Handbook of mathematics for engineers and scienteists part 45 pot

276 INTEGRALS Example 3.  dx √ 2x – x 2 =  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) e 2x dx. Taking f(x)=3x + 1 and g  (x)=e 2x , one finds that f  (x)=3 and g(x)= 1 2 e 2x . On substituting these expressions into the formula of integration by pars, one obtains  (3x + 1) e 2x dx = 1 2 (3x + 1) e 2x – 3 2  e 2x dx = 1 2 (3x + 1) e 2x – 3 4 e 2x + C =  3 2 x – 1 4  e 2x + 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 fi nitely many arithmetic operations and compositions. Here are examples of integrals that cannot be expressed via elementary functions:  dx √ x 3 + 1 ,  exp(–x 2 ) dx,  e x x dx,  dx ln x ,  cos x x dx,  sin(x 2 ) 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) Q m (x) ,(7.1.3.1) where P n (x)=a n x n + ···+ a 1 x + a 0 , Q m (x)=b m x m + ···+ b 1 x + b 0 . 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) (x 2 + β j x + γ j ) q j , j = 1, 2, , s,(7.1.3.2b) 7.1. INDEFINITE INTEGRAL 277 where the p i and q j are positive integers satisfying the condition p 1 +···+p k +2(q 1 +···+q s )= m; β 2 j – 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 (x – α i ) p i ;(7.1.3.3a) B j,1 x + D j,1 x 2 + β j x + γ j + B j,2 x + D j,2 (x 2 + β j x + γ j ) 2 + ···+ B j,q j x + D j,q j (x 2 + 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 ,andD j,r . Example 1. This is an illustration of how a proper fraction can be decomposed into partial fractions: b 5 x 5 + b 4 x 4 + b 3 x 3 + b 2 x 2 + b 1 x + b 0 (x + a)(x + c) 3 (x 2 + k 2 ) = A 1,1 x + a + A 2,1 x + c + A 2,2 (x + c) 2 + A 2,3 (x + c) 3 + Bx + D x 2 + k 2 . 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 x 2 + βx + γ dx = B 2 ln(x 2 + βx + γ)+ 2D – Bβ  4γ – β 2 arctan 2x + β  4γ – β 2 . (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 > 1 can be computed using the formula  Bx + D (x 2 + βx + γ) q dx = P (x) (x 2 + βx + γ) q–1 + λ  dx x 2 + βx + γ ,(7.1.3.5) where P (x) is a polynomial of degree 2q –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 (x 2 + βx + γ) q dx = (2D – Bβ)x + Dβ – 2Bγ (q – 1)(4γ – β 2 )(x 2 + βx + γ) q–1 + (2q – 3)(2D – Bβ) (q – 1)(4γ – β 2 )  dx (x 2 + βx + γ) q–1 . Example 2. Compute the integral  3x 2 – x – 2 x 3 + 8 dx. Let us factor the denominator of the integrand, x 3 + 8 =(x +2)(x 2 – 2x +4), and perform the partial fraction decomposition: 3x 2 – x – 2 (x + 2)(x 2 – 2x + 4) = A x + 2 + Bx + D x 2 – 2x + 4 . 278 INTEGRALS Multiplying both sides by the common denominator and collecting the coefficients of like powers of x,we obtain (A + B – 3)x 2 +(–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,andD: 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  3x 2 – x – 2 x 3 + 8 dx =  1 x + 2 dx +  2x – 3 x 2 – 2x + 4 dx =ln|x + 2| +ln  x 2 – 2x + 4  – 1 √ 3 arctan 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) m are 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 (k) n (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 + ···+ b 1 x + b 0 = b m (x – α 1 )(x – α 2 ) (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 + ···+ a 1 x + a 0 b m x m + ···+ b 1 x + b 0 = c m x n–m + ···+ c 1 x +c 0 + s m–1 x m–1 + ···+ s 1 x + s 0 b m x m + ···+ b 1 x + b 0 (n ≥ m), which are then integrated separately. 7.1. INDEFINITE INTEGRAL 279 Example 3. Evaluate the integral I =  x 2 x – 1 dx. 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 = 1 2 x 2 + 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) m are 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 (m–1) n (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)andR(x 1 , , 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 – xdx. With the change of variable √ 1 – x = z,wehavex = 1 –z 2 and dx =–2zdz. Substituting these expressions into the integral yields I =–2  (1 – z 2 )z 2 dz =– 2 3 z 3 + 2 5 z 5 + 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 . 