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INTEGRALS 225 Chapter 7 INTEGRALS v Just as a mountaineer climbs a mountain – because it is there, so a good mathematics student studies new material because...

INTEGRALS 225 Chapter 7 INTEGRALS v Just as a mountaineer climbs a mountain – because it is there, so a good mathematics student studies new material because it is there. — JAMES B. BRISTOL v 7.1 Introduction Differential Calculus is centred on the concept of the derivative. The original motivation for the derivative was the problem of defining tangent lines to the graphs of functions and calculating the slope of such lines. Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded by the graph of the functions. If a function f is differentiable in an interval I, i.e., its derivative f ′exists at each point of I, then a natural question arises that given f ′at each point of I, can we determine the function? The functions that could possibly have given function as a derivative are called anti derivatives (or G.W. Leibnitz primitive) of the function. Further, the formula that gives (1646 -1716) all these anti derivatives is called the indefinite integral of the function and such process of finding anti derivatives is called integration. Such type of problems arise in many practical situations. For instance, if we know the instantaneous velocity of an object at any instant, then there arises a natural question, i.e., can we determine the position of the object at any instant? There are several such practical and theoretical situations where the process of integration is involved. The development of integral calculus arises out of the efforts of solving the problems of the following types: (a) the problem of finding a function whenever its derivative is given, (b) the problem of finding the area bounded by the graph of a function under certain conditions. These two problems lead to the two forms of the integrals, e.g., indefinite and definite integrals, which together constitute the Integral Calculus. Reprint 2024-25 226 MATHEMATICS There is a connection, known as the Fundamental Theorem of Calculus, between indefinite integral and definite integral which makes the definite integral as a practical tool for science and engineering. The definite integral is also used to solve many interesting problems from various disciplines like economics, finance and probability. In this Chapter, we shall confine ourselves to the study of indefinite and definite integrals and their elementary properties including some techniques of integration. 7.2 Integration as an Inverse Process of Differentiation Integration is the inverse process of differentiation. Instead of differentiating a function, we are given the derivative of a function and asked to find its primitive, i.e., the original function. Such a process is called integration or anti differentiation. Let us consider the following examples: d We know that (sin x) = cos x... (1) dx d x3 ( ) = x2... (2) dx 3 d x and (e ) = e x... (3) dx We observe that in (1), the function cos x is the derived function of sin x. We say x3 that sin x is an anti derivative (or an integral) of cos x. Similarly, in (2) and (3),and 3 ex are the anti derivatives (or integrals) of x2 and ex, respectively. Again, we note that for any real number C, treated as constant function, its derivative is zero and hence, we can write (1), (2) and (3) as follows : d d x3 d x (sin x + C) = cos x , ( + C) = x 2 and (e + C) = e x dx dx 3 dx Thus, anti derivatives (or integrals) of the above cited functions are not unique. Actually, there exist infinitely many anti derivatives of each of these functions which can be obtained by choosing C arbitrarily from the set of real numbers. For this reason C is customarily referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function. d More generally, if there is a function F such that F (x) = f (x) , ∀ x ∈ I (interval), dx then for any arbitrary real number C, (also called constant of integration) d [ F (x) + C] = f (x), x ∈ I dx Reprint 2024-25  INTEGRALS 227 Thus, {F + C, C ∈ R} denotes a family of anti derivatives of f. Remark Functions with same derivatives differ by a constant. To show this, let g and h be two functions having the same derivatives on an interval I. Consider the function f = g – h defined by f (x) = g (x) – h(x), ∀ x ∈ I df Then = f′ = g′ – h′ giving f′ (x) = g′ (x) – h′ (x) ∀ x ∈ I dx or f ′ (x) = 0, ∀ x ∈ I by hypothesis, i.e., the rate of change of f with respect to x is zero on I and hence f is constant. In view of the above remark, it is justified to infer that the family {F + C, C ∈ R} provides all possible anti derivatives of f. We introduce a new symbol, namely, ∫ f (x) dx which will represent the entire class of anti derivatives read as the indefinite integral of f with respect to x. Symbolically, we write ∫ f (x) dx = F (x) + C. dy