12 derivatives and integrals of inverse trigonometric functions x

Derivatives and Integrals of Inverse Trigonometric Functions

1. y = f(x) y = f –1 (x) The graphs of f and f –1 are symmetric diagonally. y = x Derivatives and Integrals of the Inverse Trigonometric Functions

2. y = f(x) y = f –1 (x) The graphs of f and f –1 are symmetric diagonally. y = x Derivatives and Integrals of the Inverse Trigonometric Functions (a, b) Assume f(x) is differentiable and (a, b) is a point on the graph of y = f(x).

3. y = f(x) y = f –1 (x) The graphs of f and f –1 are symmetric diagonally. y = x Derivatives and Integrals of the Inverse Trigonometric Functions (a, b) Assume f(x) is differentiable and (a, b) is a point on the graph of y = f(x). Hence the slope at (a, b) is f ’(a). slope = f ’(a)

4. y = f(x) y = f –1 (x) The graphs of f and f –1 are symmetric diagonally. y = x Derivatives and Integrals of the Inverse Trigonometric Functions (a, b) Assume f(x) is differentiable and (a, b) is a point on the graph of y = f(x). Hence the slope at (a, b) is f ’(a). (b, a) slope = f ’(a) The reflection of (a, b) is (b, a) on the graph of y = f –1 (x).

5. y = f(x) y = f –1 (x) The graphs of f and f –1 are symmetric diagonally. y = x Derivatives and Integrals of the Inverse Trigonometric Functions (a, b) Assume f(x) is differentiable and (a, b) is a point on the graph of y = f(x). Hence the slope at (a, b) is f ’(a). (b, a) slope = f ’(a) The reflection of (a, b) is (b, a) on the graph of y = f –1 (x). The slope of the tangent line at (b, a) is (f –1 )’(b). slope = (f –1 )’(b)

6. The graphs of f and f –1 are symmetric diagonally. Derivatives and Integrals of the Inverse Trigonometric Functions Assume f(x) is differentiable and (a, b) is a point on the graph of y = f(x). Hence the slope at (a, b) is f ’(a). The reflection of (a, b) is (b, a) on the graph of y = f –1 (x). The slope of the tangent line at (b, a) is (f –1 )’(b). By the symmetry the tangents at (a, b) and at (b, a) are the reflection of each other. y = f(x) y = f –1 (x) y = x (a, b) (b, a) slope = f ’(a) slope = (f –1 )’(b)

7. The graphs of f and f –1 are symmetric diagonally. Derivatives and Integrals of the Inverse Trigonometric Functions Assume f(x) is differentiable and (a, b) is a point on the graph of y = f(x). Hence the slope at (a, b) is f ’(a). The reflection of (a, b) is (b, a) on the graph of y = f –1 (x). The slope of the tangent line at (b, a) is (f –1 )’(b). By the symmetry the tangents at (a, b) and at (b, a) are the reflection of each other. Diagonally-symmetric lines have reciprocal slopes (why?) y = f(x) y = f –1 (x) y = x (a, b) (b, a) slope = f ’(a) slope = (f –1 )’(b)

8. The graphs of f and f –1 are symmetric diagonally. Derivatives and Integrals of the Inverse Trigonometric Functions Assume f(x) is differentiable and (a, b) is a point on the graph of y = f(x). Hence the slope at (a, b) is f ’(a). The reflection of (a, b) is (b, a) on the graph of y = f –1 (x). The slope of the tangent line at (b, a) is (f –1 )’(b). By the symmetry the tangents at (a, b) and at (b, a) are the reflection of each other. Diagonally-symmetric lines have reciprocal slopes (why?) y = f(x) y = f –1 (x) y = x (a, b) (b, a) slope = f ’(a) slope = (f –1 )’(b) Hence of (f –1 )’(b) = 1 f ’ (a)

