More Related Content Similar to Calculus 05-6 ccccccccccccccccccccccc.ppsx Similar to Calculus 05-6 ccccccccccccccccccccccc.ppsx (20) Calculus 05-6 ccccccccccccccccccccccc.ppsx2. Exercise
Consider 𝑦 =
𝑥𝑥 ln3𝑥
𝑒 𝑥sin−1𝑥
7/2
. Find 𝑦′
ln 𝑦 =
7
2
ln
𝑥𝑥 ln3𝑥
𝑒 𝑥sin−1𝑥
ln 𝑦 =
7
2
𝑥 ln 𝑥 + 3 ln ln 𝑥 − 𝑥 − ln sin−1
𝑥
By diff.
Solution
𝑦′
𝑦
=
7
2
1 + ln 𝑥 +
3
𝑥 ln 𝑥
−
1
2 𝑥
−
1
1 − 𝑥2 sin−1𝑥
𝑦′
=
7𝑦
2
1 + ln 𝑥 +
3
𝑥 ln 𝑥
−
1
2 𝑥
−
1
1 − 𝑥2 sin−1𝑥
3. Exercise
Consider 𝑓 𝑥 = 𝑥2 + 6 𝑥 + 3, 𝑥 ≥ −3.
Find 𝑓−1
′(3). Sketch 𝑓 and 𝑓−1
. Deduce its domain
and range.
∵ 𝑓 0 = 3
Solution
𝑓−1 ′(3) =
1
𝑓′ 𝑓−1
(3)
=
1
𝑓′ 0
=
1
6
.
𝑓′ 𝑥 = 2𝑥 + 6
4. Slide 1 - 4
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
𝑓 𝑥 = 𝑥2
+ 6 𝑥 + 3 = 𝑥 + 3 2
− 6
𝑦 = 𝑥 + 3 2
− 6
𝑦 + 6 = 𝑥 + 3 2
𝑦 + 6 = 𝑥 + 3
𝑦 + 6 − 3 = 𝑥
∴ 𝑓−1
= 𝑥 + 6 − 3
Domain = [−3, ∞[ Range = [−6, ∞[
Domain = [−6, ∞[ Range = [−3, ∞[
(−3, −6)
(−6, −3)
5. Slide 1 - 5
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
1
sinh csch
2 sinh
1
cosh sech
2 cosh
sinh cosh
tanh coth
cosh sinh
x x
x x
e e
x x
x
e e
x x
x
x x
x x
x x
Hyperbolic functions
DEFINITION
6. Slide 1 - 6
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Note that sinh 𝑥 has domain and range
cosh 𝑥 has domain and range [1, ∞[.
HYPERBOLIC FUNCTIONS
7. Slide 1 - 7
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The graph of tanh 𝑥 is shown.
It has the horizontal asymptotes y = ±1.
HYPERBOLIC FUNCTIONS
8. Slide 1 - 8
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
It can be proved that, if a heavy flexible cable
is suspended between two points at the same
height, it takes the shape of a curve with
equation y = c + a cosh(x/a) called a catenary.
The Latin word
catena means
‘chain.’
APPLICATIONS
9. Slide 1 - 9
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Another application occurs in the
description of ocean waves.
The velocity of a water wave with length L moving across a
body of water with depth d is modeled by
the function
where g is the acceleration due to gravity.
2
tanh
2
gL d
v
L
APPLICATIONS
10. Slide 1 - 10
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The hyperbolic functions satisfy
a number of identities that are similar to
well-known trigonometric identities.
HYPERBOLIC IDENTITIES
11. Slide 1 - 11
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2 2
2 2
sinh( ) sinh
cosh( ) cosh
cosh sinh 1
1 tanh sech
sinh( ) sinh cosh cosh sinh
cosh( ) cosh cosh sinh sinh
x x
x x
x x
x x
x y x y x y
x y x y x y
HYPERBOLIC IDENTITIES
We list some identities here.
sinh 2𝑥 = 2 sinh 𝑥 cosh 𝑥
cosh 2𝑥 = cosh2
𝑥 + sinh2
𝑥
12. Slide 1 - 12
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Prove:
a. cosh2x – sinh2x = 1
b. 1 – tanh2 x = sech2x
HYPERBOLIC FUNCTIONS Example 1
13. Slide 1 - 13
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
2 2
2 2
2 2 2 2
cosh sinh
2 2
2 2
4 4
4
1
4
x x x x
x x x x
e e e e
x x
e e e e
HYPERBOLIC FUNCTIONS Example 1 a
14. Slide 1 - 14
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
We start with the identity proved in (a):
cosh2x – sinh2x = 1
If we divide both sides by cosh2x, we get:
2
2 2
2 2
sinh 1
1
cosh cosh
or 1 tanh sech
x
x x
x x
HYPERBOLIC FUNCTIONS Example 1 b
15. Slide 1 - 15
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The identity proved in Example 1 a
gives a clue to the reason for the name
‘hyperbolic’ functions, as follows.
