Mathematics Class 12 Part 2 Textbook PDF

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Open Knowledge Foundation Network, India : Open Education Project Help spreading the light of education. Use and share our books. It is FREE. Educate a child. Educate the economically challenged. Share and spread the word! Show your support for the cause of Openness of Knowledge. facebook: https://www.facebook.com/OKFN.India twitter: https://twitter.com/OKFNIndia Website: http://in.okfn.org/ MATHEMATICS PART II Textbook for Class XII MATHEMATICS PART II Textbook for Class XII  Foreword The National Curriculum Framework, 2005, recommends that children’s life at school must be linked to their life outside the school. This principle marks a departure from the legacy of bookish learning which continues to shape our system and causes a gap between the school, home and community. The syllabi and textbooks developed on the basis of NCF signify an attempt to implement this basic idea. They also attempt to discourage rote learning and the maintenance of sharp boundaries between different subject areas. We hope these measures will take us significantly further in the direction of a child-centred system of education outlined in the National Policy on Education (1986). The success of this effort depends on the steps that school principals and teachers will take to encourage children to reflect on their own learning and to pursue imaginative activities and questions. We must recognise that, given space, time and freedom, children generate new knowledge by engaging with the information passed on to them by adults. Treating the prescribed textbook as the sole basis of examination is one of the key reasons why other resources and sites of learning are ignored. Inculcating creativity and initiative is possible if we perceive and treat children as participants in learning, not as receivers of a fixed body of knowledge. These aims imply considerable change in school routines and mode of functioning. Flexibility in the daily time-table is as necessary as rigour in implementing the annual calendar so that the required number of teaching days are actually devoted to teaching. The methods used for teaching and evaluation will also determine how effective this textbook proves for making children’s life at school a happy experience, rather than a source of stress or boredom. Syllabus designers have tried to address the problem of curricular burden by restructuring and reorienting knowledge at different stages with greater consideration for child psychology and the time available for teaching. The textbook attempts to enhance this endeavour by giving higher priority and space to opportunities for contemplation and wondering, discussion in small groups, and activities requiring hands-on experience.  vi NCERT appreciates the hard work done by the textbook development committee responsible for this book. We wish to thank the Chairperson of the advisory group in Science and Mathematics, Professor J.V. Narlikar and the Chief Advisor for this book, Professor P.K. Jain for guiding the work of this committee. Several teachers contributed to the development of this textbook; we are grateful to their principals for making this possible. We are indebted to the institutions and organisations which have generously permitted us to draw upon their resources, material and personnel. As an organisation committed to systemic reform and continuous improvement in the quality of its products, NCERT welcomes comments and suggestions which will enable us to undertake further revision and refinement. Director New Delhi National Council of Educational 20 November 2006 Research and Training  Preface The National Council of Educational Research and Training (NCERT) had constituted 21 Focus Groups on Teaching of various subjects related to School Education, to review the National Curriculum Framework for School Education - 2000 (NCFSE - 2000) in face of new emerging challenges and transformations occurring in the fields of content and pedagogy under the contexts of National and International spectrum of school education. These Focus Groups made general and specific comments in their respective areas. Consequently, based on these reports of Focus Groups, National Curriculum Framework (NCF)-2005 was developed. NCERT designed the new syllabi and constituted Textbook Development Teams for Classes XI and XII to prepare textbooks in Mathematics under the new guidelines and new syllabi. The textbook for Class XI is