx Put sin x = u cos x dx = du cos x 1 + sin x dx = u du = log(1+u)+c = log(1+sin x)+c MATH1231 CALCULUS p.6/66

16 (0.2) Integration by parts MATH1231 CALCULUS p.7/66

17 (0.2) Integration by parts u(x) dv dx dx = u(x)v(x) v(x) du dx dx MATH1231 CALCULUS p.7/66

18 (0.2) Integration by parts u(x) dv dx dx = u(x)v(x) v(x) du dx dx Example }{{} x }{{} e x u(x) dv dx dx = x }{{} u(x) }{{} e x v(x) e x }{{} v(x). 1 }{{} du dx dx = xe x e x +C MATH1231 CALCULUS p.7/66

19 (0.2) Integration by parts u(x) dv dx dx = u(x)v(x) v(x) du dx dx Example }{{} x }{{} e x u(x) dv dx dx = x }{{} u(x) }{{} e x v(x) e x }{{} v(x). 1 }{{} du dx dx = xe x e x +C Try x 3 }{{} u(x) e x }{{} dv dx dx MATH1231 CALCULUS p.7/66

20 (1.1) Trigonometric Integrals sin 3 x cos 6 x dx cos 8 x dx tan 3 x sec 2 x dx sinh 3 x cosh 6 x dx MATH1231 CALCULUS p.8/66

21 Who Cares? Many problems in three dimensional space can be simplified using spherical polar co-ordinates z φ r (r, θ, φ) y x = r sin φ cos θ y = r sin φ sin θ z = r cos φ θ Integrals over powers of trig functions then arise, x e.g., volume calculations, charge distributions, acoustic vibrations, angular momentum of electrons, trajectories of falling space junk MATH1231 CALCULUS p.9/66

22 Example cos 3 x dx = = = cos x cos 2 x dx cos x(1 sin 2 x) dx (1 u 2 ) du u = sin x = u 1 3 u3 = cos x 1 3 cos3 x + C MATH1231 CALCULUS p.10/66

23 (1.1.1) cos m x sin n x dx MATH1231 CALCULUS p.11/66

24 (1.1.1) cos m x sin n x dx If m is odd, say m = 2k + 1, then factor cos m x sin n x = cos x(cos 2 x) k sin n x = cos x(1 sin 2 x) k sin n x The substitution u = sin x, du = cos x dx gives a polynomial to integrate. MATH1231 CALCULUS p.11/66

25 (1.1.1) cos m x sin n x dx If m is odd, say m = 2k + 1, then factor cos m x sin n x = cos x(cos 2 x) k sin n x = cos x(1 sin 2 x) k sin n x The substitution u = sin x, du = cos x dx gives a polynomial to integrate. If n is odd, use the analogous method with the roles of sin x and cos x interchanged. MATH1231 CALCULUS p.11/66

26 Examples cos 3 x sin 4 x dx = sin 3 x cos 2 x dx = sin 7 x dx = MATH1231 CALCULUS p.12/66

27 Examples cos 3 x sin 4 x dx = (cos x) cos 2 x sin 4 x dx sin 3 x cos 2 x dx = (sin x) sin 2 x cos 2 x dx sin 7 x dx = (sin x) sin 6 x dx MATH1231 CALCULUS p.12/66

28 Examples cos 3 x sin 4 x dx = (cos x)(1 sin 2 x) sin 4 x dx sin 3 x cos 2 x dx = (sin x)(1 cos 2 x) cos 2 x dx sin 7 x dx = (sin x)(1 cos 2 x) 3 dx MATH1231 CALCULUS p.12/66

29 Examples cos 3 x sin 4 x dx = sin 3 x cos 2 x dx = sin 7 x dx = (cos x) (1 sin 2 x) sin 4 x }{{} u=sin x (sin x) (1 cos 2 x) cos 2 x }{{} u=cos x (sin x) (1 cos 2 x) 3 }{{} u=cos x dx dx dx MATH1231 CALCULUS p.12/66

30 cos m x sin n x dx If both m and n are even, we use the identities cos 2 x = 1 + cos 2x 2 and sin 2 x = 1 cos 2x 2 to change the integral into a sum of integrals of the form cos j 2x dx if j is odd, we use the method of the first section if j is even, we use the method of this section again. MATH1231 CALCULUS p.13/66

