2. LESSON 1. The Indeterminate Form
• Rolle’s Theorem
• This theorem is very useful in the proof of many theorems in Calculus. It
was formulated by Michel Rolle (French Mathematician, 1652-1719).
Theorem
If a function f(x) is continuous in the closed interval [a,b];
if f’(x) exists on the open interval (a,b); and f(a) = f(b)=0,
then there is a number c in (a,b) such that f’(c)=0.
3. Example 1: Consider the function f(x)=𝑥2
− 4𝑥 + 3 and the interval [1,3]. Verify that the three
conditions of the theorem are satisfied.
Solution:
1. At [1,3] – the function is continuous because it is polynomial.
2. f’(x) = 2x-4 – the derivative of the function exists
3. f(1)=0, f(3)= 0; f(1) = f(3)= 0
Hence the three conditions are satisfied.
Therefore, there must be a number c in (1,3) which satisfies the conclusion of the theorem.
At x = c; f’(c)= 2c-4 ; f’(c)=0
2c – 4 = 0
2c = 4
C = 2
4. Mean Value Theorem
The mean value theorem (law of mean) is used to estimate the values of
functions when direct calculation is difficult. It is also used to prove the two
functions having the same derivative must differ by constant which is one of
its important uses.
Theorem
If a function f(x) is continuous on the closed interval
[a,b] and if f’(x) exist on open interval (a,b), then there is a
number c in (a,b) such that
𝑓′
𝑐 =
𝑓 𝑏 −𝑓(𝑎)
𝑏−𝑎
5. Example 2: Verify that the three conditions of the Rolle’s theorem are satisfied.
Find c.
𝑓 𝑥 = 𝑥 , [4,9]
• Exercises
a. Verify that the three conditions of the hypothesis of Rolle’s theorem are
satisfied by the given function on the indicated interval and find the value of c
which satisfies the conclusion of the theorem.
1. 𝑓 𝑥 = 𝑥2 − 𝑥 − 2, −1,2
2. 𝑓 𝑥 = 𝑥3 − 3𝑥, 0, 3
3. 𝑓 𝑥 = 𝑠𝑖𝑛𝑥. 0, 𝜋
6. Mean Value Theorem
The mean value theorem (law of mean) is used to estimate the values of
functions when direct calculation is difficult. It is also used to prove the two
functions having the same derivative must differ by constant which is one of
its important uses.
Theorem
If a function f(x) is continuous on the closed interval
[a,b] and if f’(x) exist on open interval (a,b), then there is
a number c in (a,b) such that
𝑓′
𝑐 =
𝑓 𝑏 −𝑓(𝑎)
𝑏−𝑎
7. f(x) =
𝑓 𝑏 −𝑓 𝑎
𝑏−𝑎
𝑥 − 𝑎 + 𝑓 𝑎 − 𝑓(𝑥) (1)
Note:
f(x) is continuous on [a,b], differentiable on (a,b) and f(a) = f(b) = 0.
Differentiating with respect to x;
f’(x) =
𝑓 𝑏 −𝑓 𝑎
𝑏−𝑎
− 𝑓′(𝑥) (2)
f’(c) =
𝑓 𝑏 −𝑓 𝑎
𝑏−𝑎
𝑓′ 𝑐 = 0
or equivalently
𝑓′ 𝑐 =
𝑓 𝑏 −𝑓 𝑎
𝑏−𝑎
8. Example 1:
Given 𝑓 𝑥 = 𝑥2 + 2𝑥 − 1 𝑎𝑛𝑑 0,1 . Verify that the hypothesis of the mean value
theorem is satisfied. Find c that satisfies its conclusion.
• The function is continuous since it is a polynomial.
• f’(x)=2x+2 exists on the open interval (0,1)
• Find the value of c
Since
𝑓 𝑥 = 𝑥2 + 2𝑥 − 1, 𝑎 = 0 𝑎𝑛𝑑 𝑏 = 1
f(a) = f(0) = -1
f(b)= f(1) = 2
10. Example2 : Use the mean value theorem to prove 10.77 < 117 < 11
Solution:
Let f(x)= 𝑥, 𝑎 = 100 𝑎𝑛𝑑 𝑏 = 117
𝑓 𝑎 = f(100) = 100 = 10
𝑓 𝑏 = 𝑓 117 = 117
Since f’(x) =
1
2 𝑥
, then f’(c) =
Substituting the values
1
2 𝑐
=
117−10
117−100
solving for 117 =
17
2 𝑐
+ 10
where 100 < c < 117. Since 102 = 100 𝑎𝑛𝑑 112 = 121
11. It follows that
100 < c < 121
10 < 𝑐 < 11
Taking the reciprocal
1
10
>
1
𝑐
>
1
11
Multiplying the inequality above by
17
2
and then adding by 10,
17
20
+ 10 >
17
2 𝑐
+ 10 >
17
22
+ 10
11 > 117 > 10.77
10.77< 117<11
12. If we simplify the theorem, it becomes
f(b)=f(a)+(b-a)f’(c)
In Rolle’s theorem, if b is near a, c is also near a. That is c comes
closer and closer to a as (b-a) gets smaller and smaller. Thus when (b-
a) is sufficiently small c approximates the value of a.