already in use, which was brought in 2005. The first draft of the present book (Class XII) was prepared by the team consisting of NCERT faculty, experts and practicing teachers. The draft was refined by the development team in different meetings. This draft of the book was exposed to a group of practicing teachers teaching Mathematics at higher secondary stage in different parts of the country, in a review workshop organised by the NCERT at Delhi. The teachers made useful comments and suggestions which were incorporated in the draft textbook. The draft textbook was finalised by an editorial board constituted out of the development team. Finally, the Advisory Group in Science and Mathematics and the Monitoring Committee constituted by the HRD Ministry, Government of India have approved the draft of the textbook. In the fitness of things, let us cite some of the essential features dominating the textbook. These characteristics have reflections in almost all the chapters. The existing textbook contains thirteen main chapters and two appendices. Each chapter contains the followings :  Introduction: Highlighting the importance of the topic; connection with earlier studied topics; brief mention about the new concepts to be discussed in the chapter.  Organisation of chapter into sections comprising one or more concepts/ subconcepts.  Motivating and introducing the concepts/subconcepts. Illustrations have been provided wherever possible.  viii  Proofs/problem solving involving deductive or inductive reasoning, multiplicity of approaches wherever possible have been inducted.  Geometric viewing / visualisation of concepts have been emphasized whenever needed.  Applications of mathematical concepts have also been integrated with allied subjects like Science and Social Sciences.  Adequate and variety of examples/exercises have been given in each section.  For refocusing and strengthening the understanding and skill of problem solving and applicabilities, miscellaneous types of examples/exercises have been provided involving two or more subconcepts at a time at the end of the chapter. The scope of challenging problems to talented minority have been reflected conducive to the recommendation as reflected in NCF-2005.  For more motivational purpose, brief historical background of topics have been provided at the end of the chapter and at the beginning of each chapter, relevant quotation and photograph of eminent mathematician who have contributed significantly in the development of the topic undertaken, are also provided.  Lastly, for direct recapitulation of main concepts, formulas and results, brief summary of the chapter has also been provided. I am thankful to Professor Krishan Kumar, Director, NCERT who constituted the team and invited me to join this national endeavour for the improvement of Mathematics education. He has provided us with an enlightened perspective and a very conducive environment. This made the task of preparing the book much more enjoyable and rewarding. I express my gratitude to Professor J.V. Narlikar, Chairperson of the Advisory Group in Science and Mathematics, for his specific suggestions and advice towards the improvement of the book from time to time. I, also, thank Professor G. Ravindra, Joint Director, NCERT for his help from time to time. I express my sincere thanks to Professor Hukum Singh, Chief Coordinator and Head, DESM, Dr. V. P. Singh, Coordinator and Professor, S. K. Singh Gautam who have been helping for the success of this project academically as well as administratively. Also, I would like to place on records my appreciation and thanks to all the members of the team and the teachers who have been associated with this noble cause in one or the other form. PAWAN K. JAIN Chief Advisor  Textbook Development Committee CHAIRPERSON , ADVISORY G ROUP IN SCIENCE AND MATHEMATICS J.V. Narlikar, Emeritus Professor, Inter-University Centre for Astronomy and Astrophysics (IUCAA), Ganeshkhind, Pune University, Pune CHIEF ADVISOR P.K. Jain, Professor, Department of Mathematics, University of Delhi, Delhi CHIEF COORDINATOR Hukum Singh, Professor and Head, DESM, NCERT, New Delhi M EMBERS A.K. Rajput, Reader, RIE, Bhopal, M.P. Arun Pal Singh, Sr. Lecturer, Department of Mathematics, Dayal Singh College, University of Delhi, Delhi B.S.P. Raju, Professor, RIE Mysore, Karnataka C.R. Pradeep, Assistant Professor, Department of Mathematics, Indian Institute of Science, Bangalore, Karnataka D.R. Sharma P.G.T., Jawahar Navodaya Vidyalaya, Mungeshpur, Delhi