HYPERBOLIC FUNCTIONS
16. Slide 1 - 16
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The derivatives of the hyperbolic
functions are easily computed.
For example,
(sinh ) cosh
2 2
x x x x
d d e e e e
x x
dx dx
DERIVATIVES OF HYPERBOLIC FUNCTIONS
17. Slide 1 - 17
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
We list the differentiation formulas for
the hyperbolic functions here.
DERIVATIVES
18. Slide 1 - 18
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Any of these differentiation rules can be combined with the
Chain Rule.
sinh
(cosh ) sinh
2
d d x
x x x
dx dx x
1.
𝑑
𝑑𝑥
sinh 𝑓 = cosh 𝑓 . 𝑓′
2.
𝑑
𝑑𝑥
cosh 𝑓 = sinh 𝑓 . 𝑓′
3.
𝑑
𝑑𝑥
tanh 𝑓 = sech2 𝑓 . 𝑓′
4.
𝑑
𝑑𝑥
sech 𝑓 = −sech 𝑓 tanh 𝑓 . 𝑓′
5.
𝑑
𝑑𝑥
csch 𝑓 = −csch 𝑓 coth 𝑓 . 𝑓′
6.
𝑑
𝑑𝑥
coth 𝑓 = −csch2 𝑓 . 𝑓′
Consider
𝑓 = 𝑓(𝑥)
Example:
19. Slide 1 - 19
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
You can see from the figures that sinh
and tanh are one-to-one functions.
So, they have inverse functions denoted by
sinh−1
𝑥 and tanh−1
𝑥.
INVERSE HYPERBOLIC FUNCTIONS
20. Slide 1 - 20
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
This figure shows that cosh 𝑥 is not one-to-one.
However, when restricted to the domain
[0, ∞], it becomes one-to-one.
INVERSE FUNCTIONS
21. Slide 1 - 21
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
1
1
1
sinh sinh
cosh cosh and 0
tanh tanh
y x y x
y x y x y
y x y x
Definition 2
INVERSE FUNCTIONS
The remaining inverse hyperbolic functions
are defined similarly.
22. Slide 1 - 22
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The graphs of sinh−1
𝑥,
cosh−1
𝑥 and tanh−1
𝑥 are
displayed.
INVERSE FUNCTIONS
𝑦 = sinh−1
𝑥
𝑦 = tanh−1 𝑥
𝑦 = cosh−1
𝑥
23. Slide 1 - 23
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Since the hyperbolic functions are defined
in terms of exponential functions, it’s not
surprising to learn that the inverse hyperbolic
functions can be expressed in terms of logarithms.
INVERSE FUNCTIONS
24. Slide 1 - 24
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
In particular, we have:
1 2
1 2
1 1
2
sinh ln 1
cosh ln 1 1
1
tanh ln 1 1
1
x x x x
x x x x
x
x x
x
INVERSE FUNCTIONS
25. Slide 1 - 25
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Show that .
Let y = sinh-1 x. Then,
So, e y – 2x – e -y = 0
Or, multiplying by e y, e 2y – 2xe y – 1 = 0
This is really a quadratic equation in e y:
(e y )2 – 2x (e y ) – 1 = 0
1 2
sinh ln 1
x x x
sinh
2
y y
e e
x y
INVERSE FUNCTIONS
26. Slide 1 - 26
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solving by the quadratic formula,
we get:
Note that e y > 0, but
So, the minus sign is inadmissible and we have:
Thus,
2
2
2 4 4
1
2
y x x
e x x
2
1 0
x x
2
1
y
e x x
2
ln( ) ln 1
y
y e x x
INVERSE FUNCTIONS
27. Slide 1 - 27
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
, |𝑥| < 1 , |𝑥| > 1
28. Slide 1 - 28
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Prove that .
Let y = sinh-1 x. Then, sinh y = x.