Notation Given that dx = f (x ) , we write y = ∫ f (x) dx. For the sake of convenience, we mention below the following symbols/terms/phrases with their meanings as given in the Table (7.1). Table 7.1 Symbols/Terms/Phrases Meaning ∫ f (x) dx Integral of f with respect to x f (x) in ∫ f (x) dx Integrand x in ∫ f (x) dx Variable of integration Integrate Find the integral An integral of f A function F such that F′(x) = f (x) Integration The process of finding the integral Constant of Integration Any real number C, considered as constant function Reprint 2024-25 228 MATHEMATICS We already know the formulae for the derivatives of many important functions. From these formulae, we can write down immediately the corresponding formulae (referred to as standard formulae) for the integrals of these functions, as listed below which will be used to find integrals of other functions. Derivatives Integrals (Anti derivatives) d  xn + 1  x n+1 =x ; n  ∫ x dx = + C , n ≠ –1 n (i) dx  n + 1  n +1 Particularly, we note that d ( x) = 1 ; ∫ dx = x + C dx d (ii) ( sin x ) = cos x ; ∫ cos x dx = sin x + C dx d (iii) ( – cos x ) = sin x ; ∫ sin x dx = – cos x + C dx d (iv) ( tan x ) = sec2 x ; ∫ sec 2 x dx = tan x + C dx d (v) ( – cot x ) = cosec2 x ; ∫ cosec 2 x dx = – cot x + C dx d (vi) ( sec x ) = sec x tan x ; ∫ sec x tan x dx = sec x + C dx d (vii) ( – cosec x ) = cosec x cot x ; ∫ cosec x cot x dx = – cosec x + C dx d ( (viii) dx sin x = –1 1 ) ; ∫ dx = sin – 1 x + C 1 – x2 1– x 2 d ( (ix) dx – cos x = –1 1 ; ) ∫ dx = – cos – 1 x + C 1 – x2 1– x 2 d ( (x) dx tan x = –1 1 ) 1 + x2 ; dx ∫ 1 + x 2 = tan –1 x+C d x (e ) = e x ; ∫ e dx = e +C x x (xi) dx Reprint 2024-25  INTEGRALS 229 d 1 1 (xii) ( log | x |) = ; ∫ x dx = log | x | +C dx x d  ax  ax =a ; ∫ a dx = x (xiii)  x +C dx  log a  log a A Note In practice, we normally do not mention the interval over which the various functions are defined. However, in any specific problem one has to keep it in mind. 7.2.1 Some properties of indefinite integral In this sub section, we shall derive some properties of indefinite integrals. (I) The process of differentiation and integration are inverses of each other in the sense of the following results : d dx ∫ f (x) dx = f (x) and ∫ f ′(x) dx = f (x) + C, where C is any arbitrary constant. Proof Let F be any anti derivative of f, i.e., d F(x ) = f (x) dx Then ∫ f (x) dx = F(x) + C d d Therefore ∫ f (x) dx = ( F (x) + C ) dx dx d = F (x ) = f (x) dx Similarly, we note that d f ′(x) = f (x) dx and hence ∫ f ′(x) dx = f (x) + C where C is arbitrary constant called constant of integration. (II) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. Reprint 2024-25 230 MATHEMATICS Proof Let f and g be two functions such that d d ∫ dx ∫ f (x) dx = g (x ) dx dx d  f (x) dx – ∫ g (x ) dx  = 0 dx  ∫ or Hence ∫ f (x) dx – ∫ g (x) dx = C, where C is any real number (Why?) or ∫ f (x) dx = ∫ g (x) dx + C So the families of curves {∫ f (x) dx + C , C ∈ R} 1 1 and {∫ g (x) dx + C , C ∈ R} are identical. 2 2 Hence, in this sense, ∫ f (x) dx and ∫ g (x) dx are equivalent. A Note The equivalence of the families {∫ f (x) dx + C ,C ∈ R} and 1 1 {∫ g (x) dx + C ,C ∈ R} is customarily expressed by writing ∫ f (x) dx = ∫ g (x) dx , 2 2 without mentioning the parameter. (III) ∫ [ f (x) + g (x)] dx = ∫ f (x) dx + ∫ g (x) dx Proof By Property (I), we have d  [ f (x ) + g (x )] dx  = f (x) + g (x) dx  ∫... (1) On the otherhand, we find that d  d d  ∫ f (x ) dx + ∫ g (x) dx  = ∫ f (x) dx + dx ∫ g (x) dx dx dx = f (x) + g (x)... (2) Thus, in view of Property (II), it follows by (1) and (2) that ∫ ( f (x) + g (x) ) dx = ∫ f (x) dx + ∫ g (x) dx. (IV) For any real number k, ∫ k f (x ) dx = k ∫ f (x) dx Reprint 2024-25  INTEGRALS 231 d dx ∫ Proof By the Property (I), k f (x) dx = k f (x). d  d Also dx  k ∫ f (x) dx  = k dx ∫ f (x) dx = k f (x) Therefore, using the Property (II), we have ∫k f (x ) dx = k ∫ f (x) dx. (V) Properties (III) and (IV) can be generalised to a finite number of functions f1, f2,..., fn and the real numbers, k1, k2,..., kn giving ∫ [k1 f1 (x) + k2 f 2 (x) +... + kn f n (x)] dx = k1 ∫ f1 (x) dx + k 2 ∫ f 2 (x) dx +... + kn ∫ fn (x) dx. To find an anti derivative of a given function, we search intuitively for a function whose derivative is the given function. The search for the requisite function for finding an anti derivative is known as integration by the method of inspection. We illustrate it through some examples. Example 1 Write an anti derivative for each of the following functions using the method of inspection: 1 (i) cos 2x (ii) 3x2 + 4x3 (iii) ,x≠0 x Solution (i) We look for a function whose derivative is cos 2x. Recall that d sin 2x = 2 cos 2x dx 1 d d 1  or cos 2x = (sin 2x) =  sin 2 x  2 dx dx  2  1 Therefore, an anti derivative of cos 2x is sin 2 x. 2 (ii) We look for a function whose derivative is 3x2 + 4x3. Note that d 3 dx ( ) x + x 4 = 3x2 + 4x3. Therefore, an anti derivative of 3x2 + 4x3 is x3 + x4. Reprint 2024-25 232 MATHEMATICS (iii) We know that d 1 d 1 1 (log x ) = , x > 0 and [log ( – x)] = ( – 1) = , x < 0 dx x dx –x x d 1 Combining above, we get dx ( log x ) = , x ≠ 0 x 1 1. Therefore, ∫ x dx = log x is one of the anti derivatives of x Example 2 Find the following integrals: 3 x3 – 1 2 1 (i) ∫ 2 dx (iii) ∫ (x 2 + 2 e – ) dx x x (ii) ∫ (x 3 + 1) dx x Solution (i) We have x3 – 1 ∫ x2 dx = ∫ x dx – ∫ x dx –2 (by Property V)  x1 + 1   x– 2 + 1  = + C1 –  + C 2  ; C , C are constants of integration  1+1   – 2 +1  1 2 x2 x– 1 x2 1 = + C1 – – C2 = + + C1 – C 2 2 –1 2 x x2 1 = + + C , where C = C1 – C2 is another constant of integration. 