The integral on the right-hand side is easy to take, since the integrand is the sum of power functions.. By removing the brackets, one obtains a sum of integrals like x n e ax dx, which
Trang 1are evaluated using the formulas
sin α cos β = 12[sin(α + β) + sin(α – β)], cos α cos β = 12[cos(α + β) + cos(α – β)], sin α sin β = 12[cos(α – β) – cos(α + β)].
4 Integrals of the form
sinm xcosn x dx , where m and n are integers, are evaluated
as follows:
(a) if m is odd, one uses the change of variable cos x = z, with sin x dx = –dz;
(b) if n is odd, one uses the change of variable sin x = z, with cos x dx = dz;
(c) if m and n are both even nonnegative integers, one should use the degree reduction
formulas
sin2x= 12(1– cos2x), cos2x= 12(1+ cos2x), sin x cos x = 12 sin2x
Example 3 Evaluate the integral
sin5x dx.
This integral corresponds to odd m: m =5 With simple rearrangement and the change of variable
cos x = z, we have
sin5x dx=
(sin2x)2sin x dx = –
( 1 – cos2x)2d cos x = –
( 1– z2)2dz
= 23z3–15z5– z + C = 23cos3x–15cos5x – cos x + C.
Remark In general, the integrals
sinp xcosq x dxare reduced to the integral of a differential binomial
by the substitution y = sin x.
7.1.6 Integration of Polynomials Multiplied by Elementary Functions
Throughout this section, P n (x) designates a polynomial of degree n.
7.1.6-1 Integration of the product of a polynomial by exponential functions
General formulas:
P n (x)e ax dx = e ax
P n (x)
a –P n (x)
a2 +· · · + (–1 )n P
(n)
n (x)
a n+1
+ C,
P n (x) cosh(ax) dx = sinh(ax)
P n (x)
a + P (x)
a3 +· · ·
– cosh(ax)
P
n (x)
a2 + P n (x)
a4 +· · ·
+ C,
P n (x) sinh(ax) dx = cosh(ax)
P n (x)
a + P (x)
a3 +· · ·
– sinh(ax)
P
n (x)
a2 +P n (x)
a4 +· · ·
+ C.
These formulas are obtained by repeated integration by parts; see formula 4 from
Para-graph 7.1.2-2 with f (x) = P n (x) for g(n+1)(x) = e ax , g(n+1)(x) = cosh(ax), and g(n+1)(x) = sinh(ax), respectively.
In the special case P n (x) = x n, the first formula gives
x n e ax dx = e axn
k=0
(–1)n–k
a n+1 –k
n!
k!x
k + C.
Trang 2284 INTEGRALS
7.1.6-2 Integration of the product of a polynomial by a trigonometric function
1◦ General formulas:
P n (x) cos(ax) dx = sin(ax)
P n (x)
a – P (x)
a3 +· · ·
+ cos(ax)
P
n (x)
a2 – P n (x)
a4 +· · ·
+ C,
P n (x) sin(ax) dx = sin(ax)
P
n (x)
a2 – P n (x)
a4 +· · ·
– cos(ax)
P n (x)
a – P (x)
a3 +· · ·
+ C.
These formulas are obtained by repeated integration by parts; see formula 4 from
Para-graph 7.1.2-2 with f (x) = P n (x) for g(n+1)(x) = cos(ax) and g(n+1)(x) = sin(ax), respectively.
2◦ To evaluate integrals of the form
P n (x) cos m (ax) dx,
P n (x) sin m (ax) dx, with m =2,3, , one should first use the trigonometric formulas
cos2k (ax) = 1
22k–1
k–1
i=0
C i
2kcos[2(k – i)ax] + 1
22k C2k k (m =2k),
cos2k+1(ax) = 1
22k
k
i=0
C i
2k+1cos[(2k–2i+1)ax] (m =2k+1),
sin2k (ax) = 1
22k–1
k–1
i=0
(–1)k–i C i
2kcos[2(k – i)ax] + 212k C k
2k (m =2k),
sin2k+1(ax) = 1
22k
k
i=0
(–1)k–i C i
2k+1sin[(2k–2i+1)ax] (m =2k+1),
thus reducing the above integrals to those considered in Item1◦.
3◦ Integrals of the form
P n (x) e ax sin(bx) dx,
P n (x) e ax cos(bx) dx
can be evaluated by repeated integration by parts
In particular,
x n e ax sin(bx) = e axn+1
k=1
(–1)k+1n!
(n – k +1)! (a2+ b2)k/2x n–k+1sin(bx + kθ) + C,
x n e ax cos(bx) = e axn+1
k=1
(–1)k+1n!
(n – k +1)! (a2+ b2)k/2x n–k+1cos(bx + kθ) + C, where
sin θ = – √ b
a2+ b2, cos θ =
a
√
a2+ b2.
Trang 37.1.6-3 Integrals involving power and logarithmic functions.
