- •Functions
- •The Concept of a Function
- •Trigonometric Functions
- •Inverse Trigonometric Functions
- •Logarithmic, Exponential and Hyperbolic Functions
- •Limits and Continuity
- •Introductory Examples
- •Continuity Examples
- •Linear Function Approximations
- •Limits and Sequences
- •Properties of Continuous Functions
- •The Derivative
- •The Chain Rule
- •Higher Order Derivatives
- •Mathematical Applications
- •Antidifferentiation
- •Linear Second Order Homogeneous Differential Equations
- •Linear Non-Homogeneous Second Order Differential Equations
- •Area Approximation
- •Integration by Substitution
- •Integration by Parts
- •Logarithmic, Exponential and Hyperbolic Functions
- •The Riemann Integral
- •Volumes of Revolution
- •Arc Length and Surface Area
- •Techniques of Integration
- •Integration by formulae
- •Integration by Substitution
- •Integration by Parts
- •Trigonometric Integrals
- •Trigonometric Substitutions
- •Integration by Partial Fractions
- •Fractional Power Substitutions
- •Numerical Integration
- •Integrals over Unbounded Intervals
- •Discontinuities at End Points
- •Improper Integrals
- •Sequences
- •Monotone Sequences
- •Infinite Series
- •Series with Positive Terms
- •Alternating Series
- •Power Series
- •Taylor Polynomials and Series
- •Applications
- •Parabola
- •Ellipse
- •Hyperbola
- •Polar Coordinates
- •Graphs in Polar Coordinates
- •Areas in Polar Coordinates
- •Parametric Equations
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CHAPTER 5. THE DEFINITE INTEGRAL |
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19. |
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x sinh x dx |
20. |
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x cosh x dx |
21. |
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x(ln x)3dx |
22. |
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x arctan x dx |
23. |
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xarccot x dx |
24. |
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sin3 x dx |
25. |
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cos3 x dx |
26. |
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sin4 x dx |
27. |
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cos4 x dx |
28. |
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sinh2 x dx |
29. |
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cosh2 x dx |
30. |
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sinh3 x dx |
31. |
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x2 sinh x dx |
32. |
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x2 cosh x dx |
33. |
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x3 sinh x dx |
34. |
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x3 cosh x dx |
35. |
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x2e2xdx |
36. |
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x3e−xdx |
37. |
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x sin(3x) dx |
38. |
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x cos(x + 1)dx |
39. |
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x ln(x + 1)dx |
40. |
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x 2xdx |
41. |
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x 102xdx |
42. |
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x2 103xdx |
43. |
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x2(ln x)3dx |
44. |
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arcsinh (3x)dx |
45. |
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arccosh (2x)dx |
46. |
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arctanh (2x)dx |
47. |
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arccoth (3x)dx |
48. |
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xarcsec x dx |
Z
50.xarccsc x dx
5.5Logarithmic, Exponential and Hyperbolic Functions
With the Fundamental Theorems of Calculus it is possible to rigorously develop the logarithmic, exponential and hyperbolic functions.
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS231
Definition 5.5.1 For each x > 0 we define the natural logarithm of x, denoted ln x, by the equation
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x 1 |
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ln(x) = Z1 |
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t |
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Theorem 5.5.1 |
(Natural Logarithm) |
The natural logarithm, ln x, has the |
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following properties: |
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(ln x) = |
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dx |
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The natural logarithm is an increasing, continuous and di erentiable function on (0, ∞).
(ii)If a > 0 and b > 0, then ln(ab) = ln(a) + ln(b).
(iii)If a > 0 and b > 0, then ln(a/b) = ln(a) + ln(b).
(iv)If a > 0 and n is a natural number, then ln(an) = n ln a.
(v)The range of ln x is (−∞, ∞).
(vi)ln x is one-to-one and has a unique inverse, denoted ex.
Proof.
(i)Since 1/t is continuous on (0, ∞), (i) follows from the Fundamental Theorem of Calculus, Second Form.
(ii)Suppose that a > 0 and b > 0. Then
ln(ab) = Z1 |
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t |
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= ln a + Z1 |
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adu ; u = |
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t, du = |
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au |
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= ln a + ln b.
