7. Differentiation of Trigonometric Function
RADIAN MEASURE. Let s denote the length of arc
AB intercepted by the central angle AOB on a circle of
radius r and let S denote the area of the sector AOB. (If s
AOB = 10; if s = r,
is 1/360 of the circumference,
AOB = 1 radian). Suppose AOB is measured as a
degrees; then
(i)
and
csc2 u du
cot u C
S
360
Suppose next that
r2
AOB is measured as
radian; then
(ii)
s= r
and
S = ½ r2
A comparison of (i) and (ii) will make clear one of the advantages of radian measure.
TRIGONOMETRIC FUNCTIONS. Let
be any real
number. Construct the angle whose measure is radians
with vertex at the origin of a rectangular coordinate
system and initial side along the positive x-axis. Take P(
x, y) on the terminal side of the angle a unit distance from
O; then sin = y and cos = x. The domain of definition
of both sin
of sin
is –1
tan
and cos
is the set or real number; the range
y 1 and the range of cos
sin
sec
and
cos
it follows that the range of both tan
of definition (cos
0) is
is –1 x
1
cos
and sec
1. From
is set of real numbers while the domain
2n 1
, (n = 1, 2, 3, …). It is left as an exercise for
2
the reader to consider the functions cot
In problem 1, we prove
and csc .
lim
sin
0
1
(Had the angle been measured in degrees, the limit would have been /180. For this
reason, radian measure is always used in the calculus)
This instruction material adopted of Calculus by Frank Ayres Jr
13
RULES OF DIFFERENTIATION. Let u be a differentiable function of x; then
14.
d
(sin u)
dx
15.
d
(cos u)
dx
sin u
16.
d
(tan u)
dx
sec2 u
cos u
du
dx
17.
d
(cot u)
dx
csc2 u
du
dx
du
dx
18.
d
(sec u)
dx
sec u tan u
du
dx
19.
d
(csc u)
dx
csc u cot u
du
dx
du
dx
8. Differentiation of Inverse trigonometric functions
THE INVERSE TRIGONOMETRIC FUNCTIONS. If x
function is written y
sin y , the inverse
arc sin x . The domain of definition of arc sin x is –1
x
1, the
range of sin y; the range of arc sin x is the set if real numbers, the domain of definition
of sin y. The domain if definition and the range of the remaining inverse trigonometric
functions may be established in a similar manner.
The inverse trigonometric functions are multi-valued. In order that there be
agreement on separating the graph into single-valued arcs, we define below one such
arc (called the principal branch) for each function. In the accompanying graphs, the
principal branch is indicated by a thickening of the line.
y = arc sin x
y = arc cos x
y = arc tan x
Fig. 13-1
Function
y = arc sin x
y = arc cos x
y = arc tan x
y = arc cot x
y = arc sec x
y = arc csc x
This instruction material adopted of Calculus by Frank Ayres Jr
Principal Branch
1
y 12
2
0
y
1
2
y
0
y
y
1
2
y
1
2
1
2
,0
,0
y
1
2
y
1
2
14
RULES OF DIFFERENTIATION. Let u be a differentiable function of x, then
20.
d
(arc sin u)
dx
1
du
1 u 2 dx
21.
d
(arc cos u)
dx
1
22.
d
(arc tan u)
dx
du
1 u 2 dx
1
du
1 u dx
2
23.
d
(arc cot u)
dx
24.
d
(arc sec u)
dx
25.
d
(arc csc u)
dx
1
du
1 u dx
2
1
du
1 dx
u u2
1
u u2
du
1 dx
9. DIFFERENTIATION OF EXPONENTIAL AND
LOGARITHMIC FUNCTIONS
1
THE NUMBER e = lim 1
h
h
=1 1
lim (1 k )1 / k
k
0
1 1
1

 2.178 28
2! 3!
n!
