INTEGRALS 225 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 primitive) of the function. Further, the formula that gives 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. Chapter 7 INTEGRALS G .W. Leibnitz (1646 -1716)Reprint 2026-27
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: We know that (sin ) d x dx = cos x ... (1) 3 ( ) 3 d x dx = x2 ... (2) and ( )xd e dx = ex ... (3) We observe that in (1), the function cos x is the derived function of sin x. We say that sin x is an anti derivative (or an integral) of cos x. Similarly, in (2) and (3), 3 3 x and 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 : (sin + C) cos= d x x dx , 3 2 ( + C) 3 = d x x dx and ( + C) =x xd e e 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. More generally, if there is a function F such that F ( ) = ( ) d x f x dx , ∀ x ∈ I (interval), then for any arbitrary real number C, (also called constant of integration) [ ]F ( ) + C d x dx = f (x), x ∈ IReprint 2026-27
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 Then df dx = f ′ = g ′ – h ′ giving f ′ (x) = g ′ (x) – h ′ (x) ∀ x ∈ I 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 ( ) + Cf x dx x∫ . Notation Given that ( ) dy f x dx = , 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 functionReprint 2026-27
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) (i) 1 1 n nd x x dx n + = + ; 1 C 1 n n x x dx n + = + +∫ , n ≠ –1 Particularly, we note that ( ) 1 d x dx = ; Cdx x= +∫ (ii) ( )sin cos d x x dx = ; cos sin Cx dx x= +∫ (iii) ( )– cos sin d x x dx = ; sin cos Cx dx – x= +∫ (iv) ( ) 2 tan sec d x x dx = ; 2 sec tan Cx dx x= +∫ (v) ( ) 2 – cot cosec d x x dx = ; 2 cosec cot Cx dx – x= +∫ (vi) ( )sec sec tan d x x x dx = ; sec tan sec Cx x dx x= +∫ (vii) ( )– cosec cosec cot d x x x dx = ; cosec cot – cosec Cx x dx x= +∫ (viii) ( )– 1 2 1 sin 1 d x dx – x = ; – 1 2 sin C 1 dx x – x = +∫ (ix) ( )– 1 2 1 – cos 1 d x dx – x = ; – 1 2 cos C 1 dx – x – x = +∫ (x) ( )– 1 2 1 tan 1 d x dx x = + ; – 1 2 tan C 1 dx x x = + +∫ (xi) ( )x xd e e dx = ; Cx x e dx e= +∫Reprint 2026-27
INTEGRALS 229 (xii) ( ) 1 log | | d x dx x = ; 1 log | | Cdx x x = +∫ (xiii) x xd a a dx log a = ; C x x a a dx log a = +∫ ANote 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 f x dx 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., F( ) d x dx = f (x) Then ( )f x dx∫ = F(x) + C Therefore ( ) d f x dx dx ∫ = ( )F ( ) + C d x dx = F ( ) = ( ) d x f x dx Similarly, we note that f ′(x) = ( ) d 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 2026-27
230 MATHEMATICS Proof Let f and g be two functions such that ( ) d f x dx dx ∫ = ( ) d g x dx dx ∫ or ( ) ( ) d f x dx – g x dx dx ∫ ∫ = 0 Hence ( ) ( )f x dx – g x dx∫ ∫ = C, where C is any real number (Why?) or ( )f x dx∫ = ( ) Cg x dx +∫ So the families of curves { }1 1( ) C , C Rf x dx + ∈∫ and { }2 2( ) C , C Rg x dx + ∈∫ are identical. Hence, in this sense, ( ) and ( )f x dx g x dx∫ ∫ are equivalent. A Note The equivalence of the families { }1 1( ) + C ,Cf x dx ∈∫ R and { }2 2( ) + C ,Cg x dx ∈∫ R is customarily expressed by writing ( ) = ( )f x dx g x dx∫ ∫ , 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 dx ∫ = f (x) + g (x) ... (1) On the otherhand, we find that ( ) + ( ) d f x dx g x dx dx ∫ ∫ = ( ) + ( ) d d f x 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 2026-27
INTEGRALS 231 Proof By the Property (I), ( ) ( ) d k f x dx k f x dx =∫ . Also ( ) d k f x dx dx ∫ = ( ) = ( ) d k f x dx k f x dx ∫ 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 [ ]1 1 2 2( ) ( ) ( )n nk f x k f x ... k f x dx+ + +∫ = 1 1 2 2( ) ( ) ( )n nk f x dx k f x dx ... k f 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: (i) cos 2x (ii) 3x2 + 4x3 (iii) 1 x , x ≠ 0 Solution (i) We look for a function whose derivative is cos 2x. Recall that d dx sin 2x = 2 cos 2x or cos 2x = 1 2 d dx (sin 2x) = 1 sin 2 2 d x dx Therefore, an anti derivative of cos 2x is 1 sin 2 2 x . (ii) We look for a function whose derivative is 3x2 + 4x3. Note that ( )3 4d x x dx + = 3x2 + 4x3. Therefore, an anti derivative of 3x2 + 4x3 is x3 + x4.Reprint 2026-27
232 MATHEMATICS (iii) We know that 1 1 1 (log ) 0 and [log ( )] ( 1) 0 d d x , x – x – , x dx x dx – x x = > = = < Combining above, we get ( ) 1 log 0 d x , x dx x = ≠ Therefore, 1 logdx x x =∫ is one of the anti derivatives of 1 x . Example 2 Find the following integrals: (i) 3 2 1x – dx x∫ (ii) 2 3( 1)x dx+∫ (iii) ∫ 3 2 1 ( 2 – )+ x x e dx x Solution (i) We have 3 2 2 1 –x – dx x dx – x dx x =∫ ∫ ∫ (by Property V) = 1 1 2 1 1 2C C 1 1 2 1 – x x – – + + + + + + ; C1, C2 are constants of integration = 2 1 1 2C C 2 1 – x x – – – + = 2 1 2 1 + C C 2 x – x + = 2 1 + C 2 x x + , where C = C1 – C2 is another constant of integration. ANote From now onwards, we shall write only one constant of integration in the final answer. (ii) We have 2 2 3 3( 1)x dx x dx dx+ = +∫ ∫ ∫ = 2 1 3 C 2 1 3 x x + + + + = 5 3 3 C 5 x x+ +Reprint 2026-27
INTEGRALS 233 (iii) We have 3 3 2 2 1 1 ( 2 ) 2x x x e – dx x dx e dx – dx x x + = +∫ ∫ ∫ ∫ = 3 1 2 2 – log + C 3 1 2 xx e x + + + = 5 2 2 2 – log + C 5 x x e x+ Example 3 Find the following integrals: (i) (sin cos )x x dx+∫ (ii) cosec (cosec cot )x x x dx+∫ (iii) 2 1 sin cos – x dx x∫ Solution (i) We have (sin cos ) sin cosx x dx x dx x dx+ = +∫ ∫ ∫ = – cos sin Cx x+ + (ii) We have 2 (cosec (cosec + cot ) cosec cosec cotx x x dx x dx x x dx= +∫ ∫ ∫ = – cot cosec Cx – x + (iii) We have 2 2 2 1 sin 1 sin cos cos cos – x x dx dx – dx x x x =∫ ∫ ∫ = 2 sec tan secx dx – x x dx∫ ∫ = tan sec Cx – x + 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 4 ( 6 ) d x – x dx = 4x3 – 6 Therefore, the anti derivative F is given by F(x) = x4 – 6x + C, where C is constant.Reprint 2026-27
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 2 – x e dx∫ by inspection since we can not find a function whose derivative is 2 – x e (iii) When the variable of integration is denoted by a variable other than x, the integral formulae are modified accordingly. For instance 4 1 4 51 C C 4 1 5 y y dy y + = + = + +∫ 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. e2x 4. (ax + b)2 5. sin 2x – 4 e3x Find the following integrals in Exercises 6 to 20: 6. 3 (4 + 1)x e dx∫ 7. 2 2 1 (1 – )x dx x∫ 8. 2 ( )ax bx c dx+ +∫ 9. 2 (2 )x x e dx+∫ 10. 2 1 x – dx x ∫ 11. 3 2 2 5 4x x – dx x + ∫ 12. 3 3 4x x dx x + + ∫ 13. 3 2 1 1 x x x – dx x – − + ∫ 14. (1 )– x x dx∫Reprint 2026-27
INTEGRALS 235 15. 2 ( 3 2 3)x x x dx+ +∫ 16. (2 3cos )x x – x e dx+∫ 17. 2 (2 3sin 5 )x – x x dx+∫ 18. sec (sec tan )x x x dx+∫ 19. 2 2 sec cosec x dx x∫ 20. 2 2 – 3sin cos x x∫ dx. Choose the correct answer in Exercises 21 and 22. 21. The anti derivative of 1 x x + equals (A) 1 1 3 2 1 2 C 3 x x+ + (B) 2 23 2 1 C 3 2 x x+ + (C) 3 1 2 2 2 2 C 3 x x+ + (D) 3 1 2 2 3 1 C 2 2 x x+ + 22. If 3 4 3 ( ) 4 d f x x dx x = − such that f (2) = 0. Then f (x) is (A) 4 3 1 129 8 x x + − (B) 3 4 1 129 8 x x + + (C) 4 3 1 129 8 x x + + (D) 3 4 1 129 8 x x + − 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 2026-27
236 MATHEMATICS Consider I = ( )f x dx∫ Put x = g(t) so that dx dt = g′(t). 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) (iii) 4 2 tan secx x x (iv) 1 2 sin (tan ) 1 – x x+ Solution (i) We know that derivative of mx is m. Thus, we make the substitution mx = t so that mdx = dt. Therefore, 1 sin sinmx dx t dt m =∫ ∫ = – 1 m cos t + C = – 1 m cos mx + C (ii) Derivative of x2 + 1 is 2x. Thus, we use the substitution x2 + 1 = t so that 2x dx = dt. Therefore, 2 2 sin ( 1) sinx x dx t dt+ =∫ ∫ = – cos t + C = – cos (x2 + 1) + C (iii) Derivative of x is 1 2 1 1 2 2 – x x = . Thus, we use the substitution 1 so that giving 2 x t dx dt x = = dx = 2t dt. Thus, 4 2 4 2 tan sec 2 tan secx x t t t dt dx tx =∫ ∫ = 4 2 2 tan sect t dt∫ Again, we make another substitution tan t = u so that sec2 t dt = duReprint 2026-27