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CHAPTER 5. THE DEFINITE INTEGRAL |
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(iii) If a > 0 and b > 0, then |
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ln |
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= Z1 |
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dt + Za |
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t, du = |
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=ln a − ln b.
(iv)If a > 0 and n is a natural number, then
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ln(an) = Z1 a |
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dt ; t = un, dt = nun−1du |
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as required.
(v)From the partition {1, 2, 3, 4, · · · }, we get the following inequality using upper and lower sum approximations:
graph
1312 = 12 + 31 + 41 < ln 4 < 1 + 21 + 31.
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS233
Hence, ln 4 > 1. ln(4n) = n ln 4 > n and ln 4−n = −n ln 4 < −n. By the intermediate value theorem, every interval (−n, n) is contained in the range of ln x. Therefore, the range of ln x is (−∞, ∞), since the derivative of ln x is always positive, ln x is increasing and hence one-to- one. The inverse of ln x exists.
(vi)Let e denote the number such that ln(e) = 1. Then we define y = ex if and only if x = ln(y) for x (−∞, ∞), y > 0.
This completes the proof.
Definition 5.5.2 If x is any real number, we define y = ex if and only if x = ln y.
Theorem 5.5.2 (Exponential Function) The function y = ex has the following properties:
(i)e0 = 1, ln(ex) = x for every real x and dxd (ex) = ex.
(ii)ea · eb = ea+b for all real numbers a and b.
(iii)ea = ea−b for all real numbers a and b. eb
(iv)(ea)n = ena for all real numbers a and natural numbers n.
Proof.
(i)Since ln(1) = 0, e0 = 1. By definition y = ex if and only if x = ln(y) = ln(ex). Suppose y = ex. Then x = ln y. By implicit di erentiation, we get
1 = y1 dxdy , dxdy = y = ex.
Therefore,
dxd (ex) = ex.
234 |
CHAPTER 5. THE DEFINITE INTEGRAL |
(ii) |
Since ln x is increasing and, hence, one-to-one, |
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ea · eb = ea+b ↔ |
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ln(ea · eb) = ln(ea+b) ↔ |
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ln(ea) + ln(eb) = a + b ↔ |
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a + b = a + b. |
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It follows that for all real numbers a and b, |
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ea · eb = ea+b. |
(iii)ea = ea−b ↔
eb
ea
ln eb = ln(ea−b) ↔
ln(ea) − ln(eb) = a − b ↔
a − b = a − b.
It follows that for all real numbers a and b,
ea = ea−b. eb
(iv)(ea)n = ena ↔
ln((ea)n) = ln(ena) ↔
n ln(ea) = na ↔
na = na.
Therefore, for all real numbers a and natural numbers n, we have
(ea)n = ena.
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS235
Definition 5.5.3 Suppose b > 0 and b 6= 1. Then we define the following:
(i) |
For each real number x, bx = ex ln b. |
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y = logb x = |
ln x |
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ln b |
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Theorem 5.5.3 |
(General Exponential Function) Suppose b > 0 and b 6= 1. |
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Then |
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(i)ln(bx) = x ln b, for all real numbers x.
(ii)dxd (bx) = bx ln b, for all real numbers x.
(iii)bx1 · bx2 = bx1+x2 , for all real numbers x1 and x2.
(iv)bx1 = bx1−x2 , for all real numbers x1 and x2. bx2
(v)(bx1 )x2 = bx1x2 , for all real numbers x1 and x2.
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bx dx = |
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+ c. |
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ln b |
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Proof. |
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(i) |
ln(bx) = ln(ex ln b) = x ln b |
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(bx) = |
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(ex ln b) = ex ln b · (ln b) (by the chain rule) |
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=bx ln b.
(iii)bx1 · bx2 = ex1 ln b · ex2 ln b
=e(x1 ln b+x2 ln b)
=e(x1+x2) ln b
=b(x1+x2)
236 |
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CHAPTER 5. THE DEFINITE INTEGRAL |
(iv) |
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ex1 ln b |
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=ex1 ln b−x2 ln b
=e(x1−x2) ln b
=b(x1−x2).