31 Example I = sin 2 x cos 2 x dx MATH1231 CALCULUS p.14/66

32 Example I = sin 2 x cos 2 x dx I = ( 1 cos 2x 2 )( 1 + cos 2x 2 ) dx MATH1231 CALCULUS p.14/66

33 Example I = sin 2 x cos 2 x dx I = ( 1 cos 2x 2 )( 1 + cos 2x 2 ) dx I = ( 1 cos2 2x 4 ) dx = cos 2 2x dx MATH1231 CALCULUS p.14/66

34 Example I = sin 2 x cos 2 x dx I = ( 1 cos 2x 2 )( 1 + cos 2x 2 ) dx I = ( 1 cos2 2x 4 ) dx = cos 2 2x dx I = ( 1 + cos 4x 2 ) dx MATH1231 CALCULUS p.14/66

35 (1.1.2) sin nx cos mx dx sin nx sin mx dx Use the identities cos nx cos mx dx sin A cos B = 1 2 sin A sin B = 1 2 cos A cos B = 1 2 (sin(a + B) + sin(a B)) (cos(a B) cos(a + B)) (cos(a B) + cos(a + B)) MATH1231 CALCULUS p.15/66

36 Example sin 3x sin 2x dx = 1 2 (cos(3x 2x) cos(3x + 2x)) dx MATH1231 CALCULUS p.16/66

37 (1.1.3) tan n x dx or sec n x dx MATH1231 CALCULUS p.17/66

38 (1.1.3) tan n x dx or sec n x dx Exploit the relationships sec 2 x = 1+tan 2 x, d dx tan x = sec2 x, d dx sec x = sec x tan x MATH1231 CALCULUS p.17/66

39 (1.1.3) tan n x dx or sec n x dx Exploit the relationships sec 2 x = 1+tan 2 x, d dx tan x = sec2 x, d dx sec x = sec x tan x Reduction Formula (apply repeatedly) I n = tan n xdx = = tan n 2 x(tan 2 x)dx tan n 2 x(sec 2 x 1)dx = tann 1 x n 1 I n 2 (until) I 1 = sin x cos x dx = ln sec x + C (or) I 0 = dx = x + C MATH1231 CALCULUS p.17/66

40 I n = sec n x dx = (by parts) = sec } n 2 {{ x} tan x u(x) sec n 2 x = sec n 2 x tan x (n 2) = sec n 2 x tan x (n 2) sec 2 x {}}{ d (tan x) dx dx (tan x) (n 2) sec n 3 x sec x tan x }{{} du dx sec n 2 x tan 2 x dx sec n 2 x (sec 2 x 1) dx dx = sec n 2 x tan x (n 2)I n + (n 2)I n 2 = secn 2 x tan x (n 1) + (n 2) (n 1) I n 2. I 1 = sec x dx = ln sec x + tan x + C (or) I 0 = dx = x + C MATH1231 CALCULUS p.18/66

41 Example I 4 = tan 4 x dx MATH1231 CALCULUS p.19/66

42 (1.1.4) tan m x sec n x dx MATH1231 CALCULUS p.20/66

43 (1.1.4) tan m x sec n x dx If n is even, write sec n 2 x = (sec 2 x) n 2 2 = (tan 2 x + 1) n 2 2 This yields a sum of integrals of the form tan j x sec 2 x dx. Now use the substitution u = tan x. MATH1231 CALCULUS p.20/66

44 (1.1.4) tan m x sec n x dx If n is even, write sec n 2 x = (sec 2 x) n 2 2 = (tan 2 x + 1) n 2 2 This yields a sum of integrals of the form tan j x sec 2 x dx. Now use the substitution u = tan x. If m is odd, say m = 2k + 1, write tan m x sec n x = (tan 2 x) k sec n x tan x = (sec 2 x 1) k sec n 1 x sec x tan x This yields a sum of integrals sec j x sec x tan x dx. Now use the substitution u = sec x. MATH1231 CALCULUS p.20/66

45 If m is even m = 2k then tan m x sec n x dx = (sec 2 x 1) k sec n x dx yields a sum of integrals for the form sec j x dx MATH1231 CALCULUS p.21/66

46 Examples MATH1231 CALCULUS p.22/66

47 Examples tan 2 x even {}}{ sec 4 x dx = = (tan 2 x sec 2 x) factor {}}{ sec 2 x dx tan 2 x(1 + tan 2 x) sec 2 x dx }{{} u=tan x MATH1231 CALCULUS p.22/66