2 x A Note From now onwards, we shall write only one constant of integration in the final answer. (ii) We have 2 2 ∫ (x 3 + 1) dx = ∫ x 3 dx + ∫ dx 2 +1 5 x3 = 2 + x + C = 3 x3 + x + C +1 5 3 Reprint 2024-25  INTEGRALS 233 3 3 1 1 ∫ + 2 e – ) dx = ∫ x 2 dx + ∫ 2 e x dx – ∫ dx x (iii) We have (x 2 x x 3 +1 x2 = 3 + 2 e x – log x + C +1 2 5 2 = x 2 + 2 e x – log x + C 5 Example 3 Find the following integrals: (i) ∫ (sin x + cos x) dx (ii) ∫ cosec x (cosec x + cot x) dx 1 – sin x (iii) ∫ cos2 x dx Solution (i) We have ∫ (sin x + cos x) dx = ∫ sin x dx + ∫ cos x dx = – cos x + sin x + C (ii) We have ∫ (cosec x (cosec x + cot x) dx = ∫ cosec x dx + ∫ cosec x cot x dx 2 = – cot x – cosec x + C (iii) We have 1 – sin x 1 sin x ∫ 2 cos x dx = ∫ 2 cos x dx – ∫ cos 2 x dx = ∫ sec x dx – ∫ tan x sec x dx 2 = tan x – sec x + C Example 4 Find the anti derivative F of f defined by f (x) = 4x3 – 6, where F (0) = 3 Solution One anti derivative of f (x) is x4 – 6x since d 4 (x – 6 x ) = 4x3 – 6 dx Therefore, the anti derivative F is given by F(x) = x4 – 6x + C, where C is constant. Reprint 2024-25 234 MATHEMATICS Given that F(0) = 3, which gives, 3 = 0 – 6 × 0 + C or C = 3 Hence, the required anti derivative is the unique function F defined by F(x) = x4 – 6x + 3. Remarks (i) We see that if F is an anti derivative of f, then so is F + C, where C is any constant. Thus, if we know one anti derivative F of a function f, we can write down an infinite number of anti derivatives of f by adding any constant to F expressed by F(x) + C, C ∈ R. In applications, it is often necessary to satisfy an additional condition which then determines a specific value of C giving unique anti derivative of the given function. (ii) Sometimes, F is not expressible in terms of elementary functions viz., polynomial, logarithmic, exponential, trigonometric functions and their inverses etc. We are therefore blocked for finding ∫ f (x) dx. For example, it is not possible to find ∫e – x2 2 dx by inspection since we can not find a function whose derivative is e – x (iii) When the variable of integration is denoted by a variable other than x, the integral formulae are modified accordingly. For instance y4 + 1 1 ∫ y dy = + C = y5 + C 4 4 +1 5 EXERCISE 7.1 Find an anti derivative (or integral) of the following functions by the method of inspection. 1. sin 2x 2. cos 3x 3. e 2x 2 3x 4. (ax + b) 5. sin 2x – 4 e Find the following integrals in Exercises 6 to 20: 1 ∫ (4 e ∫x ∫ (ax 2 6. 3x + 1) dx 7. (1 – ) dx 8. 2 + bx + c ) dx x2 2  1  x3 + 5x 2 – 4 ∫ (2 x + e ) dx 10. ∫  x – ∫ 2 x 9.  dx 11. dx  x x2 x3 + 3x + 4 x3 − x 2 + x – 1 12. ∫ x dx 13. ∫ x –1 dx 14. ∫ (1 – x) x dx Reprint 2024-25  INTEGRALS 235 ∫ x ( 3x + 2 x + 3) dx ∫ (2 x – 3cos x + e ) dx 2 x 15. 16. ∫ (2x – 3sin x + 5 x ) dx ∫ sec x (sec x + tan x) dx 2 17. 18. sec 2 x 2 – 3sin x 19. ∫ cosec2 x dx 20. ∫ cos2 x dx. Choose the correct answer in Exercises 21 and 22.  1  21. The anti derivative of  x +  equals  x 1 1 2 1 3 2 3 1 2 (A) x + 2x 2 + C (B) x + x +C 3 3 2 3 1 3 1 2 2 3 2 1 2 (C) x + 2x 2 + C (D) x + x +C 3 2 2 d 3 22. If f ( x) = 4 x3 − 4 such that f (2) = 0. Then f (x) is dx x 1 129 1 129 (A) x + 3 − (B) x + 4 + 4 3 x 8 x 8 1 129 1 129 (C) x4 + + (D) x3 + − x3 8 x4 8 7.3 Methods of Integration In previous section, we discussed integrals of those functions which were readily obtainable from derivatives of some functions. It was based on inspection, i.e., on the search of a function F whose derivative is f which led us to the integral of f. However, this method, which depends on inspection, is not very suitable for many functions. Hence, we need to develop additional techniques or methods for finding the integrals by reducing them into standard forms. Prominent among them are methods based on: 1. Integration by Substitution 2. Integration using Partial Fractions 3. Integration by Parts 7.3.1 Integration by substitution In this section, we consider the method of integration by substitution. The given integral ∫ f (x) dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t). Reprint 2024-25 236 MATHEMATICS Consider I= ∫ f (x) dx dx Put x = g(t) so that = g′(t). dt We write dx = g′(t) dt Thus I= ∫ f ( x) dx = ∫ f ( g (t )) g′(t ) dt This change of variable formula is one of the important tools available to us in the name of integration by substitution. It is often important to guess what will be the useful substitution. Usually, we make a substitution for a function whose derivative also occurs in the integrand as illustrated in the following examples. Example 5 Integrate the following functions w.r.t. x: (i) sin mx (ii) 2x sin (x2 + 1) tan 4 x sec 2 x sin (tan – 1 x) (iii) (iv) x 1 + x2 Solution (i) We know that derivative of mx is m. Thus, we make the substitution mx = t so that mdx = dt. 1 1 1 Therefore, ∫ sin mx dx = m ∫ sin t dt = – m cos t + C = – m cos mx + C (ii) Derivative of x2 + 1 is 2x. Thus, we use the substitution x2 + 1 = t so that 2x dx = dt. ∫ 2 x sin (x + 1) dx = ∫ sin t dt = – cos t + C = – cos (x2 + 1) + C 2 Therefore, 1 1 –2 1 (iii) Derivative of x is x =. Thus, we use the substitution 2 2 x 1 x = t so that dx = dt giving dx = 2t dt. 2 x tan 4 x sec 2 x 2t tan 4t sec 2t dt ∫ dx = ∫ = 2 ∫ tan t sec t dt 4 2 Thus, x t Again, we make another substitution tan t = u so that sec2 t dt = du Reprint 2024-25  INTEGRALS 237 u5 Therefore, 2 ∫ tan 4t sec 2t dt = 2 ∫ u 4 du = 2 +C 5 2 = tan 5 t + C (since u = tan t) 5 2 = tan x + C (since t = x ) 5 5 tan 4 x sec 2 x 2 Hence, ∫ x dx = 5 tan 5 x +C Alternatively, make the substitution tan x = t 1 (iv) Derivative of tan – 1 x =. Thus, we use the substitution 1 + x2 dx tan–1 x = t so that = dt. 1 + x2 sin (tan – 1 x) Therefore , ∫ 1 + x2 dx = ∫ sin t dt = – cos t + C = – cos (tan–1x) + C Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique. These will be used later without reference. (i) ∫ tan x dx = log sec x + C We have sin x ∫ tan x dx = ∫ cos x dx Put cos x = t so that sin x dx = – dt dt Then ∫ tan x dx = – ∫ t = – log t + C = – log cos x + C or ∫ tan x dx = log sec x + C (ii) ∫ cot x dx = log sin x + C cos x We have ∫ cot x dx = ∫ sin x dx Reprint 2024-25 238 MATHEMATICS Put sin x = t so that cos x dx = dt dt Then ∫ cot x dx = ∫ t = log t + C = log sin x + C (iii) ∫ sec x dx = log sec x + tan x + C We have sec x (sec x + tan x) ∫ sec x dx = ∫ sec x + tan x dx Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt dt Therefore, ∫ sec x dx = ∫ = log t + C = log sec x + tan x + C t (iv) ∫ cosec x dx = log cosec x – cot x + C We have cosec x (cosec x + cot x) ∫ cosec x dx = ∫ (cosec x + cot x) dx Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt dt So ∫ cosec x dx = – ∫ t = – log | t | = – log |cosec x + cot x | + C cosec 2 x − cot 2 x = – log +C cosec x − cot x = log cosec x – cot x + C Example 6 Find the following integrals: sin x 1 ∫ sin ∫ sin (x + a) dx ∫ 1 + tan x dx 3 (i) x cos 2 x dx (ii) (iii) Solution (i) We have ∫ sin x cos2 x dx = ∫ sin 2 x cos 2 x (sin x ) dx 3 = ∫ (1 – cos x) cos x (sin x ) dx 2 2 Put t = cos x so that dt = – sin x dx Reprint 2024-25  INTEGRALS 239 ∫ sin x cos 2 x (sin x ) dx = − ∫ (1 – t 2 ) t 2 dt 2 Therefore,  t3 t 5  = – ∫ (t – t ) dt = –  –  + C 2 4 3 5 1 1 = – cos x + cos x + C 3 5 3 5 (ii) Put x + a = t. Then dx = dt. Therefore sin x sin (t – a) ∫ sin (x + a) dx = ∫ sin t dt sin t cos a – cos t sin a = ∫ sin t dt = cos a ∫ dt – sin a ∫ cot t dt = (cos a ) t – (sin a) log sin t + C1  = (cos a) (x + a ) – (sin a) log sin (x + a ) + C1  = x cos a + a cos a – (sin a ) log sin (x + a) – C1 sin a sin x Hence, ∫ sin (x + a) dx = x cos a – sin a log |sin (x + a)| + C, where, C = – C1 sin a + a cos a, is another arbitrary constant. dx cos x dx (iii) ∫ 1 + tan x = ∫ cos x + sin x 1 (cos x + sin x + cos x – sin x) dx = 2 ∫ cos x + sin x 1 1 cos x – sin x = 2 ∫ dx + 2 ∫ cos x + sin x dx x C1 1 cos x – sin x 2 2 2 ∫ cos x + sin x = + + dx... (1) Reprint 2024-25 240 MATHEMATICS cos x – sin x Now, consider I = ∫ dx cos x + sin x Put cos x + sin x = t so that (cos x – sin x) dx = dt dt Therefore I=∫ = log t + C2 = log cos x + sin x + C2 t Putting it in (1), we get dx x C1 1 C ∫ 1 + tan x = 2 + + log cos x + sin x + 2 2 2 2 x 1 C C = + log cos x + sin x + 1 + 2 2 2 2 2 x 1  C C  = + log cos x + sin x + C,  C = 1 + 2  2 2  2 2  EXERCISE 7.2 Integrate the functions in Exercises 1 to 37: 1. 2x 2. ( log x )2 3. 1 1 + x2 x x + x log x 4. sin x sin (cos x) 5. sin (ax + b) cos (ax + b) 6. ax + b 7. x x + 2 8. x 1 + 2 x 2 1 x 9. (4 x + 2) x 2 + x + 1 10. x – x 11. ,x>0 x+4 3 1 5 x2 1 12. (x – 1) 3 x 13. 14. , x > 0, m ≠ 1 (2 + 3x 3 )3 x (log x ) m x x 15. 16. e2 x + 3 17. 2 9 – 4 x2 ex –1 etan x e2x – 1 e2 x – e – 2 x 18. 19. 20. 1 + x2 e2 x + 1 e2 x + e – 2 x Reprint 2024-25  INTEGRALS 241 sin – 1 x 21. tan2 (2x – 3) 22. sec2 (7 – 4x) 23. 1 – x2 2cos x – 3sin x 1 cos x 24. 6cos x + 4sin x 25. 26. cos x (1 – tan x) 2 2 x cos x 27. sin 2x cos 2 x 28. 29. cot x log sin x 1 + sin x sin x sin x 1 30. 31. 32. 1 + cos x (1 + cos x ) 2 1 + cot x 33. 1 34. tan x 35. (1 + log x)2 1 – tan x sin x cos x x 36. (x + 1) ( x + log x ) 2 37. ( x 3sin tan – 1 x 4 ) 8 x 1+ x Choose the correct answer in Exercises 38 and 39. 