9. The graphs of f and f –1 are symmetric diagonally. Derivatives and Integrals of the Inverse Trigonometric Functions Assume f(x) is differentiable and (a, b) is a point on the graph of y = f(x). Hence the slope at (a, b) is f ’(a). The reflection of (a, b) is (b, a) on the graph of y = f –1 (x). The slope of the tangent line at (b, a) is (f –1 )’(b). By the symmetry the tangents at (a, b) and at (b, a) are the reflection of each other. Diagonally-symmetric lines have reciprocal slopes (why?) y = f(x) y = f –1 (x) y = x (a, b) (b, a) slope = f ’(a) slope = (f –1 )’(b) Hence of (f –1 )’(b) = 1 = 1 f ’( f –1 (b) ) f ’ ( a )

10. Suppose f and g are a pair of inverse functions, then (f o g)(x) = x. Derivatives and Integrals of the Inverse Trigonometric Functions

11. Suppose f and g are a pair of inverse functions , then (f o g)(x) = x. Differentiate both sides with respect to x and use the chain rule: [(f o g)(x)]' = x' Derivatives and Integrals of the Inverse Trigonometric Functions

12. Suppose f and g are a pair of inverse functions , then (f o g)(x) = x. Differentiate both sides with respect to x and use the chain rule: [(f o g)(x)]' = x' f '( g(x) ) * g'(x) = 1 Derivatives and Integrals of the Inverse Trigonometric Functions

13. Suppose f and g are a pair of inverse functions , then (f o g)(x) = x. Differentiate both sides with respect to x and use the chain rule: [(f o g)(x)]' = x' f '( g(x) ) * g'(x) = 1 or g'(x) = 1 f '( g(x) ) Derivatives and Integrals of the Inverse Trigonometric Functions

14. Suppose f and g are a pair of inverse functions , then (f o g)(x) = x. Differentiate both sides with respect to x and use the chain rule: [(f o g)(x)]' = x' f '( g(x) ) * g'(x) = 1 or g'(x) = 1 f '( g(x) ) Set f = sin(x) and g = arcsin (x) Derivatives and Integrals of the Inverse Trigonometric Functions

15. Suppose f and g are a pair of inverse functions , then (f o g)(x) = x. Differentiate both sides with respect to x and use the chain rule: [(f o g)(x)]' = x' f '( g(x) ) * g'(x) = 1 or g'(x) = 1 f '( g(x) ) Set f = sin(x) and g = arcsin(x) we obtain [arcsin(x)]' = 1 dsin(y) Derivatives and Integrals of the Inverse Trigonometric Functions dy y=arcsin(x)

16. Suppose f and g are a pair of inverse functions , then (f o g)(x) = x. Differentiate both sides with respect to x and use the chain rule: [(f o g)(x)]' = x' f '( g(x) ) * g'(x) = 1 or g'(x) = 1 f '( g(x) ) Set f = sin(x) and g = arcsin(x) we obtain [arcsin(x)]' = 1 dsin(y) 1 cos( arcsin(x) ) = Derivatives and Integrals of the Inverse Trigonometric Functions dy y=arcsin(x)

17. [arcsin(x)]' 1 cos( arcsin(x) ) = Derivatives and Integrals of the Inverse Trigonometric Functions

18. θ =arcsin(x) x 1 [arcsin(x)]' 1 cos( arcsin(x) ) = Derivatives and Integrals of the Inverse Trigonometric Functions

19. θ = arcsin (x) x 1  1 – x 2 [arcsin(x)]' 1 cos( arcsin(x) ) = Derivatives and Integrals of the Inverse Trigonometric Functions

20. θ =arcsin(x) x 1  1 – x 2 [arcsin(x)]' 1 cos( arcsin(x) ) = 1  1 – x 2 [arcsin(x)]' = Derivatives and Integrals of the Inverse Trigonometric Functions