48 Examples tan 2 x even {}}{ sec 4 x dx = = (tan 2 x sec 2 x) factor {}}{ sec 2 x dx tan 2 x(1 + tan 2 x) sec 2 x dx }{{} u=tan x odd {}}{ tan 3 x sec 3 x dx = = tan 2 x sec 2 x factor {}}{ (sec x tan x) dx (sec 2 x 1) sec 2 x sec x tan x dx }{{} u=sec x MATH1231 CALCULUS p.22/66

49 (1.2) Reduction formulae Suppose we have an integrand that is a function of both an independent variable x and an integral index n e.g., (log x) n dx, x n e x dx, x n cos x dx If we now integrate by parts and obtain a similar integrand but with a smaller index n then we have a reduction formula that can be applied repeatedly. In general if [f(x)] n g(x) dx then integrate by parts with u(x) = [f(x)] n and dv dx = g(x) MATH1231 CALCULUS p.23/66

50 Example I n = x n e x dx Integrate by parts with u = x n and dv dx = ex I n = x n e x dx = x n e x n x n 1 e x dx = x n e x ni n 1 and I 0 = e x. MATH1231 CALCULUS p.24/66

51 Example I n = sin n x dx. Integrate by parts with u = sin n 1 x and dx dv I n = sin n 1 x cos x + ((n 1) sin n 2 x cos x) cos x dx = sin n 1 x cos x + (n 1) sin n 2 x(1 sin 2 x) dx = sin n 1 x cos x + (n 1)I n 2 (n 1)I n Now solve for I n I n = sinn 1 x cos x n + ( n 1 n ) In 2. MATH1231 CALCULUS p.25/66

52 Wallis s Product π/2 0 π/2 0 sin 2m x dx = sin 2m+1 x dx = ( ) ( ) ( ) 2m 1 2m 3 1 (π )... 2m 2m ( ) ( ) ( ) 2m 2m m + 1 2m 1 3 π 2 = m 2m 1 ( 2m 2m + 1 π/2 ) 0 sin 2m x dx π/2 0 sin 2m+1 x dx lim m π 2 = lim m (2m 2) (2m 1) 2 2m π = lim m (m!) 2 2 2m (2m)! m MATH1231 CALCULUS p.26/66

53 (1.2.1) Proposition I m,n = π/2 0 cos m θ sin n θ dθ = ( m 1 m+n) Im 2,n provided m 2 ( n 1 m+n) Im,n 2 provided n 2 I 1,1 = I 0,1 = π/2 0 π/2 0 cos θ sin θ dθ = 1 2 sin θ dθ = 1 I 0,0 = π/2 0 π/2 I 1,0 = 0 dθ = π 2 cos θ dθ = 1 Proof: See Notes MATH1231 CALCULUS p.27/66

54 Example π 2 0 cos 3 x sin 4 x dx = I 3,4 MATH1231 CALCULUS p.28/66

55 Sample Test Question Show that I n = 1 n I n = π 4 0 tan n θ sec θ dθ ( 2 (n 1)In 2 ) n 2 Hint sec θ tan θ = d dθ (sec θ) MATH1231 CALCULUS p.29/66

56 I n = Z π 4 0 tan n 1 θ tan θ sec θ {z } {z } u dv dθ dθ = tan n 1 θ sec θ π 4 0 Z π 4 0 sec θ(n 1) tan n 2 θ sec 2 θ dθ MATH1231 CALCULUS p.30/66

57 I n = Z π 4 0 tan n 1 θ tan θ sec θ {z } {z } u dv dθ dθ = tan n 1 θ sec θ π 4 0 Z π 4 0 sec θ(n 1) tan n 2 θ sec 2 θ dθ I n = 2 (n 1) Z π 4 0 sec θ tan n 2 θ `tan 2 θ + 1 dθ = 2 (n 1) Z π 4 0 sec θ tan n θ dθ (n 1) Z π 4 0 sec θ tan n 2 θ dθ = 2 (n 1)I n (n 1)I n 2 MATH1231 CALCULUS p.30/66

58 I n = Z π 4 0 tan n 1 θ tan θ sec θ {z } {z } u dv dθ dθ = tan n 1 θ sec θ π 4 0 Z π 4 0 sec θ(n 1) tan n 2 θ sec 2 θ dθ I n = 2 (n 1) Z π 4 0 sec θ tan n 2 θ `tan 2 θ + 1 dθ = 2 (n 1) Z π 4 0 sec θ tan n θ dθ (n 1) Z π 4 0 sec θ tan n 2 θ dθ = 2 (n 1)I n (n 1)I n 2 solve for I n MATH1231 CALCULUS p.30/66