1◦ The formula of integration by parts with g (x) = P n (x) is effective in the evaluation of
integrals of the form
P n (x) ln(ax) dx = Q n+1(x) ln(ax) – a
Q n+1(x)
x dx,
where Q n+1(x) =
P n (x) dx is a polynomial of degree n+1 The integral on the right-hand
side is easy to take, since the integrand is the sum of power functions
Example Evaluate the integral
ln x dx.
Setting f (x) = ln x and g (x) =1, we find f (x) = 1
x and g(x) = x Substituting these expressions into
the formula of integration by parts, we obtain
ln x dx = x ln x –
dx = x ln x – x + C.
2◦ The easiest way to evaluate integrals of the more general form
I =
n i=0
lni (ax)
m
j=0
b ij x β ij
dx,
where the β ij are arbitrary numbers, is to use the substitution z = ln(ax), so that
I =
n i=0
z im
j=0
b ij
a β ij+ 1e(β ij+1)z
dz
By removing the brackets, one obtains a sum of integrals like
x n e ax dx, which are easy
to evaluate by the last formula in Paragraph 7.1.6-1
7.1.6-4 Integrals involving inverse trigonometric functions
1◦ The formula of integration by parts with g (x) = P n (x) also allows the evaluation of the
following integrals involving inverse trigonometric functions:
P n (x) arcsin(ax) dx = Q n+1(x) arcsin(ax) – a
Q n+1(x)
√
1– a2x2 dx,
P n (x) arccos(ax) dx = Q n+1(x) arccos(ax) + a
Q n+1(x)
√
1– a2x2 dx,
P n (x) arctan(ax) dx = Q n+1(x) arctan(ax) – a
Q n+1(x)
a2x2+1 dx,
P n (x) arccot(ax) dx = Q n+1(x) arccot(ax) + a
Q n+1(x)
a2x2+1 dx,
where Q n+1(x) =
P n (x) dx is a polynomial of degree n+1 The integrals with radicals on
the right-hand side in the first two formulas can be evaluated using the techniques described
in Paragraph 7.1.4-2 The integrals of rational functions on the right-hand side in the last two formulas can be evaluated using the techniques described in Subsection 7.1.3
Trang 4286 INTEGRALS
Remark. The above formulas can be generalized to contain any rational functions R (x) and R(x) instead
of the polynomials P n (x) and Q n+1(x), respectively.
2◦ The following integrals are taken using a change of variable:
P n (x) arcsin m (ax) dx = 1
a
cos z P n
sin z a
z m dz, substitution z = arcsin(ax);
P n (x) arccos m (ax) dx = –1
a
sin z P n
cos z a
z m dz , substitution z = arccos(ax),
where m = 2,3, The expressions cos z sink z and sin z cos k z (k = 1, , n) in the integrals on the right-hand sides should be expressed as sums of sines and cosines with appropriate arguments Then it remains to evaluate integrals considered in Paragraph 7.1.6-2
7.2 Definite Integral
7.2.1 Basic Definitions Classes of Integrable Functions.
Geometrical Meaning of the Definite Integral
7.2.1-1 Basic definitions
Let y = f (x) be a bounded function defined on a finite closed interval [a, b] Let us partition this interval into n elementary subintervals defined by a set of points{x0, x1, , x n}such
that a = x0 < x1 <· · · < x n = b Each subinterval [x k–1, x k] will be characterized by its
lengthΔx k = x k – x k–1and an arbitrarily chosen point ξ k[x k–1, x k] Let us make up an
integral sum (a Cauchy–Riemann sum, also known as a Riemann sum)
s n=
n
k=1
f (ξ k)Δx k (x k–1 ≤ξ k ≤x k).
If, as n → ∞ and, accordingly, Δx k → 0 for all k, there exists a finite limit of the integral sums s n and it depends on neither the way the interval [a, b] was split up, nor the selection of the points ξ k, then this limit is denoted b
a f (x) dx and is called the definite
integral (also the Riemann integral) of the function y = f (x) over the interval [a, b]:
b
a f (x) dx = lim n→∞ s n
max 1≤k n Δx k →0 as n → ∞
In this case, the function f (x) is called integrable on the interval [a, b].
7.2.1-2 Classes of integrable functions
1 If a function f (x) is continuous on an interval [a, b], then it is integrable on this
interval
2 If a bounded function f (x) has finitely many jump discontinuities on [a, b], then it is integrable on [a, b].
3 A monotonic bounded function f (x) is always integrable.
Trang 57.2.1-3 Geometric meaning of the definite integral.
If f (x) ≥ 0 on [a, b], then the integral b
a f (x) dx is equal to the area of the domain
D={a≤x≤b, 0 ≤y≤f (x)}(the area of the curvilinear trapezoid shown in Fig 7.1)
D
y=f x( )
y
x
O
Figure 7.1 The integral of a nonnegative function f (x) on an interval [a, b] is equal to the area of the shaded
region.
7.2.2 Properties of Definite Integrals and Useful Formulas
7.2.2-1 Qualitative properties of integrals
1 If a function f (x) is integrable on [a, b], then the functions cf (x), with c = const, and
|f (x)| are also integrable on [a, b].
2 If two functions f (x) and g(x) are integrable on [a, b], then their sum, difference, and product are also integrable on [a, b].
3 If a function f (x) is integrable on [a, b] and its values lie within an interval [c, d], where a function g(y) is defined and continuous, then the composite function g(f (x)) is also integrable on [a, b].
4 If a function f (x) is integrable on [a, b], then it is also integrable and on any subin-terval [α, β] ⊂ [a, b] Conversely, if an interval [a, b] is partitioned into a number of subintervals and f (x) is integrable on each of the subintervals, then it is integrable on the whole interval [a, b].
5 If the values of a function are changed at finitely many points, this will not affect the integrability of the function and will not change the value of the integral
7.2.2-2 Properties of integrals in terms of identities
1 The integral over a zero-length interval is zero:
a
a f (x) dx =0
2 Antisymmetry under the swap of the integration limits:
b
a f (x) dx = –
a
b f (x) dx.
This property can be taken as the definition of a definite integral with a > b.
Trang 6288 INTEGRALS
3 Linearity If functions f (x) and g(x) are integrable on an interval [a, b], then
b
a
Af (x) Bg (x)
dx = A
b
a f (x) dx B
b
a g (x) dx for any numbers A and B.
4 Additivity If c[a, b] and f (x) is integrable on [a, b], then
b
a f (x) dx =
c
a f (x) dx +
b
c f (x) dx.
5 Differentiation with respect the variable upper limit If f (x) is continuous on [a, b],
then the function Φ(x) = x
a f (t) dt is differentiable on [a, b], andΦ (x) = f (x) This can
be written in one relation:
d dx
x
a f (t) dt
= f (x).
6 Newton–Leibniz formula:
b
a f (x) dx = F (x)
b
a = F (b) – F (a), where F (x) is an antiderivative of f (x) on [a, b].
7 Integration by parts If functions f (x) and g(x) have continuous derivatives on [a, b],
a f (x)g
(x) dx =
f (x)g(x)b
a–
b
a f
(x)g(x) dx.
8 Repeated integration by parts:
b
a f (x)g
(n+1 )(x) dx =
*
f (x)g(n) (x) – f (x)g(n–1)(x) + · · · + (–1)n f(n) (x)g(x)+b
a
+ (–1)n+1 b
a f
(n+1 )(x)g(x) dx, n=0,1,
9 Change of variable (substitution) in a definite integral Let f (x) be a continuous function on [a, b] and let x(t) be a continuously differentiable function on [α, β] Suppose also that the range of values of x(t) coincides with [a, b], with x(α) = a and x(β) = b Then
b
a f (x) dx =
β
α f x (t)
x (t) dt.
Example Evaluate the integral
3 0
dx
(x –8 )√
x+ 1.
Perform the substitution x +1= t2, with dx =2t dt We have t =1at x =0and t =2at x =3 Therefore
3
0
dx
(x –8 )√
x+ 1 =
2 1
2t dt
(t2 – 9)t =2
2 1
dt
t2– 9 =
1
3lnt– 3
t+ 3
2 1
= 1
3ln
2
5.
10 Differentiation with respect to a parameter Let f (x, λ) be a continuous function in
a domain a≤x≤b , λ1≤λ≤λ2and let it has a continuous partial derivative ∂λ ∂ f (x, λ) in the
Trang 7same domain Also let u(λ) and v(λ) be differentiable functions on the interval λ1≤λ≤λ2
such that a≤u (λ)≤b and a≤v (λ)≤b Then
d
dλ
u(λ)
v(λ) f (x, λ) dx = f u (λ), λ
du(λ)
dλ – f v (λ), λ dv(λ)
dλ +
u(λ)
v(λ)
∂
∂λ f (x, λ) dx.
11 Cauchy’s formula for multiple integration:
x
a dx1
x1
a dx2 .
x n–1
a f (x n ) dx n=
1
(n –1)!
x
a (x – t)
n–1f (t) dt.
7.2.3 General Reduction Formulas for the Evaluation of Integrals
Below are some general formulas, involving arbitrary functions and parameters, that could facilitate the evaluation of integrals
7.2.3-1 Integrals involving functions of a linear or rational argument
b
a f (a + b – x) dx =
b
a f (x) dx;
a
0 [f (x) + f (a – x)] dx =2
a
0 f (x) dx;
a
0 [f (x) – f (a – x)] dx =0;
a –a f (x) dx =0 if f (x) is odd;
a –a f (x) dx =2
a
0 f (x) dx if f (x) is even;
b
a f (x, a + b – x) dx =0 if f (x, y) = –f (y, x);
1
0 f 2x √
1– x2
dx=
1
0 f 1– x2) dx.
7.2.3-2 Integrals involving functions with trigonometric argument
π
0 f (sin x) dx =2
π/2
0 f (sin x) dx;
π/2
0 f (sin x) dx =
π/2
0 f (cos x) dx;
π/2
0 f (sin x, cos x) dx =0 if f (x, y) = –f (y, x);
π/2
0 f(sin2x ) cos x dx =
π/2
0 f(cos
2x ) cos x dx;