This instruction material adopted of Calculus by Frank Ayres Jr 13
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 is 1/360 of the circumference, AOB = 10; if s = r,
AOB = 1 radian). Suppose AOB is measured as a degrees; then
(i) csc2udu cotu C and
2 360 r S
Suppose next that 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 and cos is the set or real number; the range of sin is –1 y 1 and the range of cos is –1 x 1. From
cos sin
tan and
cos 1 sec
it follows that the range of both tan and sec is set of real numbers while the domain of definition (cos 0) is
2 1 2n
, (n = 1, 2, 3, …). It is left as an exercise for
the reader to consider the functions cot and csc . In problem 1, we prove
1 sin lim
0
This instruction material adopted of Calculus by Frank Ayres Jr 14
RULES OF DIFFERENTIATION. Let u be a differentiable function of x; then 14.
dx du u cos ) u (sin dx
d
17.
dx du u csc )
u (cot dx
d 2
15.
dx du u sin ) u (cos dx
d
18.
dx du u tan u sec ) u (sec dx
d
16.
dx du u sec ) u (tan dx
d 2 19.
dx du u cot u csc )
u (csc dx
d
8. Differentiation of Inverse trigonometric functions
THE INVERSE TRIGONOMETRIC FUNCTIONS. If x sin , the inverse y
function is written y arcsin . The domain of definition of arc sin x 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 Principal Branch
y = arc sin x 2
1 2
1 y
y = arc cos x 0 y
y = arc tan x 2
1 2
1 y
y = arc cot x 0 y
y = arc sec x 2
1
y , 0 y 21
y = arc csc x 2
1
This instruction material adopted of Calculus by Frank Ayres Jr 15
RULES OF DIFFERENTIATION. Let u be a differentiable function of x, then 20.
9. DIFFERENTIATION OF EXPONENTIAL AND
LOGARITHMIC FUNCTIONS
THE NUMBER e= k
Rules of differentiation. If u is a differentiable function of x,
This instruction material adopted of Calculus by Frank Ayres Jr 16
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. lny
10. DIFFERENTIATION OF HYPERBOLIC FUNCTIONS
DEFINITIONS OF HYPERBOLIC FUNCTION. For u any real number, except where noted:
2
DIFFERENTIATION FORMULAS. If u is a differentiable function of x, 31.
DEFINITIONS OF INVERSE HYPERBOLIC FUNCTIONS.
This instruction material adopted of Calculus by Frank Ayres Jr 17
Differentiation formulas. If u is a differentiable function of x, 37.
dx du
u 1
1 )
u (sinh dx
d
2 1
38. , (u 1)
dx du
u 1
1 )
u (cosh dx
d
2 1
39. , (u 1)
dx du
u 1
1 ) u (tanh dx
d 2
2 1
40. , (u 1)
dx du
u 1
1 ) u (coth dx
d 2
2 1
41. , (0 u 1)
dx du
u 1 u
1 )
u h (sec dx
d
2 1
42. , (u 0)
dx du
u 1 u
1 )
u h (csc dx
d
2 1
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 4sin2 are parametric equations, with parameter , of the parabola 4x + y = 4, since 4x2 y 4cos2 4sin2 4
(b) 2 2
1t, y 4 t
x is another parametric representation, with parameter t, of the same curve.
(a) (b)
This instruction material adopted of Calculus by Frank Ayres Jr 18 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
dx dy
is given by
du dx
du dy dx dy
/ /
The second derivative 2
2
dx y d
is given by
dx du dx dy du
d dx
y d
2 2
SOLVED PROBLEMS
1. Find dx dy
and 2 2
dx y d
, given x = - sin , y = 1- cos
sin ,
cos 1
d dy d
dx
, and
cos 1
sin /
/ d dx
d dy dx dy
2 2
2 2
cos 1
1 cos
1 1 cos
1
1 cos cos
1 sin
dx d d
d dx
y d
12. FUNDAMENTAL INTEGRATION FORMULAS
IF F(x) IS A FUNCTION Whose derivative F`(x)=f(x) on a certain interval of the x -axis, 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
integral of f(x) = 2x since x x
dx d x
dx d x dx
d
2 ) 4 ( ) 5 ( )
( 2 2 2 . All indefinite
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)dxis used to indicate that the indefinite integral of f(x) is to be found. Thus we write 2xdx 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
2 2 2
2 1 2 2 2
1 arcsin C a u
a u a
u a u du
d
Absolute value signs appear in certain of the formulas. For example, we write
5. u C
du d
This instruction material adopted of Calculus by Frank Ayres Jr 19 instead of
5(a) lnu C, u 0
u du
5(b). ln( u) C, u 0
u du
and
10. tanu du lnsecu C
instead of
10(a) tanudu lnsecu C, all u such that sec u 1 10(b) tanudu ln ( secu) C, all u such that sec u -1
Fundamental Integration Formulas
1. f(x) dx f(x) C dx
d
18. cscucotudu csc u C
2. (u v)dx udx v dx 19. C
a u tan arc a 1
u a
du
2 2
3. audx a udx, a any constant 20. C
a u sin arc u
a du
2 2
4. C, m 1
1 m u du u
1 m m
5. lnu C
u du
6. C, a 0,a 1
a ln
a du a
u u
7. eu du eu C,
8. sinudu cosu C
9. cosudu sinu C
10. tanudu ln secu C
11. cotudu ln sin u C
12. secudu ln secu tanu C
13. cscudu ln csc u cotu C
14. sec2 udu tanu C
15. csc2udu cotu C
This instruction material adopted of Calculus by Frank Ayres Jr 20
21. C
a u sec arc a 1
a u u
du
2 2
22. C
a u
a u ln a 2
1
a u
du
2 2
23. C
u a
u a ln a 2
1
u a
du
2 2
24. ln u u a C
a u
du 2 2
2 2
25. lnu u a C
a u
du 2 2
2 2
26. C
a u arcsin a
u a u du u
a 2
2 1 2 2 2 1 2 2
27. u a du u u a a2ln u u2 a2 C
2 1 2 2 2 1 2 2
28. u a du u u a a2ln u u2 a2 C
2 1 2 2 2 1 2 2
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)
(i) udv uv vdu
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) vdu must not be more complex than udv
Example 1: Find x3ex2dx
Take u = x2and dv ex2xdx; then du = 2x dx and 21 x2 e
v . Now by the rule C
e e x dx e x e
x dx e
x x2 x2 x2 2 x2 21 x2
2 1 2
This instruction material adopted of Calculus by Frank Ayres Jr 21
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:
This instruction material adopted of Calculus by Frank Ayres Jr 22
14. TRIGONOMETRIC INTEGRALS
THE FOLLOWING IDENTITIES are employed to find the trigonometric integrals of this chapter.