NOTATION. If a > 0 and a
y
h
1, and if ay = x, then y = log ax
log e x
ln x
y
log 10 x
log x
The domain of definition is x > 0; the range is the set of real numbers.
y = eax
Fig. 14-1
Rules of differentiation. If u is a differentiable function of x,
y = ln x
d a
1 du
( log u)
, a
dx
u ln a dx
d
1 du
27.
(ln u)
dx
u dx
26.
d u
(a )
dx
d u
29.
(e )
dx
28.
au ln a
eu
du
, a
dx
y = e-ax
0, a 1
0
du
dx
This instruction material adopted of Calculus by Frank Ayres Jr
15
LOGARITHMIC DIFFERENTIATION. If a differentiable function y = f(x) is the
product of several factors, the process of differentiation may be simplified by taking the
natural logarithm of the function before differentiating or, what is the same thing, by
using the formula
30.
d
( y)
dx
y
du
ln y
dx
10. DIFFERENTIATION OF HYPERBOLIC FUNCTIONS
DEFINITIONS OF HYPERBOLIC FUNCTION. For u any real number, except
where noted:
sinh u
cosh u
tanh u
eu
e
u
2
eu
e
coth u
1
tanh u
sec h u
1
coshu
csch u
1
sinh u
u
2
sinh u
cosh u
eu
eu
e
e
u
u
eu
eu
e
e
u
u
, (u
0)
, (u
0)
2
e
u
e
u
e
u
2
e
u
DIFFERENTIATION FORMULAS. If u is a differentiable function of x,
31.
d
(sinh u)
dx
coshu
du
dx
34.
d
(coth u)
dx
csc h2 u
32.
d
(cosh u)
dx
sinh u
du
dx
35.
d
(sec h u)
dx
sec h u tanh u
33.
d
(tanh u)
dx
sec h2 u
36.
d
(csc h u)
dx
csc h u coth u
du
dx
du
dx
DEFINITIONS OF INVERSE HYPERBOLIC FUNCTIONS.
u 1
coth 1 u 12 ln
, (u 2
sinh 1 u ln(u
1 u 2 ) ,all u
u 1
1
cosh u
ln(u
tanh 1 u
1
2
ln
u
2
1), (u 1)
1 u
, (u 2
1 u
1)
sec h 1u
ln
csc h 1u
ln
This instruction material adopted of Calculus by Frank Ayres Jr
1
1
u
du
dx
du
dx
1)
1 u2
, (0
u
u
1 u2
, (u
u
1)
0)
16
Differentiation formulas. If u is a differentiable function of x,
37.
d
(sinh 1 u)
dx
1
du
1 u 2 dx
38.
d
(cosh 1 u)
dx
39.
d
(tanh 1 u)
dx
40.
d
(coth 1 u)
dx
du
,
1 u dx
41.
d
(sec h 1 u)
dx
1
42.
d
(csc h 1 u)
dx
1
du
,
2 dx
1 u
1
du
,
1 u dx
2
1
2
(u
1)
(u 2
1)
(u 2
1)
du
,
2 dx
u 1 u
1
du
,
u 1 u2 dx
(0
u 1)
(u
0)
11. PARAMETRIC REPRESENTATION OF CURVES
PARAMETRIC EQUATIONS. If the coordinates (x, y) of a point P on a curve are
given as functions x = f(u), y = g(u) of a third variable or parameters u, the equations x
= f(u), y = g(u) are called parametric equations of the curve.
Example:
(a) x
cos , y
4 sin 2
4x + y = 4, since 4 x 2
(b) x
1
2
t, y
are parametric equations, with parameter , of the parabola
y
4 cos 2
4 sin 2
4
4 t 2 is another parametric representation, with parameter t, of the
same curve.
(a)
(b)
Fig. 16-1
This instruction material adopted of Calculus by Frank Ayres Jr
17
It should be noted that the first set of parametric equations represents only a
portion of the parabola, whereas the second represents the entire curve.