INTEGRALS 237 Therefore, 4 2 4 2 tan sec 2t t dt u du=∫ ∫ = 5 2 C 5 u + = 52 tan C 5 t + (since u = tan t) = 52 tan C (since ) 5 x t x+ = Hence, 4 2 tan secx x dx x ∫ = 52 tan C 5 x + Alternatively, make the substitution tan x t= (iv) Derivative of 1 2 1 tan 1 – x x = + . Thus, we use the substitution tan–1 x = t so that 2 1 dx x+ = dt. Therefore , 1 2 sin (tan ) sin 1 – x dx t dt x = +∫ ∫ = – 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 = log sec + Cx dx x We have sin tan cos x x dx dx x =∫ ∫ Put cos x = t so that sin x dx = – dt Then tan log C log cos C dt x dx – – t – x t = = + = +∫ ∫ or tan log sec Cx dx x= +∫ (ii) ∫cot = log sin + Cx dx x We have cos cot sin x x dx dx x =∫ ∫Reprint 2026-27
238 MATHEMATICS Put sin x = t so that cos x dx = dt Then cot dt x dx t =∫ ∫ = log Ct + = log sin Cx + (iii) ∫sec = log sec + tan + Cx dx x x We have sec (sec tan ) sec sec + tan x x x x dx dx x x + =∫ ∫ Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt Therefore, sec log + C = log sec tan C dt x dx t x x t = = + +∫ ∫ (iv) ∫cosec = log cosec – cot + Cx dx x x We have cosec (cosec cot ) cosec (cosec cot ) x x x x dx dx x x + = +∫ ∫ Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt So cosec – – log | | – log |cosec cot | C dt x dx t x x t = = = + +∫ ∫ = 2 2 cosec cot – log C cosec cot x x x x − + − = log cosec cot Cx – x + Example 6 Find the following integrals: (i) 3 2 sin cosx x dx∫ (ii) sin sin ( ) x dx x a+∫ (iii) 1 1 tan dx x+∫ Solution (i) We have 3 2 2 2 sin cos sin cos (sin )x x dx x x x dx=∫ ∫ = 2 2 (1 – cos ) cos (sin )x x x dx∫ Put t = cos x so that dt = – sin x dxReprint 2026-27
INTEGRALS 239 Therefore, 2 2 sin cos (sin )x x x dx∫ = 2 2 (1 – )t t dt− ∫ = 3 5 2 4 ( – ) C 3 5 t t – t t dt – – = + ∫ = 3 51 1 cos cos C 3 5 – x x+ + (ii) Put x + a = t. Then dx = dt. Therefore sin sin ( ) sin ( ) sin x t – a dx dt x a t = +∫ ∫ = sin cos cos sin sin t a – t a dt t∫ = cos – sin cota dt a t dt∫ ∫ = 1(cos ) (sin ) log sin Ca t – a t + = 1(cos ) ( ) (sin ) log sin ( ) Ca x a – a x a + + + = 1cos cos (sin ) log sin ( ) C sinx a a a – a x a – a+ + Hence, sin sin ( ) x dx x a+∫ = x cos a – sin a log |sin (x + a)| + C, where, C = – C1 sin a + a cos a, is another arbitrary constant. (iii) cos 1 tan cos sin dx x dx x x x = + +∫ ∫ = 1 (cos + sin + cos – sin ) 2 cos sin x x x x dx x x+∫ = 1 1 cos – sin 2 2 cos sin x x dx dx x x + +∫ ∫ = 1C 1 cos sin 2 2 2 cos sin x x – x dx x x + + +∫ ... (1)Reprint 2026-27
240 MATHEMATICS Now, consider cos sin I cos sin x – x dx x x = +∫ Put cos x + sin x = t so that (cos x – sin x) dx = dt Therefore 2I log C dt t t = = +∫ = 2log cos sin Cx x+ + Putting it in (1), we get 1 2C C1 + + log cos sin 1 tan 2 2 2 2 dx x x x x = + + +∫ = 1 2C C1 + log cos sin 2 2 2 2 x x x+ + + = 1 2C C1 + log cos sin C C 2 2 2 2 x x x , + + = + EXERCISE 7.2 Integrate the functions in Exercises 1 to 37: 1. 2 2 1 x x+ 2. ( )2 log x x 3. 1 logx x x+ 4. sin sin (cos )x x 5. sin ( ) cos ( )ax b ax b+ + 6. ax b+ 7. 2x x + 8. 2 1 2x x+ 9. 2 (4 2) 1x x x+ + + 10. 1 x – x 11. 4 x x + , x > 0 12. 1 3 53( 1)x – x 13. 2 3 3 (2 3 ) x x+ 14. 1 (log )m x x , x > 0, 1≠m 15. 2 9 4 x – x 16. 2 3x e + 17. 2 x x e 18. 1 2 1 – tan x e x+ 19. 2 2 1 1 x x e – e + 20. 2 2 2 2 x – x x – x e – e e e+Reprint 2026-27
INTEGRALS 241 21. tan2 (2x – 3) 22. sec2 (7 – 4x) 23. 1 2 sin 1 – x – x 24. 2cos 3sin 6cos 4sin x – x x x+ 25. 2 2 1 cos (1 tan )x – x 26. cos x x 27. sin 2 cos 2x x 28. cos 1 sin x x+ 29. cot x log sin x 30. sin 1 cos x x+ 31. ( )2 sin 1 cos x x+ 32. 1 1 cot x+ 33. 1 1 tan– x 34. tan sin cos x x x 35. ( )2 1 log x x + 36. ( )2 ( 1) logx x x x + + 37. ( )3 1 4 sin tan 1 – x x x8 + Choose the correct answer in Exercises 38 and 39. 38. 9 10 10 10 log 10 10 x e x x dx x + +∫ equals (A) 10x – x10 + C (B) 10x + x10 + C (C) (10x – x10)–1 + C (D) log (10x + x10) + C 39. 2 2 equals sin cos dx x x∫ (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) 2 cos x dx∫ (ii) sin 2 cos 3x x dx∫ (iii) 3 sin x dx∫Reprint 2026-27
242 MATHEMATICS Solution (i) Recall the identity cos 2x = 2 cos2 x – 1, which gives cos2 x = 1 cos 2 2 x+ Therefore, = 1 (1 + cos 2 ) 2 x dx∫ = 1 1 cos 2 2 2 dx x dx+∫ ∫ = 1 sin 2 C 2 4 x x+ + (ii) Recall the identity sin x cos y = 1 2 [sin (x + y) + sin (x – y)] (Why?) Then = = 1 1 cos 5 cos C 2 5 – x x + + = 1 1 cos 5 cos C 10 2 – x x+ + (iii) From the identity sin 3x = 3 sin x – 4 sin3 x, we find that sin3 x = 3sin sin 3 4 x – x Therefore, 3 sin x dx∫ = 3 1 sin sin 3 4 4 x dx – x dx∫ ∫ = 3 1 – cos cos 3 C 4 12 x x+ + Alternatively, 3 2 sin sin sinx dx x x dx=∫ ∫ = 2 (1 – cos ) sinx x dx∫ Put cos x = t so that – sin x dx = dt Therefore, 3 sin x dx∫ = ( )2 1 – t dt− ∫ = 3 2 C 3 t – dt t dt – t+ = + +∫ ∫ = 31 cos cos C 3 – x x+ + Remark It can be shown using trigonometric identities that both answers are equivalent.Reprint 2026-27
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 4. sin3 (2x + 1) 5. sin3 x cos3 x 6. sin x sin 2x sin 3x 7. sin 4x sin 8x 8. 1 cos 1 cos – x x+ 9. cos 1 cos x x+ 10. sin4 x 11. cos4 2x 12. 2 sin 1 cos x x+ 13. cos 2 cos 2 cos cos x – x – α α 14. cos sin 1 sin 2 x – x x+ 15. tan3 2x sec 2x 16. tan4x 17. 3 3 2 2 sin cos sin cos x x x x + 18. 2 2 cos 2 2sin cos x x x + 19. 3 1 sin cosx x 20. ( )2 cos 2 cos sin x x x+ 21. sin – 1 (cos x) 22. 1 cos ( ) cos ( )x – a x – b Choose the correct answer in Exercises 23 and 24. 23. 2 2 2 2 sin cos is equal to sin cos x x dx x x − ∫ (A) tan x + cot x + C (B) tan x + cosec x + C (C) – tan x + cot x + C (D) tan x + sec x + C 24. 2 (1 ) equals cos ( ) x x e x dx e x + ∫ (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: (1) ∫ 2 2 1 – = log + C 2 +– dx x a a x ax aReprint 2026-27
244 MATHEMATICS (2) ∫ 2 2 1 + = log + C 2 –– dx a x a a xa x (3) ∫ – 1 2 2 1 tan C dx x = + a ax + a (4) ∫ 2 2 2 2 = log + – + C – dx x x a x a (5) ∫ – 1 2 2 = sin + C – dx x aa x (6) ∫ 2 2 2 2 = log + + + C + dx x x a x a We now prove the above results: (1) We have 2 2 1 1 ( ) ( )x – a x ax – a = + = 1 ( ) – ( ) 1 1 1 2 ( ) ( ) 2 x a x – a – a x – a x a a x – a x a + = + + Therefore, 2 2 1 2 dx dx dx – a x – a x ax – a = + ∫ ∫ ∫ = [ ] 1 log ( )| log ( )| C 2 | x – a – | x a a + + = 1 log C 2 x – a a x a + + (2) In view of (1) above, we have 2 2 1 1 ( ) ( ) 2 ( ) ( )– a x a x a a x a xa x + + − = + − = 1 1 1 2a a x a x + − + Reprint 2026-27
INTEGRALS 245 Therefore, 2 2 – dx a x∫ = 1 2 dx dx a a x a x + − + ∫ ∫ = 1 [ log | | log | |] C 2 a x a x a − − + + + = 1 log C 2 a x a 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. Therefore, 2 2 dx x a+∫ = = 11 1 1 θ θ C tan C– x d a a a a = + = +∫ (4) Let x = a secθ. Then dx = a secθ tan θ d θ. Therefore, 2 2 dx x a− ∫ = 2 2 2 secθ tanθ θ sec θ a d a a− ∫ = 1secθ θ log secθ + tanθ + Cd =∫ = 2 12 log 1 C x x – a a + + = 2 2 1log log Cx x – a a+ − + = 2 2 log + Cx x – a+ , where C = C1 – log |a| (5) Let x = a sinθ. Then dx = a cosθ dθ. Therefore, 2 2 dx a x− ∫ = 2 2 2 θ θ θ cos sin a d a – a ∫ = 1 θ = θ + C = sin C– x d a +∫ (6) Let x = a tan θ. Then dx = a sec2 θ dθ. Therefore, 2 2 dx x a+ ∫ = 2 2 2 2 θ θ θ sec tan a d a a+ ∫ = 1θ θsecθ θ = log (sec tan ) Cd + +∫Reprint 2026-27