If we differentiate this equation implicitly
with respect to x, we get:
As cosh2 y - sinh2 y = 1 and cosh y ≥ 0, we have:
So,
1
2
1
(sinh )
1
d
x
dx x
cosh 1
dy
y
dx
2
cosh 1 sinh
y y
2 2
1 1 1
cosh 1 sinh 1
dy
dx y y x
DERIVATIVES
29. Slide 1 - 29
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
By another way, we have:
1 2
2
2
2 2
2
2
2 2
sinh ln 1
1
1
1
1
1
1 1
1 1
1
1 1
d d
x x x
dx dx
d
x x
dx
x x
x
x x x
x x
x
x x x
DERIVATIVES
31. Slide 1 - 31
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Implicit Differentiation
So far, all the equations and functions we
looked at were all stated explicitly in terms
of one variable:
x
y
1
In this function, y is defined
explicitly in terms of x. If we re-
wrote it as 𝑥𝑦 = 1, y is now
defined implicitly in terms of x.
It is easy to find the derivative of an explicit
function, but what about:
5
3
4 2
3
y
y
y
x
32. Slide 1 - 32
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
2 2
1
x y
This is not a function,
but it would still be nice
to be able to find the
slope.
2 2
1
d d d
x y
dx dx dx
Do the same thing to both sides.
2 2 0
dy
x y
dx
Note use of chain rule.
2 2
dy
y x
dx
2
2
dy x
dx y
dy x
dx y
33. Slide 1 - 33
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
2
2 sin
y x y
2
2 sin
d d d
y x y
dx dx dx
This can’t be solved for y.
2 2 cos
dy dy
x y
dx dx
2 cos 2
dy dy
y x
dx dx
2
2 cos
dy
x
y
dx
2
2 cos
dy x
dx y
This technique is called
implicit differentiation.
1 Differentiate both sides w.r.t. x.
2 Solve for .
dy
dx
34. Slide 1 - 34
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
4
5
2
3
3
4
i) x
x
y
y
Find dy/dx if:
3
4
2
12
5
8
3 x
x
dx
dy
y
dx
dy
y
3
4
2
12
5
8
3 x
x
y
y
dx
dy
y
y
x
x
dx
dy
8
3
12
5
2
3
4
Examples.
35. Slide 1 - 35
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Find
dy/dx
if:
2
2
2
2
sin
cos
cos
sin
ii) x
y
x
y
)
2
(
cos
2
sin
)
2
(
sin
2
cos 2
2
2
2
x
x
dx
dy
y
y
x
x
dx
dy
y
y
2
2
2
2
sin
2
cos
2
sin
2
cos
2 x
x
x
x
dx
dy
y
y
dx
dy
y
y
2
2
2
2
sin
2
cos
2
sin
2
cos
2 x
x
x
x
y
y
y
y
dx
dy
2
2
2
2
sin
2
cos
2
sin
2
cos
2
y
y
y
y
x
x
x
x
dx
dy
2
2
2
2
sin
cos
sin
cos
y
y
y
x
x
x
dx
dy
36. Slide 1 - 36
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Find
dy/dx
if:
8
4
5
3
iii) 3
2
2
y
xy
x
0
12
10
5
6 2
2
dx
dy
y
dx
dy
xy
y
x
Product
Rule!
2
2
5
6
12
10 y
x
y
xy
dx
dy
2
2
12
10
5
6
y
xy
y
x
dx
dy
37. Slide 1 - 37
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
We need the slope. Since we can’t solve for y, we use implicit
differentiation to solve for .
dy
dx
2 2
7
x xy y
( 1,2)
2 2
7
x xy y
2 2 0
dy
dy
x y
x y
dx
dx
Note product rule.
2 2 0
dy dy
x x y y
dx dx
2
2
dy
y x
y x
dx
2
2
dy y x
dx y x
2 2 1
2 2 1
m
2 2
4 1
4
5
Find the equations of the lines tangent and normal to the curve
at .
38. Slide 1 - 38
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
2 2
7
x xy y
( 1,2)
4
5
m
tangent:
4
2 1
5
y x
4 4
2
5 5
y x
4 14
5 5
y x
Find the equations of the lines tangent and normal to the curve
at .
normal:
5
2 1
4
y x
5 5
2
4 4
y x
5 3
4 4
y x
Normal line is
perpendicular
to tangent
)
( 1
1 x
x
m
y
y
Eq. of straight line
39. Slide 1 - 39
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example. Find derivative at (1, 1)
x
x
y
x
y
2
3
3
2
Product Rule is
easier than quotient
rule, so let’s cross
multiply!