(v)By Definition 5.5.3 (i), we get
(b |
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x2 ln(bx1 ) |
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=ex2 ln(ex1 ln b)
=ex2·x1 ln b
=e(x1x2) ln b
=bx1x2 .
(vi) Since |
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(bx) = bx ln b, |
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we get |
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bx(ln b) dx = bx + c, |
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ln b Z bx dx = bx + c, |
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ex dx = |
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+ D, |
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ln b |
where D is some constant. This completes the proof.
Theorem 5.5.4 If u(x) > 0 for all x, and u(x) and v(x) are di erentiable functions, then we define
y = (u(x))v(x) = ev(x) ln(u(x)).
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS237
Then y is a di erentiable function of x and |
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= dx (u(x))v(x) = (u(x))v(x) v0 |
(x) ln(u(x)) + v(x) |
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Proof. This theorem follows by the chain rule and the product rule as follows
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[uv] = dx [ev ln u] = ev ln u v0 |
ln u + v |
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= uv v0 ln u + v |
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Theorem 5.5.5 The following di erentiation formulas for the hyperbolic functions are valid.
(i) |
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(sinh x) = cosh x |
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(iii) |
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(coth x) = −csch2x |
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(csch x) = −csch x coth x |
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Proof. We use the definitions and properties of hyperbolic functions given in Chapter 1 and the di erentiation formulas of this chapter.
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(ii) |
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(cosh x) = |
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(iii) |
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(tanh x) = |
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sinh x |
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cosh x |
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(cosh x)2 |
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cosh x)2 |
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238 |
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CHAPTER 5. THE DEFINITE INTEGRAL |
(iv) |
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(coth x) = |
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=−sinh2 x · cosh2 x = −sinh2 x
=−csch2x.
(v)dxd (sech x) = dxd (cosh x)−1 = −1(cosh x)−2 · sinh x
=− sech x tanh x.
(vi)dxd (csch x) = dxd (sinh x)−1 = −1(sinh x)−2 · cosh x
=− coth x csch x.cosh2 x
This completes the proof.
Theorem 5.5.6 The following integration formulas are valid:
(i) |
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sinh x dx = cosh x + c |
(ii) |
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cosh x dx = sinh x + c |
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(iii) |
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tanh x dx = ln(cosh x) + c |
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coth xdx = ln | sinh x| + c |
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(v) |
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sech x dx = 2 arctan(ex) + c |
(vi) |
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csch x dx = ln |
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tanh |
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Proof. Each formula can be easily verified by di erentiating the right-hand side to get the integrands on the left-hand side. This proof is left as an exercise.
Theorem 5.5.7 The following di erentiation and integration formulas are valid:
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS239
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(arcsinh x) = |
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√ |
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arcsinh x + c |
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1 + x2 |
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(arccosh x) = |
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(iv) |
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arccosh x + c |
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(arctanh x) = 1 − x2 , |x| < 1 |
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Proof. This theorem follows directly from the following definitions: |
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arcsinh x = ln(x + √ |
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(2) arccosh x = ln(x + √ |
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1 + x2 |
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x2 − 1) |
(3) arctanh x = |
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The proof is left as an exercise.
Exercises 5.5
1.Prove Theorem 5.5.6.
2.Prove Theorem 5.5.7.
3.Show that sinh mx and cosh mx are linearly independent if m 6= 0. (Hint:
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Show that the Wronskian W (sinh mx, cosh mx) is not zero if m 6= 0.) |
4. |
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Show that emx and e−mx are linearly independent if m = 0. |
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Show that solution of the equation y00 − m2y = 0 can be expressed as |
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y = c1emx + c2e−mx. |
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Show that every solution of y00 − m2y = 0 can be written as y = |
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A sinh mx + B cosh mx. |
7.Determine the relation between c1 and c2 in problem 5 with A and B in problem 6.
8.Prove the basic identities for hyperbolic functions:
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CHAPTER 5. THE DEFINITE INTEGRAL |
(i)sinh(x + y) = sinh x cosh y + cosh x sinh y.
