Mathematics-16.Unit-12 Indefinite Integration

INDEFINITE INTEGRAL BASIC CONCEPT Let F(x) be a differentiable function of x such that . Then F(x) is called the integral of f(x). Symbolically, it is written as  f(x) dx = F(x). f(x), the function to be integrated, is called the integrand. F(x) is also called the anti-derivative (or primitive function) of f(x). Constant of Integration: As the differential coefficient of a constant is zero, we have . Therefore, f(x) dx = F(x) + c.This constant c is called the constant of integration and can take any real value. INTEGRATION AS THE INVERSE PROCESS OF DIFFERENTIATION Basic formulae: Anti-derivatives or integrals of some of the widely used functions (integrands) are given below. , n  – 1 ( a > 0) Standard Formulae: Example - 1: Evaluate . Solution: = 1 – = 1 – = 1 – sec2 x – sec x tan x Now (x – tan x – sec x) = 1 – sec2 x – sec x tan x = x – tan x – sec x + c. Example - 2: Evaluate dx. Solution: dx = + =  tanx secx dx +  cot x cosecx dx = secx – cosecx + c . METHODS OF INTEGRATION If the integrand is not a derivative of a known function, then the corresponding integrals cannot be found directly. In order to find the integral of complex problems, generally three rules of integration are used. (1) Integration by substitution or by change of the independent variable. (2) Integration by parts. (3) Integration by partial fractions. INTEGRATION BY SUBSTITUTION Direct Substitution If integral is of the form  f(g(x)) g(x) dx, then put g(x) = t, provided exists. (i). = ln |f (x)| + c Put f (x) = t  f (x) dx = dt  = ln |t| + c = ln |f (x)| + c. (ii). Put f (x) = t = . Example -3: Evaluate . Solution: Let lnx = t. Then dt = dx Hence I = sint dt = -cost + c = -cos(lnx) + c Example -4: Evaluate . Solution: Put tan–1 x4 = t  I = = = (tan–1 x4)2 + A. Example -5: Evaluate . Solution: Let z = 2x3 + 3x dz= (6x2 +3)dx = 3( 2x2 +1)dx = = + c. Standard Substitutions: • For terms of the form x2 + a2 or , put x = a tan or a cot For terms of the form x2 - a2 or , put x = a sec  or a cosec (A) For terms of the form a2 - x2 or , put x = a sin  or a cos • If both , are present, then put x = a cos . • For the type , put x = a cos2 + b sin2 • For the type , put the expression within the bracket = t. • For the type (n  N, n> 1), put . • For , n1,n2  N (and > 1), again put (x + a) = t (x + b) Example -6: Evaluate . Solution: = I2 Put 1 –x = t2 dx = 2t dt I = – Example -7: Evaluate . Solution: I = = Put = t  dt = dx Hence I = Indirect Substitution If the integrand is of the form f(x)g(x), where g(x) is a function of the integral of f(x), then put integral of f(x) = t. Example -8: Evaluate Solution: Integral of the numerator = Put x3/2 = t. We get I = = . Derived Substitution: Some time it is useful to write the integral as a sum of two related integrals which can be evaluated by making suitable substitutions. Examples of such integrals are: A. Algebraic Twins , , . B. Trigonometric twins , , , . Method of evaluating these integral are illustrated by mean of the following examples: Example -9: Evaluate . Solution: I = = = = = = For I1, we write x - = t  (1 + )dx = dt I1 = = = For I2 , we write x + = t  (1 - )dx = dt  I2= = = Combining the two integrals, we get I = + Example -10: Evaluate . Solution: Put tanx = t2  sec2x dx = 2t dt  dx =  I = = = = . This can be solved by the method used in example (9). INTEGRATION BY PARTS If u and v be two functions of x, then integral of product of these two functions is given by: Note: In applying the above rule care has to be taken in the selection of the first function(u) and the second function (v). Normally we use the following methods: If in the product of the two functions, one of the functions is not directly integrable (e.g. lnx, sin-1x, cos-1x, tan-1x etc.) then we take it as the first function and the remaining function is taken as the second function. e.g. In the integration of x tan-1x dx, tan-1x is taken as the first function and x as the second function. If there is no other function, then unity is taken as the second function e.g. In the integration of tan-1x dx , tan-1x is taken as the first function and 1 as the second function. If both of the function are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable. Usually we use the following preference order for the first function (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent) In the above stated order, the function on the left is always chosen as the first function. This rule is called as ILATE e.g. In the integration of , x is taken as the first function and sinx is taken as the second function. Example -11: Evaluate . Solution: = = = +c Example -12: Evaluate . Solution: I = = sec = sec tan - = sectan - I + ln|sec + tan|  I = Example -13: Evaluate . Solution: Here u = lnx  du = and v = x3  dv = 3x2 dx I = = = . An important result: In the integral if g(x) can be expressed as g(x) = f(x) + f(x) then = ex f(x) + c Example -13: Evaluate . Solution: Let I = Here f(x) = 1/x, f(x) = -1/x2 Hence I = Example -14: Evaluate . Solution: dx = = INTEGRATION BY PARTIAL FRACTIONS A function of the form P(x)/Q(x), where P(x) and Q(x) are polynomials, is called a rational function. Consider the rational function The two fractions on the RHS are called partial fractions. To integrate the rational function on the LHS, it is enough to integrate the two fractions on the RHS, which are easily integrable. This is known as method of partial fractions. In case the degree of P(x) (numerator) is not less than that of Q(x) (denominator), we carry out the division of P(x) by Q(x) and reduce the degree of the numerator. In order to write P(x)/Q(x) in partial fractions, first of all we write Q(x) = (x - a)k ... (x2 + x + )r ... where binomials are different, and then set where A1, A2, ..., Ak, M1, M2, ......, Mr, N1, N2, ...... ,Nr are real constants to be determined. These are determined by reducing both sides of the above identity to integral form and equating the coefficients of equal powers of x, which gives a system of linear equations in the coefficient. (This method is called the method of comparison of coefficients). The constants can also be obtained by substituting suitably chosen numerical values of x in both sides of the identity. Note: Before proceeding to write a rational function as a sum of partial fractions, we should be taken to ascertain that it is either a proper rational fraction or is rewritten as one. A rational function is proper if the degree of polynomial Q(x) is greater than the degree of the polynomial P(x). In case the degree of P(x) greater than or equal to the degree of Q(x), we first write , where h(x) is a polynomial and p(x) is a polynomial of degree less than the degree of polynomial Q(x). Example -15: Evaluate . Solution: Put sinx = t  cosx dx = dt = = log (1+ t) – log (2 +t ) +c = log . Example -16: Evaluate . Solution: Put tan = t  sec2 . dx = dt  dx =  dx = dt, then we have I = = Expands into simple fractions After solve the coefficients, A = ; B = ; C = -1 Hence I = = = . Example -17: Evaluate . Solution: Since x3 + 1 = (x + 1) (x2 - x + 1) (the second factor is not a product of linear factors), the partial fractions of the given integer