21. θ = arcsin (x) x 1  1 – x 2 [arcsin(x)]' 1 cos( arcsin(x) ) = 1  1 – x 2 [arcsin(x)]' = Derivatives and Integrals of the Inverse Trigonometric Functions –1 1 – π /2 π /2 x y= arcsin (x )

22. θ =arcsin(x) x 1  1 – x 2 [arcsin(x)]' 1 cos( arcsin(x) ) = 1  1 – x 2 [arcsin(x)]' = We use the same technique to obtain the derivatives of the other inverse trig-functions. Derivatives and Integrals of the Inverse Trigonometric Functions –1 1 – π /2 π /2 x y= arcsin (x )

23. θ =arcsin(x) x 1  1 – x 2 [arcsin(x)]' 1 cos( arcsin(x) ) = 1  1 – x 2 [arcsin(x)]' = We use the same technique to obtain the derivatives of the other inverse trig-functions. Derivatives and Integrals of the Inverse Trigonometric Functions –1 1 – π /2 π /2 x y= arcsin (x ) We display the graphs of each inverse–trig and list each of their derivatives below.

24. Derivatives of the Inverse Trig–Functions 1  1 – x 2 [sin –1 (x)] ' = y =sin –1 (x)

25. Derivatives of the Inverse Trig–Functions [cos –1 (x)]' = 1  1 – x 2 [sin –1 (x)] ' = –1  1 – x 2 y =sin –1 (x) y =cos –1 (x)

26. Derivatives of the Inverse Trig–Functions [cos –1 (x)]' = 1 1 + x 2 [tan –1 (x)]' = 1  1 – x 2 [sin –1 (x)] ' = –1  1 – x 2 y =tan –1 (x) y =sin –1 (x) y =cos –1 (x)

27. Derivatives of the Inverse Trig–Functions [cos –1 (x)]' = 1 1 + x 2 [tan –1 (x)]' = –1 1 + x 2 [cot –1 (x)]' = 1  1 – x 2 [sin –1 (x)] ' = –1  1 – x 2 y =tan –1 (x) y =sin –1 (x) y =cos –1 (x) y =cot –1 (x)

28. Derivatives of the Inverse Trig–Functions [cos –1 (x)]' = 1 1 + x 2 [tan –1 (x)]' = |x|  x 2 – 1 [sec –1 (x)]' = –1 1 + x 2 [cot –1 (x)]' = y =sec –1 (x) 1  1 – x 2 [sin –1 (x)] ' = –1  1 – x 2 1 y =tan –1 (x) y =sin –1 (x) y =cos –1 (x) y =cot –1 (x)

29. Derivatives of the Inverse Trig–Functions [cos –1 (x)]' = 1 1 + x 2 [tan –1 (x)]' = |x|  x 2 – 1 [sec –1 (x)]' = –1 1 + x 2 [cot –1 (x)]' = –1 |x|  x 2 – 1 [csc –1 (x)]' = y =sec –1 (x) y =csc –1 (x) 1  1 – x 2 [sin –1 (x)] ' = –1  1 – x 2 1 y =tan –1 (x) y =sin –1 (x) y =cos –1 (x) y =cot –1 (x)

30. Derivatives of the Inverse Trig–Functions u'  1 – u 2 [sin –1 (u )]' = –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = –u ' 1 + u 2 [cot –1 (u )]' = –u ' |u|  u 2 – 1 [csc –1 (u )]' = dsin -1 (u) dx = 1  1 – u 2 du dx –1  1 – u 2 du dx 1 1 + u 2 du dx 1 |u|  u 2 – 1 du dx dcos -1 (u) dx = dtan -1 (u) dx = dsec -1 (u) dx = –1 1 + u 2 du dx dtan -1 (u) dx = –1 |u|  u 2 – 1 du dx dcsc -1 (u) dx = Below are the chain–rule versions where u = u(x).

31. Derivatives and Integrals of the Inverse Trigonometric Functions –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example.