10 x 9 + 10 x log e 10 dx 38. ∫ x10 + 10 x equals (A) 10x – x10 + C (B) 10x + x10 + C (C) (10x – x10)–1 + C (D) log (10x + x10) + C dx 39. ∫ sin 2 x cos2 x equals (A) tan x + cot x + C (B) tan x – cot x + C (C) tan x cot x + C (D) tan x – cot 2x + C 7.3.2 Integration using trigonometric identities When the integrand involves some trigonometric functions, we use some known identities to find the integral as illustrated through the following example. Example 7 Find (i) ∫ cos x dx (ii) ∫ sin 2 x cos 3 x dx (iii) ∫ sin x dx 2 3 Reprint 2024-25 242 MATHEMATICS Solution (i) Recall the identity cos 2x = 2 cos2 x – 1, which gives 1 + cos 2 x cos2 x = 2 1 1 1 Therefore, = 2 ∫ (1 + cos 2x ) dx = ∫ dx + ∫ cos 2 x dx 2 2 x 1 = + sin 2 x + C 2 4 1 (ii) Recall the identity sin x cos y = [sin (x + y) + sin (x – y)] (Why?) 2 Then = 1  1  = 2  – 5 cos 5 x + cos x  + C 1 1 cos 5 x + cos x + C = – 10 2 (iii) From the identity sin 3x = 3 sin x – 4 sin3 x, we find that 3sin x – sin 3x sin3 x = 4 3 1 ∫ sin ∫ sin x dx – ∫ sin 3x dx 3 Therefore, x dx = 4 4 3 1 = – cos x + cos 3 x + C 4 12 ∫ sin x dx = ∫ sin 2 x sin x dx = ∫ (1 – cos 3 2 Alternatively, x ) sin x dx Put cos x = t so that – sin x dx = dt t3 Therefore, 2 ( ) ∫ sin x dx = − ∫ 1 – t dt = – ∫ dt + ∫ t dt = – t + 3 2 3 +C 1 = – cos x + cos3 x + C 3 Remark It can be shown using trigonometric identities that both answers are equivalent. Reprint 2024-25  INTEGRALS 243 EXERCISE 7.3 Find the integrals of the functions in Exercises 1 to 22: 1. sin2 (2x + 5) 2. sin 3x cos 4x 3. cos 2x cos 4x cos 6x 3 3 3 4. sin (2x + 1) 5. sin x cos x 6. sin x sin 2x sin 3x 1 – cos x cos x 7. sin 4x sin 8x 8. 9. 1 + cos x 1 + cos x sin 2 x 10. sin4 x 11. cos4 2x 12. 1 + cos x cos 2 x – cos 2α cos x – sin x 13. 14. 15. tan3 2x sec 2x cos x – cos α 1 + sin 2 x 4 sin 3 x + cos3 x cos 2 x + 2sin 2 x 16. tan x 17. 18. sin 2 x cos 2 x cos 2 x 1 cos 2 x 19. 20. 21. sin – 1 (cos x) sin x cos3 x ( cos x + sin x )2 1 22. cos (x – a ) cos (x – b) Choose the correct answer in Exercises 23 and 24. sin 2 x − cos 2 x 23. ∫ sin 2 x cos 2 x dx is equal to (A) tan x + cot x + C (B) tan x + cosec x + C (C) – tan x + cot x + C (D) tan x + sec x + C e x (1 + x) 24. ∫ cos2 (e x x) dx equals (A) – cot (exx) + C (B) tan (xex) + C (C) tan (ex) + C (D) cot (ex) + C 7.4 Integrals of Some Particular Functions In this section, we mention below some important formulae of integrals and apply them for integrating many other related standard integrals: dx 1 x–a (1) ∫ 2 2 = log +C x –a 2 a x +a Reprint 2024-25 244 MATHEMATICS dx 1 a+x (2) ∫ a 2 – x 2 = 2a log a–x +C dx 1 x ∫ x 2 + a 2 = a tan –1 (3) +C a dx (4) ∫ x –a2 2 = log x + x 2 – a 2 + C dx x (5) ∫ a2 – x2 = sin – 1 a +C dx (6) ∫ x +a2 2 = log x + x 2 + a 2 + C We now prove the above results: 1 1 (1) We have = x –a22 (x – a ) (x + a ) 1  (x + a) – (x – a)  1  1 1   = – 2a  (x – a) (x + a)  2a  x – a x + a   = dx 1  dx dx  Therefore, ∫ x2 – a 2 = 2a  ∫ x – a – ∫ x + a  1 = [log | (x – a)| – log | (x + a)|] + C 2a 1 x–a = log +C 2a x+a (2) In view of (1) above, we have 1 1  (a + x) + (a − x)  1  1 1  =   = + 2 a –x 2 2a  (a + x) (a − x)  2a  a − x a + x   Reprint 2024-25  INTEGRALS 245 dx 1  dx dx  Therefore, ∫ a2 – x2 =  ∫ 2a  a − x +∫ a + x  1 = [ − log | a − x | + log | a + x |] + C 2a 1 a+ x = log +C 2a a−x ANote The technique used in (1) will be explained in Section 7.5. (3) Put x = a tan θ. Then dx = a sec2 θ dθ. dx Therefore, ∫ x2 + a2 = 1 1 1 x a ∫ dθ = θ + C = tan – 1 + C = a a a (4) Let x = a sec θ. Then dx = a sec θ tan θ d θ. dx a secθ tanθ dθ Therefore, ∫ x 2 − a 2 = ∫ a2 sec2θ − a2 = ∫ secθ dθ = log secθ + tanθ + C1 x x2 = log + – 1 + C1 a a2 = log x + x – a − log a + C1 2 2 = log x + x – a + C , where C = C1 – log |a| 2 2 (5) Let x = a sinθ. Then dx = a cosθ dθ. dx a cosθ dθ Therefore, ∫ a −x 2 2 = ∫ a 2 – a 2 sin 2θ x ∫ dθ = θ + C = sin +C –1 = a (6) Let x = a tan θ. Then dx = a sec2 θ dθ. dx a sec 2θ dθ Therefore, ∫ x2 + a2 = ∫ a 2 tan 2θ + a 2 = ∫ secθ dθ = log (secθ + tanθ) + C1 Reprint 2024-25 246 MATHEMATICS x x2 = log + + 1 + C1 a a2 2 = log x + x + a − log | a | + C1 2 2 = log x + x + a + C, where C = C1 – log |a| 2 Applying these standard formulae, we now obtain some more formulae which are useful from applications point of view and can be applied directly to evaluate other integrals. dx (7) To find the integral ∫ ax 2 + bx + c , we write  2 b c  b   c b2  2 2 a ax + bx + c =  x + x + = a   x +  + –   a a   2a   a 4a 2   b c b2 = t so that dx = dt and writing – 2 = ± k. We find the 2 Now, put x + 2a a 4a 1 dt  c b2  integral reduced to the form ∫ 2 depending upon the sign of  – 2 a t ± k2  a 4a  and hence can be evaluated. (8) To find the integral of the type , proceeding as in (7), we obtain the integral using the standard formulae. px + q (9) To find the integral of the type ∫ ax 2 + bx + c dx , where p, q, a, b, c are constants, we are to find real numbers A, B such that d px + q = A (ax 2 + bx + c) + B = A (2ax + b) + B dx To determine A and B, we equate from both sides the coefficients of x and the constant terms. A and B are thus obtained and hence the integral is reduced to one of the known forms. Reprint 2024-25  INTEGRALS 247 ( px + q) dx (10) For the evaluation of the integral of the type ax 2 + bx + c ∫ , we proceed as in (9) and transform the integral into known standard forms. Let us illustrate the above methods by some examples. Example 8 Find the following integrals: dx dx (i) ∫ x 2 − 16 (ii) ∫ 2x − x2 Solution dx dx 1 x–4 (i) We have ∫ x2 − 16 = ∫ x2 – 42 = 8 log x+4 + C [by 7.4 (1)] (ii) Put x – 1 = t. Then dx = dt. dx dt Therefore, ∫ 2x − x 2 = ∫ 1– t 2 = sin (t ) + C –1 [by 7.4 (5)] = sin (x – 1) + C –1 Example 9 Find the following integrals : dx dx dx (i) ∫ x2 − 6 x + 13 (ii) ∫ 3x 2 + 13x − 10 (iii) ∫ 5x2 − 2 x Solution (i) We have x2 – 6x + 13 = x2 – 6x + 32 – 32 + 13 = (x – 3)2 + 4 dx 1 So, ∫ x2 − 6 x + 13 = ∫ ( x – 3)2 + 22 dx Let x – 3 = t. Then dx = dt dx dt 1 t ∫ x2 − 6 x + 13 = ∫ t 2 + 22 = 2 tan +C –1 Therefore, [by 7.4 (3)] 2 1 x–3 = tan – 1 +C 2 2 Reprint 2024-25 248 MATHEMATICS (ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand,  2 13x 10  3 x 2 + 13x – 10 = 3  x + –   3 3  13   17   2 2 = 3  x + 6  –  6   (completing the square)      dx 1 dx Thus ∫ 3x 2 + 13x − 10 = ∫ 3  2 13   17  2  x + −     6  6 13 Put x + = t. Then dx = dt. 6 dx 1 dt ∫ 3x 2 + 13x − 10 3∫ Therefore, = 2  17  t2 −    6 17 t– 1 6 +C = log 1 [by 7.4 (i)] 17 17 3× 2× t+ 6 6 13 17 x+ – 1 6 6 +C = log 1 17 13 17 x+ + 6 6 1 6x − 4 = log + C1 17 6 x + 30 1 3x − 2 1 1 = log + C1 + log 17 x+5 17 3 1 3x − 2 1 1 = log + C , where C = C1 + log 17 x+5 17 3 Reprint 2024-25  INTEGRALS 249 dx dx (iii) We have ∫ 2 5x − 2 x =∫  2x  5 x2 –   5  1 dx = 5 ∫ 2 2 (completing the square)  1 1 x–  –   5 5 1 Put x – = t. Then dx = dt. 5 dx 1 dt Therefore, ∫ 5x2 − 2 x = 5 ∫ 2 1 t2 –   5 2 1 1 = log t + t 2 –   +C [by 7.4 (4)] 5 5 1 1 2x = log x – + x2 – +C 5 5 5 Example 10 Find the following integrals: x+2 x+3 (i) ∫ 2 x 2 + 6 x + 5 dx (ii) ∫ 5 − 4x – x2 dx Solution (i) Using the formula 7.4 (9), we express x+2= A d dx ( ) 2 x 2 + 6 x + 5 + B = A (4 x + 6) + B Equating the coefficients of x and the constant terms from both sides, we get 1 1 4A = 1 and 6A + B = 2 or A= and B =. 4 2 x+2 1 4x + 6 1 dx Therefore, ∫ 2x2 + 6x + 5 = ∫ 2 4 2x + 6x + 5 dx + ∫ 2 2 2x + 6 x + 5 1 1 = I1 + I2 (say)... (1) 4 2 Reprint 2024-25 250 MATHEMATICS In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt dt Therefore, I1 = ∫ t = log t + C1 = log | 2 x + 6 x + 5 | + C1 2... (2) dx 1 dx and I2 = ∫ 2x2 + 6x + 5 = 2 ∫ 5 x 2 + 3x + 2 1 dx = ∫ 2  2 3 1 2  x + +     2 2 3 Put x + = t , so that dx = dt, we get 2 1 dt 1 I2 = 2 ∫ 1 2 = 1 tan –1 2t + C 2 [by 7.4 (3)] t2 +   2× 2 2 –1  3 = tan 2  x +  + C2 = tan ( 2 x + 3 ) + C 2 –1... (3)  2  Using (2) and (3) in (1), we get x+2 1 1 ∫ 2 x 2 + 6 x + 5 dx = 4 log 2 x 2 + 6x + 5 + tan – 1 ( 2 x + 3) + C 2 C1 C2 where, C=+ 4 2 (ii) This integral is of the form given in 7.4 (10). Let us express d x+3= A (5 – 4 x – x 2 ) + B = A (– 4 – 2x) + B dx Equating the coefficients of x and the constant terms from both sides, we get 1 – 2A = 1 and – 4 A + B = 3, i.e., A = – and B = 1 2 Reprint 2024-25  INTEGRALS 251 x+3 1 ( – 4 – 2 x ) dx + dx Therefore, ∫ dx = – 2 ∫ 5 − 4 x − x2 ∫ 5 − 4 x − x2 5 − 4x − x2 1 = – I +I... (1) 2 1 2 In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt. ( – 4 − 2 x ) dx = dt Therefore, I1 = ∫ 5 − 4x − x 2 ∫ t = 2 t + C1 = 2 5 – 4 x – x 2 + C1... (2) dx dx Now consider I2 = ∫ 5 − 4x − x 2 =∫ 9 – (x + 2) 2 Put x + 2 = t, so that dx = dt. dt t Therefore, I2 = ∫ 3 −t 2 = sin – 1 + C 2 2 3 [by 7.4 (5)] x+2 = sin –1 + C2... (3) 3 Substituting (2) and (3) in (1), we obtain x+3 x+2 C ∫ 5 – 4x – x2 = – 5 – 4x – x 2 + sin – 1 3 + C , where C = C2 – 1 2 EXERCISE 7.4 Integrate the functions in Exercises 1 to 23. 3x 2 1 1 1. 2. 3. x6 + 1 1 + 4x 2 (2 – x) 2 +1 1 3x x2 4. 5. 6. 9 – 25 x 2 1 + 2x4 1 − x6 x –1 x2 sec 2 x 7. 8. 9. x2 – 1 x6 + a6 tan 2 x + 4 Reprint 2024-25 252 MATHEMATICS 1 1 1 10. 11. 12. x + 2x + 2 2 9x + 6x + 5 2 7 – 6 x – x2 1 1 1 13. 14. 15. ( x – 1)( x – 2 ) 8 + 3x – x 2 ( x – a )( x – b ) 4x + 1 x+2 5x − 2 16. 17. 18. 2x + x – 3 2 2 x –1 1 + 2x + 3x2 6x + 7 x+2 x+2 19. 20. 21. ( x – 5 )( x – 4 ) 4x – x 2 x + 2x + 3 2 x+3 5x + 3 22. 