59 (1.3) Trigonometric and hyperbolic substitution Substitution is inspired guesswork. This table is for inspiration. integral factor substitutions identities a 2 x 2 x = a sin θ x = a cos θ cos 2 θ + sin 2 θ = 1 dx = a cos θ dθ dx = a sin θ dθ a 2 + x 2 x = a sinh θ x = a tan θ sec 2 θ = tan 2 θ + 1 dx = a cosh θ dθ dx = a sec 2 θ dθ x 2 a 2 x = a sec θ x = a cosh θ cosh 2 θ = sinh 2 θ + 1 dx = a sec θ tan θ dθ dx = a sinh θ dθ MATH1231 CALCULUS p.31/66

60 Example I = 1 x 2 dx MATH1231 CALCULUS p.32/66

61 Example I = 1 x 2 dx substitution x = sin θ, dx dθ = cos θ MATH1231 CALCULUS p.32/66

62 Example I = 1 x 2 dx substitution x = sin θ, dx dθ = cos θ I = cos θ cos θ dθ = 1 2 (1 + cos 2θ) dθ = θ 2 + sin 2θ 4 = sin 1 x 2 + C = θ 2 + x 1 x sin θ cos θ 2 + C + C MATH1231 CALCULUS p.32/66

63 Example Find the area of the ellipse y (0,b) (a,0) x A = 4 a 0 b a a 2 x 2 dx A = πab exercise MATH1231 CALCULUS p.33/66

64 Example I = x 3 x2 16 dx substitution x = 4 sec θ MATH1231 CALCULUS p.34/66

65 Example I = x 2 a 2 dx x a 0 substitution x = a sec θ (see notes) I = a2 2 ( x x 2 a 2 a 2 ln x a + ) x 2 a 2 a + C compare the substitution x = a cosh θ MATH1231 CALCULUS p.35/66

66 Example Planetary Motion Kepler (1609,1619) The motion of a planet around the sun is described by r θ θ(r) = l r 2 dr 2µ(E + k r 2µr l2 ) 2 + C E, l, µ, k, C are constants What are the possible motions of the planets? Substitute u = 1 r then du = 1 r 2 dr θ(r) = l du 2µ(E + ku 2µ l2 u2 ) + C MATH1231 CALCULUS p.36/66

67 Example ctd. Complete the square in the denominator Substitution (u kµ l 2 ) = θ(r) = du ( k2 µ 2 l + 2µE 4 l ) (u kµ 2 l ) 2 2 ( k2 µ 2 l 4 + 2µE l 2 ) cos φ, du = ( k2 µ 2 + 2µE l 4 l 2 ) sin φ dφ θ(r) = ( k2 µ 2 ) sin φ dφ ( k2 µ 2 l + 2µE 4 l ) sin φ 2 l + 2µE 4 l 2 = φ MATH1231 CALCULUS p.37/66

68 Example ctd. parabola, ε = 1 ellipse, 0 < ε < 1 hyperbola, ε > 1 circle, ε = 0 focus α r = 1 + ɛ cos θ α = l2 µk ɛ = 1 + 2El2 µk 2 MATH1231 CALCULUS p.38/66

69 (1.4) Rational functions and partial fractions A rational function has the form p(x) q(x) where p(x), q(x) are polynomial functions Example x 3 (x 2)(x 2 4x + 8) dx Any rational function can be integrated by systematic reduction with partial fractions MATH1231 CALCULUS p.39/66

70 Step 1 SIMPLIFY using long division p(x) q(x) = s(x) + r(x) q(x) deg r(x) < deg q(x) MATH1231 CALCULUS p.40/66

71 Step 1 SIMPLIFY using long division p(x) q(x) = s(x) + r(x) q(x) deg r(x) < deg q(x) Step 2 FACTORIZE the denominator q(x) as product of linear (x + a) and irreducible quadratic (x 2 + bx + c), b 2 4c < 0 factors. MATH1231 CALCULUS p.40/66

72 Step 1 SIMPLIFY using long division p(x) q(x) = s(x) + r(x) q(x) deg r(x) < deg q(x) Step 2 FACTORIZE the denominator q(x) as product of linear (x + a) and irreducible quadratic (x 2 + bx + c), b 2 4c < 0 factors. Step 3 DECOMPOSE r(x) q(x) as a sum of rational functions whose denominators contain only one of the factors of q(x) from step 2. MATH1231 CALCULUS p.40/66