246 MATHEMATICS = 2 12 log 1 C x x a a + + + = 2 1log log Cx x a | a |2 + + − + = 2 log Cx x a2 + + + , where C = C1 – log |a| 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. (7) To find the integral 2 dx ax bx c+ +∫ , we write ax2 + bx + c = 2 2 2 2 2 4 b c b c b a x x a x – a a a a a + + = + + Now, put 2 b x t a + = so that dx = dt and writing 2 2 2 4 c b – k a a = ± . We find the integral reduced to the form 2 2 1 dt a t k±∫ depending upon the sign of 2 2 4 c b – a a 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. (9) To find the integral of the type 2 px q dx ax bx c + + +∫ , where p, q, a, b, c are constants, we are to find real numbers A, B such that 2 + = A ( ) + B = A (2 ) + B d px q ax bx c ax 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 2026-27
INTEGRALS 247 (10) For the evaluation of the integral of the type 2 ( )px q dx ax 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: (i) 2 16 dx x −∫ (ii) 2 2 dx x x− ∫ Solution (i) We have 2 2 2 16 4 dx dx x x – = −∫ ∫ = 4 log C 8 4 x – x 1 + + [by 7.4 (1)] (ii) Put x – 1 = t. Then dx = dt. Therefore, 2 2 dx x x− ∫ = 2 1 dt – t ∫ = 1 sin ( ) C– t + [by 7.4 (5)] = 1 sin ( – 1) C– x + Example 9 Find the following integrals : (i) 2 6 13 dx x x− +∫ (ii) 2 3 13 10 dx x x+ −∫ (iii) 2 5 2 dx x x− ∫ Solution (i) We have x2 – 6x + 13 = x2 – 6x + 32 – 32 + 13 = (x – 3)2 + 4 So, 6 13 dx x x2 − +∫ = ( )2 2 1 3 2 dx x – + ∫ Let x – 3 = t. Then dx = dt Therefore, 6 13 dx x x2 − +∫ = 1 2 2 1 tan C 2 22 –dt t t = + +∫ [by 7.4 (3)] = 11 3 tan C 2 2 – x – +Reprint 2026-27
248 MATHEMATICS (ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand, 2 3 13 10x x –+ = 2 13 10 3 3 3 x x – + = 2 2 13 17 3 6 6 x – + (completing the square) Thus 3 13 10 dx x x2 + −∫ = 2 2 1 3 13 17 6 6 dx x + − ∫ Put 13 6 x t+ = . Then dx = dt. Therefore, 3 13 10 dx x x2 + −∫ = 2 2 1 3 17 6 dt t − ∫ = 1 17 1 6log C 17 17 3 2 6 6 t – t + × × + [by 7.4 (i)] = 1 13 17 1 6 6log C 13 1717 6 6 x – x + + + + = 1 1 6 4 log C 17 6 30 x x − + + = 1 1 3 2 1 1 log C log 17 5 17 3 x x − + + + = 1 3 2 log C 17 5 x x − + + , where C = 1 1 1 C log 17 3 +Reprint 2026-27
INTEGRALS 249 (iii) We have 2 25 2 5 5 dx dx xx x x – 2 = − ∫ ∫ = 2 2 1 5 1 1 5 5 dx x – – ∫ (completing the square) Put 1 5 x – t= . Then dx = dt. Therefore, 5 2 dx x x2 − ∫ = 2 2 1 5 1 5 dt t – ∫ = 2 21 1 log C 55 t t – + + [by 7.4 (4)] = 21 1 2 log C 5 55 x x – x –+ + Example 10 Find the following integrals: (i) 2 2 6 5 x dx x x2 + + +∫ (ii) 2 3 5 4 x dx x – x + − ∫ Solution (i) Using the formula 7.4 (9), we express x + 2 = ( )2 A 2 6 5 B d x x dx + + + = A (4 6) Bx + + Equating the coefficients of x and the constant terms from both sides, we get 4A = 1 and 6A + B = 2 or A = 1 4 and B = 1 2 . Therefore, 2 2 6 5 x x x2 + + +∫ = 1 4 6 1 4 22 6 5 2 6 5 x dx dx x x x x2 2 + + + + + +∫ ∫ = 1 2 1 1 I I 4 2 + (say) ... (1)Reprint 2026-27
250 MATHEMATICS In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt Therefore, I1 = 1log C dt t t = +∫ = 2 1log | 2 6 5 | Cx x+ + + ... (2) and I2 = 2 2 1 522 6 5 3 2 dx dx x x x x = + + + + ∫ ∫ = 2 2 1 2 3 1 2 2 dx x + + ∫ Put 3 2 x t+ = , so that dx = dt, we get I2 = 2 2 1 2 1 2 dt t + ∫ = 1 2 1 tan 2 C 1 2 2 – t + × [by 7.4 (3)] = 1 2 3 tan 2 + C 2 – x + = ( )1 2tan 2 3 + C– x + ... (3) Using (2) and (3) in (1), we get ( )2 12 1 1 log 2 6 5 tan 2 3 C 4 22 6 5 –x dx x x x x x2 + = + + + + + + +∫ where, C = 1 2C C 4 2 + (ii) This integral is of the form given in 7.4 (10). Let us express x + 3 = 2 A (5 4 ) + B d – x – x dx = A (– 4 – 2x) + B Equating the coefficients of x and the constant terms from both sides, we get – 2A = 1 and – 4 A + B = 3, i.e., A = 1 2 – and B = 1Reprint 2026-27
INTEGRALS 251 Therefore, 2 3 5 4 x dx x x + − − ∫ = ( ) 2 2 4 21 2 5 4 5 4 – – x dx dx – x x x x + − − − − ∫ ∫ = 1 2 – I1 + I2 ... (1) In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt. Therefore, I1= ( ) 2 4 2 5 4 – x dx dt tx x − = − − ∫ ∫ = 12 Ct + = 2 12 5 4 C– x – x + ... (2) Now consider I2 = 2 2 5 4 9 ( 2) dx dx x x – x = − − + ∫ ∫ Put x + 2 = t, so that dx = dt. Therefore, I2 = 1 2 2 2 sin + C 33 –dt t t = − ∫ [by 7.4 (5)] = 1 2 2 sin C 3 – x + + ... (3) Substituting (2) and (3) in (1), we obtain 2 1 2 3 2 5 – 4 – + sin C 35 4 –x x – x x – x – x + + = +∫ , where 1 2 C C C 2 –= EXERCISE 7.4 Integrate the functions in Exercises 1 to 23. 1. 2 6 3 1 x x + 2. 2 1 1 4x+ 3. ( )2 1 2 1– x + 4. 2 1 9 25– x 5. 4 3 1 2 x x+ 6. 2 6 1 x x− 7. 2 1 1 x – x – 8. 2 6 6 x x a+ 9. 2 2 sec tan 4 x x +Reprint 2026-27
252 MATHEMATICS 10. 2 1 2 2x x+ + 11. 2 1 9 6 5x x+ + 12. 2 1 7 6– x – x 13. ( )( ) 1 1 2x – x – 14. 2 1 8 3x – x+ 15. ( )( ) 1 x – a x – b 16. 2 4 1 2 3 x x x – + + 17. 2 2 1 x x – + 18. 2 5 2 1 2 3 x x x − + + 19. ( )( ) 6 7 5 4 x x – x – + 20. 2 2 4 x x – x + 21. 2 2 2 3 x x x + + + 22. 2 3 2 5 x x – x + − 23. 2 5 3 4 10 x x x + + + . Choose the correct answer in Exercises 24 and 25. 24. 2 equals 2 2 dx x x+ +∫ (A) x tan–1 (x + 1) + C (B) tan–1 (x + 1) + C (C) (x + 1) tan–1x + C (D) tan–1x + C 25. 2 equals 9 4 dx x x− ∫ (A) –11 9 8 sin C 9 8 x − + (B) –11 8 9 sin C 2 9 x − + (C) –11 9 8 sin C 3 8 x − + (D) –11 9 8 sin C 2 9 x − + 7.5 Integration by Partial Fractions Recall that a rational function is defined as the ratio of two polynomials in the form P( ) Q( ) x x , where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0. If the degree of P(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 rationalReprint 2026-27
INTEGRALS 253 functions by long division process. Thus, if P( ) Q( ) x x is improper, then 1P ( )P( ) T( ) Q( ) Q( ) xx x x x = + , where T(x) is a polynomial in x and 1P ( ) Q( ) x x is a proper rational function. As we know 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 linear and quadratic factors. Assume that we want to evaluate P( ) Q( ) x dx x∫ , where P( ) Q( ) x 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 1. ( – ) ( – ) px q x a x b + , a ≠ b A B x – a x – b + 2. 2 ( – ) px q x a + ( )2 A B x – a x – a + 3. 2 ( – ) ( ) ( ) px qx r x a x – b x – c + + A B C x – a x – b x – c + + 4. 2 2 ( – ) ( ) px qx r x a x – b + + 2 A B C ( )x – a x – bx – a + + 5. 2 2 ( – ) ( ) px qx r x a x bx c + + + + 2 A B + Cx 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 2026-27
254 MATHEMATICS Example 11 Find ( 1) ( 2) dx x x+ +∫ Solution The integrand is a proper rational function. Therefore, by using the form of partial fraction [Table 7.2 (i)], we write 1 ( 1) ( 2)x x+ + = A B 1 2x x + + + ... (1) 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) ( 2)x x+ + = 1 – 1 1 2x x + + + Therefore, ( 1) ( 2) dx x x+ +∫ = 1 2 dx dx – x x+ +∫ ∫ = log 1 log 2 Cx x+ − + + = 1 log C 2 x x + + + 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. Example 12 Find 2 2 1 5 6 x dx x x + − +∫ Solution Here the integrand 2 2 1 5 6 x x – x + + is not proper rational function, so we divide x2 + 1 by x2 – 5x + 6 and find thatReprint 2026-27
INTEGRALS 255 2 2 1 5 6 x x – x + + = 2 5 5 5 5 1 1 ( 2) ( 3)5 6 x – x – x – x –x – x + = + + Let 5 5 ( 2) ( 3) x – x – x – = A B 2 3x – x – + 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 Thus, 2 2 1 5 6 x x – x + + = 5 10 1 2 3x – x – − + Therefore, 2 2 1 5 6 x dx x – x + +∫ = 1 5 10 2 3 dx dx dx x – x – − +∫ ∫ ∫ = x – 5 log | x – 2 | + 10 log | x – 3 | + C. Example 13 Find 2 3 2 ( 1) ( 3) x dx x x − + +∫ Solution The integrand is of the type as given in Table 7.2 (4). We write 2 3 2 ( 1) ( 3) x – x x+ + = 2 A B C 1 3( 1)x xx + + + ++ 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 x2, 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 A B and C 4 2 4 – – ,= = = . Thus the integrand is given by 2 3 2 ( 1) ( 3) x x x − + + = 2 11 5 11 4 ( 1) 4 ( 3)2 ( 1) – – x xx+ ++ Therefore, 2 3 2 ( 1) ( 3) x x x − + +∫ = 2 11 5 11 4 1 2 4 3( 1) dx dx dx – x xx − + ++∫ ∫ ∫ = 11 5 11 log +1 log 3 C 4 2 ( + 1) 4 x x x + − + + = 11 + 1 5 log + C 4 + 3 2 ( + 1) x x x +Reprint 2026-27