It follows that
f’(c ) = f’(a)
f(b) = f(a)+ (b-a)f’(a)
15. L’Hospital’s Rule
In Calculus I we learned how to find the limit of the quotient of two
functions when the numerator and denominator approaches zero. In
evaluating such function, we changed it into a form where the limit
theorems can be applied. That is where we employed algebraic
manipulation such as conjugation, factoring or rationalizing the
denominator.
However, there are functions having indeterminate forms whose limit
can not be evaluated by the mentioned methods. So the purpose of this
topic is to introduced a systematic way for evaluating the limits of such
function and it is the L’Hospital’s Rule or for convenience LHR.
16. “ LHR states that to evaluate the limit of the fraction
𝑓(𝑥)
𝑔(𝑥)
that takes
the form
0
0
at x = a, differentiate the numerator and denominator
separately and then take the limit of the new fraction
𝑓′(𝑥)
𝑔′(𝑥)
.” In case,
the limit of the new fraction is again
0
0
, repeat the process, that is
reapply LHR. However, before reapplying LHR, simplify first the new
quotient whenever possible.
18. 3. Evaluate lim
𝑥→0
𝑥𝑐𝑜𝑡𝑥
• If the lim
𝑥→𝑎
𝑓 𝑥 = ∞ and lim
𝑥→𝑎
𝑔 𝑥 = ∞, then the function defined by their
difference is said to have the indeterminate form ∞ − ∞. To apply LHR
change it into the form 0/0 or
∞
∞
using algebraic manipulation.
4. Evaluate lim
𝑥→
𝜋
2
(𝑠𝑒𝑐𝑥 − 𝑡𝑎𝑛𝑥).
5. Evaluate lim
𝑥→
𝜋
2
𝜋
2
− 𝑥 𝑡𝑎𝑛𝑥
20. • LESSON 2. Differentiation of The Transcendental Functions
Functions which are not algebraic are transcendental
functions. The trigonometric functions and their inverses,
together with logarithmic and exponential functions are the
simplest transcendental functions. Though some of which
were already discussed in Calculus I, but we need to review
their values for applications purposes.
22. Differentiation of Inverse Trigonometric Function
In trigonometry, we recall that
y = Arcsinx iff x = sin y and −
𝜋
2
< 𝑦 ≤
𝜋
2
y = Arccosx iff x = cos y and 0 ≤ 𝑦 ≤ 𝜋
y = Arctanx iff x = tany and −
𝜋
2
< 𝑦 <
𝜋
2
y = Arccotx iff x = coty and 0 < 𝑦 <π
y = Arcsecx iff x = sec y and −𝜋 ≤ 𝑦 < −
𝜋
2
𝑓𝑜𝑟 𝑥 ≤ −1
0 ≤ 𝑦 <
𝜋
2
𝑓𝑜𝑟 𝑥 ≥ −1
Y = Arccscx iff x = csc y and −𝜋 < 𝑦 ≤ −
𝜋
2
𝑓𝑜𝑟 𝑥 ≤ −1
0 < 𝑦 ≤
𝜋
2
𝑓𝑜𝑟 𝑥 ≥ −1
23. The following formulas are used for differentiating inverse trigonometric functions. The
symbol u denotes an arbitrary differentiable function of x.
𝑑
𝑑𝑥
𝐴𝑟𝑐𝑠𝑖𝑛𝑢 =
1
1−𝑢2
𝑑𝑢
𝑑𝑥
•
𝒅
𝒅𝒙
𝑨𝒓𝒄𝒄𝒐𝒔𝒖 =
−𝟏
𝟏−𝒖𝟐
𝒅𝒖
𝒅𝒙
•
𝑑
𝑑𝑥
𝐴𝑟𝑐𝑡𝑎𝑛𝑢 =
1
1+𝑢2
𝑑𝑢
𝑑𝑥
•
𝑑
𝑑𝑥
𝐴𝑟𝑐𝑐𝑜𝑡𝑢 =
−1
1+𝑢2
𝑑𝑢
𝑑𝑥
•
𝑑
𝑑𝑥
𝐴𝑟𝑐𝑠𝑒𝑐𝑢 =
1
𝑢 𝑢2−1
𝑑𝑢
𝑑𝑥
•
𝑑
𝑑𝑥
𝐴𝑟𝑐𝑐𝑠𝑐𝑢 =
−1
𝑢 𝑢2−1
𝑑𝑢
𝑑𝑥
24. Proof:
Let y = Arcsinu
u = sin y
Differentiating with respect to x;
𝑑𝑢
𝑑𝑥
= cos 𝑦
𝑑𝑦
𝑑𝑥
Solving for
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
1
cos 𝑦
𝑑𝑢
𝑑𝑥
In trigonometry, cos y = 1 − 𝑠𝑖𝑛2𝑦 = 1 − 𝑢2 . (The + sign of the radical is chosen
since cosy > 0 for −
𝜋
2
≤ 𝑦 ≤
𝜋
2
. Hence
𝑑𝑦
𝑑𝑥
=
1
1−𝑢2
𝑑𝑢
𝑑𝑥
𝑑
𝑑𝑥
(𝐴𝑟𝑐𝑠𝑖𝑛𝑢) =
1
1−𝑢2
𝑑𝑢
𝑑𝑥
26. • Example3. A ladder 25 ft. long leans against a vertical wall. If the lower end is
pulled away at the rate of 6ft/sec, how fast is the angle between the ladder and
the floor changing when the lower end is 7ft from the wall?