73 (1.4.1) q(x) = α(x + a) linear deg r(x) < deg q(x) r(x) = r = const r(x) q(x) dx = r dx α(x + a) = r α ln x + a + C (x a) MATH1231 CALCULUS p.41/66

74 Example I = 2x 3 +7x x+3 dx 2x + 3 x 2 + 2x 3 2x 3 + 7x x 3 3x 2 4x 2 4x 2 6x 6x + 3 6x + 9 I = 12 (x 2 + 2x 3) x + 3 dx = x3 3 + x2 3x + 6 ln 2x C MATH1231 CALCULUS p.42/66

75 (1.4.2) q(x) = α(x + a 1 )(x + a 2 )... (x + a n ) We can write distinct linear factors r(x) q(x) = A 1 x + a 1 + A 2 x + a A n x + a n Then r(x) q(x) dx = j A j x + a j dx How to find A j? MATH1231 CALCULUS p.43/66

76 To find the constants A j For example, to find A 2, multiply both sides of ( ) by x + a 2 then r(x) α(x + a 1 )(x + a 3 ) (x + a n ) = A x + a 2 x + a 2 1 +A 2 + +A n x + a 1 x + a n Now substitute x = a 2. In effect, we find A 2 by covering up the factor (x + a 2 ) in the denominator of r(x) q(x) and then substituting x = a 2: A 2 = r( a 2 ) α( a 2 + a 1 ) (x + a 2 ) ( a 2 + a 3 ) ( a 2 + a n ) MATH1231 CALCULUS p.44/66

77 Example I = dx (x 2)(x 3) 1 (x 2)(x 3) = A x 2 + B x 3 MATH1231 CALCULUS p.45/66

78 Example I = dx (x 2)(x 3) 1 (x 2)(x 3) = A x 2 + B x 3 A = 1 (x 2) (2 3) = 1, B = 1 (3 2) (x 3) = 1 MATH1231 CALCULUS p.45/66

79 Example I = dx (x 2)(x 3) 1 (x 2)(x 3) = A x 2 + B x 3 A = 1 (x 2) (2 3) = 1, B = 1 (3 2) (x 3) = 1 Hence I = dx x 2 + dx x 3 = ln x 3 ln x 2 + C = ln x 3 x 2 + C. MATH1231 CALCULUS p.45/66

80 (1.4.3) q(x) = α(x + a) k a power of a linear factor We can write r(x) q(x) = A 1 x + a + A 2 (x + a) A k (x + a) k Multiply both sides of ( ) by (x + a) k then substitute x = a to find A k Multiply both sides of ( ) by (x + a) k then differentiate and substitute x = a to find A k 1 Multiply both sides of ( ) by (x + a) k then differentiate twice and substitute x = a to find A k 2... MATH1231 CALCULUS p.46/66

81 Example x 2 3 (x 1) 3 dx MATH1231 CALCULUS p.47/66

82 (1.4.4) q(x) = α(x + a 1 ) k 1 (x + a 2 ) k 2... (x + a n ) k n powers of distinct linear factors r(x) q(x) = A 11 x + a 1 + A 12 (x + a 1 ) A 1k 1 (x + a 1 ) k A 21 x + a A 2k 2 (x + a 2 ) k A n1 x + a n + + A nk n (x + a n ) k n Finding the constants A jk involves solving simultaneous equations. MATH1231 CALCULUS p.48/66

83 Example I = dx x(x+1) 3 1 x(x + 1) 3 = A x + B x C (x + 1) 2 + D (x + 1) 3 MATH1231 CALCULUS p.49/66

84 Example I = dx x(x+1) 3 1 x(x + 1) 3 = A x + B x C (x + 1) 2 + D (x + 1) 3 multiply by x(x + 1) 3 1 = A(x + 1) 3 + Bx(x + 1) 2 + Cx(x + 1) + Dx MATH1231 CALCULUS p.49/66

85 Example I = dx x(x+1) 3 1 x(x + 1) 3 = A x + B x C (x + 1) 2 + D (x + 1) 3 multiply by x(x + 1) 3 1 = A(x + 1) 3 + Bx(x + 1) 2 + Cx(x + 1) + Dx substitute substitute x = 0 1 = A x = 1 1 = D 1 = (x + 1) 3 + Bx(x + 1) 2 + Cx(x + 1) x MATH1231 CALCULUS p.49/66