256 MATHEMATICS Example 14 Find 2 2 2 ( 1) ( 4) x dx x x+ +∫ Solution Consider 2 2 2 ( 1) ( 4) x x x+ + and put x2 = y. Then 2 2 2 ( 1) ( 4) x x x+ + = ( 1) ( 4) y y y+ + Write ( 1) ( 4) y y y+ + = A B 1 4y y + + + 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 A = 1 4 and B 3 3 − = Thus, 2 2 2 ( 1) ( 4) x x x+ + = 2 2 1 4 3 ( 1) 3 ( 4) – x x + + + Therefore, 2 2 2 ( 1) ( 4) x dx x x+ +∫ = 2 2 1 4 3 31 4 dx dx – x x + + +∫ ∫ = 1 11 4 1 tan tan C 3 3 2 2 – – x – x + × + = 1 11 2 tan tan C 3 3 2 – – x – x + + 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. Example 15 Find ( ) 2 3 sin 2 cos 5 cos 4 sin – d – – φ φ φ φ φ∫ Solution Let y = sinφ Then dy = cosφ dφReprint 2026-27
INTEGRALS 257 Therefore, ( ) 2 3 sin 2 cos 5 cos 4 sin – d – – φ φ φ φ φ∫ = 2 (3 – 2) 5 (1 ) 4 y dy – – y – y∫ = 2 3 2 4 4 y – dy y – y +∫ = ( )2 3 2 I (say) 2 y – y – =∫ Now, we write ( )2 3 2 2 y – y – = 2 A B 2 ( 2)y y + − − [by Table 7.2 (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 I = 2 3 4 [ + ] 2 ( 2) dy y – y –∫ = 2 3 + 4 2 ( 2) dy dy y – y –∫ ∫ = 1 3 log 2 4 C 2 y – y − + + − = 4 3 log sin 2 C 2 sin– φ − + + φ = 4 3 log (2 sin ) + C 2 sin − φ + − φ (since, 2 – sin φ is always positive) Example 16 Find 2 2 1 ( 2) ( 1) x x dx x x + + + +∫ Solution The integrand is a proper rational function. Decompose the rational function into partial fraction [Table 2.2(5)]. Write 2 2 1 ( 1) ( 2) x x x x + + + + = 2 A B + C 2 ( 1) x x x + + + Therefore, x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2)Reprint 2026-27
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 2 1 ( 1) ( 2) x x x x + + + + = 2 2 1 3 5 5 5 ( 2) 1 x x x + + + + = 2 3 1 2 1 5 ( 2) 5 1 x x x + + + + Therefore, 2 2 1 ( +1) ( 2) x x dx x x + + +∫ = 2 2 3 1 2 1 1 5 2 5 51 1 dx x dx dx x x x + + + + +∫ ∫ ∫ = 2 13 1 1 log 2 log 1 tan C 5 5 5 – x x x+ + + + + EXERCISE 7.5 Integrate the rational functions in Exercises 1 to 21. 1. ( 1) ( 2) x x x+ + 2. 2 1 9x – 3. 3 1 ( 1) ( 2) ( 3) x – x – x – x – 4. ( 1) ( 2) ( 3) x x – x – x – 5. 2 2 3 2 x x x+ + 6. 2 1 (1 2 ) – x x – x 7. 2 ( 1) ( – 1) x x x+ 8. 2 ( 1) ( 2) x x – x + 9. 3 2 3 5 1 x x – x x + − + 10. 2 2 3 ( 1) (2 3) x x – x − + 11. 2 5 ( 1) ( 4) x x x+ − 12. 3 2 1 1 x x x + + − 13. 2 2 (1 ) (1 )x x− + 14. 2 3 1 ( 2) x – x + 15. 4 1 1x − 16. 1 ( 1)n x x + [Hint: multiply numerator and denominator by x n – 1 and put xn = t ] 17. cos (1 – sin ) (2 – sin ) x x x [Hint : Put sin x = t]Reprint 2026-27
INTEGRALS 259 18. 2 2 2 2 ( 1) ( 2) ( 3) ( 4) x x x x + + + + 19. 2 2 2 ( 1) ( 3) x x x+ + 20. 4 1 ( 1)x x – 21. 1 ( 1)x e – [Hint : Put ex = t] Choose the correct answer in each of the Exercises 22 and 23. 22. ( 1) ( 2) x dx x x− −∫ equals (A) 2 ( 1) log C 2 x x − + − (B) 2 ( 2) log C 1 x x − + − (C) 2 1 log C 2 x x − + − (D) log ( 1) ( 2) Cx x− − + 23. 2 ( 1) dx x x +∫ equals (A) 21 log log ( +1) + C 2 x x− (B) 21 log log ( +1) + C 2 x x+ (C) 21 log log ( +1) + C 2 x x− + (D) 21 log log ( +1) + C 2 x x+ 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 uv dx = dv du u v dx dx + Integrating both sides, we get uv = dv du u dx v dx dx dx +∫ ∫ or dv u dx dx∫ = du uv – v dx dx∫ ... (1) Let u = f (x) and dv dx = g (x). Then du dx = f ′(x) and v = ( )g x dx∫Reprint 2026-27
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 cosx x dx∫ Solution Put f (x) = x (first function) and g (x) = cos x (second function). Then, integration by parts gives cosx x dx∫ = cos [ ( ) cos ] d x x dx – x x dx dx dx∫ ∫ ∫ = sin sinx x – x dx∫ = x sin x + cos x + C Suppose, we take f (x) = cos x and g (x) = x. Then cosx x dx∫ = cos [ (cos ) ] d x x dx – x x dx dx dx∫ ∫ ∫ = ( ) 2 2 cos sin 2 2 x x x x dx+ ∫ Thus, it shows that the integral cosx 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 sinx 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 xReprint 2026-27
INTEGRALS 261 as sin x + k, where k is any constant, then cosx x dx∫ = (sin ) (sin )x x k x k dx+ − +∫ = (sin ) (sinx x k x dx k dx+ − −∫ ∫ = (sin ) cos Cx x k x – kx+ − + = sin cos Cx x x+ + This shows that adding a constant to the integral of the second function is superfluous so far as the final result is concerned while applying the method of integration by parts. (iii) Usually, if any function is a power of x or a polynomial in x, then we take it as the first function. However, in cases where other function is inverse trigonometric function or logarithmic function, then we take them as first function. Example 18 Find log x dx∫ Solution To start with, we are unable to guess a function whose derivative is log x. We take log x as the first function and the constant function 1 as the second function. Then, the integral of the second function is x. Hence, (log .1)x dx∫ = log 1 [ (log ) 1 ] d x dx x dx dx dx −∫ ∫ ∫ = 1 (log ) – log Cx x x dx x x – x x ⋅ = +∫ . Example 19 Find x x e dx∫ Solution Take first function as x and second function as ex. The integral of the second function is ex. Therefore, x x e dx∫ = 1x x x e e dx− ⋅∫ = xex – ex + C. Example 20 Find 1 2 sin 1 – x x dx x− ∫ Solution Let first function be sin – 1x and second function be 2 1 x x− . First we find the integral of the second function, i.e., 2 1 x dx x− ∫ . Put t =1 – x2. Then dt = – 2x dxReprint 2026-27
262 MATHEMATICS Therefore, 2 1 x dx x− ∫ = 1 2 dt – t ∫ = 2 – 1t x= − − Hence, 1 2 sin 1 – x x dx x− ∫ = ( )1 2 2 2 1 (sin ) 1 ( 1 ) 1 – x – x – x dx x − − − − ∫ = 2 1 1 sin C– x x x− − + + = 2 1 1 sin Cx – x x− − + Alternatively, this integral can also be worked out by making substitution sin–1 x = θ and then integrating by parts. Example 21 Find sinx e x dx∫ Solution Take ex as the first function and sin x as second function. Then, integrating by parts, we have I sin ( cos ) cosx x x e x dx e – x e x dx= = +∫ ∫ = – ex cos x + I1 (say) ... (1) Taking ex and cos x as the first and second functions, respectively, in I1, we get I1 = sin sinx x e x – e x dx∫ Substituting the value of I1 in (1), we get I = – ex cos x + ex sin x – I or 2I = ex (sin x – cos x) Hence, I = sin (sin cos ) + C 2 x x e e x dx x – x=∫ Alternatively, above integral can also be determined by taking sin x as the first function and ex the second function. 7.6.1 Integral of the type [ ( ) + ( )]x e f x f x dx′∫ We have I = [ ( ) + ( )]x e f x f x dx′∫ = ( ) + ( )x x e f x dx e f x dx′∫ ∫ = 1 1I ( ) , where I = ( )x x e f x dx e f x dx′+ ∫ ∫ ... (1) Taking f (x) and ex as the first function and second function, respectively, in I1 and integrating it by parts, we have I1 = f (x) ex – ( ) Cx f x e dx′ +∫ Substituting I1 in (1), we get I = ( ) ( ) ( ) Cx x x e f x f x e dx e f x dx′ ′− + +∫ ∫ = ex f (x) + CReprint 2026-27