Solution: Let
x = distance between the lower end of the ladder AB from the
wall BC
𝜃 = angle between the ladder and the floor AC
Find 𝑑𝜃
𝑑𝑡
𝑤ℎ𝑒𝑛
𝑑𝑥
𝑑𝑡
= 6𝑓𝑡./ sec 𝑎𝑛𝑑 𝑥 = 7
27. • Differentiating with respect to t,
𝑑𝜃
𝑑𝑡
=
−1
1−
𝑥2
625
∙
1
25
𝑑𝑥
𝑑𝑡
Substituting x = 7 and
𝑑𝑥
𝑑𝑡
=6 , we get
𝑑𝜃
𝑑𝑡
=
−1
625−𝑥2
625
∙
1
25
∙
𝑑𝑥
𝑑𝑡
𝑑𝜃
𝑑𝑡
=
−1
625−𝑥2
25
∙
1
25
∙
𝑑𝑥
𝑑𝑡
𝑑𝜃
𝑑𝑡
=
−1
625−72
∙ 6
𝑑𝜃
𝑑𝑡
= -1/4 rad/sec ( the minus sign indicates that θ is decreasing)
28. • Differentiation of Exponential Functions
The following formulas are used to find the derivatives of exponential functions.
𝑑
𝑑𝑥
𝑎𝑢
= 𝑎𝑢
𝑙𝑛𝑎
𝑑𝑢
𝑑𝑥
𝑑
𝑑𝑥
𝑒𝑢 = 𝑒𝑢 𝑑𝑢
𝑑𝑥
Example
1. 42𝑥 𝑑𝑢
𝑑𝑥
, find dy/dx
2. 𝑦 = 𝑒𝑠𝑖𝑛𝑥
3. y =
1+2𝑥
1−2𝑥
4. 𝑦 = 4𝑥
𝑙𝑛4𝑥
29. • The Hyperbolic Functions
• Functions which are combinations of 𝑒𝑥 𝑎𝑛𝑑 𝑒−𝑥 are hyperbolic functions and
frequently occurs in mathematics, sciences and engineering. They are defined as
follows:
𝑠𝑖𝑛ℎ𝑥 =
𝑒𝑥−𝑒−𝑥
2
𝑐𝑜𝑡ℎ𝑥 =
𝑐𝑜𝑠ℎ𝑥
𝑠𝑖𝑛ℎ𝑥
𝑐𝑜𝑠ℎ𝑥 =
𝑒𝑥+𝑒−𝑥
2
𝑐𝑠𝑐ℎ𝑥 =
1
𝑠𝑖𝑛ℎ𝑥
𝑡𝑎𝑛ℎ𝑥 =
𝑠𝑖𝑛ℎ𝑥
𝑐𝑜𝑠ℎ𝑥
𝑠𝑒𝑐ℎ𝑥 =
1
𝑐𝑜𝑠ℎ𝑥
30. The following identities can be deduced directly from the definitions of the
hyperbolic functions.
• 𝑐𝑜𝑠ℎ2𝑥 − 𝑠𝑖𝑛ℎ2𝑥 = 1
• 𝑡𝑎𝑛ℎ2
𝑥 + 𝑠𝑒𝑐ℎ2
𝑥 = 1
• 𝑐𝑜𝑡ℎ2𝑥 − 𝑐𝑠𝑐ℎ2𝑥 = 1
• 𝑠𝑖𝑛ℎ2𝑥 = 2𝑠𝑖𝑛ℎ𝑥𝑐𝑜𝑠ℎ𝑥
• 𝑐𝑜𝑠ℎ2𝑥 = 𝑐𝑜𝑠ℎ2𝑥 + 𝑠𝑖𝑛ℎ2𝑥
= 1 + 2𝑠𝑖𝑛ℎ2𝑥
= 2𝑐𝑜𝑠ℎ2𝑥 − 1
36. • Differentiation of Inverse Hyperbolic Functions
The inverse of hyperbolic of sinx is written 𝑠𝑖𝑛ℎ−1𝑥 and it is define as follows:
𝑦 = 𝑠𝑖𝑛ℎ−1𝑥 𝑖𝑓 𝑥 = 𝑠𝑖𝑛ℎ𝑦
This can also be defined in like manner in terms of logarithms as
𝑠𝑖𝑛ℎ−1𝑥 = ln(𝑥 + 1 + 𝑥2) for all x.
37. Inverse hyperbolic functions can be expressed in terms of logarithmic functions.
Following are the expressions for the inverse hyperbolic functions:
𝑠𝑖𝑛ℎ−1
𝑥 = 𝑙𝑛 𝑥 + 𝑥2 + 1 ; for x is any real number
𝑐𝑜𝑠ℎ−1𝑥 = 𝑙𝑛 𝑥 + 𝑥2 − 1 ; x ≥ 1
𝑡𝑎𝑛ℎ−1𝑥 =
1
2
ln
1+𝑥
1−𝑥
; 𝑥 < 1
𝑐𝑜𝑡ℎ−1
𝑥 =
1
2
ln
𝑥+1
𝑥−1
; 𝑥 > 1
• Express the following in terms of logarithms
1. 𝑠𝑖𝑛ℎ−1(2) 2. 𝑡𝑎𝑛ℎ−1(4/5)
41. PARTIAL DIFFERENTIATION
Let x =f(x0 ) be a function of two independent variables x and y. If y is
held constant, then x temporarily a function of single variable. From this
point of view, we can compute the derivative of z with respect to x by
employing the rules of ordinary differentiation of function with single
variable. The derivative found in this manner is partial differentiation.