will have the form. . Hence, x = A(x2 - x + 1) + (Bx + D)(x + 1) = (A + B)x2+ (-A + B + D)x + (A + D). Equating the coefficients of equal powers of x, we get A = -1/3, B = 1/3, D = 1/3. express (x + 1) = l(d.c. of x2 - x + 1) + m  (x + 1) = l(2x - 1) + m = 2xl - l + m Comparing the coefficients of like powers of x, we get l =1/2 and m=3/2  ALGEBRAIC INTEGRALS Integrals of the form: In these type of integrals we write px + q = (diff coefficient of ax2 + bx + c) + m. Find and m by comparing the coefficient of x and constant term on both sides of the identity. In this way the question will reduce to the sum of two integrals which can be integrated easily. Integral of the type In this case substitute ax2 + bx + c = M (px2 + qx + r) + N (2px + q) + R Find M, N, & R. The integration reduces to integration of three independent functions. Example -18: Evaluate . Solution: Let x + 1 = (differential Coefficient of 2x2 + x – 3) + B x +1 = A(4x + 1) + B = 4ax + A + B equating the coefficients, A = , B = Now I = = Let I1 = and I2 = Put 2x2 + x-3 =z  (4x +1) dx =dz I1 = = = I2 = = = Hence I = . Example -19: Evaluate . Solution: Express 3x + 2 = (d.c. of 4x2 + 4x + 5) + m or, 3x + 2 = (8x + 4) + m = 8 x + 4 + m. Comparing the coefficients, we get 8 = 3 and 4 + m = 2 = 3/8 and m = 2 - 4 = 1/2  = . Integration of Irrational Algebraic Fractions: 1. Irrational functions of (ax+b)1/n and x can be easily evaluated by the substitution tn = ax+ b. Thus . 2. . Here we substitute, x –k = 1/t. This substitution will reduce the given integral to - . 3. , so that . Now the substitution C + Dt2 = u2 reduces it to the form . 4. Here, we write, ax2 +bx +c = A1 (dx +e) ( 2fx +g) +B1( dx +e) +C1 where A1, B1 and C1 are constants which can be obtained by comparing the coefficient of like terms on both sides. And given integral will reduce to the form A1 Example -20: Evaluate . Solution: Put x + 1 = t2, we get TRIGONOMETRIC INTEGRALS Integrals of the form  R ( sinx, cosx) dx: Here R is a rational function of sin x and cos x. This can be translated into integrals of a rational function by the substitution: tan(x/2) = t. This is the so called universal substitution. In this case . Sometimes, instead of the substitution tanx/2 = t, it is more advantageous to make the substitution cot x/2 = t Universal substitution often leads to very cumbersome calculations. Indicated below are those cases where the aim can be achieved with the aid of simpler substitutions. (a) If R(-sin x, cos x) = -R(sin x, cos x), substitute cos x = t (b) If R(sin x, -cos x) = -R(sin x, cos x), substitute sin x = t (c) If R(-sin x, - cos x) = R(sin x, cos x), substitute tan x = t Integrals of the form: Rule for (i) : In this integral express numerator as l (Denominator) + m(d.c. of denominator) + n. Find l, m, n by comparing the coefficients of sinx, cosx and constant term and split the integral into sum of three integrals. Rule for (ii) : Express numerator as l (denominator) + m(d.c. of denominator) and find l and m as above Example -21: Evaluate . Solution: If in expression we substitute -sinx for sin x, then the integrand will change its sign. Hence, we take advantage of the substitution t = cosx; dt = - sinx dx. This gives Integration of the Type: (sinMxcosNx)dx M & N  natural numbers. If one of them is odd, then substitute for term of even power. If both are odd, substitute either of the term. If both are even, use trigonometric identities only. Example -22: Evaluate . Solution: Put sinx = t = . Example -23: Evaluate Solution: Assignment 1. = (A) (B) (C) (D) none of these Solution: Put x = acos2 + bsin2 the given integral becomes. = = 2 + c = Hence (A) is the correct answer. 