INTEGRALS 263 Thus, ′∫ [ ( ) ( )]x e f x + f x dx = ( ) Cx e f x + Example 22 Find (i) 1 2 1 (tan ) 1 x – e x x + +∫ dx (ii) 2 2 ( + 1) ( + 1) x x e x∫ dx Solution (i) We have I = 1 2 1 (tan ) 1 x – e x dx x + +∫ Consider f (x) = tan– 1x, then f ′(x) = 2 1 1 x+ Thus, the given integrand is of the form ex [ f (x) + f ′(x)]. Therefore, 1 2 1 I (tan ) 1 x – e x dx x = + +∫ = ex tan– 1x + C (ii) We have 2 2 ( + 1) I ( + 1) x x e x = ∫ dx 2 2 1 + 1+1) [ ] ( + 1) x x – e dx x = ∫ 2 2 2 1 2 [ ] ( + 1) ( +1) x x – e dx x x = +∫ 2 1 2 [ + ] + 1 ( +1) x x – e dx x x = ∫ Consider 1 ( ) 1 x f x x − = + , then 2 2 ( ) ( 1) f x x ′ = + Thus, the given integrand is of the form ex [f (x) + f ′(x)]. Therefore, 2 2 1 1 C 1( 1) x xx x e dx e xx + − = + ++∫ EXERCISE 7.6 Integrate the functions in Exercises 1 to 22. 1. x sin x 2. x sin 3x 3. x2 ex 4. x log x 5. x log 2x 6. x2 log x 7. x sin– 1x 8. x tan–1 x 9. x cos–1 x 10. (sin–1x)2 11. 1 2 cos 1 x x x − − 12. x sec2 x 13. tan–1x 14. x (log x)2 15. (x2 + 1) log xReprint 2026-27
264 MATHEMATICS 16. ex (sinx + cosx) 17. 2 (1 ) x x e x+ 18. 1 sin 1 cos x x e x + + 19. 2 1 1 –x e x x 20. 3 ( 3) ( 1) x x e x − − 21. e2x sin x 22. 1 2 2 sin 1 – x x + Choose the correct answer in Exercises 23 and 24. 23. 3 2 x x e dx∫ equals (A) 31 C 3 x e + (B) 21 C 3 x e + (C) 31 C 2 x e + (D) 21 C 2 x e + 24. sec (1 tan )x e x x dx+∫ equals (A) ex cos x + C (B) ex sec x + C (C) ex sin x + C (D) ex tan x + C 7.6.2 Integrals of some more types Here, we discuss some special types of standard integrals based on the technique of integration by parts : (i) 2 2 x a dx−∫ (ii) 2 2 x a dx+∫ (iii) 2 2 a x dx−∫ (i) Let 2 2 I x a dx= −∫ Taking constant function 1 as the second function and integrating by parts, we have I = 2 2 2 2 1 2 2 x x x a x dx x a − − − ∫ = 2 2 2 2 2 x x x a dx x a − − − ∫ = 2 2 2 2 2 2 2 x a a x x a dx x a − + − − − ∫Reprint 2026-27
INTEGRALS 265 = 2 2 2 2 2 2 2 dx x x a x a dx a x a − − − − − ∫ ∫ = 2 2 2 2 2 I dx x x a a x a − − − − ∫ or 2I = 2 2 2 2 2 dx x x a a x a − − − ∫ or I = ∫ 2 2 x – a dx = 2 2 2 2 2 – – log + – + C 2 2 x a x a x x a Similarly, integrating other two integrals by parts, taking constant function 1 as the second function, we get (ii) ∫ 2 2 2 2 2 2 21 + = + + log + + + C 2 2 a x a dx x x a x x a (iii) Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric substitution x = a secθ in (i), x = a tanθ in (ii) and x = a sinθ in (iii) respectively. Example 23 Find 2 2 5x x dx+ +∫ Solution Note that 2 2 5x x dx+ +∫ = 2 ( 1) 4x dx+ +∫ Put x + 1 = y, so that dx = dy. Then 2 2 5x x dx+ +∫ = 2 2 2y dy+∫ = 2 21 4 4 log 4 C 2 2 y y y y+ + + + + [using 7.6.2 (ii)] = 2 21 ( 1) 2 5 2 log 1 2 5 C 2 x x x x x x+ + + + + + + + + Example 24 Find 2 3 2x x dx− −∫ Solution Note that 2 2 3 2 4 ( 1)x x dx x dx− − = − +∫ ∫Reprint 2026-27
266 MATHEMATICS Put x + 1 = y so that dx = dy. Thus 2 3 2x x dx− −∫ = 2 4 y dy−∫ = 2 11 4 4 sin C 2 2 2 – y y y− + + [using 7.6.2 (iii)] = 2 11 1 ( 1) 3 2 2 sin C 2 2 – x x x x + + − − + + EXERCISE 7.7 Integrate the functions in Exercises 1 to 9. 1. 2 4 x− 2. 2 1 4x− 3. 2 4 6x x+ + 4. 2 4 1x x+ + 5. 2 1 4x x− − 6. 2 4 5x x+ − 7. 2 1 3x x+ − 8. 2 3x x+ 9. 2 1 9 x + Choose the correct answer in Exercises 10 to 11. 10. 2 1 x dx+∫ is equal to (A) ( )2 21 1 log 1 C 2 2 x x x x+ + + + + (B) 3 2 2 2 (1 ) C 3 x+ + (C) 3 2 2 2 (1 ) C 3 x x+ + (D) 2 2 2 21 1 log 1 C 2 2 x x x x x+ + + + + 11. 2 8 7x x dx− +∫ is equal to (A) 2 21 ( 4) 8 7 9log 4 8 7 C 2 x x x x x x− − + + − + − + + (B) 2 21 ( 4) 8 7 9log 4 8 7 C 2 x x x x x x+ − + + + + − + + (C) 2 21 ( 4) 8 7 3 2 log 4 8 7 C 2 x x x x x x− − + − − + − + + (D) 2 21 9 ( 4) 8 7 log 4 8 7 C 2 2 x x x x x x− − + − − + − + +Reprint 2026-27
INTEGRALS 267 7.7 Definite Integral In the previous sections, we have studied about the indefinite integrals and discussed few methods of finding them including integrals of some special functions. In this section, we shall study what is called definite integral of a function. The definite integral has a unique value. A definite integral is denoted by ( ) b a f x dx∫ , where a is called the lower limit of the integral and b is called the upper limit of the integral. The definite integral is introduced either as the limit of a sum or if it has an anti derivative F in the interval [a, b], then its value is the difference between the values of F at the end points, i.e., F(b) – F(a). 7.8 Fundamental Theorem of Calculus 7.8.1 Area function We have defined ( ) b a f x dx∫ as the area of the region bounded by the curve y = f (x), the ordinates x = a and x = b and x-axis. Let x be a given point in [a, b]. Then ( ) x a f x dx∫ represents the area of the light shaded region in Fig 7.1 [Here it is assumed that f (x) > 0 for x ∈ [a, b], the assertion made below is equally true for other functions as well]. The area of this shaded region depends upon the value of x. In other words, the area of this shaded region is a function of x. We denote this function of x by A(x). We call the function A(x) as Area function and is given by A (x) = ∫ ( ) x a f x dx ... (1) Based on this definition, the two basic fundamental theorems have been given. However, we only state them as their proofs are beyond the scope of this text book. 7.8.2 First fundamental theorem of integral calculus Theorem 1 Let f be a continuous function on the closed interval [a, b] and let A (x) be the area function. Then A′′′′′(x) = f (x), for all x ∈∈∈∈∈ [a, b]. Fig 7.1Reprint 2026-27
268 MATHEMATICS 7.8.3 Second fundamental theorem of integral calculus We state below an important theorem which enables us to evaluate definite integrals by making use of anti derivative. Theorem 2 Let f be continuous function defined on the closed interval [a, b] and F be an anti derivative of f. Then ∫ ( ) b a f x dx = [F( )] =b ax F (b) – F(a). Remarks (i) In words, the Theorem 2 tells us that ( ) b a f x dx∫ = (value of the anti derivative F of f at the upper limit b – value of the same anti derivative at the lower limit a). (ii) This theorem is very useful, because it gives us a method of calculating the definite integral more easily, without calculating the limit of a sum. (iii) The crucial operation in evaluating a definite integral is that of finding a function whose derivative is equal to the integrand. This strengthens the relationship between differentiation and integration. (iv) In ( ) b a f x dx∫ , the function f needs to be well defined and continuous in [a, b]. For instance, the consideration of definite integral 1 3 2 2 2 ( – 1)x x dx −∫ is erroneous since the function f expressed by f (x) = 1 2 2( – 1)x x is not defined in a portion – 1 < x < 1 of the closed interval [– 2, 3]. Steps for calculating ( ) b a f x dx∫ . (i) Find the indefinite integral ( )f x dx∫ . Let this be F(x). There is no need to keep integration constant C because if we consider F(x) + C instead of F(x), we get ( ) [F ( ) C] [F( ) C] – [F( ) C] F( ) – F( ) b b aa f x dx x b a b a= + = + + =∫ . Thus, the arbitrary constant disappears in evaluating the value of the definite integral. (ii) Evaluate F(b) – F(a) = [F ( )]b ax , which is the value of ( ) b a f x dx∫ . We now consider some examplesReprint 2026-27
INTEGRALS 269 Example 25 Evaluate the following integrals: (i) 3 2 2 x dx∫ (ii) 9 34 22(30 – ) x dx x ∫ (iii) 2 1 ( 1) ( 2) x dx x x+ +∫ (iv) 34 0 sin 2 cos 2t t dt π ∫ Solution (i) Let 3 2 2 I x dx= ∫ . Since 3 2 F ( ) 3 x x dx x= =∫ , Therefore, by the second fundamental theorem, we get I = 27 8 19 F (3) – F (2) – 3 3 3 = = (ii) Let 9 34 22 I (30 – ) x dx x = ∫ . We first find the anti derivative of the integrand. Put 3 2 3 30 – . Then – 2 x t x dx dt= = or 2 – 3 x dx dt= Thus, 3 2 22 2 – 3 (30 – ) x dt dx t x =∫ ∫ = 2 1 3 t = 3 2 2 1 F ( ) 3 (30 – ) x x = Therefore, by the second fundamental theorem of calculus, we have I = 9 3 2 4 2 1 F(9) – F(4) 3 (30 – )x = = 2 1 1 3 (30 – 27) 30 – 8 − = 2 1 1 19 3 3 22 99 − = (iii) Let 2 1 I ( 1) ( 2) x dx x x = + +∫Reprint 2026-27