𝜕𝑧
𝜕𝑥
,
𝜕𝑓
𝜕𝑥
, 𝑧𝑥, 𝑓𝑥 𝑥, 𝑦
42. • Definition of partial derivative:
If z=f(x,y), then the partial derivative of z with respect to x is symbolically
defined as
𝜕𝑧
𝜕𝑥
= lim
∆𝑥→0
𝑓 𝑥+∆𝑥,𝑦 −𝑓(𝑥,𝑦)
∆𝑥
and
𝜕𝑧
𝜕𝑦
= lim
∆𝑦→0
𝑓 𝑥, 𝑦+∆𝑦 −𝑓(𝑥,𝑦)
∆𝑦
46. Geometric Interpretation of Partial Derivative
• Let the graph of the surface be defined by the equation z = f(x,y). Let the point
𝑃(𝑥0,𝑦0, 𝑧0) be a point on the surface. So the plane passing through P and parallel to xz
plane has the equation y = 𝑦0 . The intersection of the surface z = f(x,y) and the plane y
= 𝑦0 is the curve APB. As a point moves along the curve APB, the coordinates of x and
z vary while y remains constant. The slope of the tangent line at P represents the rate at
which z changes with respect to x. Hence, the partial derivative
𝜕𝑧
𝜕𝑥
is the slope of the tangent
to the curve intersecting APB at the point P.
47. The equations of the tangent at P are:
𝑧 − 𝑧0 = 𝑚0 𝑥 − 𝑥0 , 𝑦 = 𝑦0 where 𝑚0 = 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓
𝜕𝑧
𝜕𝑥
at P.
If the curve of intersection of the surface z = f(x,y) and the plane x = 𝑥0, the
point moves along the curve where y and z vary and x remains constant. So
𝜕𝑧
𝜕𝑦
is
the slope of the tangent to the curve at point P. The equations of the tangent at P
are:
𝑧 − 𝑧0 = 𝑚0 𝑦 − 𝑦0 , 𝑥 = 𝑥0 where 𝑚0 = 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓
𝜕𝑧
𝜕𝑦
at P.
48. If the curve of intersection of the surface z=f(x,y) and the plane x= 𝑥0.
The equations are
b. 𝑧 − 𝑧0 = 𝑚0 𝑦 − 𝑦0 , 𝑥 = 𝑥0. where 𝑚0 =
𝜕𝑧
𝜕𝑦
𝑎𝑡 𝑃.
If
𝜕𝑧
𝜕𝑦
𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝑃
49. Example:
1. Find the equations of the tangent to the parabola 𝑧 = 𝑥2
+ 3𝑦2
, 𝑦 =
1 𝑎𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 2,1,7
50. .
Since y is constant, we differentiate partially the equation with
respect to x.
𝜕𝑧
𝜕𝑥
= 2𝑥
At point (2,1,7), we have
𝜕𝑧
𝜕𝑥
= 2 2 = 4, 𝑠𝑜 𝑚0 = 4.
𝑆𝑖𝑛𝑐𝑒 𝑥0 = 2 , 𝑎𝑛𝑑 𝑧0 = 7
𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑡 𝑃 𝑎𝑟𝑒
z – 7 = 4(x-2). y = 1
z = 4x – 1, y = 1
51. • Example2. Find the equations of the tangent to the ellipse
𝑥2 + 4𝑦 + 𝑧2 = 0 , 𝑧 = −1 at the point (2, -1, 1).
Example:3
Find the slope of the tangent line to the curve of intersection
of the surface
𝑧 =
1
2
24 − 𝑥2 − 2𝑦2 with the plane y=2 at the point
(2,2, 3). Find also the equation of the line tangent to the
curve.
52. Partial Derivatives of Higher Order
Continued differentiation of function of two (or more variables give rise to partial differentiation
of higher order. If z=f(x,y), then
𝜕𝑧
𝜕𝑥
is the first partial derivative of z with respect to x. Then the partial
derivative of
𝜕𝑧
𝜕𝑥
with respect to x is the second partial derivative of z. Hence, for a function of to
variables, there are four possible ways in which second partial derivative may arise. The following
notations for the second partial derivatives are used:
1)
𝜕
𝜕𝑥
𝜕𝑧
𝜕𝑥
=
𝜕2𝑧
𝜕𝑥2 =
𝜕2𝑓
𝜕𝑥2 = 𝑧𝑥𝑥 = 𝑓𝑥𝑥
2)
𝜕
𝜕𝑦
𝜕𝑧
𝜕𝑥
=
𝜕2𝑧
𝜕𝑦𝜕𝑥
=
𝜕2𝑓
𝜕𝑦𝜕𝑥
= 𝑧𝑥𝑦 = 𝑓𝑥𝑦
3)
𝜕
𝜕𝑥
𝜕𝑧
𝜕𝑦
=
𝜕2𝑧
𝜕𝑥𝜕𝑦
=
𝜕2𝑓
𝜕𝑥𝜕𝑦
= 𝑧𝑦𝑥 = 𝑓𝑦𝑥
4)
𝜕
𝜕𝑦
𝜕𝑧
𝜕𝑦
=
𝜕2𝑧
𝜕𝑦2 =
𝜕2𝑓
𝜕𝑦2 = 𝑧𝑦𝑦 = 𝑓𝑦𝑦
58. • Evaluate the following functions if you still remember the process.