2. If , then f (x) is equal to (A) (B) (C) (D) none of these Solution: I = real part of = = = e(1 + i)x = ex (cos x + i sin x) = [icos x – sin x][(x – 1) + ix] I = [(1 – x) sin x – x cos x] + c. Hence (A) is the correct answer. 3. I = = f (x) + c then f (x) is equal to (A) 2 tan–1 (2 ex) (B) (C) (D) Solution: I = = f (x) + c  I = Let ex = t  ex dx = dt  I = + c = + c. Hence (B) is the correct answer. 4. is equal to : (A) tan-1 (B) loge (C) loge (D) tan-1 Solution: Substitute and solve by partial fractions. Hence (A) is the correct answer. 5. is equal to : (A) (B) (C) (D) none of these Solution: Let sin x = A( sin x – cos x) + B. d.c of (sin x – cos x) or sin x = A ( sin x – cos x) + B ( cos x + sin x) or sin x= (A + B ) sin x + (B –A) cos x equating the coefficient of sin x and cos x, we get A + B = 1 and B – A = 0 A = 1/2, B = 1/2 I = = = . Hence (B) is the correct answer. 6. I = is equal to : (A) ln (B) ln (C) ln (D) ln Solution: Let I = = (1+ xex = p, ex (1+x) dx = dp) I = 1= Ap2 + B (p) (p-1) + C (p – 1) For p = 1, p = 0, and p = -1, A = 1, C= -1 and B= -1.  I = = ln + c = ln Hence (D) is the correct answer. 7. is equal to : (A) (B) (C) (D) Solution: = = = – = . Hence (C) is the correct answer. 8. (A) (B) (C) (D) none of these Solution: Put x ex = t  I = = tan t + c = tan(xex) + c; Hence (B) is the correct answer. 9. (A) sec3 2x + 3 sec 2x (B) (C) (D) none of these Solution: I = = Put sec 2x = t  2 sec 2x tan 2x dx = dt  I = = Hence (B) is the correct answer. 10. (A) log (tan x) (B) cot (log x) (C) log log (tan x) (D) tan (log x) Solution: put log tan x = t sec x cosec x dx = dt = log t + c = log (log tan x) Hence (C) is the correct answer. 11. is equal to (A) (B) (C) (D) none of these Solution:  Put cosx = t  - = Hence (B) is the correct answer. 12. dx is equal to (A) (B) (C) (D) none of these Solution: I = dx . Put 1 - = t  ,  I = = Hence (A) is the correct answer. 13. equals (A) (B) (C) (D) Solution: I = Now put t =  dt =  I = = Hence (C) is the correct answer. 14. is equal to (A) (B) (C) (D) Solution: I = as 1 –sin 2x = (sin x –cos x)2 Hence Put sinx = t  cosxdx = dt = e sinx + c . Hence (A) is the correct answer. 15. If then a is equal to (A) (B) (C) (D) Solution: Multiplying above and below by x2 and put 1 –x3 = t2  -3x2dx = 2t dt  I = + c I = + c  a = Hence (B) is the correct answer. 16. dx is equal to (A) (B) (C) (D) none of these Solution: Put ax = t  I = again put at = z  I = = Hence (C) is the correct answer. 17. is equal to (A) 2 sin x + log (sec x –tan x) + c (B) 2 sin x - log (sec x –tan x) + c (C) 2 sin x + log (sec x + tan x) + c (D) 2 sin x - log (sec x + tan x) + c Solution: I = I = = 2 = 2 sinx – log| secx + tanx| + c Hence (D) is the correct answer. 18. (A) (B) (C) (D) none of these Solution: (by property of exponential function) = = = Hence (B) is the correct answer. 19. (A) x2 sin x – 2x cos x + c (B) x2 sin x + c (C) –x2 cos x + 2 x sin x + 2 cos x + c (D) –x2 sin x – 2x cos x + sin x + c Solution: By parts I = –x2 cos x + 2 x sin x + 2cos x + c with alternate +, – sign. Hence (C) is the correct answer. 20. = (A) log (sec x + tan x) –2 tan (B) log (sec x + tan x) +2 tan (C) log (sec x - tan x) –2 tan (D) none of these Solution: Here we have cos x but itsd.c.i.e., -sin x is not present in the numeratior and as such we cannot make the substitution of cos x = t. but we simply put cos x = t to split the integrand into partial fractions. =  I = dx = dx = log (sec x + tan x) –2 tan Hence (A) is the correct answer. 