270 MATHEMATICS Using partial fraction, we get –1 2 ( 1) ( 2) 1 2 x x x x x = + + + + + So ( 1) ( 2) x dx x x+ +∫ = – log 1 2log 2 F( )x x x+ + + = Therefore, by the second fundamental theorem of calculus, we have I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3] = – 3 log 3 + log 2 + 2 log 4 = 32 log 27 (iv) Let 34 0 I sin 2 cos 2t t dt π = ∫ . Consider 3 sin 2 cos 2∫ t t dt Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = 1 2 du So 3 sin 2 cos 2∫ t t dt = 31 2 u du∫ = 4 41 1 [ ] sin 2 F ( ) say 8 8 u t t= = Therefore, by the second fundamental theorem of integral calculus I = 4 41 1 F ( ) – F (0) [sin – sin 0] 4 8 2 8 π π = = EXERCISE 7.8 Evaluate the definite integrals in Exercises 1 to 20. 1. 1 1( 1)x dx − +∫ 2. 3 2 1 dx x∫ 3. 2 3 2 1 (4 – 5 6 9)x x x dx+ +∫ 4. sin 2 0 4 x dx π ∫ 5. cos 2 0 2 x dx π ∫ 6. 5 4 x e dx∫ 7. 4 0 tan x dx π ∫ 8. 4 6 cosec x dx π π∫ 9. 1 0 2 1 – dx x ∫ 10. 1 20 1 dx x+∫ 11. 3 22 1 dx x −∫Reprint 2026-27
INTEGRALS 271 12. 22 0 cos x dx π ∫ 13. 3 22 1 x dx x +∫ 14. 1 20 2 3 5 1 x dx x + +∫ 15. 21 0 x x e dx∫ 16. 2 2 21 5 4 3 x x x+ +∫ 17. 2 34 0 (2sec 2)x x dx π + +∫ 18. 2 2 0 (sin – cos ) 2 2 x x dx π ∫ 19. 2 20 6 3 4 x dx x + +∫ 20. 1 0 ( sin ) 4 x x x e dx π +∫ Choose the correct answer in Exercises 21 and 22. 21. 3 21 1 dx x+∫ equals (A) 3 π (B) 2 3 π (C) 6 π (D) 12 π 22. 2 3 20 4 9 dx x+∫ equals (A) 6 π (B) 12 π (C) 24 π (D) 4 π 7.9 Evaluation of Definite Integrals by Substitution In the previous sections, we have discussed several methods for finding the indefinite integral. One of the important methods for finding the indefinite integral is the method of substitution. To evaluate ( ) b a f x dx∫ , by substitution, the steps could be as follows: 1. Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce the given integral to a known form. 2. Integrate the new integrand with respect to the new variable without mentioning the constant of integration. 3. Resubstitute for the new variable and write the answer in terms of the original variable. 4. Find the values of answers obtained in (3) at the given limits of integral and find the difference of the values at the upper and lower limits.Reprint 2026-27
272 MATHEMATICS ANote In order to quicken this method, we can proceed as follows: After performing steps 1, and 2, there is no need of step 3. Here, the integral will be kept in the new variable itself, and the limits of the integral will accordingly be changed, so that we can perform the last step. Let us illustrate this by examples. Example 26 Evaluate 1 4 5 15 1x x dx − +∫ . Solution Put t = x5 + 1, then dt = 5x4 dx. Therefore, 4 5 5 1x x dx+∫ = t dt∫ = 3 2 2 3 t = 3 5 2 2 ( 1) 3 x + Hence, 1 4 5 15 1x x dx − +∫ = 13 5 2 – 1 2 ( 1) 3 x + = ( ) 3 3 5 52 2 2 (1 1) – (– 1) 1 3 + + = 3 3 2 2 2 2 0 3 − = 2 4 2 (2 2) 3 3 = Alternatively, first we transform the integral and then evaluate the transformed integral with new limits. Let t = x5 + 1. Then dt = 5 x4 dx. Note that, when x = – 1, t = 0 and when x = 1, t = 2 Thus, as x varies from – 1 to 1, t varies from 0 to 2 Therefore 1 4 5 15 1x x dx − +∫ = 2 0 t dt∫ = 2 3 3 3 2 2 2 0 2 2 2 – 0 3 3 t = = 2 4 2 (2 2) 3 3 = Example 27 Evaluate – 1 1 20 tan 1 x dx x+∫Reprint 2026-27
INTEGRALS 273 Solution Let t = tan – 1x, then 2 1 1 dt dx x = + . The new limits are, when x = 0, t = 0 and when x = 1, 4 t π = . Thus, as x varies from 0 to 1, t varies from 0 to 4 π . Therefore –1 1 20 tan 1 x dx x+∫ = 2 4 4 0 0 2 t t dt π π ∫ = 2 2 1 – 0 2 16 32 π π = EXERCISE 7.9 Evaluate the integrals in Exercises 1 to 8 using substitution. 1. 1 20 1 x dx x +∫ 2. 52 0 sin cos d π φ φ φ∫ 3. 1 – 1 20 2 sin 1 x dx x + ∫ 4. 2 0 2x x +∫ (Put x + 2 = t2) 5. 2 20 sin 1 cos x dx x π +∫ 6. 2 20 4 – dx x x+∫ 7. 1 21 2 5 dx x x− + +∫ 8. 2 2 21 1 1 – 2 x e dx x x ∫ Choose the correct answer in Exercises 9 and 10. 9. The value of the integral 1 3 31 1 4 3 ( )x x dx x − ∫ is (A) 6 (B) 0 (C) 3 (D) 4 10. If f (x) = 0 sin x t t dt∫ , then f ′(x) is (A) cosx + x sin x (B) x sinx (C) x cosx (D) sinx + x cosx 7.10 Some Properties of Definite Integrals We list below some important properties of definite integrals. These will be useful in evaluating the definite integrals more easily. P0 : ( ) ( ) b b a a f x dx f t dt=∫ ∫ P1 : ( ) – ( ) b a a b f x dx f x dx=∫ ∫ . In particular, ( ) 0 a a f x dx =∫ P2 : ( ) ( ) ( ) b c b a a c f x dx f x dx f x dx= +∫ ∫ ∫Reprint 2026-27
274 MATHEMATICS P3 : ( ) ( ) b b a a f x dx f a b x dx= + −∫ ∫ P4 : 0 0 ( ) ( ) a a f x dx f a x dx= −∫ ∫ (Note that P4 is a particular case of P3) P5 : 2 0 0 0 ( ) ( ) (2 ) a a a f x dx f x dx f a x dx= + −∫ ∫ ∫ P6 : 2 0 0 ( ) 2 ( ) , if (2 ) ( ) a a f x dx f x dx f a x f x= − =∫ ∫ and 0 if f (2a – x) = – f (x) P7 : (i) 0 ( ) 2 ( ) a a a f x dx f x dx − =∫ ∫ , if f is an even function, i.e., if f (– x) = f (x). (ii) ( ) 0 a a f x dx − =∫ , if f is an odd function, i.e., if f (– x) = – f (x). We give the proofs of these properties one by one. Proof of P0 It follows directly by making the substitution x = t. Proof of P1 Let F be anti derivative of f. Then, by the second fundamental theorem of calculus, we have ( ) F ( ) – F ( ) – [F ( ) F ( )] ( ) b a a b f x dx b a a b f x dx= = − = −∫ ∫ Here, we observe that, if a = b, then ( ) 0 a a f x dx =∫ . Proof of P2 Let F be anti derivative of f. Then ( ) b a f x dx∫ = F(b) – F(a) ... (1) ( ) c a f x dx∫ = F(c) – F(a) ... (2) and ( ) b c f x dx∫ = F(b) – F(c) ... (3) Adding (2) and (3), we get ( ) ( ) F( ) – F( ) ( ) c b b a c a f x dx f x dx b a f x dx+ = =∫ ∫ ∫ This proves the property P2. Proof of P3 Let t = a + b – x. Then dt = – dx. When x = a, t = b and when x = b, t = a. Therefore ( ) b a f x dx∫ = ( – ) a b f a b t dt− +∫Reprint 2026-27
INTEGRALS 275 = ( – ) b a f a b t dt+∫ (by P1) = ( – ) b a f a b x+∫ dx by P0 Proof of P4 Put t = a – x. Then dt = – dx. When x = 0, t = a and when x = a, t = 0. Now proceed as in P3. Proof of P5 Using P2, we have 2 2 0 0 ( ) ( ) ( ) a a a a f x dx f x dx f x dx= +∫ ∫ ∫ . Let t = 2a – x in the second integral on the right hand side. Then dt = – dx. When x = a, t = a and when x = 2a, t = 0. Also x = 2a – t. Therefore, the second integral becomes 2 ( ) a a f x dx∫ = 0 – (2 – ) a f a t dt∫ = 0 (2 – ) a f a t dt∫ = 0 (2 – ) a f a x dx∫ Hence 2 0 ( ) a f x dx∫ = 0 0 ( ) (2 ) a a f x dx f a x dx+ −∫ ∫ Proof of P6 Using P5, we have 2 0 0 0 ( ) ( ) (2 ) a a a f x dx f x dx f a x dx= + −∫ ∫ ∫ ... (1) Now, if f (2a – x) = f (x), then (1) becomes 2 0 ( ) a f x dx∫ = 0 0 0 ( ) ( ) 2 ( ) , a a a f x dx f x dx f x dx+ =∫ ∫ ∫ and if f (2a – x) = – f (x), then (1) becomes 2 0 ( ) a f x dx∫ = 0 0 ( ) ( ) 0 a a f x dx f x dx− =∫ ∫ Proof of P7 Using P2, we have ( ) a a f x dx −∫ = 0 0 ( ) ( ) a a f x dx f x dx − +∫ ∫ . Then Let t = – x in the first integral on the right hand side. dt = – dx. When x = – a, t = a and when x = 0, t = 0. Also x = – t. Therefore ( ) a a f x dx −∫ = 0 0 – (– ) ( ) a a f t dt f x dx+∫ ∫ = 0 0 (– ) ( ) a a f x dx f x dx+∫ ∫ (by P0) ... (1)Reprint 2026-27
276 MATHEMATICS (i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes 0 0 0 ( ) ( ) ( ) 2 ( ) a a a a a f x dx f x dx f x dx f x dx − = + =∫ ∫ ∫ ∫ (ii) If f is an odd function, then f (–x) = – f (x) and so (1) becomes 0 0 ( ) ( ) ( ) 0 a a a a f x dx f x dx f x dx − = − + =∫ ∫ ∫ Example 28 Evaluate 2 3 1 –x x dx −∫ Solution We note that x3 – x ≥ 0 on [– 1, 0] and x3 – x ≤ 0 on [0, 1] and that x3 – x ≥ 0 on [1, 2]. So by P2 we write 2 3 1 –x x dx −∫ = 0 1 23 3 3 1 0 1 ( – ) – ( – ) ( – )x x dx x x dx x x dx − + +∫ ∫ ∫ = 0 1 23 3 3 1 0 1 ( – ) ( – ) ( – )x x dx x x dx x x dx − + +∫ ∫ ∫ = 0 1 24 2 2 4 4 2 – 1 0 1 – – – 4 2 2 4 4 2 x x x x x x + + = ( ) 1 1 1 1 1 1 – – – 4 – 2 – – 4 2 2 4 4 2 + + = 1 1 1 1 1 1 – 2 4 2 2 4 4 2 + + − + − + = 3 3 11 2 2 4 4 − + = Example 29 Evaluate 24 – 4 sin x dx π π∫ Solution We observe that sin2 x is an even function. Therefore, by P7 (i), we get 24 – 4 sin x dx π π∫ = 24 0 2 sin x dx π ∫ = 4 0 (1 cos 2 ) 2 2 x dx π − ∫ = 4 0 (1 cos 2 )x dx π −∫Reprint 2026-27