1. 𝑠𝑒𝑐2 4𝑥 − 3 𝑑𝑥
2.
𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
𝑠𝑖𝑛2𝑥
𝑑𝑥
3. 𝑥2
𝑐𝑜𝑠4𝑥3
𝑑𝑥
4. (𝑒3𝑥 + 1)2𝑑𝑥
5. 24𝑥𝑑𝑥
59. • Product of Sines and Cosines
• Trigonometric integrals are those whose integrands involve trigonometric functions. Some
trigonometric integrals cannot be evaluated directly from their given forms but they can be
reduced to standard forms by use of appropriate trigonometric identities. In this lesson, we
shall evaluate trigonometric integrals of the following types:
𝑠𝑖𝑛𝑢𝑐𝑜𝑠𝑣𝑑𝑥
𝑐𝑜𝑠𝑢𝑐𝑜𝑠𝑣𝑑𝑥
𝒔𝑖𝑛𝑢𝑠𝑖𝑛𝑣𝑑𝑥
Where u and v are differentiable function of x and u ≠ v. These functions can be evaluated
using the following formula in trigonometry:
2 sinucosv = sin (u + v) + sin ( u – v)
2 cosucosv = cos (u +v) + cos (u – v)
2 sinusinv = cos (u – v) – cos (u + v)
60. Note:
• The integral of a product of a sine and cosine can be reduced to an integral of a sum of two
sines.
The integral of a product of two cosines (or two sines) can be reduced to an integral of a sum (
or difference) of two cosines.
2𝑠𝑖𝑛𝑢𝑐𝑜𝑠𝑣𝑑𝑥 = [ sin (u + v) + sin ( u – v)]dx
2𝑐𝑜𝑠𝑢𝑐𝑜𝑠𝑣𝑑𝑥 = [ cos (u +v) + cos (u – v)]dx
𝟐𝒔𝑖𝑛𝑢𝑠𝑖𝑛𝑣𝑑𝑥 = [ cos (u – v) – cos (u + v)]dx
61. • Example1. Evaluate 𝑐𝑜𝑠6𝑥𝑐𝑜𝑠2𝑥𝑑𝑥
Solution: u = 6x ; v = 2x
𝑐𝑜𝑠6𝑥𝑐𝑜𝑠2𝑥𝑑𝑥 =
1
2
2𝑐𝑜𝑠6𝑥𝑐𝑜𝑠2𝑥𝑑𝑥
=
1
2
[cos 6𝑥 + 2𝑥 + cos(6𝑥 − 2𝑥)]𝑑𝑥
=
1
2
(
1
8
𝑠𝑖𝑛8𝑥 +
1
4
sin 4x) + C
=
1
16
𝑠𝑖𝑛8𝑥 +
1
8
sin 4x + C
63. • Evaluate each of the following:
1. 𝑠𝑖𝑛5𝑥𝑠𝑖𝑛𝑥𝑑𝑥
2. 𝑐𝑜𝑠2𝑥𝑐𝑜𝑠7𝑥𝑑𝑥
3. sin 3𝑥 − 2𝜋 cos 𝑥 + 𝜋 𝑑𝑥
4. 4𝑠𝑖𝑛8𝑥𝑐𝑜𝑠4𝑥𝑑𝑥
5. 8 sin 3𝑥 −
𝜋
4
cos 𝑥 +
𝜋
2
𝑑𝑥
64. Powers of
Sines and
Cosines
In integrating power of sine or cosine or
product of such powers, we shall consider the
trigonometric integral of the form:
𝑠𝑖𝑛𝑚
𝑣 𝑐𝑜𝑠𝑛
𝑣𝑑𝑥
Where v is differentiable function
of x and m, n, are real numbers.
If m = n = 1; or m=1, n≠1; or
m≠1, n=1, the integral can be
easily evaluated by chain rule.
It can easily be substituted as:
𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥𝑑𝑥 =
𝑠𝑖𝑛2𝑥
2
+ 𝐶
𝑠𝑖𝑛𝑥𝑐𝑜𝑠2
𝑥𝑑𝑥 = −
𝑐𝑜𝑠3𝑥
3
+ 𝐶
𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥𝑑𝑥 =
2𝑐𝑜𝑠3/2
3
+ 𝐶
65. In this part, we
will be dealing
with integral of
general form but
with m, n ≠ 1.
Consider the
following cases:
Case I: When m is positive odd integer and n is any number, we may
write
𝑠𝑖𝑛𝑚𝑥𝑐𝑜𝑠𝑛𝑥 = (𝑠𝑖𝑛𝑚−1𝑥𝑐𝑜𝑠𝑛𝑥)𝑠𝑖𝑛𝑥
Since m is odd, m - 1 is even and
we can use Pythagorean theorem
for trigonometric identity.