3
3
3
2
x
xy
x
y
2
3
2
2
3
3
3
2 x
y
dx
dy
xy
x
dx
dy
y
2
3
2
6
3
2 x
y
xy
y
dx
dy
2
2
3
3
2
6
xy
y
x
y
dx
dy
2
2
3
)
1
)(
1
(
3
)
1
(
2
)
1
(
6
)
1
(
dx
dy
5
1
5
dx
dy
40. Slide 1 - 40
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Higher Order Derivatives
Find if .
2
2
d y
dx
3 2
2 3 7
x y
3 2
2 3 7
x y
2
6 6 0
x y y
2
6 6
y y x
2
6
6
x
y
y
2
x
y
y
2
2
2
y x x y
y
y
2
2
2x x
y y
y y
2 2
2
2x x
y
y
x
y
y
4
3
2x x
y
y y
Substitute
back into the
equation.
y
42. Derivative of Parametric Equations
Consider the graph of
x = 2 sin t, y = cos t
We seek the slope, that
is
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
.
𝑑𝑡
𝑑𝑥
For parametric equations:
𝑑𝑦
𝑑𝑥
=
𝑑𝑦/𝑑𝑡
𝑑𝑥/𝑑𝑡
For our example :
𝑑𝑦
𝑑𝑥
=
− sin 𝑡
2 cos 𝑡
=
− tan 𝑡
2
43. Second Derivatives
The second derivative is the derivative of the
first derivative
But the first derivative is a function of t
We seek the derivative with respect to x
We must use the chain rule
𝑑2𝑦
𝑑𝑥2 =
𝑑
𝑑𝑥
𝑦′
=
𝑑
𝑑𝑡
𝑦′
.
𝑑𝑡
𝑑𝑥
=
𝑑
𝑑𝑡
𝑦′
𝑑𝑥
𝑑𝑡
44. Second Derivatives
Find the second derivative of the parametric
equations
x = 3 + 4cos t
y = 1 – sin t
First derivative:
𝑑𝑦
𝑑𝑥
=
𝑦.
𝑥. =
− 𝑐𝑜𝑠 𝑡
−4 sin 𝑡
=
cot 𝑡
4
Second derivative:
𝑑2𝑦
𝑑𝑥2 =
𝑑
𝑑𝑡
𝑦′
.
𝑑𝑡
𝑑𝑥
=
−1
4
csc2𝑡
−4 sin 𝑡
∴ 𝑦′′
=
1
16 sin3𝑡
45. Try This!
Where does the curve described by the
parametric equations have a horizontal tangent?
x = t – 4
y = (t 2 + t)2
Find the derivative
For what value of t does dy/dx = 0?
46. Exercise
Consider 𝑥 = 2 sin
𝑡+1
𝑡−1
, 𝑦 = cos
𝑡+1
𝑡−1
. Prove 𝑦′′ =
−1
4 𝑦3
∵ cos2
𝜃 + sin2
𝜃 = 1
∴
𝑥2
4
+ 𝑦2
= 1
𝑥
2
+ 2𝑦𝑦′ = 0
∴ 𝑦′
=
−𝑥
4𝑦
By diff.
4𝑦′′
=
−𝑦+𝑥𝑦′
𝑦2
4𝑦′′
=
−𝑦+𝑥
−𝑥
4𝑦
𝑦2
By diff.
4𝑦′′
=
−𝑦2−
𝑥2
4
𝑦3
𝑦′′
=
−1
4𝑦3
Solution
47. Exercise
i) Consider
∵ cos 2𝛽 = 1 − 2 sin2
𝛽
∴ 𝑦 = 1 − 2𝑥2
Solution
𝑦′
= −4 𝑥 → 𝑦′ 2
= 16𝑥2
𝑦′′
= −4
𝑦′ 2
𝑦′′
= −64𝑥2
= 32 −2𝑥2
= 32(𝑦 − 1)
𝑦 = cos 2
𝑡
1−𝑡
, 𝑥 = sin
𝑡
1−𝑡
. Prove
𝑦′ 2𝑦′′ = 32(𝑦 − 1)
48. Slide 1 - 48
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Find
dy/dx
if: .
tanh
sin
5 2
1
y
xy
xy
y
h
y
y
x
x
yy
y
x
y
x
y
e x
y 2
4
2
2
ln
sec
'
1
1
'
2
5
ln
'
.