INTEGRALS 277 = 4 0 1 – sin 2 2 x x π = 1 1 – sin – 0 – 4 2 2 4 2 π π π = Example 30 Evaluate 20 sin 1 cos x x dx x π +∫ Solution Let I = 20 sin 1 cos x x dx x π +∫ . Then, by P4, we have I = 20 ( ) sin ( ) 1 cos ( ) x x dx x π π − π − + π −∫ = 20 ( ) sin 1 cos x x dx x π π − +∫ = 20 sin I 1 cos x dx x π π − +∫ or 2 I = π π sin cos x dx x1 20 +∫ or I = 20 sin 2 1 cos x dx x ππ +∫ Put cos x = t so that – sin x dx = dt. When x = 0, t = 1 and when x = π, t = – 1. Therefore, (by P1) we get I = 1 21 – 2 1 dt t −π +∫ = 1 212 1 dt t− π +∫ = 1 20 1 dt t π +∫ (by P7, 2 1 since 1 t+ is even function) = 2 1– 1 – 1 1 0 tan tan 1 – tan 0 – 0 4 4 t − π π π = π = π = Example 31 Evaluate 1 5 4 1 sin cosx x dx −∫ Solution Let I = 1 5 4 1sin cosx x dx −∫ . Let f(x) = sin5 x cos4 x. Then f (– x) = sin5 (– x) cos4 (– x) = – sin5 x cos4 x = – f (x), i.e., f is an odd function. Therefore, by P7 (ii), I = 0Reprint 2026-27
278 MATHEMATICS Example 32 Evaluate 4 2 4 40 sin sin cos x dx x x π +∫ Solution Let I = 4 2 4 40 sin sin cos x dx x x π +∫ ... (1) Then, by P4 I = 4 2 0 4 4 sin ( ) 2 sin ( ) cos ( ) 2 2 x dx x x π π − π π − + − ∫ = 4 2 4 40 cos cos sin x dx x x π +∫ ... (2) Adding (1) and (2), we get 2I = 4 4 22 2 4 40 0 0 sin cos [ ] 2sin cos x x dx dx x x x ππ π + π = = = +∫ ∫ Hence I = 4 π Example 33 Evaluate 3 6 1 tan dx x π π + ∫ Solution Let I = 3 3 6 6 cos 1 tan cos sin x dxdx x x x π π π π = + + ∫ ∫ ... (1) Then, by P3 I = 3 6 cos 3 6 cos sin 3 6 3 6 x dx x x π π π π + − π π π π + − + + − ∫ = 3 6 sin sin cos x dx x x π π + ∫ ... (2) Adding (1) and (2), we get 2I = [ ]3 3 6 6 3 6 6 dx x π π π π π π π = = − =∫ . Hence I 12 π =Reprint 2026-27
INTEGRALS 279 Example 34 Evaluate 2 0 log sin x dx π ∫ Solution Let I = 2 0 log sin x dx π ∫ Then, by P4 I = 2 2 0 0 log sin log cos 2 x dx x dx π π π − = ∫ ∫ Adding the two values of I, we get 2I = ( )2 0 log sin log cosx x dx π +∫ = ( )2 0 log sin cos log 2 log 2x x dx π + −∫ (by adding and subtracting log 2) = 2 2 0 0 log sin 2 log 2x dx dx π π −∫ ∫ (Why?) Put 2x = t in the first integral. Then 2 dx = dt, when x = 0, t = 0 and when 2 x π = , t = π. Therefore 2I = 0 1 log sin log 2 2 2 t dt π π −∫ = 2 0 2 log sin log 2 2 2 t dt π π −∫ [by P6 as sin (π – t) = sin t) = 2 0 log sin log 2 2 x dx π π −∫ (by changing variable t to x) = I log 2 2 π − Hence 2 0 log sin x dx π ∫ = – log 2 2 π .Reprint 2026-27
280 MATHEMATICS EXERCISE 7.10 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 1. 22 0 cos x dx π ∫ 2. 2 0 sin sin cos x dx x x π +∫ 3. 3 2 2 3 30 2 2 sin sin cos x dx x x π + ∫ 4. 5 2 5 50 cos sin cos x dx x x π +∫ 5. 5 5 | 2 |x dx − +∫ 6. 8 2 5x dx−∫ 7. 1 0 (1 )n x x dx−∫ 8. 4 0 log (1 tan )x dx π +∫ 9. 2 0 2x x dx−∫ 10. 2 0 (2log sin log sin 2 )x x dx π −∫ 11. 22 – 2 sin x dx π π∫ 12. 0 1 sin x dx x π +∫ 13. 72 – 2 sin x dx π π∫ 14. 2 5 0 cos x dx π ∫ 15. 2 0 sin cos 1 sin cos x x dx x x π − +∫ 16. 0 log (1 cos )x dx π +∫ 17. 0 a x dx x a x+ − ∫ 18. 4 0 1x dx−∫ 19. Show that 0 0 ( ) ( ) 2 ( ) a a f x g x dx f x dx=∫ ∫ , if f and g are defined as f(x) = f (a – x) and g(x) + g(a – x) = 4 Choose the correct answer in Exercises 20 and 21. 20. The value of 3 52 2 ( cos tan 1)x x x x dx π −π + + +∫ is (A) 0 (B) 2 (C) π (D) 1 21. The value of 2 0 4 3 sin log 4 3 cos x dx x π + + ∫ is (A) 2 (B) 3 4 (C) 0 (D) –2Reprint 2026-27
INTEGRALS 281 Miscellaneous Examples Example 35 Find cos 6 1 sin 6x x dx+∫ Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx Therefore 1 2 1 cos 6 1 sin 6 6 x x dx t dt+ =∫ ∫ = 3 3 2 2 1 2 1 ( ) C = (1 sin 6 ) C 6 3 9 t x× + + + Example 36 Find 1 4 4 5 ( )x x dx x − ∫ Solution We have 1 1 4 4 4 3 5 4 1 (1 ) ( )x x xdx dx x x − − =∫ ∫ Put – 3 3 4 1 3 1 1 – , so thatx t dx dt x x − = = = Therefore 1 14 4 4 5 ( ) 1 3 x x dx t dt x − =∫ ∫ = 5 5 4 4 3 1 4 4 1 C = 1 C 3 5 15 t x × + − + Example 37 Find 4 2 ( 1) ( 1) x dx x x− +∫ Solution We have 4 2 ( 1) ( 1) x x x− + = 3 2 1 ( 1) 1 x x x x + + − + − = 2 1 ( 1) ( 1) ( 1) x x x + + − + ... (1) Now express 2 1 ( 1)( 1)x x− + = 2 A B C ( 1) ( 1) x x x + + − + ... (2)Reprint 2026-27
282 MATHEMATICS So 1 = A (x2 + 1) + (Bx + C) (x – 1) = (A + B) x2 + (C – B) x + A – C Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1, which give 1 1 A , B C – 2 2 = = = . Substituting values of A, B and C in (2), we get 2 1 ( 1) ( 1)x x− + = 2 2 1 1 1 2( 1) 2 ( 1) 2( 1) x x x x − − − + + ... (3) Again, substituting (3) in (1), we have 4 2 ( 1) ( 1) x x x x− + + = 2 2 1 1 1 ( 1) 2( 1) 2 ( 1) 2( 1) x x x x x + + − − − + + Therefore 4 2 2 – 1 2 1 1 1 log 1 – log ( 1) – tan C 2 2 4 2( 1) ( 1) x x dx x x x x x x x = + + − + + − + +∫ Example 38 Find 2 1 log (log ) (log ) x dx x + ∫ Solution Let 2 1 I log (log ) (log ) x dx x = + ∫ = 2 1 log (log ) (log ) x dx dx x +∫ ∫ In the first integral, let us take 1 as the second function. Then integrating it by parts, we get I = 2 1 log (log ) log (log ) dx x x x dx x x x − +∫ ∫ = 2 log (log ) log (log ) dx dx x x x x − +∫ ∫ ... (1) Again, consider log dx x∫ , take 1 as the second function and integrate it by parts, we have 2 1 1 – – log log (log ) dx x x dx x x xx = ∫ ∫ ... (2)Reprint 2026-27
INTEGRALS 283 Putting (2) in (1), we get 2 2 I log (log ) log (log ) (log ) x dx dx x x x x x = − − +∫ ∫ = log (log ) C log x x x x − + Example 39 Find cot tanx x dx + ∫ Solution We have I = cot tanx x dx + ∫ tan (1 cot )x x dx= +∫ Put tan x = t2, so that sec2 x dx = 2t dt or dx = 4 2 1 t dt t+ Then I = 2 4 1 2 1 (1 ) t t dt t t + + ∫ = 2 2 2 4 2 2 2 1 1 1 1 ( 1) 2 = 2 = 2 11 1 2 dt dt t t t dt t t t t t + + + + + − + ∫ ∫ ∫ Put 1 t t − = y, so that 2 1 1 t + dt = dy. Then I = ( ) – 1 – 1 2 2 1 2 2 tan C = 2 tan C 2 22 t dy y t y − = + + + ∫ = 2 – 1 – 11 tan 1 2 tan C = 2 tan C 2 2 tan t x t x − − + + Example 40 Find 4 sin 2 cos 2 9 – cos (2 ) x x dx x ∫ Solution Let 4 sin 2 cos 2 I 9 – cos 2 x x dx x = ∫Reprint 2026-27
284 MATHEMATICS Put cos2 (2x) = t so that 4 sin 2x cos 2x dx = – dt Therefore –1 1 2 2 1 1 1 1 I – – sin C sin cos 2 C 4 4 3 4 39 – dt t x t − = = + = − + ∫ Example 41 Evaluate 3 2 1 sin ( )x x dx − π∫ Solution Here f (x) = | x sin πx | = sin for 1 1 3 sin for 1 2 x x x x x x π − ≤ ≤ − π ≤ ≤ Therefore 3 2 1 | sin |x x dx − π∫ = 3 1 2 1 1 sin sinx x dx x x dx − π + − π∫ ∫ = 3 1 2 1 1 sin sinx x dx x x dx − π − π∫ ∫ Integrating both integrals on righthand side, we get 3 2 1 | sin |x x dx − π∫ = = 2 2 2 1 1 3 1 − − − = + π π ππ π Example 42 Evaluate 2 2 2 20 cos sin x dx a x b x π +∫ Solution Let I = 2 2 2 2 2 2 2 20 0 ( ) cos sin cos ( ) sin ( ) x dx x dx a x b x a x b x π π π − = + π − + π −∫ ∫ (using P4) = 2 2 2 2 2 2 2 20 0cos sin cos sin dx x dx a x b x a x b x π π π − + +∫ ∫ = 2 2 2 20 I cos sin dx a x b x π π − +∫ Thus 2I = 2 2 2 20 cos sin dx a x b x π π +∫Reprint 2026-27