21. (A) (B) (C) (D) none of these Solution: Put (1 –x) = t –dx = dt = = + c = . Hence (C) is the correct answer. 22. (A) x sin (log x) + c (B) x cos (log x) + c (C) x log (sin x) + c (D) x log (cos x) + c Solution: Put log x = t  dx = et dt x = et I = = et sin t = x sin (log x) + c. Hence (A) is the correct answer. 23. is equal to (A) loge (tan x – cot x) + c (B) loge (cot x – tan x) + c (C) tan-1 (tan x – cot x) +c (D) tan-1 (2 cot 2 x) + c Solution: Let I = = = If tan x = p, then sec2 x dx = dp = = = If , then = = = Hence (C) is the correct answers. 24. is equal to (A) (B) (C) (D) Solution: Let I = Multiplying Nr. & Dr. by cosec2x  I = dx = - dx = = . Hence (A) is the correct answer. 25. is equal to (A) (B) (C) (D) none of these Solution: Substituting x = p6, dx = 6 p5 dp . Hence (A) is the correct answer. 26. is equal to (A) ln + c (B) tan ln + c (C) ln + c (D) none of these Solution: ex = t ex dx = dt , put sin–1 t = z = = ln + k = ln + k Hence (C) is the correct answer. 27. is equal to (A) x sin x + cos x –sin2 x + c (B) x cos x –sin2 x + c (C) x sin x + cos x –(cos2 x)/2 + c (D) x2 sin x + cos x –sin3 x + c Solution: elog x = x I = Hence (C) is the correct answer. 28. dx is equal to (A) (B) (C) (D) Solution: I = I = = = x tanx/2 +c Hence (A) is the correct answer. 29. is equal to : (A) ln|sec 3x – tan 3x| + c (B) ln|sec 3x + tan 3x| + c (C) ln|sec 3x + tan 3x| + c (D) ln|sec 3x – tan 3x| + c Solution: Let I = = = = ln|sec 3x + tan 3x| + c. Hence (B) is the correct answer. 30. is equal to : (A) [sin (ln x) + cos (ln x)] + c (B) [cos (ln x) – sin (ln x)] + c (C) [sin (ln x) – cos (ln x)] + c (D) x[sin (ln x) – cos (ln x)] + c Solution: Let I = Let ln x = t  x = et  dx = et dt  I = = sin t . et – = sin t . et – 2I = et(sin t – cos t)  I = et(sin t – cos t) = elnx [sin (ln x) – cos (ln x)] + c = [sin (ln x) – cos (ln x)] + c. Hence (C) is the correct answer. 31. is equal to : (A) + c (B) + c (C) + c (D) + c Solution: Let I = Let ln x = t  dx = dt  I = = = + c I = + c. Hence (D) is the correct answer. 32. dx is equal to : (A) xx – c (B) xx + c (C) xlogx + c (D) None of these Solution: Let I = dx Let xx = t  x ln x = ln t   dx(1 + ln x)xx = dt  I =  I = t + c I = xx + c. Hence (B) is the correct answer. 33. is equal to : (A) + c (B) + c (C) + c (D) + c Solution: Let I = Let t = sin x – cos x  t2 = 1 – sin 2x  sin 2x = (1 – t2)  I = = = + c = + c = + c. Hence (A) is the correct answer. 34. For what value of a and b, the equation (sin2x – cos 2x) dx = holds good? (A) a = – , b is any arbitrary constant (B) a = , b any arbitrary constant (C) a = – , b any arbitrary constant (D) a = , b any arbitrary constant Solution: = - = = a = – , b is any arbitrary constant Hence (A) is the correct answer. 35. For what value of a and b, the equation  holds good? (A) a = – , b is any arbitrary constant (B) a = , b any arbitrary constant (C) a = – , b any arbitrary constant (D) a = , b any arbitrary constant Solution: = dx = =  a = – , b any arbitrary constant Hence (C) is the correct answer. 36. is equal to : (A) + c (B) + c (C) + c (D) + c Solution: Let I = Let ln = t  x = a.et  dx = aet dt = a = [cos t + sin t] + c = + c. Hence (B) is the correct answer. 37. is equal to : (A) + c (B) + c (C) + c (D) + c Solution: I = Let ln (x + ) = t   dx = dt  I = = + c I = + c. Hence (C) is the correct answer. 38. is equal to (A) + k (B) – + k (C) + k (D) + k Solution: Let I = Let = t    I = – = – + c. Hence (B) is the correct answer. 39. dx is equal to : (A) sin–1 (sin x + cos x) + c (B) sin–1 (sin x – cos x) + c (C) sin–1 (sin x – cos x) + c (D) sin–1 (sin x + cos x) + c Solution: Let I = dx = dx = dx Let t = sin x – cos x  t2 = 1 – sin 2x  dt = (cos x + sin x) dx  I = = sin–1 (t) + c I = sin–1 (sin x – cos x) + c. Hence (C) is the correct answer. 