280 INTEGRALS 3 ◦ . The integrals of the more general form  R  x,  ax + b cx + d  q 1 , ,  ax + b cx + d  q k  dx, where q 1 , , q k are rational numbers, are reduced to integrals of rational functions using the change of variable z m = ax + b cx + d ,wherem is the common denominator of the fractions q 1 , , 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 (k) n (a) is its kth derivative at x = a,andβ 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, √ ax 2 + bx + c  dx, where b 2 ≠ 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) √ ax 2 + bx + c = t x √ a if a > 0; 2) √ ax 2 + bx + c = xt √ c if c > 0; 3) √ ax 2 + bx + c = t(x – x 1 )if4ac – b 2 < 0, where x 1 is a root of the quadratic equation ax 2 +bx+c = 0. In all three cases, the variable x and the radical √ ax 2 + 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 √ at+ b , √ ax 2 + bx + c = √ at 2 + bt + c √ a 2 √ at+ b , dx = 2 √ at 2 + bt + c √ a (2 √ at+ b) 2 dt; 2) x = 2 √ ct– b a –t 2 , √ ax 2 + bx + c = √ ct 2 – bt + c √ a a –t 2 , dx = 2 √ ct 2 – bt + c √ a (a –t 2 ) 2 dt; 3) x = (t 2 + a)x 1 + b t 2 – a , √ ax 2 + bx + c = (2ax 1 + b)t t 2 – a , dx =–2 (2ax 1 + b)t (t 2 – a) 2 dt. 2 ◦ . Trigonometric substitutions. The function √ ax 2 + bx + c can be reduced, by making a perfect square in the radicand, to one of the three forms: 1) √ a  (x – p) 2 + q 2 if a > 0; 2) √ a  (x – p) 2 – q 2 if a > 0; 3) √ –a  q 2 –(x – p) 2 if a < 0, 7.1. INDEFINITE INTEGRAL 281 where p =– 1 2 b/a. Different trigonometric substitutions are further used in each case to evaluate the integral: 1) x – p = q tan t,  (x – p) 2 + q 2 = q cos t , dx = qdt cos 2 t ; 2) x – p = q cos t ,  (x – p) 2 – q 2 = q tan t, dx = q sin tdt cos 2 t ; 3) x – p = q sin t,  q 2 –(x – p) 2 = q cos t, dx = q cos tdt. Example 2. Evaluate the integral  √ 6 + 4x – 2x 2 dx. This integral corresponds to case 3 with a =–2, p = 1,andq = 2. The integrand can be rewritten in the form: √ 6 + 4x – 2x 2 = √ 2 √ 3 + 2x – x 2 = √ 2  4 –(x – 1) 2 . Using the trigonometric substitution x–1 =2 sin t and the formulas √ 3 + 2x – x 2 = 2 cos t and dx = 2 cos tdt, we obtain  √ 6 + 4x – 2x 2 dx = 4 √ 2  cos 2 tdt = 2 √ 2  (1 +cos2t) dt = 2 √ 2t + √ 2 sin 2t + C = 2 √ 2 arcsin x – 1 2 + √ 2 sin  2 arcsin x – 1 2  + C = 2 √ 2 arcsin x – 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 ,wherer 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 ,wherek 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 ,wherek is the denominator of the fraction p. Remark. In cases 2 and 3, the substitution z = x n leads 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,whereR(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,wehavem = 1, and the substitution becomes z = e x . 282 INTEGRALS Example 1. Evaluate the integral  e 3x dx e x + 2 . This integral corresponds to integer p and q: p = 1 and q = 3. So we use the substitution z = e x .Then x =lnz and dx = dz z . Therefore,  e 3x dx e x + 2 =  z 2 dz z + 2 =   z – 2 + 4 z + 2  dz = 1 2 z 2 – 2z + 4 ln |z + 2| + C = 1 2 e 2x – 2e x + 4 ln(e x + 2)+C. 2. Integrals of the form  R(sinh ax,coshax) dx are evaluated by converting the hyperbolic functions to exponentials, using the formulas sinh ax = 1 2 (e ax – e –ax )and cosh ax = 1 2 (e ax + e –ax ), and performing the substitution z = e ax .Then  R(sinh ax,coshax) dx = 1 a  R  z 2 – 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,coshax) dx = 2 a  R  2t 1 – t 2 , 1 + t 2 1 – t 2  dt 1 – t 2 . 7.1.5-2. Integration of trigonometric functions. 1. Integrals of the form  R(sin ax,cosax) dx can be converted to integrals of rational functions using the versatile trigonometric substitution t =tan  ax 2  :  R(sin ax,cosax) dx = 2 a  R  2t 1 + t 2 , 1 – t 2 1 + t 2  dt 1 + t 2 . Example 2. Evaluate the integral  dx 2 +sinx . Using the versatile trigonometric substitution t =tan x 2 ,wehave  dx 2 +sinx = 2  dt  2 + 2t 1 + t 2  (1 + t 2 ) =  dt t 2 + t + 1 = 2  d(2t + 1) (2t + 1) 2 + 3 = 2 √ 3 arctan 2t + 1 √ 3 + C = 2 √ 3 arctan  2 √ 3 tan x 2 + 1 √ 3  + C. 2. Integrals of the form  R(sin 2 ax,cos 2 ax,tanax) dx are converted to integrals of rational functions with the change of variable z =tanax:  R(sin 2 ax,cos 2 ax,tanax) dx = 1 a  R  z 2 1 + z 2 , 1 1 + z 2 , z  dz 1 + z 2 . 3. Integrals of the form  sin ax cosbx dx,  cos ax cos bx dx,  sin ax sinbx dx . 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. 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 ,. 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 ,andD j,r . Example 1. This is an illustration of how a