THE FIRST DERIVATIVE
dy
dy
is given by
dx
dx
d2y
d2y
The second derivative
is given by
dx 2
dx 2
dy / du
dx / du
d dy
du dx
du
dx
SOLVED PROBLEMS
1. Find
d2y
dy
and
, given x =
dx
dx 2
dx
dy
1 cos ,
d
d
d2y
dx 2
d
d
sin
1 cos
- sin , y = 1- cos
sin ,
d
dx
dy
dx
and
cos
1
1 cos
2
dy / d
dx / d
sin
1 cos
1
1 cos
1
1 cos
2
12. FUNDAMENTAL INTEGRATION FORMULAS
IF F(x) IS A FUNCTION Whose derivative F`(x)=f(x) on a certain interval of the xaxis, then F(x) is called an anti-derivative or indefinite integral of f(x). The indefinite
integral of a given function is not unique; for example, x2, x2 + 5, x2 – 4 are indefinite
d 2
d 2
d 2
integral of f(x) = 2x since
(x )
( x 5)
( x 4) 2 x . All indefinite
dx
dx
dx
integrals of f(x) = 2x are then included in x2 + C where C, called the constant of
integration, is an arbitrary constant.
The symbol
f ( x)dx is used to indicate that the indefinite integral of f(x) is to be
found. Thus we write 2 x dx
x2
C
FUNDAMENTAL INTEGRATION FORMULAS. A number of the formulas below
follow immediately from the standard differentiation formulas of earlier chapters while
Formula 25, for example, may be checked by showing that
d
du
5.
1
2
u a2
u2
1
2
a 2 arcsin
u
a
C
a2
u2
Absolute value signs appear in certain of the formulas. For example, we write
d
ln u C
du
This instruction material adopted of Calculus by Frank Ayres Jr
18
instead of
du
u
5(a)
ln u C , u
5(b).
0
du
u
ln( u ) C , u
0
and
tan u du
10.
ln secu
C
instead of
10(a)
tan u du
ln sec u C ,
all u such that sec u
1
10(b)
tan u du
ln ( sec u) C ,
all u such that sec u
-1
Fundamental Integration Formulas
1.
d
f (x) dx
dx
2.
(u
3.
au dx
4.
um du
5.
du
u
6.
au du
au
ln a
7.
e u du
eu
8.
sin u du
9.
cos u du
f ( x)
v ) dx
18.
C
u dx
19.
v dx
a u dx , a any constant
um 1
m 1
ln u
C,
m
20.
csc u cot u du
du
a
2
u
2
du
a2
u2
csc u C
1
u
arc tan
a
a
C
u
a
C
arc sin
1
C
C,
a
0, a
1
C,
cos u C
sin u C
10. tan u du ln sec u
C
11. cot u du
C
ln sin u
12. sec u du ln sec u tan u
C
13. csc u du ln csc u cot u
C
14. sec 2 u du tan u C
15. csc 2 u du
cot u C
16. sec u tan u du sec u C
This instruction material adopted of Calculus by Frank Ayres Jr
19
du
21.
u u
22.
23.
a
2
du
u2
a2
du
a2
u2
u
2
a
C
1
u
ln
2a u
a
a
C
1
a
ln
2a a
u
u
C
ln u
u2
a2
C
ln u
u2
a2
C
du
24.
25.
2
1
u
arc sec
a
a
2
du
u2
a2
26.
a2
u2 du
1u
2
a2
u2
1 a2
2
27.
u2
a 2 du
1u
2
u2
a2
1 a 2 ln
2
u
u2
a2
C
28.
u2
a 2 du
1u
2
u2
a2
1 a 2 ln
2
u
u2
a2
C
arcsin
u
a
C
13. INTEGRATION BY PARTS
INTEGRATION BY PARTS. When u and v are differentiable function of
d (uv) = u dv + v du
u dv = d(uv) – v(du)
u dv
(i)
uv
v du
To use (i) in effecting a required integration, the given integral must be
separated into two parts, one part being u and the other part, together with dx, being dv.