40. dx is equal to : (A) + c (B) – + c (C) – + c (D) + c Solution: Let I = dx = dx Let 1 + x–4 = t  dx = dt  dx = – dt  I = – + c  I = – + c. Hence (B) is the correct answer. 41. dx is equal to : (A) – + c (B) – + c (C) + c (D) + c Solution: Let I = dx Let 1 + = t  5 dx = dt  dx = – dt  I = – = –  t3/2 = – + c. Hence (A) is the correct answer. 42. is equal to (A) –cot x – + k (B) + 2cot–3 x + k (C) + 2 tan3 x + k (D) none of these Solution: Let I = = = =  I = – cot x – cot3 x + c. Hence (A) is the correct answer. 43. dx is equal to : (A) + c (B) + c (C) + c (D) + c Solution: Let I = dx Let = t  – dx = dt  I = – = – = t3/2[2 – 3 ln t] + c = + c. Hence (D) is the correct answer. 44. dx is equal to : (A) + c (B) + c (C) – + c (D) – + c Solution: Let I = dx = dx Let 1 + = t  – dx = dt  I = – dt = – . t3/2 + c = – + c. Hence (C) is the correct answer. 45. is equal to : (A) tan–1 + c (B) – tan–1 + c (C) tan–1 + c (D) None of these Solution: Let I = Let t = tan  dx =  I = = 2 = 2. .tan–1 + c  I = tan–1 + c. Hence (A) is the correct answer. 46. is equal to : (A) tan–1 + c (B) tan–1 + c (C) tan–1 + c (D) tan–1 + c Solution: Let I = Put 4x + 5 = t  x = dx =  I = = Let = u  I =  I = tan–1 + c. Hence (B) is the correct answer. 47. is equal to (A) real part of (B) imaginary part of (C) neither real nor Imaginary part of (D) none of these Solution: Let I = = real part of = real part of Hence (A) is the correct answer. 48. is equal to : (A) (B) (C) (D) Solution: Let I = Let x2 = t  2x dx = dt  I = = + c = + c. Hence (A) is the correct answer. 49. is equal to (A) (B) (C) (D) none of these Solution: = = + c = . Hence (B) is the correct answer. 50. If In = then In + nIn –1 is equal to (A) x (ln x)n –1 (B) x (ln x)n (C) nx (ln x)n (D) none of these Solution: In = In = x(ln x)n – n In = x(ln x)n – n In –1 In + nIn –1 = x(ln x)n. Hence (B) is the correct answer. 51. then A is equal to (A) 2 (B) (C) (D) Solution: I = Let = t  – dx = dt  I = – ln |t – 1| + c = – ln + c = ln + c  A = . Hence (B) is the correct answer. 52. is equal to (A) (B) (C) (D) + c Solution: I = (If p = tan-1x  x = tanp  dx = sec2p dp) = ep tanp = x + c Hence (A) is the correct answer. 53. is equal to (A) (B) (C) (D) Solution: = = + ln |x| – 2x + c Hence (D) is the correct answer. 54. is equal to (A) sin x – cos x + c (B) – sin x + cos x + c (C) sin x + cos x + c (D) none of these Solution: = = sin x + cos x + c Hence (C) is the correct answer. 55. If = A + B, then A is equal to (A) 1 (B) 2 (C) –1 (D) – 2 Solution: = . cosec2x dx = = - 2 Hence A = –2 . Hence (D) is the correct answer. 56. The value of is : (A) + c (B) + c (C) + c (D) + c Solution: Let I = = Put ln x = t  x = et  dx = etdt I = = . Hence (C) is the correct answer. 57. I = is equal to : (A) 10x + x10 + c (B) 10x – x10 + c (C) 10x + x10 + c (D) loge (10x + x10) + c Solution: If 10x +x10 = p  (10xln 10 + 10 x9 ) dx = dp ; I = ln ( x10 +10x) + c Hence (D) is the correct answer. 58. The value of the integral sinx dx is (A) (B) (C) (D) Solution: Put t = sin2 x The integral reduces to I = = = ( option A) = ( option B ) Hence (A) & (B) are the correct answers. 59. is equal to : (A) (B) (C) (D) Solution: = x – = = Hence (B) is the correct answer. 60. is equal to (A) ln (xe + ex) + c (B) ln (x + e) + c (C) ln (x–e + e–x) + c (D) none of these Solution: Let I = Let ex + xe = t  (ex + e.xe – 1) dx = dt  e(ex – 1 + xe – 1)dx = dt  I = = ln(ex + xe) + c. Hence (A) is the correct answer.