23.. x – 2x − 5 2 x + 4 x + 10 2 Choose the correct answer in Exercises 24 and 25. dx 24. ∫ x 2 + 2 x + 2 equals (A) x tan–1 (x + 1) + C (B) tan–1 (x + 1) + C (C) (x + 1) tan–1x + C (D) tan–1x + C dx 25. ∫ 9x − 4x2 equals 1 –1  9 x − 8  1  8x − 9  (A) sin  +C (B) sin –1  +C 9  8  2  9  1 –1  9 x − 8  1  9x − 8  (C) sin  +C (D) sin –1  +C 3  8  2  9  7.5 Integration by Partial Fractions Recall that a rational function is defined as the ratio of two polynomials in the form P(x) , where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0. If the degree of P(x) Q(x) is less than the degree of Q(x), then the rational function is called proper, otherwise, it is called improper. The improper rational functions can be reduced to the proper rational Reprint 2024-25  INTEGRALS 253 P(x) P(x) P (x) functions by long division process. Thus, if is improper, then = T(x) + 1 , Q(x) Q(x) Q(x) P1 (x) where T(x) is a polynomial in x and is a proper rational function. As we know Q(x) how to integrate polynomials, the integration of any rational function is reduced to the integration of a proper rational function. The rational functions which we shall consider here for integration purposes will be those whose denominators can be factorised into P(x ) P(x) linear and quadratic factors. Assume that we want to evaluate ∫ Q(x) dx , where Q(x) is proper rational function. It is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. After this, the integration can be carried out easily using the already known methods. The following Table 7.2 indicates the types of simpler partial fractions that are to be associated with various kind of rational functions. Table 7.2 S.No. Form of the rational function Form of the partial fraction px + q A + B 1. ,a≠b (x –a) (x –b) x–a x–b px + q A B 2. + (x – a ) 2 x – a ( x – a )2 px 2 + qx + r A B C 3. + + (x – a) (x – b) (x – c) x –a x –b x –c px 2 + qx + r A B C 4. + 2 + (x – a) 2 (x – b) x – a (x – a) x–b px 2 + qx + r A Bx + C , 5. + 2 (x – a) (x 2 + bx + c) x – a x + bx + c where x2 + bx + c cannot be factorised further In the above table, A, B and C are real numbers to be determined suitably. Reprint 2024-25 254 MATHEMATICS dx Example 11 Find ∫ (x + 1) (x + 2) Solution The integrand is a proper rational function. Therefore, by using the form of partial fraction [Table 7.2 (i)], we write 1 A B = +... (1) (x + 1) (x + 2) x +1 x + 2 where, real numbers A and B are to be determined suitably. This gives 1 = A (x + 2) + B (x + 1). Equating the coefficients of x and the constant term, we get A+B=0 and 2A + B = 1 Solving these equations, we get A =1 and B = – 1. Thus, the integrand is given by 1 1 –1 = + (x + 1) (x + 2) x +1 x + 2 dx dx dx Therefore, ∫ (x + 1) (x + 2) = ∫ x + 1 – ∫ x + 2 = log x + 1 − log x + 2 + C x +1 = log +C x+2 Remark The equation (1) above is an identity, i.e. a statement true for all (permissible) values of x. Some authors use the symbol ‘≡’ to indicate that the statement is an identity and use the symbol ‘=’ to indicate that the statement is an equation, i.e., to indicate that the statement is true only for certain values of x. x2 + 1 Example 12 Find ∫ x 2 − 5x + 6 dx x2 + 1 Solution Here the integrand 2 is not proper rational function, so we divide x – 5x + 6 x2 + 1 by x2 – 5x + 6 and find that Reprint 2024-25  INTEGRALS 255 x2 + 1 5x – 5 5x – 5 = 1+ 2 =1+ x – 5x + 6 2 x – 5x + 6 (x – 2) (x – 3) 5x – 5 A B Let = + (x – 2) (x – 3) x–2 x–3 So that 5x – 5 = A (x – 3) + B (x – 2) Equating the coefficients of x and constant terms on both sides, we get A + B = 5 and 3A + 2B = 5. Solving these equations, we get A = – 5 and B = 10 x2 + 1 5 10 Thus, = 1− + x – 5x + 6 2 x –2 x –3 x2 + 1 1 dx Therefore, ∫ x 2 – 5 x + 6 dx = ∫ dx − 5 ∫ x – 2 dx + 10∫ x – 3 = x – 5 log | x – 2 | + 10 log | x – 3 | + C. 3x − 2 Example 13 Find ∫ dx (x + 1)2 (x + 3) Solution The integrand is of the type as given in Table 7.2 (4). We write 3x – 2 A B C = + + (x + 1) (x + 3) 2 x + 1 (x + 1) 2 x+3 So that 3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2 = A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1 ) Comparing coefficient of x 2 , x and constant term on both sides, we get A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2. Solving these equations, we get 11 –5 –11. Thus the integrand is given by A= ,B = and C = 4 2 4 3x − 2 11 5 11 = 4 (x + 1) – – 4 (x + 3) (x + 1) (x + 3) 2 2 (x + 1) 2 3x − 2 11 dx 5 dx 11 dx Therefore, ∫ (x + 1)2 (x + 3) = ∫ – ∫ 4 x + 1 2 (x + 1) 2 − ∫ 4 x+3 11 5 11 = log x +1 + − log x + 3 + C 4 2 (x + 1) 4 11 x +1 5 = log + +C 4 x + 3 2 (x + 1) Reprint 2024-25 256 MATHEMATICS x2 Example 14 Find ∫ (x2 + 1) (x 2 + 4) dx x2 Solution Consider 2 and put x2 = y. ( x + 1) ( x 2 + 4) x2 y Then = (x + 1) (x + 4) 2 2 (y + 1) (y + 4) y A B Write = + (y + 1) (y + 4) y +1 y + 4 So that y = A (y + 4) + B (y + 1) Comparing coefficients of y and constant terms on both sides, we get A + B = 1 and 4A + B = 0, which give 1 4 A= − and B = 3 3 x2 1 4 Thus, = – + (x + 1) (x + 4) 2 2 3 (x + 1) 3 (x + 4) 2 2 x 2 dx 1 dx 4 dx Therefore, ∫ (x2 + 1) (x 2 + 4) = – 3 ∫ x 2 + 1 + 3 ∫ x2 + 4 1 –1 4 1 –1 x = – tan x + × tan +C 3 3 2 2 1 –1 2 –1 x = – tan x + tan +C 3 3 2 In the above example, the substitution was made only for the partial fraction part and not for the integration part. Now, we consider an example, where the integration involves