32. Derivatives and Integrals of the Inverse Trigonometric Functions b. cos –1 (e x ) Example A. Find the following derivatives. a. tan –1 (x 3 ) –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example. 2 c. sec –1 (ln(x ))

33. Derivatives and Integrals of the Inverse Trigonometric Functions b. cos –1 (e x ) 2 Example A. Find the following derivatives. Set u = x 3 , a. tan –1 (x 3 ) –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example. c. sec –1 (ln(x ))

34. Derivatives and Integrals of the Inverse Trigonometric Functions b. cos –1 (e x ) 2 Example A. Find the following derivatives. Set u = x 3 , so [ tan –1 (x 3 )]' = (x 3 ) ' 1 + (x 3 ) 2 a. tan –1 (x 3 ) –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example. c. sec –1 (ln(x ))

35. Derivatives and Integrals of the Inverse Trigonometric Functions b. cos –1 (e x ) 2 Example A. Find the following derivatives. Set u = x 3 , so [ tan –1 (x 3 )]' = (x 3 ) ' 1 + (x 3 ) 2 = 3x 2 1 + x 6 a. tan –1 (x 3 ) –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example. c. sec –1 (ln(x ))

36. Derivatives and Integrals of the Inverse Trigonometric Functions b. cos –1 (e x ) Set u = e x , 2 Example A. Find the following derivatives. Set u = x 3 , so [ tan –1 (x 3 )]' = (x 3 ) ' 1 + (x 3 ) 2 = 3x 2 1 + x 6 a. tan –1 (x 3 ) –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example. c. sec –1 (ln(x ))

37. Derivatives and Integrals of the Inverse Trigonometric Functions b. cos –1 (e x ) Set u = e x , so [cos -1 (e x )]' = 2 2  1 – (e x ) 2 –( e x )' 2 2 Example A. Find the following derivatives. Set u = x 3 , so [ tan –1 (x 3 )]' = (x 3 ) ' 1 + (x 3 ) 2 = 3x 2 1 + x 6 a. tan –1 (x 3 ) –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example. c. sec –1 (ln(x ))

38. Derivatives and Integrals of the Inverse Trigonometric Functions b. cos –1 (e x ) Set u = e x , so [cos -1 (e x )]' = 2 2  1 – (e x ) 2 –( e x )' 2 2 = –2xe x  1 – e 2x 2 2 Example A. Find the following derivatives. Set u = x 3 , so [ tan –1 (x 3 )]' = (x 3 ) ' 1 + (x 3 ) 2 = 3x 2 1 + x 6 a. tan –1 (x 3 ) –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example. c. sec –1 (ln(x ))

39. Derivatives and Integrals of the Inverse Trigonometric Functions b. cos –1 (e x ) Set u = e x , so [cos -1 (e x )]' = 2 2  1 – (e x ) 2 –( e x )' 2 2 = –2xe x  1 – e 2x 2 2 Set u = ln (x ), so [sec –1 ( ln (x)]' = Example A. Find the following derivatives. Set u = x 3 , so [ tan –1 (x 3 )]' = (x 3 ) ' 1 + (x 3 ) 2 = 3x 2 1 + x 6 a. tan –1 (x 3 ) –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example. c. sec –1 (ln(x ))

40. Derivatives and Integrals of the Inverse Trigonometric Functions b. cos –1 (e x ) Set u = e x , so [cos -1 (e x )]' = 2 2  1 – (e x ) 2 –( e x )' 2 2 = –2xe x  1 – e 2x 2 2 c. sec –1 (ln(x )) Set u = ln (x), so [sec –1 ( ln (x )]' = 1/x |ln(x)|  ln 2 (x) – 1 Example A. Find the following derivatives. Set u = x 3 , so [ tan –1 (x 3 )]' = (x 3 ) ' 1 + (x 3 ) 2 = 3x 2 1 + x 6 a. tan –1 (x 3 ) –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example.