1
10
ln
sec
1
5
'
4
2
ln
2
4
2
2
ln
y
x
yx
x
e
y
h
y
x
y
e
x
y
y
x
y
x
y
ii) Find 𝑦’ for:
49. Slide 1 - 49
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Introduction to integrtion
In differentiation, the differential coefficient
𝑑𝑦
𝑑𝑥
indicates
that a function of x is being differentiated with respect
to x, the dx indicating that it is “with respect to x”.
In integration, the variable of integration is shown by
adding d(the variable) after the function to be
integrated. When we want to integrate a function, we
use a special notation: 𝒇 𝒙 𝒅𝒙.
Thus, to integrate 4x, we will write it as follows:
49
50. Slide 1 - 50
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Introduction
4𝑥 𝒅𝒙 = 2x2
+ c , c ∈ ℝ.
Integral
sign
This term is called
the integrand
There must always be
a term of the form dx
Constant of
integration
50
51. Slide 1 - 51
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Indefinite integral
Note that along with the integral sign ( 𝑑𝑥), there is a term of the form dx,
which must always be written, and which indicates the variable involved, in
our example x.
We say that 4x is integrated with respect x, i.e: 𝟒𝒙 𝒅𝒙
The function being integrated is called the integrand.
Technically, integrals of this type are called indefinite integrals, to
distinguish them from definite integrals, which we will deal with later.
When you are required to evaluate an indefinite integral, your answer must
always include a constant of integration.; i.e:
𝟒𝒙 𝒅𝒙 = 2𝑥2 + c; where c ∈ ℝ
51
52. Slide 1 - 52
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Definition of anti-derivative
Formally, we define the anti-derivative
as: If f(x) is a continuous function and F(x)
is the function whose derivative is f(x), i.e.:
𝑭′
(x) = f(x) , then:
𝒇 𝒙 𝒅𝒙 = F(x) + c;
where c is any arbitrary constant.
52
53. Slide 1 - 53
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Table of
INTEGRATION
54. Slide 1 - 54
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
i)
𝑑𝑥
𝑥 1+ln 𝑥 3 •
𝑓′𝑑𝑥
𝑓 𝑛 =
𝑓−𝑛+1
−𝑛+1
+ 𝑐
=
1+ln 𝑥 −2
−2
+c
ii)
1+tan 𝑥 3𝑑𝑥
1−𝑠𝑖𝑛2 𝑥
• 𝑓𝑛
𝑓′
𝑑𝑥 =
𝑓𝑛+1
𝑛+1
+ 𝑐
=
1+tan 𝑥 4
4
+c
=
1 + tan 𝑥 3𝑑𝑥
𝑐𝑜𝑠2 𝑥
= sec2 𝑥 1 + tan 𝑥 3𝑑𝑥
Examples. Find the following integrals:
55. Slide 1 - 55
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
iii)
1−2𝑥3 𝑑𝑥
2𝑥−𝑥4+1
•
𝑓′𝑑𝑥
𝑓
= 2 𝑓 + 𝑐
=
1
2
2 1−2𝑥3 𝑑𝑥
2𝑥−𝑥4+1
=
1
2
. 2 2𝑥 − 𝑥4 + 1 + 𝐶
•
𝑓′𝑑𝑥
𝑎2−𝑓2
= sin−1 𝑓
𝑎
+ 𝑐
iv)
𝑑𝑥
𝑥 4−9 ln2 𝑥
=
1
3
. sin−1 3 ln 𝑥
2
+ 𝐶
=
1
3
3𝑑𝑥
𝑥 4− 3 ln 𝑥 2
56. Slide 1 - 56
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
iv)
3+𝑥 𝑑𝑥
𝑥−5
=
𝑥−5 +3+5
𝑥−5
𝑑𝑥
= 1 +
8
𝑥−5
𝑑𝑥
= 𝑥 + 8 ln |𝑥 − 5| + 𝑐
v)
𝑑𝑥
sin−1 𝑥 1−𝑥2
•
𝑓′𝑑𝑥
𝑓
= ln |𝑓| + 𝑐
= ln | sin−1
𝑥 | + 𝑐
57. Slide 1 - 57
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
vi)
1+sin 𝑥 𝑑𝑥
cos2𝑥
= sec2
𝑥 + tan 𝑥 sec 𝑥 𝑑𝑥
= tan 𝑥 + sec 𝑥 + 𝑐
vii)
1−cos 2𝑥 𝑑𝑥
sin2𝑥
=
1−cos 2𝑥 𝑑𝑥
1
2 1−cos 2𝑥
= 2 𝑑𝑥 = 2𝑥 + 𝑐.