86 Example ctd. We need two more equations for B and C MATH1231 CALCULUS p.50/66

87 Example ctd. We need two more equations for B and C substitute x = 1 1 = 8 + 4B + 2C 1 3 = C + 2B substitute x = 2 4 = C + 3B solve simultaneously then B = 1, C = 1 MATH1231 CALCULUS p.50/66

88 Example ctd. We need two more equations for B and C substitute x = 1 1 = 8 + 4B + 2C 1 3 = C + 2B substitute x = 2 4 = C + 3B solve simultaneously then B = 1, C = 1 Finally 1 1 x(x + 1) 3 = x 1 x (x + 1) 2 = ln x x x (x + 1) 2 + C 1 (x + 1) 3 MATH1231 CALCULUS p.50/66

89 Sample Class Test Problem x + 3 I = (x + 1)(x + 2) 2 dx MATH1231 CALCULUS p.51/66

90 Sample Class Test Problem x + 3 I = (x + 1)(x + 2) 2 dx x + 3 (x + 1)(x + 2) 2 = A x B x C (x + 2) 2 MATH1231 CALCULUS p.51/66

91 Sample Class Test Problem x + 3 I = (x + 1)(x + 2) 2 dx x + 3 (x + 1)(x + 2) 2 = A x B x C (x + 2) 2 (x + 3)(x + 1) A(x + 1) = (x + 1)(x + 2) 2 x B(x + 1) x C(x + 1) (x + 2) 2 MATH1231 CALCULUS p.51/66

92 Sample Class Test Problem x + 3 I = (x + 1)(x + 2) 2 dx x + 3 (x + 1)(x + 2) 2 = A x B x C (x + 2) 2 (x + 3)(x + 1) A(x + 1) = (x + 1)(x + 2) 2 x B(x + 1) x C(x + 1) (x + 2) 2 x 1 A = 2 MATH1231 CALCULUS p.51/66

93 (x + 3)(x + 2) 2 A(x + 2)2 = (x + 1)(x + 2) 2 x B(x + 2)2 x C(x + 2)2 (x + 2) 2 MATH1231 CALCULUS p.52/66

94 (x + 3)(x + 2) 2 A(x + 2)2 = (x + 1)(x + 2) 2 x B(x + 2)2 x C(x + 2)2 (x + 2) 2 x 2 C = 1 MATH1231 CALCULUS p.52/66

95 (x + 3)(x + 2) 2 A(x + 2)2 = (x + 1)(x + 2) 2 x B(x + 2)2 x C(x + 2)2 (x + 2) 2 x 2 C = 1 x + 3 (x + 1)(x + 2) 2 = 2 x B x (x + 2) 2 MATH1231 CALCULUS p.52/66

96 (x + 3)(x + 2) 2 A(x + 2)2 = (x + 1)(x + 2) 2 x B(x + 2)2 x C(x + 2)2 (x + 2) 2 x 2 C = 1 x + 3 (x + 1)(x + 2) 2 = 2 x B x (x + 2) 2 x = 0 B = 2 MATH1231 CALCULUS p.52/66

97 I = 2 x + 1 dx 2 x + 2 dx 1 (x + 2) 2 dx = 2 log x log x x C MATH1231 CALCULUS p.53/66

98 (1.4.5) q(x) = α(x 2 + bx + c), b 2 4c < 0 an irreducible quadratic If b 2 4c 0 then the quadratic factorizes so an earlier case applies w.l.o.g. write r(x) = L(2x + b) + M r(x) q(x) dx = L α 2x + b x 2 + bx + c dx + M α dx x 2 + bx + c = L α ln x2 + bx + c + M α c b2 4 tan 1 ( x + 2 b c b2 4 ) + C MATH1231 CALCULUS p.54/66

99 I = dx x 2 +bx+c x 2 + bx + c = (x + b 2 )2 + (c b2 ) complete the square 4 = (c b2 4 ) (x + b 2 2 ) + 1 I = 1 (c b2 4 ) " dx (x+ 2 b ) c b2 4! 2 +1# substitute tan θ = q (x+ b ) 2 c b2 4 c b2 4 then sec 2 θ dθ = q 1 c b2 4 dx I = 1 (c b2 4 ) q c b2 4 sec2 θ dθ = q θ tan 2 θ+1 c b2 4 = q 1 c b2 4 tan 1 q (x+ b ) 2 c b2 4 MATH1231 CALCULUS p.55/66