(For this reason, integration by the use of (i) is called integration by parts.) Two
general rules can be stated:
(a) the part selected as dv must be readily integrable
(b)
v du must not be more complex than u dv
Example 1: Find
2
x 3 e x dx
Take u = x2 and dv
2
x 2 e x dx
2
e x x dx ; then du = 2x dx and v
1
2
x 2e x
2
2
x e x dx
This instruction material adopted of Calculus by Frank Ayres Jr
1
2
x 2e x
2
1
2
1
2
ex
2
e x . Now by the rule
2
C
20
ln( x 2
Example 2: Find
2)dx
Take u = ln (x2 + 2) and dv = dx; then du
ln( x 2
x ln( x 2
2)dx
2 x 2 dx
x2 2
2)
x ln ( x 2
2 x dx
and v = x. By the rule.
x2 2
x ln( x 2
2)
2) 2 x 2 2 arc tan x / 2
4
2
x
2
2
dx
C
REDUCTION FORMULAS. The labour involved in successive applications of
integration by parts to evaluate an integral may be materially reduced by the use of
reduction formulas. In general, a reduction formula yields a new integral of the same
form as the original but with an exponent increased or reduced. A reduction formula
succeeds if ultimately it produces an integral which can be evaluated. Among the
reduction formulas are:
(A).
du
(a u 2 ) m
(B). (a 2
(C).
1
u
2
a (2m 2)(a 2
2
u 2 ) m du
du
(u a 2 ) m
u )
2ma 2
(a 2
2m 1
1
u
2
a (2m 2)(u 2
2
(D). (u 2
u (a 2 u 2 ) m
2m 1
2 m 1
a )
u (u 2 a 2 ) m 2ma 2
(u 2
2m 1
2m 1
1 m au m m 1 au
u e
u e du
2
a
a 2 ) m du
(E). u m e au du
(F). sin m u du
(G). cos m u du
sin m 1 u cos u
m
cos m 1 u sin u
m
(H). sin m u cos m u du
m
(I). n sin bu du
(J). n cos bu du
u 2 ) m 1 du, m
2m 3
du
2
2m 2 (u a 2 ) m
a 2 ) m 1 du, m
, m 1
1
1/ 2
1
, m 1
1/ 2
m 1
sin m 2 u du
m
m 1
cos m 2 u du
m
sin m 1 u cos n 1 u
m n
sin m 1 u cos n 1 u
m n
m
2m 3
du
2
2m 2 ( a u 2 ) m
2 m 1
n 1
sin m u cos n 2 u du
m n
m 1
sin m 2 u cos n u du, m
m n
um
cos bu
b
m m1
u cos bu du
b
um
sin bu
b
m m1
u sin bu du
b
This instruction material adopted of Calculus by Frank Ayres Jr
n
21
14. TRIGONOMETRIC INTEGRALS
THE FOLLOWING IDENTITIES are employed to find the trigonometric integrals
of this chapter.
1. sin 2 x cos 2 x 1
7. sin x cos y
2. 1 tan2 x
sec 2 x
8. sin x sin y
3. 1 cot 2 x
csc 2 x
2
4. sin x
5. cos 2 x
1 (1
2
6. sin x cos x
1
2
cos 2x)
1 sin 2x
2
sin(x
y) sin(x
y)
cos(x
y) cos(x
y)
1
2
9. cos x cos y
cos 2x)
1 (1
2
1
2
cos(x
10. 1 cos x
2 sin 2 1 x
11. 1 cos x
2 cos 2 1 x
y) cos(x
y)
2
2
12. 1 sin x 1 cos(12
x)
SOLVED PROBLEMS
SINES AND COSINES
1.
sin 2 x dx
2.
cos2 3x dx
3.
sin 3 x dx
sin 2 x sin x dx
(1 cos2 x) sin x dx
4.
cos3 x dx
cos4 x cos x dx
(1 sin 2 x) 2 cos x dx
1
2
1
2
(1 cos2 x) dx
1
2
(1 cos6 x) dx
1
4
x
1
2
x
sin 2 x C
1
12
sin 6 x C
cos x dx 2 sin 2 x cos x dx
sin x
5.