a combination of the substitution method and the partial fraction method. ( 3 sin φ – 2 ) cos φ Example 15 Find ∫ 5 – cos2 φ – 4 sin φ d φ Solution Let y = sinφ Then dy = cosφ dφ Reprint 2024-25  INTEGRALS 257 ( 3 sinφ – 2 ) cosφ (3y – 2) dy Therefore, ∫ 5 – cos 2φ – 4 sinφ d φ = ∫ 5 – (1 – y 2 ) – 4 y 3y – 2 = ∫ y 2 – 4 y + 4 dy 3y – 2 = ∫ ( y – 2 )2 = I (say) 3y – 2 A B Now, we write = + [by Table 7.2 (2)] ( y – 2) 2 y − 2 (y − 2) 2 Therefore, 3y – 2 = A (y – 2) + B Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2, which gives A = 3 and B = 4. Therefore, the required integral is given by 3 4 dy dy I = ∫[ + ] dy = 3 ∫ +4∫ y – 2 (y – 2) 2 y–2 (y – 2) 2  1  = 3 log y − 2 + 4  – +C  y−2 4 = 3 log sin φ − 2 + +C 2 – sin φ 4 = 3 log (2 − sin φ) + + C (since, 2 – sin φ is always positive) 2 − sin φ x 2 + x + 1 dx Example 16 Find ∫ (x + 2) (x 2 + 1) Solution The integrand is a proper rational function. Decompose the rational function into partial fraction [Table 2.2(5)]. Write x2 + x + 1 A Bx + C = + 2 (x + 1) (x + 2) 2 x + 2 (x + 1) Therefore, x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2) Reprint 2024-25 258 MATHEMATICS Equating the coefficients of x2, x and of constant term of both sides, we get A + B =1, 2B + C = 1 and A + 2C = 1. Solving these equations, we get 3 2 1 A = , B = and C = 5 5 5 Thus, the integrand is given by 2 1 x+ x2 + x + 1 3 3 1  2x + 1  = + 2 5 = 5 +  2  (x + 1) (x + 2) 2 5 (x + 2) x + 1 5 (x + 2) 5  x + 1  x2 + x + 1 3 dx 1 2x 1 1 Therefore, ∫ (x2 +1) (x + 2) dx = 5 ∫ x + 2 + 5 ∫ x2 + 1 dx + 5 ∫ x2 + 1 dx 3 1 1 = log x + 2 + log x 2 + 1 + tan –1 x + C 5 5 5 EXERCISE 7.5 Integrate the rational functions in Exercises 1 to 21. x 1 3x – 1 1. 2. 3. (x + 1) (x + 2) 2 x –9 (x – 1) (x – 2) (x – 3) x 2x 1 – x2 4. 5. 2 6. (x – 1) (x – 2) (x – 3) x + 3x + 2 x (1 – 2 x) x x 3x + 5 7. 8. 9. (x + 1) (x – 1) 2 (x – 1) (x + 2) 2 x – x2 − x + 1 3 2x − 3 5x x3 + x + 1 10. 11. 12. (x – 1) (2x + 3) 2 (x + 1) (x 2 − 4) x2 − 1 2 3x – 1 1 13. 14. 15. (1 − x) (1 + x 2 ) (x + 2) 2 x −1 4 1 16. [Hint: multiply numerator and denominator by x n – 1 and put xn = t ] x (x n + 1) cos x 17. [Hint : Put sin x = t] (1 – sin x) (2 – sin x) Reprint 2024-25  INTEGRALS 259 (x 2 + 1) (x 2 + 2) 2x 1 18. 19. 20. (x 2 + 3) (x 2 + 4) (x + 1) (x 2 + 3) 2 x (x 4 – 1) 1 21. x [Hint : Put ex = t] (e – 1) Choose the correct answer in each of the Exercises 22 and 23. x dx 22. ∫ equals ( x − 1) ( x − 2) ( x − 1)2 ( x − 2) 2 (A) log +C (B) log +C x−2 x −1 2  x −1  (C) log   +C (D) log ( x − 1) ( x − 2) + C  x−2 dx 23. ∫ x ( x 2 + 1) equals 1 1 (A) log x − log (x +1) + C (B) log x + log (x +1) + C 2 2 2 2 1 1 (D) log x + log (x +1) + C 2 (C) − log x + log (x 2 +1) + C 2 2 7.6 Integration by Parts In this section, we describe one more method of integration, that is found quite useful in integrating products of functions. If u and v are any two differentiable functions of a single variable x (say). Then, by the product rule of differentiation, we have d dv du (uv ) = u + v dx dx dx Integrating both sides, we get dv du uv = ∫ u dx + ∫ v dx dx dx dv du or ∫ u dx dx = uv – ∫ v dx dx... (1) dv Let u = f (x) and = g (x). Then dx du = f ′(x) and v = ∫ g (x ) dx dx Reprint 2024-25 260 MATHEMATICS Therefore, expression (1) can be rewritten as ∫ f (x) g (x) dx = f (x) ∫ g (x) dx – ∫ [ ∫ g (x) dx] f ′(x ) dx i.e., ∫ f (x) g (x) dx = f (x) ∫ g (x ) dx – ∫ [ f ′ (x) ∫ g (x ) dx ] dx If we take f as the first function and g as the second function, then this formula may be stated as follows: “The integral of the product of two functions = (first function) × (integral of the second function) – Integral of [(differential coefficient of the first function) × (integral of the second function)]” Example 17 Find ∫ x cos x dx Solution Put f (x) = x (first function) and g (x) = cos x (second function). Then, integration by parts gives d ∫ x cos x dx = x ∫ cos x dx – ∫ [ (x ) ∫ cos x dx ] dx dx = x sin x – ∫ sin x dx = x sin x + cos x + C Suppose, we take f (x) = cos x and g (x) = x. Then d ∫ x cos x dx = cos x ∫ x dx – ∫ [ (cos x) ∫ x dx] dx dx x2 x2 = ( cos x ) + ∫ sin x dx 2 2 Thus, it shows that the integral ∫ x cos x dx is reduced to the comparatively more complicated integral having more power of x. Therefore, the proper choice of the first function and the second function is significant. Remarks (i) It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for ∫ x sin x dx. The reason is that there does not exist any function whose derivative is x sin x. (ii) Observe that while finding the integral of the second function, we did not add any constant of integration. If we write the integral of the second function cos x Reprint 2024-25  INTEGRALS 261 as sin x + k, where k is any constant, then ∫ x cos x dx = x (sin x + k ) − ∫ (sin x + k ) dx = x (sin x + k ) − ∫ (sin x dx − ∫ k dx = x (sin x + k ) − cos x – kx + C = x sin x + cos x + C This shows t