41. Derivatives and Integrals of the Inverse Trigonometric Functions b. cos –1 (e x ) Set u = e x , so [cos -1 (e x )]' = 2 2  1 – (e x ) 2 –( e x )' 2 2 = –2xe x  1 – e 2x 2 2 Set u = ln (x), so [sec –1 ( ln (x )]' = 1/x |ln(x)|  ln 2 (x) – 1 = 1 x|ln(x)||  ln 2 (x) – 1 Example A. Find the following derivatives. Set u = x 3 , so [ tan –1 (x 3 )]' = (x 3 ) ' 1 + (x 3 ) 2 = 3x 2 1 + x 6 a. tan –1 (x 3 ) –u '  1 – u 2 [cos –1 ( u )]' = u' 1 + u 2 [tan –1 (u )]' = u' |u|  u 2 – 1 [sec –1 (u )]' = We’ll use the following formulas for the next example. c. sec –1 (ln(x ))

42. Derivatives and Integrals of the Inverse Trigonometric Functions = sin -1 (u) + C  1 – u 2 du Expressing the relations in integrals: ∫

43. Derivatives and Integrals of the Inverse Trigonometric Functions = sin -1 (u) + C  1 – u 2 du Expressing the relations in integrals: ∫ = cos -1 (u) + C  1 – u 2 - du ∫

44. Derivatives and Integrals of the Inverse Trigonometric Functions = sin -1 (u) + C  1 – u 2 du Expressing the relations in integrals: ∫ = cos -1 (u) + C  1 – u 2 - du ∫ = tan -1 (u) + C du ∫ 1 + u 2

45. Derivatives and Integrals of the Inverse Trigonometric Functions = sin -1 (u) + C  1 – u 2 du |u|  u 2 – 1 Expressing the relations in integrals: ∫ = cos -1 (u) + C  1 – u 2 - du ∫ = tan -1 (u) + C du ∫ 1 + u 2 = sec -1 (u) + C ∫ du

46. Derivatives and Integrals of the Inverse Trigonometric Functions = sin -1 (u) + C  1 – u 2 du Expressing the relations in integrals: ∫ = cos -1 (u) + C  1 – u 2 - du ∫ = tan -1 (u) + C du ∫ 1 + u 2 = sec -1 (u) + C ∫ Example B. Find the integral ∫ dx 9 + 4x 2 |u|  u 2 – 1 du

47. Derivatives and Integrals of the Inverse Trigonometric Functions = sin -1 (u) + C  1 – u 2 du Expressing the relations in integrals: ∫ = cos -1 (u) + C  1 – u 2 - du ∫ = tan -1 (u) + C du ∫ 1 + u 2 = sec -1 (u) + C ∫ Match the form of the integral to the one for tan -1 (u). |u|  u 2 – 1 du Example B. Find the integral ∫ dx 9 + 4x 2

48. Derivatives and Integrals of the Inverse Trigonometric Functions = sin -1 (u) + C  1 – u 2 du Expressing the relations in integrals: ∫ = cos -1 (u) + C  1 – u 2 - du ∫ = tan -1 (u) + C du ∫ 1 + u 2 = sec -1 (u) + C ∫ Match the form of the integral to the one for tan -1 (u). Write 9 + 4x 2 = 9 (1 + x 2 ) 4 9 |u|  u 2 – 1 du Example B. Find the integral ∫ dx 9 + 4x 2

49. Derivatives and Integrals of the Inverse Trigonometric Functions = sin -1 (u) + C  1 – u 2 du Expressing the relations in integrals: ∫ = cos -1 (u) + C  1 – u 2 - du ∫ = tan -1 (u) + C du ∫ 1 + u 2 = sec -1 (u) + C ∫ Match the form of the integral to the one for tan -1 (u). Write 9 + 4x 2 = 9 (1 + x 2 ) = 9 [1 + ( x) 2 ] 2 3 4 9 |u|  u 2 – 1 du Example B. Find the integral ∫ dx 9 + 4x 2