sin 2 x cos3 x dx
2
3
sin 3 x
1
5
cos x
1
8
cos3 x C
sin 4 x cos x dx
sin 3 x C
sin 2 x cos2 x cos x dx
sin 2 x cos x dx
sin 2 x(1 sin 2 x) cos x dx
sin 4 x cos x dx
This instruction material adopted of Calculus by Frank Ayres Jr
1
3
sin 3 x
1
5
sin 5 x C
22
15. TRIGONOMETRIC SUBSTITUTIONS
a2
AN INTEGRAND, which contains one of the forms
b 2u 2 ,
a2
b 2 u 2 , or
b 2 u 2 a 2 but no other irrational factor, may be transformed into another involving
trigonometric functions of a new variable as follows:
For
Use
To obtain
a
a 1 sin 2 z a cos z
sin z
b
a
a 1 tan 2 z a sec z
a 2 b 2u 2
u
tan z
b
a
a sec2 z 1 a tan z
b 2u 2 a 2
u
sec z
b
In each case, integration yields an expression in the variable z. The corresponding
expression in the original variable may be obtained by the use of a right triangle as
shown in the solved problems below.
a2
b 2u 2
u
SOLVED PROBLEMS
dx
1. Find
x
2
4
x2
Let x = 2 tan z; then dx = 2 sec2z dz and
2 sec2 z dz
(4 tan 2 z )(2 sec z )
dx
x2 4 x2
1
sin
4
2
4 x2
2 sec z
1 sec z
dz
4 tan 2 z
1
4 sin z
z cos z dz
C
4 x2
4x
C
x2
2. Find
dx
x2 4
Let x = 2 sec z; then dx = 2 sec z tan z dz and
x2
4
2 tan z
x2
x
2
4 sec2 z
(2 sec z tan z dz)
2 tan z
dx
4
2 sec z tan z 2 ln sec z tan z
1
2
x x2
4 2 ln x
4 sec3 z dz
C
x2
4
This instruction material adopted of Calculus by Frank Ayres Jr
C
23
3. Find
9 4x 2
dx
x
Let
x
9 4x 2
3
2
sin z ;
then
dx
3
2
cos z dz
and
3 cos z
9 4x 2
dx
x
3 cos z
3
2 sin z
3
3
2
cos z dz
1 sin 2 z
dz
sin z
3 ln csc z cot z
3 ln
3
cos2 z
3
dz
sin z
9 4x 2
x
3 csc z dz 3 sin z dz
3 cos z C
9 4x 2
C
16. INTEGRATION BY PARTIAL FRACTIONS
A POLYNOMIAL IN x is a function of the form a0xn + a1xn-1 + … + an-1x + an, where
the a’s are constants, a0 0, and n is a positive integer including zero.
If two polynomials of the same degree are equal for all values of the variable, the
coefficients of the like powers of the variable in the two polynomials are equal.
Every polynomial with real coefficients can be expressed (at least, theoretically)
as a product of real linear factors of the form ax + b and real irreducible quadratic factors
of the form ax2 + bx + c.
A function F ( x)
f ( x)
, where f(x) and g(x) are polynomials, is called a rational
g ( x)
fraction.
If the degree of f(x) is less than the degree of g(x), F(x) is called proper;
otherwise. F(x) is called improper.
An improper rational fraction can be expressed as the sum of a polynomial and a
proper rational fraction. Thus,
x3
x
x
.
x2 1
x2 1
Every proper rational fraction can be expressed (at least, theoretically) as a sum of
simpler fractions (partial fractions) whose denominators are of the form (ax + b)n and
(ax2 + bx + c)n, n being a positive integer. Four cases, depending upon the nature of the
factors of the denominator, arise.
This instruction material adopted of Calculus by Frank Ayres Jr
24
CASE I. DISTINCT LINEAR FACTORS
To each linear factor ax + b occurring once in the denominator of a proper
A
rational fraction, there corresponds a single partial fraction of the form
, where A
ax b
is a constant to be determined.