• sin2 𝑥 = 1
2 1 − cos 2𝑥
• cos2
𝑥 = 1
2 1 + cos 2𝑥
Note:
58. Slide 1 - 58
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
viii) 1 + cos 𝑥 2
𝑑𝑥
= 1 + 2 cos 𝑥 + cos2
𝑥 𝑑𝑥
= 1 + 2 cos 𝑥 + 1
2 1 + cos 2𝑥 𝑑𝑥
= 𝑥 + 2 sin 𝑥 + 1
2 𝑥 +
sin 2𝑥
2
+ 𝑐
ix) cos3
𝑥 𝑑𝑥
= cos2
𝑥 cos 𝑥 𝑑𝑥
= 1 − sin2
𝑥 cos 𝑥 𝑑𝑥
= cos 𝑥 − sin2
𝑥 cos 𝑥 𝑑𝑥 = sin 𝑥 −
sin3𝑥
3
+ 𝑐
59. Slide 1 - 59
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
x) cos 𝑥 + sin 𝑥 2
𝑑𝑥
= cos2
𝑥 + sin2
𝑥 + 2 sin 𝑥 cos 𝑥 𝑑𝑥
= 1 + sin 2𝑥 𝑑𝑥
= 𝑥 −
cos 2𝑥
2
+ 𝑐
60. Slide 1 - 60
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
xi) tan 𝑥 ln | cos 𝑥 | 𝑑𝑥
=−
ln2| cos 𝑥|
2
+ 𝑐
xii)
𝑒𝑥+𝑒2𝑥
𝑒2𝑥+1
𝑑𝑥
=
1
2
ln 𝑒2𝑥
+ 1 + tan−1
𝑒𝑥
+ 1
•
𝑓′𝑑𝑥
𝑓
= ln |𝑓| + 𝑐
•
𝑓′𝑑𝑥
𝑎2+𝑓2 =
1
𝑎
tan−1 𝑓
𝑎
+ 𝑐
=
1
2
2𝑒2𝑥
𝑒2𝑥+1
+
𝑒𝑥
(𝑒𝑥)2+1
𝑑𝑥
61. Slide 1 - 61
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
xiii)
sin 𝑥 cos 𝑥
sin4 𝑥+ cos4 𝑥
𝑑𝑥
÷ cos4
𝑥
=
sin 𝑥
cos3𝑥
tan4 𝑥+1
𝑑𝑥
=
1
2
2 tan 𝑥sec2𝑥
tan2𝑥 2+1
𝑑𝑥
=
1
2
tan−1
tan2
𝑥 + 𝑐
•
𝑓′𝑑𝑥
𝑎2+𝑓2 =
1
𝑎
tan−1 𝑓
𝑎
+ 𝑐
62. Slide 1 - 62
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
xiv)
sec 𝑥 cos 2𝑥
sin 𝑥+sec 𝑥
𝑑𝑥
÷ sec 𝑥
=
cos 2𝑥
sin 𝑥 cos 𝑥+1
𝑑𝑥
=
cos 2𝑥
sin 2𝑥
2+1
𝑑𝑥
= ln |
sin 2𝑥
2
+ 1| + 𝑐
63. Slide 1 - 63
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
xv)
sec 𝑥+ tan 𝑥
sec 𝑥− tan 𝑥
𝑑𝑥
=
sec 𝑥+ tan 𝑥
sec 𝑥− tan 𝑥
.
sec 𝑥+ tan 𝑥
sec 𝑥+ tan 𝑥
𝑑𝑥
=
sec 𝑥+ tan 𝑥 2
sec2 𝑥−tan2 𝑥
𝑑𝑥
= sec2
𝑥 + sec2
𝑥 − 1 + 2 sec 𝑥 tan 𝑥 𝑑𝑥
tan2
𝑥
= 2 tan 𝑥 − 𝑥 + 2 sec 𝑥 + 𝑐.
𝑡𝑎𝑛2𝑥 + 1 = 𝑠𝑒𝑐2𝑥
64. Slide 1 - 64
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
65. Slide 1 - 65
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
66. Slide 1 - 66
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley