Handbook of mathematics for engineers and scienteists part 45 pot

The integration of a polynomial multiplied by an exponential function can be accom-plished by using the formula of integration by parts or repeated integration by parts given in Paragra

Example 3.



dx

√

2x– x2 =



dx

1– (x –1 )2 =



d (x –1 )

1– (x –1 )2 = arcsin(x –1) + C.

4◦ The integration of a polynomial multiplied by an exponential function can be

accom-plished by using the formula of integration by parts (or repeated integration by parts) given

in Paragraph 7.1.2-2

Example 4 Compute the integral



( 3x + 1) e2x dx.

Taking f (x) =3x + 1and g  (x) = e2x , one finds that f  (x) =3and g(x) = 12e2x On substituting these expressions into the formula of integration by pars, one obtains



( 3x + 1) e2x dx= 1

2(3x+1) e2x–

3 2



e2x dx= 1

2(3x+1) e2x–

3

4e2x + C =

3

2x–

1 4



e2x + C.

Remark 1 More complex examples of the application of integration by parts or repeated integration by parts can be found in Subsection 7.1.6.

Remark 2 Examples of using a change of variables (see Item 5 in Paragraph 7.1.2-2) for the computation

of integrals can be found in Subsections 7.1.4 and 7.1.5.

7.1.2-4 Remark on uncomputable integrals

The differentiation of elementary functions is known to always result in elementary func-tions However, this is not the case with integration, which is the reverse of differentiation The integrals of elementary functions are often impossible to express in terms elementary functions using finitely many arithmetic operations and compositions

Here are examples of integrals that cannot be expressed via elementary functions:



dx

√

x3+1,



exp(–x2) dx,



e x

x dx,



dx

ln x,



cos x

x dx,



sin(x2) dx.

Such integrals are sometimes called intractable It is significant that all these integrals exist; they generate nonelementary (special) functions

7.1.3 Integration of Rational Functions

7.1.3-1 Partial fraction decomposition of a rational function

A rational function (also know as a rational polynomial function) is a quotient of

polyno-mials:

R(x) = P n (x)

where

P n (x) = a n x n+· · · + a1x + a0,

Q m (x) = b m x m+· · · + b1x + b0

The fraction (7.1.3.1) is called proper if m > n and improper if m≤n.

Every proper fraction (7.1.3.1) can be decomposed into a sum of partial fractions To

this end, one should factorize the denominator Q m (x) into irreducible multipliers of the

form

(x – α i)p i, i=1,2, , k; (7.1.3.2a) (x2+ β j x + γ j q j, j =1,2, , s, (7.1.3.2b)

where the p i and q j are positive integers satisfying the condition p1+· · ·+p k+2(q1+· · ·+q s) = m; β j2–4γ j<0 The rational function (7.1.3.1) can be represented as a sum of irreducibles and to each irreducible of the form (7.1.3.2) there correspond as many terms as the power

p i or q i:

A i,1

x – α i +

A i,2

(x – α i)2 +· · · + A i,p i

B j,1x + D j,1

x2+ β j x + γ j +

B j,2x + D j,2 (x2+ β j x + γ j 2 +· · · + B j,q j x + D j,q j

(x2+ β j x + γ j q j (7.1.3.3b) The constants A i,l , B j,r , D j,r are found by the method of undetermined coefficients.

To that end, one should equate the original rational fraction (7.1.3.1) with the sum of the above partial fractions (7.1.3.3) and reduce both sides of the resulting equation to a common

denominator Then, one collects the coefficients of like powers of x and equates them with zero, thus arriving at a system of linear algebraic equations for the A i,l , B j,r , and D j,r

Example 1 This is an illustration of how a proper fraction can be decomposed into partial fractions:

b5x5+ b4x4+ b3x3+ b2x2+ b1x + b0

(x + a)(x + c)3(x2+ k2 ) =

A1, 1

x + a +

A2, 1

x + c+

A2, 2

(x + c)2 + A2,3

(x + c)3 +Bx + D

x2+ k2

7.1.3-2 Integration of a proper fraction

1◦ To integrate a proper fraction, one should first rewrite the integrand (7.1.3.1) in the

form of a sum of partial fractions Below are the integrals of most common partial fractions

(7.1.3.3a) and (7.1.3.3b) (with q j =1):



A

x – α dx = A ln|x – α|,



A (x – α) p dx= –

A (p –1)(x – α) p–1,



Bx + D

x2+ βx + γ dx=

B

2 ln(x2+ βx + γ) +

2D – Bβ

4γ – β2 arctan

2x + β

4γ – β2.

(7.1.3.4)

The constant of integration C has been omitted here More complex integrals of partial fractions (7.1.3.3b) with q j >1can be computed using the formula



Bx + D (x2+ βx + γ) q dx=

P (x) (x2+ βx + γ) q–1 + λ



dx

x2+ βx + γ, (7.1.3.5)

where P (x) is a polynomial of degree2q–3 The coefficients of P (x) and the constant λ can

be found by the method of undetermined coefficients by differentiating formula (7.1.3.5)

Remark The following recurrence relation may be used in order to compute the integrals on the left-hand side in (7.1.3.5):



Bx + D

(x2+ βx + γ) q dx= (2D– Bβ)x + Dβ –2Bγ

(q –1 )( 4γ– β2)(x2+ βx + γ) q–1 +(2q – 3 )( 2D– Bβ)

(q –1 )( 4γ– β2)



dx

(x2+ βx + γ) q–1

Example 2 Compute the integral

 3x2

– x –2

x3+ 8 dx.

Let us factor the denominator of the integrand, x3+ 8= (x +2)(x2– 2x + 4 ), and perform the partial fraction decomposition:

3x 2– x –2

(x +2)(x2– 2x + 4 ) =

A

x+ 2 +

Bx + D

x2– 2x + 4.

Multiplying both sides by the common denominator and collecting the coefficients of like powers of x, we

obtain

(A + B –3)x2+ (– 2A + 2B+ D +1)x +4A + 2D + 2 = 0

Now equating the coefficients of the different powers of x with zero, we arrive at a system of algebraic equations for A, B, and D:

A + B –3 = 0 , – 2A + 2B+ D +1 = 0 , 4A + 2D + 2 = 0

Its solution is: A =1, B =2, D = –3 Hence, we have

 3x2

– x –2

x3+ 8 dx=

x+ 2dx+

 2x

– 3

x2– 2x + 4dx

= ln |x + 2| + ln x2– 2x + 4–√1

3arctan

x √– 1

3 + C.

Here, the last integral of (7.1.3.4) has been used.

2◦ The integrals of proper rational functions defined as the ratio of a polynomial to a power function (x – α) mare given by the formulas



P n (x)

(x – α) m dx= –

n



k=0

P(k)

n (α) k! (m – k –1)(x – α) m–k–1 + C, m > n +1;



P n (x)

(x – α) n+1 dx= –

n–1



k=0

P(k)

n (α) k! (n – k)(x – α) n–k +

P(n)

n (α) n! ln|x – α|+ C,

where P n (x) is a polynomial of degree n and P n(k) (α) is its kth derivative at x = α.

3◦ Suppose the roots in the factorization of the denominator of the fraction (7.1.3.1) are

all real and distinct:

Q m (x) = b m x m+· · · + b1x + b0= b m (x – α1)(x – α2) (x – α m), α i ≠α j.

Then the following formula holds:



P n (x)

Q m (x) dx=

m



k=1

P n (α k)

Q  m (α k) ln|x – α k|+ C, where m > n and the prime denotes a derivative.

7.1.3-3 Integration of improper fractions

1◦ In order to integrate an improper fraction, one should first isolate a proper fraction by

division with remainder As a result, the improper fraction is represented as the sum of a polynomial and a proper fraction,

a n x n+· · · + a1x + a0

b m x m+· · · + b1x + b0 = c m x

n–m+· · · + c1x + c0+ s m–1x

m–1+· · · + s1x + s0

b m x m+· · · + b1x + b0 (n≥m),

which are then integrated separately

Example 3 Evaluate the integral I =



x2

x– 1dx.

Let us rewrite the integrand (improper fraction) as the sum of a polynomial and a proper fraction: x

2

x– 1=

x+ 1 + 1

x– 1 Hence, I =

 

x+ 1 + 1

x– 1



dx= 12x2+ x + ln|x – 1|+ C.

2◦ The integrals of improper rational functions defined as the ratio of a polynomial to a simple power function (x – α) mare evaluated by the formula



P n (x)

(x – α) m dx=

n



k=m

P(k)

n (α) k! (k – m +1)(x – α)

k–m+1+ P n(m–1)(α)

(m –1)! ln|x – α|

–

m–2



k=0

P(k)

n (α) k! (m – k –1)(x – α) m–k–1 + C, where n≥m.

Remark 1 The indefinite integrals of rational functions are always expressed in terms of elementary functions.

Remark 2 Some of the integrals reducible to integrals of rational functions are considered in Subsections 7.1.5 and 7.1.6.

7.1.4 Integration of Irrational Functions

The integration of some irrational functions can be reduced to that of rational functions using

a suitable change of variables In what follows, the functions R(x, y) and R(x1, , x k) are assumed to be rational functions in each of the arguments

7.1.4-1 Integration of expressions involving radicals of linear-fractional functions

1◦ The integrals with roots of linear functions



R x, √ n

ax + b

dx are reduced to integrals of rational functions by the change of variable z = √ n

ax + b.

Example 1 Evaluate the integral I =



x √

1– x dx.

With the change of variable√

1– x = z, we have x =1– z2and dx = –2z dz Substituting these expressions into the integral yields

I= – 2 ( 1– z2)z2dz= – 2

3z3+

2

5z5+ C = –

2 3

( 1– x)3+ 2

5

( 1– x)5+ C.

2◦ The integrals with roots of linear-fractional functions



R



x, n ax + b

cx + d



dx

are reduced to integrals of rational functions by the substitution z = n ax + b

cx + d.

3◦ The integrals of the more general form



R



x, ax + b

cx + d

q1

, ,ax + b

cx + d

q k

dx, where q1, , q k are rational numbers, are reduced to integrals of rational functions using

the change of variable z m= ax + b

cx + d , where m is the common denominator of the fractions

q1, , q k.

4◦ Integrals containing the product of a polynomial by a simple power function of the form (x – a) β are evaluated by the formula



P n (x)(x – a) β dx=

n



k=0

P(k)

n (a) k! (k + β +1)(x – a)

k+β+1,

where P n (x) is a polynomial of degree n, P n(k) (a) is its kth derivative at x = a, and β is any positive or negative proper fraction (to be more precise, β≠–1, –2, , –n –1)

7.1.4-2 Euler substitutions Trigonometric substitutions

We will be considering integrals involving the radical of a quadratic trinomial:



R x, √

ax2+ bx + c

dx, where b2≠ 4ac Such integrals are expressible in terms of elementary functions.

1◦ Euler substitutions The given integral is reduced to the integral of a rational fraction

by one of the following three Euler substitutions:

1) √

ax2+ bx + c = t

x √

a if a>0;

2) √

ax2+ bx + c = xt √

c if c>0;

3) √

ax2+ bx + c = t(x – x1) if 4ac – b2<0,

where x1is a root of the quadratic equation ax2+ bx + c =0 In all three cases, the variable x

and the radical√

ax2+ bx + c are expressible in terms of the new variable t as (the formulas

correspond to the upper signs in the substitutions):

1) x = t

2– c

2√ a t + b,

√

ax2+ bx + c = √

a t2+ bt + c √

a

2√ a t + b , dx =2√ a t2+ bt + c √ a

(2√ a t + b)2 dt;

2) x = 2√ c t – b

a – t2 ,

√

ax2+ bx + c =

√

c t2– bt + c √

a

a – t2 , dx =2

√

c t2– bt + c √

a (a – t2)2 dt;

3) x = (t2+ a)x1+ b

t2– a ,

√

ax2+ bx + c = (2ax1+ b)t

t2– a , dx= –2(2ax1+ b)t

(t2– a)2 dt.

2◦ Trigonometric substitutions The function √

ax2+ bx + c can be reduced, by making a

perfect square in the radicand, to one of the three forms:

1) √

a

(x – p)2+ q2 if a>0;

2) √

a

(x – p)2– q2 if a>0;

3) √ –a

q2– (x – p)2 if a<0,

where p = –12b/a Different trigonometric substitutions are further used in each case to

evaluate the integral:

1) x – p = q tan t,

(x – p)2+ q2 = q

cos t, dx=

q dt

cos2t; 2) x – p = q

cos t,

(x – p)2– q2 = q tan t, dx= q sin t dt

cos2t ; 3) x – p = q sin t,

q2– (x – p)2 = q cos t, dx = q cos t dt.

Example 2 Evaluate the integral √

6 + 4x – 2x 2dx.

This integral corresponds to case 3 with a = –2, p =1, and q =2 The integrand can be rewritten in the

6 + 4x – 2x 2 =√

2√3 + 2x– x2=√

24– (x –1 )2.

Using the trigonometric substitution x–1 = 2sin t and the formulas √

3 + 2x– x2= 2cos t and dx =2cos t dt,

we obtain

 √

6 + 4x – 2x 2dx= 4√2 cos2t dt= 2√2 ( 1 + cos 2t) dt

= 2√2t +√

2 sin 2t+ C =2√2 arcsinx–1

2 +

√

2 sin



2 arcsin x–1

2



+ C

= 2√2 arcsinx–1

2 +

√

2

2 (x –1)

4– (x –1 )2+ C.

7.1.4-3 Integral of a differential binomial

The integral of a differential binomial,



x m (a + bx n)p dx,

where a and b are constants, and n, m, p are rational numbers, is expressible in terms of

elementary functions in the following three cases only:

1) If p is an integer For p≥ 0, removing the brackets gives the sum of power functions

For p <0, the substitution x = t r , where r is the common denominator of the fractions

m and n, leads to the integral of a rational function.

2) If m+1

n is an integer One uses the substitution a + bx n = t k , where k is the denominator of the fraction p.

3) If m+1

n + p is an integer One uses the substitution ax–n + b = t k , where k is the denominator of the fraction p.

Remark. In cases 2 and 3, the substitution z = x nleads to integrals of the form 3◦from Paragraph 7.1.4-1.

7.1.5 Integration of Exponential and Trigonometric Functions

7.1.5-1 Integration of exponential and hyperbolic functions

1 Integrals of the form



R(e px , e qx ) dx, where R(x, y) is a rational function of its arguments and p, q are rational numbers, may be evaluated using the substitution z m = e x,

where m is the common denominator of the fractions p and q In the special case of integer

p and q, we have m =1, and the substitution becomes z = e x

Example 1 Evaluate the integral



e3x dx

e + 2.

This integral corresponds to integer p and q: p =1and q =3 So we use the substitution z = e x Then

x = ln z and dx = dz

z Therefore,



e3x dx

e + 2 =



z2dz

z+ 2 =

 

z– 2 + 4

z+ 2



dz= 1

2z2–2z+4ln|z+2|+ C =

1

2e2x–2e +4ln(e x+2) + C.

2 Integrals of the form



R(sinh ax, cosh ax) dx are evaluated by converting the hyperbolic functions to exponentials, using the formulas sinh ax = 12(e ax – e–ax) and cosh ax = 12(e ax + e–ax ), and performing the substitution z = e ax Then



R(sinh ax, cosh ax) dx = 1

a



R



z2–1

2z , z

2+1

2z



dz

z

Alternatively, the substitution t = tanh

ax

2

 can also be used to evaluate integrals of the above form Then



R(sinh ax, cosh ax) dx = 2

a



R

 2

t

1– t2,

1+ t2

1– t2



dt

1– t2.

7.1.5-2 Integration of trigonometric functions

1 Integrals of the form



R(sin ax, cos ax) dx can be converted to integrals of rational functions using the versatile trigonometric substitution t = tanax

2

 :



R(sin ax, cos ax) dx = 2

a



R

 2

t

1+ t2,

1– t2

1+ t2



dt

1+ t2.

Example 2 Evaluate the integral



dx

2+ sin x. Using the versatile trigonometric substitution t = tan x

2, we have



dx

2+ sin x =2  dt

2 + 2t

1+ t2



( 1+ t2)

=



dt

t2+ t +1 =2



d( 2t + 1 ) ( 2t + 1 )2+ 3

= √2

3arctan

2t√+ 1

3 + C =

2

√

3arctan

 2

√

3tan

x

2 +√13



+ C.

2 Integrals of the form



R(sin2ax, cos2ax, tan ax) dx are converted to integrals of rational functions with the change of variable z = tan ax:



R(sin2ax, cos2ax, tan ax) dx = 1

a



R



z2

1+ z2,

1

1+ z2, z



dz

1+ z2.

3 Integrals of the form



sin ax cos bx dx,



cos ax cos bx dx,



sin ax sin bx dx

Remark Examples of using a change of variables... collects the coefficients of like powers of x and equates them with zero, thus arriving at a system of linear algebraic equations for the A i,l , B j,r , and D j,r... Integration of a proper fraction

1◦ To integrate a proper fraction, one should first rewrite the integrand (7.1.3.1) in the

form of a sum of partial