𝑠𝑖𝑛2𝑥 = 1 − 𝑐𝑜𝑠2𝑥
66. • Example1. Evaluate 𝑠𝑖𝑛34𝑥𝑐𝑜𝑠24𝑥 𝑑𝑥
Case II. When m is any number and n is a positive odd, we may write
𝑠𝑖𝑛𝑚𝑥𝑐𝑜𝑠𝑛𝑥 = (𝑠𝑖𝑛𝑚𝑥𝑐𝑜𝑠𝑛−1𝑥)𝑐𝑜𝑠𝑥
and then the identity
𝑐𝑜𝑠2𝑥 = 1 − 𝑠𝑖𝑛2𝑥
to reduce the integral to the form
𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 𝑠𝑖𝑛𝑥 cos 𝑑𝑥
which can be evaluated by letting
u = sinx
67. • Example2. Evaluate 𝑠𝑖𝑛2𝑥𝑐𝑜𝑠3𝑥 𝑑𝑥
• Example3. Evaluate 𝑐𝑜𝑠5𝑥 𝑑𝑥
Note:
If m and n are both positive odd integer, then 𝑠𝑖𝑛𝑚𝑥𝑐𝑜𝑠𝑛𝑥 𝑑𝑥 can be
evaluated by the method Case I or Case II.
68. Case III. When m and n are both even integers (either both positive or
one positive and other 0),
we may write 𝑠𝑖𝑛𝑚𝑥𝑐𝑜𝑠𝑛𝑥 = (𝑠𝑖𝑛2𝑥)2 𝑜𝑟 (𝑐𝑜𝑠𝑥)2
• 𝑠𝑖𝑛2𝑥 =
1−𝑐𝑜𝑠2𝑥
2
, 𝑐𝑜𝑠2𝑥 =
1+𝑐𝑜𝑠2𝑥
2
(these can be use
repeatedly when m or n or both are greater than 2).
and then use one or both of the following identities:
70. Exercises: Evaluate each of the following.
1. 𝑠𝑖𝑛5𝑥𝑐𝑜𝑠4𝑥𝑑𝑥
2. 𝑠𝑖𝑛32𝑥 cos 2𝑥 𝑑𝑥
3. 𝑠𝑖𝑛3𝑥 + cos 2𝑥 2𝑑𝑥
4. 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥
2
𝑑𝑥
5. 𝑠𝑖𝑛32𝑥𝑑𝑥
71. Power of Tangents and Secants
Consider the trigonometric integral of the form
𝑡𝑎𝑛𝑚𝑥𝑠𝑒𝑐𝑛𝑥𝑑𝑥
If m = n ≠ 1, we evaluate the integral by the definition
𝑠𝑒𝑐𝑢𝑡𝑎𝑛𝑢𝑑𝑢 = 𝑠𝑒𝑐𝑢 + 𝑐.
If m is any number and n = 2, the integral is evaluated by the ordinary method
of substitution used as
𝑡𝑎𝑛3
𝑥𝑠𝑒𝑐2
𝑥𝑑𝑥 =
𝑡𝑎𝑛4𝑥
4
+ 𝐶
the use of letting u = tanx ; du = 𝑠𝑒𝑐2𝑥𝑑𝑥
72. Case I: When m is any number and n is a
positive even integer greater than 2, we may write
𝑡𝑎𝑛𝑚
𝑥𝑠𝑒𝑐𝑛
= 𝑡𝑎𝑛𝑚
𝑥𝑠𝑒𝑐𝑛−2
𝑥 𝑠𝑒𝑐2
𝑥
then use the identity
𝑠𝑒𝑐2𝑥 = 1 + 𝑡𝑎𝑛2𝑥
to reduce the given integral to the form
(𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 tan 𝑥)𝑠𝑒𝑐2𝑥𝑑𝑥
74. Case II: When m is a positive odd integer and n
is any number, we may write
𝑡𝑎𝑛𝑚𝑥𝑠𝑒𝑐𝑛𝑥 = (𝑡𝑎𝑛𝑚−1𝑥𝑠𝑒𝑐𝑛−1𝑥)𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
and then use the identity
𝑡𝑎𝑛2
𝑥 = 𝑠𝑒𝑐2
𝑥 − 1
to reduce the integral to the form
(𝑠𝑢𝑚 𝑜𝑓 𝑝𝑜𝑤𝑒𝑟𝑠 𝑜𝑓 sec 𝑥)sec𝑥𝑡𝑎𝑛𝑥𝑑𝑥
which can be integrated with u = secx
76. Case III: When m is a positive odd (or even)
integer and n is zero, we may write
𝑡𝑎𝑛𝑚𝑥 = 𝑡𝑎𝑛𝑚−2𝑥𝑡𝑎𝑛2𝑥
And then used the identity
𝑡𝑎𝑛2𝑥 = 𝑠𝑒𝑐2 𝑥 − 1
to reduce the integral to integrable form.
79. Powers of Cotangents and Cosecants
𝑐𝑜𝑡𝑚
𝑥𝑐𝑠𝑐𝑛
𝑥𝑑𝑥
The technique involved in evaluating the integral
where x is a differentiable function of x, is similar for that evaluating
the integral
𝑡𝑎𝑛𝑚𝑥𝑠𝑒𝑐𝑛𝑥𝑑𝑥
The identity 𝑐𝑠𝑐2
𝑥 = 1 + 𝑐𝑜𝑡2
𝑥 𝑜𝑟 𝑐𝑜𝑡2
𝑥 = 𝑐𝑠𝑐2
𝑥-1 is used to reduce
the original equation into an integrable form.