CASE II. REPEATED LINEAR FACTORS
To each linear factor ax + b occurring n times in the denominator of a proper
rational fraction, there corresponds a sum of n partial fractions of the form
A1
ax b
A2
(ax b)
2
An

(ax b) n
where the A’s are constants to be determined.
CASE III. DISTINCT QUADRATIC FACTORS
To each irreducible quadratic factor ax2 + bx + c occurring once in the
denominator of a proper rational fraction, there corresponds a single partial fraction of
Ax B
the form
, where A and B are constants to be determined.
2
ax bx c
CASE IV. REPEATED QUADRATIC FACTORS
To each irreducible quadratic factor ax2 + bx + c occurring n times in the
denominator of a proper rational fraction, there corresponds a sum of n partial fraction of
the form
A1 x B1
ax 2 bx c
A2 x B2
(ax 2 bx c) 2

An x Bn
(ax 2 bx c) n
where the A’s and B’s are constants to be determined.
SOLVED PROBLEMS
1. Find
dx
2
x 4
(a) Factor the denominator: x2 – 4 = (x-2)(x+2)
1
A
B
Write 2
and clear of fraction to obtain
x 4 x 2 x 2
(1) 1 = A(x + 2) + B(x – 2) or (2) 1 = (A + B)x + (2A – 2B)
(b) Determine the constants
General method. Equate coefficients of like powers of x in (2) and solve
simultaneously for the constants. Thus, A + B = 0 and 2A – 2B = 1; A 14 , and
B
1
4
.
This instruction material adopted of Calculus by Frank Ayres Jr
25
Short method. Substitute in (1) the values x = 2 and x = -2 to obtain 1 =
1
4A and 1 = -4B; then A 14 and B
4 , as before. (Note that the values of x
used are those for which the denominator of the partial fractions become 0).
1
1
1
4
4
(c) By either method: 2
and
x 4 x 2 x 2
dx
x
2. Find
2
4
1 dx
4 x 2
1 dx
4 x 2
1
ln x 2
4
1
ln x 2
4
1 x 2
ln
4 x 2
C
C
( x 1)dx
x x 2 6x
3
x 1
A
B
C
and
2
x x 6x x x 2 x 3
(1) x + 1 = A(x – 2)(x + 3) + Bx(x + 3) + Cx (x – 2) or
(2) x + 1 = (A + B + C)x2 + (A + 3B – 2C)x – 6A
(b) General method. Solve simultaneously the system of equation
A + B + C = 0, A + 3B – 2C = 1, and -6A = 1
To obtain A = -1/6, B = 3/10, and C = -2/15.
Short method. Substitute in (1) the values x = 0, x = 2, and x = -3 to obtain 1 = 6A or A = -1/6, 3 = 10B or B = 3/10, and –2 = 15C or C = -2/15
( x 1)dx
1 dx 3
dx
2
dx
(c)
3
2
6 x 10 x 2 15 x 3
x x 6x
(a) x3+ x2 – 6x = x(x – 2)(x + 3). Then
1
ln x
6
3. Find
3
3
ln x 2
10
2
ln x 3
15
C
x 2
ln
x
1/ 6
3 / 10
x 3
2 / 15
C
(3x 5)dx
x x2 x 1
3
x3 – x2 – x + 1 = (x+1)(x-1)2. Then
3x 5
x3
x2
A
x 1
x 1
B
C
x 1 ( x 1) 2
and
3x + 5 = A(x – 1)2 + B(x + 1)(x – 1) + C (x+1).
For x = -1, 2 = 4A
and A = ½.
For x = 1, 8 = 2C and C = 4. To determine the
remaining constant, use any other value of x, say x = 0; for x = 0,
5 = A – B + C and B = -½ . Thus
(3 x 5)dx
x x2 x 1
3
1 dx
2 x 1
1 dx
dx
4
2 x 1
( x 1) 2
1
1
ln x 1
ln x 1
2
2
4
1
x 1
ln
C
x -1 2
x -1
This instruction material adopted of Calculus by Frank Ayres Jr
4
x -1
C
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