INTEGRALS 285 or I = 2 2 2 2 2 2 2 2 20 0 2 2 2cos sin cos sin dx dx a x b x a x b x π ππ π = ⋅ + +∫ ∫ (using P6) = 24 2 2 2 2 2 2 2 2 0 4 cos sin cos sin ππ π π + + + ∫ ∫ dx dx a x b x a x b x = 2 2 24 2 2 2 2 2 2 0 4 sec cosec tan cot ππ π π + + + ∫ ∫ x dx x dx a b x a x b = ( ) 01 2 2 2 2 2 2 0 1 tan t cot π − = = + + ∫ ∫ dt du put x and x u a b t a u b = 1 0 –1 –1 0 1 tan – tan π π bt au ab a ab b = –1 –1 tan tan π + b a ab a b = 2 2 π ab Miscellaneous Exercise on Chapter 7 Integrate the functions in Exercises 1 to 23. 1. 3 1 x x− 2. 1 x a x b+ + + 3. 2 1 x ax x− [Hint: Put x = a t ] 4. 3 2 4 4 1 ( 1)x x + 5. 11 32 1 x x+ [Hint: 11 1 1 32 3 6 1 1 1x x x x = + + , put x = t6] 6. 2 5 ( 1) ( 9) x x x+ + 7. sin sin ( ) x x a− 8. 5 log 4 log 3 log 2 log x x x x e e e e − − 9. 2 cos 4 sin x x− 10. 8 8 2 2 sin cos 1 2sin cos x x x − − 11. 1 cos ( ) cos ( )x a x b+ + 12. 3 8 1 x x− 13. (1 ) (2 ) x x x e e e+ + 14. 2 2 1 ( 1) ( 4)x x+ + 15. cos3 x elog sinx 16. e3 logx (x4 + 1)– 1 17. f ′ (ax + b) [f (ax + b)]nReprint 2026-27
286 MATHEMATICS 18. 3 1 sin sin ( )x x + α 19. 1 1 x x − + 20. 2 sin 2 1 cos 2 xx e x + + 21. 2 2 1 ( 1) ( 2) x x x x + + + + 22. – 1 1 tan 1 x x − + 23. 2 2 4 1 log ( 1) 2 logx x x x + + − Evaluate the definite integrals in Exercises 24 to 31. 24. 2 1 sin 1 cos π π − − ∫ x x e dx x 25. 4 4 40 sin cos cos sin x x dx x x π +∫ 26. 2 2 2 20 cos cos 4 sin x dx x x π +∫ 27. 3 6 sin cos sin 2 x x dx x π π + ∫ 28. 1 0 1 dx x x+ − ∫ 29. 4 0 sin cos 9 16 sin 2 x x dx x π + +∫ 30. 12 0 sin 2 tan (sin )x x dx π − ∫ 31. 4 1 [| 1| | 2 | | 3 |]x x x dx− + − + −∫ Prove the following (Exercises 32 to 37) 32. 3 21 2 2 log 3 3( 1) dx x x = + +∫ 33. 1 0 1x x e dx =∫ 34. 1 17 4 1 cos 0x x dx − =∫ 35. 32 0 2 sin 3 x dx π =∫ 36. 34 0 2 tan 1 log 2x dx π = −∫ 37. 1 1 0 sin 1 2 x dx− π = −∫ Choose the correct answers in Exercises 38 to 40 38. x x dx e e− +∫ is equal to (A) tan–1 (ex) + C (B) tan–1 (e–x) + C (C) log (ex – e–x) + C (D) log (ex + e–x) + C 39. 2 cos 2 (sin cos ) x dx x x+∫ is equal toReprint 2026-27
INTEGRALS 287 (A) –1 C sin cosx x + + (B) log |sin cos | Cx x+ + (C) log |sin cos | Cx x− + (D) 2 1 (sin cos )+x x 40. If f (a + b – x) = f (x), then ( ) b a x f x dx∫ is equal to (A) ( ) 2 b a a b f b x dx + −∫ (B) ( ) 2 b a a b f b x dx + +∫ (C) ( ) 2 b a b a f x dx − ∫ (D) ( ) 2 b a a b f x dx + ∫ Summary ® Integration is the inverse process of differentiation. In the differential calculus, we are given a function and we have to find the derivative or differential of this function, but in the integral calculus, we are to find a function whose differential is given. Thus, integration is a process which is the inverse of differentiation. Let F( ) ( ) d x f x dx = . Then we write ( ) F ( ) Cf x dx x= +∫ . These integrals are called indefinite integrals or general integrals, C is called constant of integration. All these integrals differ by a constant. ® Some properties of indefinite integrals are as follows: 1. [ ( ) ( )] ( ) ( )f x g x dx f x dx g x dx+ = +∫ ∫ ∫ 2. For any real number k, ( ) ( )k f x dx k f x dx=∫ ∫ More generally, if f1, f2, f3, ... , fn are functions and k1, k2, ... ,kn are real numbers. Then 1 1 2 2[ ( ) ( ) ... ( )]n nk f x k f x k f x dx+ + +∫ = 1 1 2 2( ) ( ) ... ( )n nk f x dx k f x dx k f x dx+ + +∫ ∫ ∫ ® Some standard integrals (i) 1 C 1 n n x x dx n + = + +∫ , n ≠ – 1. Particularly, Cdx x= +∫Reprint 2026-27
288 MATHEMATICS (ii) cos sin Cx dx x= +∫ (iii) sin – cos Cx dx x= +∫ (iv) 2 sec tan Cx dx x= +∫ (v) 2 cosec – cot Cx dx x= +∫ (vi) sec tan sec Cx x dx x= +∫ (vii) cosec cot – cosec Cx x dx x= +∫ (viii) 1 2 sin C 1 dx x x − = + − ∫ (ix) 1 2 cos C 1 dx x x − = − + − ∫ (x) 1 2 tan C 1 dx x x − = + +∫ (xi) 1 2 cot C 1 dx x x − = − + +∫ (xii) Cx x e dx e= +∫ (xiii) C log x x a a dx a = +∫ (xiv) 1 log | | Cdx x x = +∫ ® Integration by partial fractions Recall that a rational function is ratio of two polynomials of the form P( ) Q( ) x x , where P(x) and Q (x) are polynomials in x and Q (x) ≠ 0. If degree of the polynomial P (x) is greater than the degree of the polynomial Q (x), then we may divide P (x) by Q (x) so that 1P ( )P( ) T ( ) Q( ) Q( ) xx x x x = + , where T(x) is a polynomial in x and degree of P1 (x) is less than the degree of Q(x). T(x) being polynomial can be easily integrated. 1P ( ) Q( ) x x can be integrated by expressing 1P ( ) Q( ) x x as the sum of partial fractions of the following type: 1. ( ) ( ) px q x a x b + − − = A B x a x b + − − , a ≠ b 2. 2 ( ) px q x a + − = 2 A B ( )x a x a + − −Reprint 2026-27
INTEGRALS 289 3. 2 ( ) ( ) ( ) px qx r x a x b x c + + − − − = A B C x a x b x c + + − − − 4. 2 2 ( ) ( ) px qx r x a x b + + − − = 2 A B C ( )x a x bx a + + − −− 5. 2 2 ( ) ( ) px qx r x a x bx c + + − + + = 2 A B + Cx x a x bx c + − + + where x2 + bx + c can not be factorised further. ® Integration by substitution A change in the variable of integration often reduces an integral to one of the fundamental integrals. The method in which we change the variable to some other variable is called the method of substitution. When the integrand involves some trigonometric functions, we use some well known identities to find the integrals. Using substitution technique, we obtain the following standard integrals. (i) tan log sec Cx dx x= +∫ (ii) cot log sin Cx dx x= +∫ (iii) sec log sec tan Cx dx x x= + +∫ (iv) cosec log cosec cot Cx dx x x= − +∫ ® Integrals of some special functions (i) 2 2 1 log C 2 dx x a a x ax a − = + +−∫ (ii) 2 2 1 log C 2 dx a x a a xa x + = + −−∫ (iii) 1 2 2 1 tan C dx x a ax a − = + +∫ (iv) 2 2 2 2 log C dx x x a x a = + − + − ∫ (v) 1 2 2 sin C dx x aa x − = + − ∫ (vi) 2 2 2 2 log | | C dx x x a x a = + + + + ∫ ® Integration by parts For given functions f1 and f2, we haveReprint 2026-27
290 MATHEMATICS , i.e., 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}. Care must be taken in choosing the first function and the second function. Obviously, we must take that function as the second function whose integral is well known to us. ® [ ( ) ( )] ( ) Cx x e f x f x dx e f x dx′+ = +∫ ∫ ® Some special types of integrals (i) 2 2 2 2 2 2 2 log C 2 2 x a x a dx x a x x a− = − − + − +∫ (ii) 2 2 2 2 2 2 2 log C 2 2 x a x a dx x a x x a+ = + + + + +∫ (iii) 2 2 2 2 2 1 sin C 2 2 x a x a x dx a x a − − = − + +∫ (iv) Integrals of the types 2 2 or dx dx ax bx c ax bx c+ + + + ∫ ∫ can be transformed into standard form by expressing ax2 + bx + c = 2 2 2 2 2 4 b c b c b a x x a x a a a a a + + = + + − (v) Integrals of the types 2 2 or px q dx px q dx ax bx c ax bx c + + + + + + ∫ ∫ can be transformed into standard form by expressing 2 A ( ) B A (2 ) B d px q ax bx c ax b dx + = + + + = + + , where A and B are determined by comparing coefficients on both sides. ® We have defined ( ) b a f x dx∫ as the area of the region bounded by the curve y = f (x), a ≤ x ≤ b, the x-axis and the ordinates x = a and x = b. Let x be aReprint 2026-27
INTEGRALS 291 given point in [a, b]. Then ( ) x a f x dx∫ represents the Area function A (x). This concept of area function leads to the Fundamental Theorems of Integral Calculus. ® First fundamental theorem of integral calculus Let the area function be defined by A(x) = ( ) x a f x dx∫ for all x ≥ a, where the function f is assumed to be continuous on [a, b]. Then A′ (x) = f (x) for all x ∈ [a, b]. ® Second fundamental theorem of integral calculus Let f be a continuous function of x defined on the closed interval [a, b] and let F be another function such that F( ) ( ) d x f x dx = for all x in the domain of f, then [ ]( ) F( ) C F ( ) F ( ) b b aa f x dx x b a= + = −∫ . This is called the definite integral of f over the range [a, b], where a and b are called the limits of integration, a being the lower limit and b the upper limit. — vvvvv—Reprint 2026-27