80. • Example1. Evaluate 𝑐𝑜𝑡4𝑥𝑐𝑠𝑐4𝑥𝑑𝑥
Solution:
𝑐𝑜𝑡4𝑥𝑐𝑠𝑐4𝑥𝑑𝑥 = 𝑐𝑜𝑡4𝑥𝑐𝑠𝑐2𝑥𝑐𝑠𝑐2𝑥𝑑𝑥
= 𝑐𝑜𝑡4
𝑥𝑐𝑠𝑐2
𝑥(1 + 𝑐𝑜𝑡2
𝑥)𝑑𝑥
= (𝑐𝑜𝑡4𝑥𝑐𝑠𝑐2𝑥 + 𝑐𝑜𝑡6𝑥𝑐𝑠𝑐2𝑥)𝑑𝑥
Let u = cot x
du = −𝑐𝑠𝑐2
𝑥𝑑𝑥
= −
𝑐𝑜𝑡5𝑥
5
−
𝑐𝑜𝑡7𝑥
7
+ 𝐶
83. • Walli’s Formula Surface
The integral of powers of trigonometric functions can be evaluated in a more
neatly fashion and quite easily too by a formula called Walli’s Formula.
84. Conditions:
0
𝜋
2 𝑠𝑖𝑛𝑚𝑥𝑐𝑜𝑠𝑛𝑥 𝑑𝑥 =
𝑚−1 𝑚−3 …. 2 𝑜𝑟 1 ∙ 𝑛−1 𝑛−3 ….(2 𝑜𝑟 1)
𝑚+𝑛 𝑚+𝑛−2 …….( 2 𝑜𝑟 1 )
∙ α
Where: m and n are nonnegative integers
𝛼 =
𝜋
2
if both m and n are even
𝛼 = 1 if either one or both are odd
85.
86.
87.
88. • If n = 0, then it can be reduced to
0
𝜋
2 𝑠𝑖𝑛𝑚
𝑥 𝑑𝑥 =
𝑚−1 𝑚−3 ……(2 𝑜𝑟 1)
𝑚 𝑚−2 ……( 2 𝑜𝑟 1)
∙ α
Where
𝛼 =
𝜋
2
if m is even
𝛼 = 1 if m is odd
• If m = 0, then it can be simplified to
0
𝜋
2 𝑐𝑜𝑠𝑛𝑥 𝑑𝑥 =
𝑛−1 𝑛−3 ……(2 𝑜𝑟 1)
𝑛 𝑛−2 ……( 2 𝑜𝑟 1)
∙ α
Where
𝛼 =
𝜋
2
if n is even
𝛼 = 1 if n is odd
90. • Example4. 0
𝜋
6 𝑐𝑜𝑠83𝑥 𝑑𝑥
Solution:
n = 4 is even then 𝛼 =
𝜋
2
0
𝜋
6 𝑐𝑜𝑠83𝑥 𝑑𝑥 =
1
3
∙
7∙5∙3∙1
8∙6∙4∙2
∙
𝜋
2
=
35𝜋
768
91. • Trigonometric Substitutions
The problem of evaluating certain types of integrals involving algebraic
expressions maybe transformed into problems of evaluating trigonometric integrals.
The transformation is effected by appropriate trigonometric substitutions for the
original variable of integration.
These are the trigonometric substitutions which lead to integrable forms:
TS1. When the integrand contains 𝑎2
− 𝑢2
, use the substitution u = asinθ
TS2. When the integrand contains 𝑢2 + 𝑎2 , use the substitution u = a tanθ
TS3. When the integrand contains 𝑢2
- 𝑎2
, use the substitution 𝑢 = 𝑎𝑠𝑒𝑐𝜃.
We just assume that 𝜃 is an acute angle, u is a differentiable function of x and a is any number.
92. • Example 1. Evaluate
𝑥𝑑𝑥
4−𝑥2
Solution:
The integrand contain 4 − 𝑥 2
𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑎 2 − 𝑢 2 , that is a = 2 and u = x.
Let 𝑥 = 2𝑠𝑖𝑛𝜃 (1)
𝑑𝑥 = 2𝑐𝑜𝑠𝜃𝑑𝜃 (2)
From (1) 𝑠𝑖𝑛𝜃 =
𝑥
2
(3)
Using (3), draw a right triangle. Use values of trigonometric identities in relation to the sides of
a right triangle and Pythagorean theorem to determine the parts of the triangle.
93.
94. • Example 2. Evaluate
𝑑𝑥
4𝑥2+9
Solution:
Apply TS 2 since 4𝑥2 + 9 𝑡𝑎𝑘𝑒𝑠 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑢2 + 𝑎2, where u = 2x and a = 3.