100 Sample Class Test Problem I = x x 2 4x + 13 dx You are given that du u 2 + a 2 = 1 a tan 1 u a + C, a 0 MATH1231 CALCULUS p.56/66

101 Sample Class Test Problem I = x x 2 4x + 13 dx You are given that du u 2 + a 2 = 1 a tan 1 u a + C, a 0 First step, write as L(2x 4) x 2 4x + 13 dx + M x 2 4x + 13 dx L = 1 2 and M = 2 MATH1231 CALCULUS p.56/66

102 I = 1 2 log x2 4x x 2 4x + 13 dx complete the square 2 x 2 4x + 13 dx = 2 (x 2) dx = 2 (x 2) dx put (x 2) = 3 tan θ or put (x 2) = u and use table result du u 2 + a 2 = 1 a tan 1 u a + C, a 0 MATH1231 CALCULUS p.57/66

103 (1.4.6) q(x) = α(x 2 + bx + c) k, b 2 4c < 0 irreducible quadratic powers r(x) q(x) = B 1x + C 1 x 2 + bx + c + B 2x + C 2 (x 2 + bx + c) B kx + C k (x 2 + bx + c) k MATH1231 CALCULUS p.58/66

104 (1.4.6) q(x) = α(x 2 + bx + c) k, b 2 4c < 0 irreducible quadratic powers r(x) q(x) = B 1x + C 1 x 2 + bx + c + B 2x + C 2 (x 2 + bx + c) B kx + C k (x 2 + bx + c) k to find constants B j, C j multiply by (x 2 + bx + c) k and equate coefs of equal powers of x MATH1231 CALCULUS p.58/66

105 (1.4.6) q(x) = α(x 2 + bx + c) k, b 2 4c < 0 irreducible quadratic powers r(x) q(x) = B 1x + C 1 x 2 + bx + c + B 2x + C 2 (x 2 + bx + c) B kx + C k (x 2 + bx + c) k to find constants B j, C j multiply by (x 2 + bx + c) k and equate coefs of equal powers of x This yields a sum of integrals B j x + C j = L(2x + b) + M B j x + C j (x 2 + bx + c) j dx = L 2x + b (x 2 + bx + c) j dx+m dx (x 2 + bx + c) j the first integral is easy with the substitution u = x 2 + bx + c. MATH1231 CALCULUS p.58/66

106 I = dx (x 2 + bx + c) j = 1 (c b2 4 )j ( (x+ b 2 ) q c b2 4 dx ) j substitute tan θ = q (x+ b ) 2 c b2 4 then sec 2 θ dθ = q 1 c b2 4 dx I = c b2 4 (c b2 4 )j sec 2 θ dθ (tan 2 θ + 1) j = c b2 4 (c b2 4 )j cos 2j 2 θ dθ MATH1231 CALCULUS p.59/66

107 (1.4.7) q(x) = α(x + a 1 ) k1 (x + a m ) k m (x 2 + b 1 x + c 1 ) l1 (x 2 + b n x + c n ) l n general case r(x) q(x) = A 11 x + a A m1 x + a m + A 12 (x + a 1 ) A 1k1 (x + a 1 ) k 1 + A m2 (x + a m ) A mkm (x + a m ) k m + + B 11x + C 11 x 2 + b 1 x + c 1 + B 12x + C 12 (x 2 + b 1 x + c 1 ) B 1l 1 x + C 1l1 (x 2 + b 1 x + c 1 ) l B n1x + C n1 x 2 + b n x + c n + B n2x + C n2 (x 2 + b n x + c n ) B nl n x + C nln (x 2 + b n x + c n ) l n MATH1231 CALCULUS p.60/66

108 Example 3x 2 5x+3 (x 1)(x 2 2x+2) dx MATH1231 CALCULUS p.61/66

109 (1.5) Further Substitution Techniques Look for a substitution that will convert the integrand to a rational function and then use the method of partial fractions. Example I = dt 1 + t 1/4 Substitute x = t 1/4 (or x 4 = t) then dt = 4x 3 dx I = 4x x dx MATH1231 CALCULUS p.62/66

110 Example ctd. I = 4x 3 1+x dx x + 1 4x 3 4x 3 4x 2 4x 2 4x + 4 4x 2 4x 2 + 4x I = 4x 4x 4 4 4x 2 4x + 4 dx x dx = 4 3 x3 2x 2 + 4x 4 log 1 + x = 4 3 t3/4 2t 1/2 + 4t 1/4 4 log 1 + t 1/4 MATH1231 CALCULUS p.63/66