50. Derivatives and Integrals of the Inverse Trigonometric Functions = sin -1 (u) + C  1 – u 2 du Expressing the relations in integrals: ∫ = cos -1 (u) + C  1 – u 2 - du ∫ = tan -1 (u) + C du ∫ 1 + u 2 = sec -1 (u) + C ∫ Match the form of the integral to the one for tan -1 (u). Write 9 + 4x 2 = 9 (1 + x 2 ) = 9 [1 + ( x) 2 ] 2 3 4 9 Hence dx 9 + 4x 2 ∫ = dx 1 + ( x) 2 ∫ 1 9 2 3 |u|  u 2 – 1 du Example B. Find the integral ∫ dx 9 + 4x 2

51. Derivatives and Integrals of the Inverse Trigonometric Functions dx 1 + ( x) 2 ∫ 1 9 2 3 substitution method

52. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = dx 1 + ( x) 2 ∫ 1 9 2 3 2 3 x substitution method

53. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = dx 1 + ( x) 2 ∫ 1 9 2 3 2 3 x  du dx = 2 3 substitution method

54. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = dx 1 + ( x) 2 ∫ 1 9 2 3 2 3 x  du dx = 2 3 So dx = 3 2 du substitution method

55. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = dx 1 + ( x) 2 ∫ 1 9 2 3 2 3 x  du dx = 2 3 So dx = 3 2 du = ∫ 1 9 1 1 + u 2 3 2 du substitution method

56. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = dx 1 + ( x) 2 ∫ 1 9 2 3 2 3 x  du dx = 2 3 So dx = 3 2 du = ∫ 1 9 1 1 + u 2 3 2 du = ∫ 1 6 1 1 + u 2 du substitution method

57. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = dx 1 + ( x) 2 ∫ 1 9 2 3 2 3 x  du dx = 2 3 So dx = 3 2 du = ∫ 1 9 1 1 + u 2 3 2 du = ∫ 1 6 1 1 + u 2 du = tan -1 (u) + C 1 6 substitution method

58. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = dx 1 + ( x) 2 ∫ 1 9 2 3 2 3 x  du dx = 2 3 So dx = 3 2 du = ∫ 1 9 1 1 + u 2 3 2 du = ∫ 1 6 1 1 + u 2 du = tan -1 (u) + C 1 6 = tan -1 ( x) + C 1 6 2 3 substitution method

59. Derivatives and Integrals of the Inverse Trigonometric Functions e x ∫ Example C. Find the definite integral  1 – e 2x dx ln(1/2)

60. Derivatives and Integrals of the Inverse Trigonometric Functions e x ∫ Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2)

61. Derivatives and Integrals of the Inverse Trigonometric Functions e x ∫ substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2)

62. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫ substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x e x ∫  1 – e 2x dx ln(1/2)

63. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x e x e x ∫  1 – e 2x dx ln(1/2)

64. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x e x e x ∫  1 – e 2x dx ln(1/2)

65. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x e x for x = ln(1/2)  u = 1/2 e x ∫  1 – e 2x dx ln(1/2)

66. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2) e x e x for x = ln(1/2)  u = 1/2 x = 0  u = 1

67. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x = substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2) e x e x for x = ln(1/2)  u = 1/2 x = 0  u = 1 e x ∫  1 – u 2

68. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x = substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2) e x e x for x = ln(1/2)  u = 1/2 x = 0  u = 1 e x ∫  1 – u 2 du e x

69. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x = substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2) e x e x for x = ln(1/2)  u = 1/2 x = 0  u = 1 e x ∫  1 – u 2 du 1 1/2 e x

70. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x = substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2) e x e x for x = ln(1/2)  u = 1/2 x = 0  u = 1 e x ∫  1 – u 2 du 1 1/2 e x = ∫  1 – u 2 du 1 1/2

71. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x = substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2) e x e x for x = ln(1/2)  u = 1/2 x = 0  u = 1 e x ∫  1 – u 2 du 1 1/2 e x = ∫  1 – u 2 du 1 1/2 = sin -1 (u) | 1/2 1

72. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x = substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2) e x e x for x = ln(1/2)  u = 1/2 x = 0  u = 1 e x ∫  1 – u 2 du 1 1/2 e x = ∫  1 – u 2 du 1 1/2 = sin -1 (u) | 1/2 1 = sin -1 (1) – sin -1 (1/2)

73. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x = substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2) e x e x for x = ln(1/2)  u = 1/2 x = 0  u = 1 e x ∫  1 – u 2 du 1 1/2 e x = ∫  1 – u 2 du 1 1/2 = sin -1 (u) | 1/2 1 = sin -1 (1) – sin -1 (1/2) = π /2 – π /6

74. Derivatives and Integrals of the Inverse Trigonometric Functions Set u = e x ∫  du dx = So dx = du/e x = substitution method Example C. Find the definite integral  1 – e 2x dx ln(1/2) e x ∫  1 – e 2x dx ln(1/2) e x e x for x = ln(1/2)  u = 1/2 x = 0  u = 1 e x ∫  1 – u 2 du 1 1/2 e x = ∫  1 – u 2 du 1 1/2 = sin -1 (u) | 1/2 1 = sin -1 (1) – sin -1 (1/2) = π /2 – π /6 = π /3

75. Lastly, we have the hyperbolic trigonometric functions. Derivatives and Integrals of the Inverse Trigonometric Functions

76. Lastly, we have the hyperbolic trigonometric functions. These functions are made from the exponential functions with relations among them that are similar to the trig-family. Derivatives and Integrals of the Inverse Trigonometric Functions

77. Derivatives and Integrals of the Inverse Trigonometric Functions We define the hyperbolic sine as sinh (x) = (pronounced as " sinch of x "). e x – e -x 2 Lastly, we have the hyperbolic trigonometric functions. These functions are made from the exponential functions with relations among them that are similar to the trig-family.

78. Derivatives and Integrals of the Inverse Trigonometric Functions We define the hyperbolic sine as sinh (x) = (pronounced as " sinch of x") We define the hyperbolic cosine as cosh(x) = (pronounced as "cosh of x") e x – e -x 2 e x + e -x 2 Lastly, we have the hyperbolic trigonometric functions. These functions are made from the exponential functions with relations among them that are similar to the trig-family.

79. Derivatives and Integrals of the Inverse Trigonometric Functions We define the hyperbolic sine as sinh (x) = (pronounced as " sinch of x") We define the hyperbolic cosine as cosh(x) = (pronounced as "cosh of x") e x – e -x 2 e x + e -x 2 We define the hyperbolic tangent, cotangent, secant and cosecant as in the trig-family. Lastly, we have the hyperbolic trigonometric functions. These functions are made from the exponential functions with relations among them that are similar to the trig-family.

80. Derivatives and Integrals of the Inverse Trigonometric Functions The hyperbolic tangent: tanh(x) = sinh(x) cosh(x) = e x – e -x e x + e -x The hyperbolic cotangent: coth(x) = cosh(x) sinh(x) = e x – e -x e x + e -x The hyperbolic secant: sech(x) = 1 cosh(x) = e x + e -x 2 The hyperbolic cosecant: csch(x) = 1 sinh(x) = e x – e -x 2

81. Derivatives and Integrals of the Inverse Trigonometric Functions 3,11,12,33,77,78,82,89,90.

82. As with the trig-family, we have the hyperbolic-trig hexagram to help us with their relations. Derivatives and Integrals of the Inverse Trigonometric Functions

83. Hyperbolic Trig Hexagram Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 co-side As with the trig-family, we have the hyperbolic-trig hexagram to help us with their relations.

84. Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 Starting from any position, take two steps without turning, we have I = II III

85. Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 Starting from any position, take two steps without turning, we have I = II III For example, staring at cosh(x), go to sinh(x) then to tanh(x),

86. Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 Starting from any position, take two steps without turning, we have I = II III For example, staring at cosh(x), go to sinh(x) then to tanh(x), we get the relation cosh(x) = sinh(x) tanh(x)

87. Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 Starting from any position, take two steps without turning, we have I = II III For example, staring at cosh(x), go to sinh(x) then to tanh(x), we get the relation cosh(x) = , similarly sech(x) = sinh(x) tanh(x) csch(x) coth(x)

88. Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 Square-difference Relations:

89. Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 Square-difference Relations: The three upside down triangles give the sq-differernce relations.

90. Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 Square-difference Relations: The three upside down triangles give the sq-differernce relations. The difference of the squares on top is the square of the bottom one.

91. Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 Square-difference Relations: The three upside down triangles give the sq-differernce relations. The difference of the squares on top is the square of the bottom one. cosh 2 (x) – sinh 2 (x) = 1

92. Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 Square-difference Relations: The three upside down triangles give the sq-differernce relations. The difference of the squares on top is the square of the bottom one. cosh 2 (x) – sinh 2 (x) = 1 coth 2 (x) – 1 = csch 2 (x)

93. Derivatives and Integrals of the Inverse Trigonometric Functions sinh(x) cosh(x) coth(x) csch(x) tanh(x) sech(x) 1 Square-difference Relations: The three upside down triangles give the sq-differernce relations. The difference of the squares on top is the square of the bottom one. cosh 2 (x) – sinh 2 (x) = 1 coth 2 (x) – 1 = csch 2 (x) 1 – tanh 2 (x) = sech 2 (x)

94. Derivatives and Integrals of the Inverse Trigonometric Functions Hyperbolic trig-functions show up in engineering.

95. Derivatives and Integrals of the Inverse Trigonometric Functions Hyperbolic trig-functions show up in engineering. Specifically the graph y = cosh(x) gives the shape of a hanging cable.

96. Derivatives and Integrals of the Inverse Trigonometric Functions Graph of y = cosh(x) Hyperbolic trig-functions show up in engineering. Specifically the graph y = cosh(x) gives the shape of a hanging cable. (0, 1)

97. Derivatives and Integrals of the Inverse Trigonometric Functions The derivatives of the hyperbolic trig-functions are similar, but not the same as, the trig-family.

98. Derivatives and Integrals of the Inverse Trigonometric Functions The derivatives of the hyperbolic trig-functions are similar, but not the same as, the trig-family. One may easily check that: [sinh(x)]' = cosh(x) [cosh(x)]' = sinh(x)

99. Derivatives and Integrals of the Inverse Trigonometric Functions The derivatives of the hyperbolic trig-functions are similar, but not the same as, the trig-family. One may easily check that: [sinh(x)]' = cosh(x) [cosh(x)]' = sinh(x) [tanh(x)]' = sech 2 (x) [coth(x)]' = -csch 2 (x)

100. Derivatives and Integrals of the Inverse Trigonometric Functions The derivatives of the hyperbolic trig-functions are similar, but not the same as, the trig-family. One may easily check that: [sinh(x)]' = cosh(x) [cosh(x)]' = sinh(x) [tanh(x)]' = sech 2 (x) [coth(x)]' = -csch 2 (x) [sech(x)]' = -sech(x)tanh(x) [csch(x)]' = -csch(x)coth(x)  Frank Ma 2006

101. Derivatives and Integrals of the Inverse Trigonometric Functions The derivatives of the hyperbolic trig-functions are similar, but not the same as, the trig-family. One may easily check that: [sinh(x)]' = cosh(x) [cosh(x)]' = sinh(x) [tanh(x)]' = sech 2 (x) [coth(x)]' = -csch 2 (x) [sech(x)]' = -sech(x)tanh(x) [csch(x)]' = -csch(x)coth(x) HW. Write down the chain–rule versions of the derivatives of the hyperbolic trig-functions.