2𝑥 = 3𝑡𝑎𝑛𝜃 (1) or
𝑥 =
3
2
𝑡𝑎𝑛𝜃 (2) and
𝑑𝑥 =
3
2
𝑠𝑒𝑐2𝜃𝑑𝜃 (3)
From ( 1) 𝑡𝑎𝑛𝜃 =
2𝑥
3
(4)
95. Using (4) draw the triangle and name the sides
From the figure, we read 𝑠𝑒𝑐𝜃 =
4𝑥2+9
3
(5)
4𝑥 2+ 9 = 9𝑠𝑒𝑐2𝜃 (6)
Substituting the value of (3) and (6) in the given integral
𝑑𝑥
4𝑥2
+9
=
3
2
𝑠𝑒𝑐 2𝜃𝑑𝜃
9𝑠𝑒𝑐2θ
=
1
6
𝑑𝜃
=
1
6
𝜃 + 𝐶
=
1
6
𝐴𝑟𝑐𝑡𝑎𝑛
2𝑥
3
+ C
96. Exercise: Find the integral of the following:
1. ∫
𝑥2𝑑𝑥
4−𝑥2
2.
9−𝑥2𝑑𝑥
𝑥2
3. ∫ 𝑥2
9 − 𝑥2 𝑑𝑥
97. • Additional Standard Formula
There are lots of integral formulas found at the back of some calculus
book. But these standard formulas maybe obtained by the method of
trigonometric substitutions. These formulas can be used instead of applying
trigonometric substitutions which is too long, if possible.
SF1.
𝑑𝑢
𝑢2+𝑎2 =
1
𝑎
𝐴𝑟𝑐𝑡𝑎𝑛
𝑢
𝑎
+ 𝐶
SF2.
𝑑𝑢
𝑢2−𝑎2 =
1
2𝑎
𝑙𝑛
𝑢−𝑎
𝑢+𝑎
+ 𝐶∗∗
SF3.
𝑑𝑢
𝑢2+𝑎2
= 𝑙𝑛 𝑢 + 𝑢2 + 𝑎2 + 𝐶
SF4.
𝑑𝑢
𝑢2−𝑎2
= 𝑙𝑛 𝑢 + 𝑢2 − 𝑎2 + 𝐶
99. • Example1. Evaluate ∫
𝑑𝑥
4x2 +9
Solution:
This can be evaluated using SF1 where u =2x and a =3.
Since u = 2x, then du = 2dx.
∫
𝑑𝑥
4x2 +9
=
1
2
∙
1
3
𝐴𝑟𝑐𝑡𝑎𝑛
2𝑥
3
+ 𝐶
=
1
6
𝐴𝑟𝑐𝑡𝑎𝑛
2𝑥
3
+ 𝐶
100. • Example2. Evaluate ∫
𝑑𝑥
4−𝑥2
Solution:
This resembles SF5 with u =x and a = 2.
Since u = x, then du = dx.
∫
𝑑𝑥
4−𝑥2
= 𝐴𝑟𝑐𝑠𝑖𝑛
𝑥
2
+ 𝐶
101. • Example3. Evaluate ∫
𝑑𝑥
9x2 −16
Solution:
This resembles SF2 with u =3x and a = 4.
Since u = 3x, then du = 3dx
and nf = 1/3
∫
𝑑𝑥
9x2 −16
=
1
3
∙
1
2 4
𝑙𝑛
3𝑥−4
3𝑥+4
+ 𝐶
=
1
24
𝑙𝑛
3𝑥−4
3𝑥+4
+ 𝐶
103. • Integrands Involving Quadratic Expressions
Consider the problem of evaluating an integral which takes the general form
∫ 𝑓(𝑥) 𝑑x ,
𝑑𝑥
𝑓(𝑥)
, or
𝑑𝑥
𝑓(𝑥)
where f(x) is a quadratic expression in x. The process of completing the square is
an important aid in evaluating this type of integral. Using completing the square, the
quadratic equation can be transformed into the sum (or difference) of two squares.
Then the appropriate standard formula can be used to evaluate the integrals.
106. • Next, consider the integral of the form
∫
𝑔(𝑥)
𝑓(𝑥)
𝑑𝑥 or ∫
𝑔(𝑥)
𝑓(𝑥)
𝑑𝑥
where g(x) is a linear in x and f(x) is quadratic in x. To evaluate this type of integral,
reduce it to a form for which a standard integration formula can be applied.
Example3. Evaluate ∫
2𝑥+9
x2 +2x+5
𝑑x
Solution:
Note that if we let u = 𝑥 2+ 2𝑥 +5, du = (2x+2)dx. This leads us to the “trick” of
writing the numerator 2x+9 into the form (2x+2) +7. Thus,
∫
2𝑥+9
x2 +2x+5
𝑑x =
2𝑥+2 +7
x2 +2x+5
𝑑𝑥
=
2𝑥+2
x2 +2x+5
𝑑𝑥 +
7
x2 +2x+5
𝑑𝑥 (SF1)
= ln 𝑥2
+ 2𝑥 + 5 +
7
2
𝐴𝑟𝑐𝑡𝑎𝑛
𝑥+1
2
+ 𝐶
107. • Directions. Evaluate each of the following.
1. ∫
𝑑𝑥
x 2−3x+2
2. ∫
3+2𝑥
x2 +9
𝑑𝑥
3. ∫
𝑑𝑥
2x 2−2x+1
![LESSON 1. The Indeterminate Form
• Rolle’s Theorem
• This theorem is very useful in the proof of many theorems in Calculus. It
was formulated by Michel Rolle (French Mathematician, 1652-1719).
Theorem
If a function f(x) is continuous in the closed interval [a,b];
if f’(x) exists on the open interval (a,b); and f(a) = f(b)=0,
then there is a number c in (a,b) such that f’(c)=0.](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)