111 (1.5.1) I = dθ a+b sin θ+c cos θ Half angle substitution t = tan θ 2 MATH1231 CALCULUS p.64/66

112 (1.5.1) I = dθ a+b sin θ+c cos θ Half angle substitution t = tan θ 2 sin θ = 2 sin θ 2 cos θ 2 = 2 tan θ 2 cos2 θ 2 = 2 tan θ 2 sec 2 θ 2 = 2 tan θ 2 tan 2 θ = 2t 1 + t 2 MATH1231 CALCULUS p.64/66

113 (1.5.1) I = dθ a+b sin θ+c cos θ Half angle substitution t = tan θ 2 sin θ = 2 sin θ 2 cos θ 2 = 2 tan θ 2 cos2 θ 2 = 2 tan θ 2 sec 2 θ 2 = 2 tan θ 2 tan 2 θ = 2t 1 + t 2 cos θ = p 1 sin 2 θ = s 1 4t 2 (1 + t 2 ) 2 = 1 + 2t 2 + t 4 4t t 2 = 1 t2 1 + t 2 MATH1231 CALCULUS p.64/66

114 (1.5.1) I = dθ a+b sin θ+c cos θ Half angle substitution t = tan θ 2 sin θ = 2 sin θ 2 cos θ 2 = 2 tan θ 2 cos2 θ 2 = 2 tan θ 2 sec 2 θ 2 = 2 tan θ 2 tan 2 θ = 2t 1 + t 2 cos θ = p 1 sin 2 θ = s 1 4t 2 (1 + t 2 ) 2 = 1 + 2t 2 + t 4 4t t 2 = 1 t2 1 + t 2 dt = 1 2 sec2 θ 2 dθ = 1 + tan2 θ 2 2 dθ dθ = 2dt 1 + t 2 MATH1231 CALCULUS p.64/66

115 (1.5.1) I = dθ a+b sin θ+c cos θ Half angle substitution t = tan θ 2 sin θ = 2 sin θ 2 cos θ 2 = 2 tan θ 2 cos2 θ 2 = 2 tan θ 2 sec 2 θ 2 = 2 tan θ 2 tan 2 θ = 2t 1 + t 2 cos θ = p 1 sin 2 θ = s 1 4t 2 (1 + t 2 ) 2 = 1 + 2t 2 + t 4 4t t 2 = 1 t2 1 + t 2 I = dt = 1 2 sec2 θ 2 dθ = 1 + tan2 θ 2 2 a+b 2dt 1+t 2 2t 1 t2 1+t2 +c dθ dθ = 2dt 1 + t 2 = 2 a(1+t 2 )+2bt+c(1 t 2 ) 1+t 2 MATH1231 CALCULUS p.64/66

116 Example π 2 0 dx 2+cos x MATH1231 CALCULUS p.65/66

117 (1.6) MAPLE excells at integration and partial fractions but it isn t omniscient > # Example 1 > g:=x->x^3/((x+1)^2*(x^2+1)); x x 3 (x+1) 2 (x 2 +1) := x x 3 (x+1) 2 (x 2 +1) > convert(g(x),parfrac,x); > int(g(x),x); > # Example 2 > f:=x->x*(x^2+5)^8; (x + 1) 1 1/2 (x + 1) 2 1/2 ( x ) 1 ln (x + 1) 1/2 arctan (x) + 1/2 (x + 1) 1 and it doesn t always simplify well x x ( x ) 8 ( := x x x ) 8 > int(f(x),x); > <br> 1/18 x /2 x x x x x x x > simplify(%); 2 x 2 1/18 x /2 x x x x x x x x 2 > h:=x->1/18*(x^2+5)^9; > diff(h(x),x); x 1/18 ( x ) 9 := x 1/18 ( x ) 9 x ( x ) 8 MATH1231 CALCULUS p.66/66

MATH 1231 MATHEMATICS 1B Calculus Section 1: - Integration.

MATH 1231 MATHEMATICS 1B Calculus Section 1: - Integration. MATH 1231 MATHEMATICS 1B 2007. For use in Dr Chris Tisdell s lectures: Tues 11 + Thur 10 in KBT Calculus Section 1: - Integration. 1. Motivation 2. What you should already know 3. Useful integrals 4. Integrals