1. sin2x cos2 x 1
2. 1 tan2x sec2 x
3. 1 cot2x csc2x
4. sin x (1 cos2x)
2 1 2
5. cos2x 12(1 cos2x)
6. sinxcosx sin2x
2 1
7. sinxcosy sin(x y) sin(x y)
2 1
8. sinxsiny cos(x y) cos(x y)
2 1
9. cosxcosy 21 cos(x y) cos(x y)
10.1 cosx 2sin x
2 1 2
11.1 cosx 2cos212x
12.1 sinx 1 cos( x)
2 1
SOLVED PROBLEMS
SINES AND COSINES
1. sin xdx (1 cos2x)dx x 41sin2x C
2 1 2
1 2
2. cos 3xdx (1 cos6x)dx x 121sin6x C
2 1 2
1 2
3. xdx x xdx x xdx x 3x C
8 1 2
2 3
cos cos
sin ) cos 1 ( sin
sin sin
4. cos3xdx cos4xcosxdx (1 sin2x)2cosxdx
dx x x dx
x x dx
x 2 sin cos sin cos
cos 2 4
C x x
x 3 51 3
3
2sin sin
sin
5. sin2xcos3xdx sin2xcos2xcosxdx sin2x(1 sin2x)cosxdx
C x x
dx x x dx
x
x 3 51 5
3 1 4
2
sin sin
cos sin
This instruction material adopted of Calculus by Frank Ayres Jr 23
15. TRIGONOMETRIC SUBSTITUTIONS
AN INTEGRAND, which contains one of the forms a2 b2u2 , a2 b2u2 , or 2
2 2
a u
b but no other irrational factor, may be transformed into another involving trigonometric functions of a new variable as follows:
For Use To obtain
2 2 2
u b
a z
b a
u sin a 1 sin2z acosz
2 2 2
u b
a z
b a
u tan a 1 tan2z asecz
2 2 2
a u
b z
b a
u sec a sec2z 1 atanz
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.
SOLVED PROBLEMS
1. Find
2 2
4 x x
dx
Let x = 2 tan z; then dx = 2 sec2z dz and 4 x2 2secz
dz z z z
z dz z
x x
dx
2 2
2
2
2 tan
sec 4 1 ) sec 2 )( tan 4 (
sec 2 4
C
x x C
z dz
z z
4 4 sin
4 1 cos
sin 4
1 2 2
2. Find dx
x x
4 2
2
Let x = 2 sec z; then dx = 2 sec z tan z dz and z
x2 4 2tan
dz z dz
z z z
z dx
x
x 2 3
2 2
sec 4 ) tan sec 2 ( tan 2
sec 4 4
C z z
z
ztan 2lnsec tan sec
2
This instruction material adopted of Calculus by Frank Ayres Jr 24
3. Find dx
x x2
4 9
Let x 23sinz; then dx z dz
cos
2
3 and
z x 3cos 4
9 2
dz z
z dz
z z
z dx
x x
sin cos 3 cos
sin cos 3 4
9 2
2 3
2 3 2
dz zdz zdz
z z
sin 3 csc
3 sin
sin 1 3
2
3lncscz cotz 3cosz C
x C
x
x 2
2
4 9 4
9 3 ln 3
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
) (
) ( ) (
x g
x f x
F , where f(x) and g(x) are polynomials, is called a rational 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,
1 x
x x 1 x
x
2 2
3
.
This instruction material adopted of Calculus by Frank Ayres Jr 25
CASE I. DISTINCT LINEAR FACTORS
To each linear factor ax + b occurring once in the denominator of a proper rational fraction, there corresponds a single partial fraction of the form
b ax
A
, where A 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
n n
b ax
A b
ax A b
ax A
) ( )
( 2
2
1
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 the form
c bx ax
B Ax
2 , where A and B are constants to be determined.
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
n n n
c bx ax
B x A c
bx ax
B x A c
bx ax
B x A
) (
)
( 2 2 2
2 2 2
1
1
where the A’s and B’s are constants to be determined.
SOLVED PROBLEMS
1. Find
4
2 x
dx
(a) Factor the denominator: x2– 4 = (x-2)(x+2) Write
2 2
4 1
2 x
B x
A
x and clear of fraction to obtain
(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 41, and
4 1
This instruction material adopted of Calculus by Frank Ayres Jr 26 used are those for which the denominator of the partial fractions become 0). (c) By either method:
2 (b) General method. Solve simultaneously the system of equation