(ii)sinh(x − y) = sinh x cosh y − cosh x sinh y.
(iii)cosh(x + y) = cosh x cosh y + sinh x sinh y.
(iv)cosh(x − y) = cosh x cosh y − sinh x sinh y.
(v)sinh 2x = 2 sinh x cosh x.
(vi)cosh2 x + sinh2 x = 2 cosh2 x − 1 = 1 + 2 sinh2 x = cosh 2x.
(vii) cosh2 x − sinh2 x = 1, 1 − tanh2 x = sech2x, coth2 x − 1 = csch2x.
9. Eliminate the |
radical sign using the given substitution: |
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√ |
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a2 + x2 |
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x = a sinh t |
(ii) |
a2 − x2 |
√
(iii)x2 − a2, x = a cosh t.
10.Compute y0 in each of the following:
(i) |
y = 2 sinh(3x) + 4 cosh(2x) |
(ii) |
y = 4 tanh(5x) − 6 coth(3x) |
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y = x sech (2x) + x2 csch (5x) |
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y = 3 sinh2(4x + 1) |
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y = 4 cosh2(2x − 1) |
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(vi) |
y = sinh(2x) cosh(3x) |
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11. Compute y0 in each of the following: |
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(i) |
y = x2e−x3 |
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(ii) |
y = 2x2 |
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(iii) |
y = (x2 + 1)sin(2x) |
(iv) |
y = log10(x |
2 |
+ 1) |
(v) |
y = log2 |
(sec x + tan x)(vi) |
(x3+1) |
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y = 10 |
12. Compute y0 in each of the following:
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√ |
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√ |
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(i) y = x ln x − x |
x |
2 |
− 4) |
2 |
) |
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(ii) y = ln(x + |
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(iii) y = ln(x + |
4 + x |
(iv) y = |
1 |
ln |
1 + x |
(v) y = arcsinh (3x) |
(vi) y = arccosh (3x) |
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2 |
1 − x |
5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS241
13. |
Evaluate each of the following integrals: |
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(i) |
Z |
sinh(3x) dx |
(ii) |
Z |
x3ex2 dx |
(iii) |
Z |
x2 ln(x + 1) dx |
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(iv) |
Z |
x sinh 2x dx |
(v) |
Z |
x cosh 3x dx |
(vi) |
Z |
x4x2 |
dx |
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14. |
Evaluate each of the following integrals: |
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(i) |
Z |
arcsinh x dx |
(ii) |
Z |
arccosh x dx |
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(iii) |
Z |
arctanh x dx |
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(iv) |
Z |
√4 − x2 |
(v) |
Z |
√4 + x2 |
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(vi) |
Z |
√x2 − 4 |
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dx |
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dx |
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dx |
15.Logarithmic Di erentiation is a process of computing derivatives by first taking logarithms and then using implicit di erentiation. Find y0 in each of the following, using logarithmic di erentiation.
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(x2 + 1)3(x2 + 4)10 |
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2 |
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(x3+1) |
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(i) |
y = |
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(ii) |
y = (x |
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+ 4) |
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(x2 + 2)5(x2 + 3)4 |
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(iii) |
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(4 cos x+7) |
(iv) |
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(x3 |
+1) |
y = (sin x + 3) |
y = (3 sinh x + cos x + 5) |
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(v) |
y = (e |
x2 |
(2x+1) |
(vi) |
2 |
(x |
2 |
(x3 |
+1) |
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+ 1) |
y = x |
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+ 1) |
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In problems 16–30, compute f0(x) each f(x).
16. |
f(x) = Z1x sinh3(t)dt |
17. |
f(x) = Zxx2 |
cosh5(t)dt |
|
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cosh x |
|
|
sech x |
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18. |
f(x) = Zsinh x |
(1 + t2)3/2dt |
19. |
f(x) = Ztanh x (1 + t3)1/2dt |
|
|
(ln x)2 |
|
ex2 |
||
20. |
f(x) = Zln x |
(4 + t2)5/2dt |
21. |
f(x) = Zex2 |
(1 + 4t2)πdt |