R.P. Maurya, Reader, DESM, NCERT, New Delhi Ram Avtar, Professor (Retd.) and Consultant, DESM, NCERT, New Delhi S.K. Kaushik, Reader, Department of Mathematics, Kirori Mal College, University of Delhi, Delhi S.K.S. Gautam, Professor, DESM, NCERT, New Delhi S.S. Khare, Pro-Vice-Chancellor, NEHU, Tura Campus, Meghalaya Sangeeta Arora, P.G.T., Apeejay School, Saket, New Delhi Shailja Tewari, P.G.T., Kendriya Vidyalaya, Barkakana, Hazaribagh, Jharkhand Sunil Bajaj, Sr. Specialist, SCERT, Gurgaon, Haryana Vinayak Bujade, Lecturer, Vidarbha Buniyadi Junior College, Sakkardara Chowk, Nagpur, Maharashtra M EMBER-COORDINATOR V.P. Singh, Reader, DESM, NCERT, New Delhi ;  Acknowledgements The Council gratefully acknowledges the valuable contributions of the following participants of the Textbook Review Workshop: Jagdish Saran, Professor, Deptt. of Statistics, University of Delhi; Quddus Khan, Lecturer, Shibli National P.G. College, Azamgarh (U.P.); P.K. Tewari, Assistant Commissioner (Retd.), Kendriya Vidyalaya Sangathan; S.B. Tripathi, Lecturer, R.P.V.V., Surajmal Vihar, Delhi; O.N. Singh, Reader, RIE, Bhubaneswar, Orissa; Miss Saroj, Lecturer, Govt. Girls Senior Secondary School No.1, Roop Nagar, Delhi; P. Bhaskar Kumar, P.G.T., Jawahar Navodaya Vidyalaya, Lepakshi, Anantapur, (A.P.); Mrs. S. Kalpagam, P.G.T., K.V. NAL Campus, Bangalore; Rahul Sofat, Lecturer, Air Force Golden Jubilee Institute, Subroto Park, New Delhi; Vandita Kalra, Lecturer, Sarvodaya Kanya Vidyalaya, Vikaspuri, District Centre, New Delhi; Janardan Tripathi, Lecturer, Govt. R.H.S.S., Aizawl, Mizoram and Ms. Sushma Jaireth, Reader, DWS, NCERT, New Delhi. The Council acknowledges the efforts of Deepak Kapoor, Incharge,Computer Station; Sajjad Haider Ansari, Rakesh Kumar and Nargis Islam, D.T.P. Operators; Monika Saxena, Copy Editor; and Abhimanu Mohanty, Proof Reader. The contribution of APC-Office, administration of DESM and Publication Department is also duly acknowledged.  Contents of MATHEMATICS P ART I For Class XII Chapter 1 Relations and Functions 1 - 32 Chapter 2 Inverse Trigonometric Functions 33 - 55 Chapter 3 Matrices 56 - 102 Chapter 4 Determinants 103 - 146 Chapter 5 Continuity and Differentiability 147 - 193 Chapter 6 Application of Derivatives 194 - 246 Appendix 1: Proofs in Mathematics 247 - 255 Appendix 2: Mathematical Modelling 256 - 267 Answers 268 - 286  Contents PART II Foreword v Preface vii 7. Integrals 287 7.1 Introduction 288 7.2 Integration as an Inverse Process of Differentiation 288 7.3 Methods of Integration 300 7.4 Integrals of some Particular Functions 307 7.5 Integration by Partial Fractions 316 7.6 Integration by Parts 323 7.7 Definite Integral 331 7.8 Fundamental Theorem of Calculus 334 7.9 Evaluation of Definite Integrals by Substitution 338 7.10 Some Properties of Definite Integrals 341 8. Application of Integrals 359 8.1 Introduction 359 8.2 Area under Simple Curves 359 8.3 Area between Two Curves 366 9. Differential Equations 379 9.1 Introduction 379 9.2 Basic Concepts 379 9.3 General and Particular Solutions of a 383 Differential Equation 9.4 Formation of a Differential Equation whose 385 General Solution is given 9.5 Methods of Solving First order, First Degree 391 Differential Equations 10. Vector Algebra 424 10.1 Introduction 424 10.2 Some Basic Concepts 424 10.3 Types of Vectors 427 10.4 Addition of Vectors 429  xiv 10.5 Multiplication of a Vector by a Scalar 432 10.6 Product of Two Vectors 441 11. Three Dimensional Geometry 463 11.1 Introduction 463 11.2 Direction Cosines and Direction Ratios of a Line 463 11.3 Equation of a Line in Space 468 11.4 Angle between Two Lines 471 11.5 Shortest Distance between Two Lines 473 11.6 Plane 479 11.7 Coplanarity of Two Lines 487 11.8 Angle between Two Planes 488 11.9 Distance of a Point from a Plane 490 11.10 Angle between a Line and a Plane 492 12. Linear Programming 504 12.1 Introduction 504 12.2 Linear Programming Problem and its Mathematical Formulation 505 12.3 Different Types of Linear Programming Problems 514 13. Probability 531 13.1 Introduction 531 13.2 Conditional Probability 531 13.3 Multiplication Theorem on Probability 540 13.4 Independent Events 542 13.5 Bayes' Theorem 548 13.6 Random Variables and its Probability Distributions 557 13.7 Bernoulli Trials and Binomial Distribution 572 Answers 588  INTEGRALS 287 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. 288 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 ) = ex... (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) = x2 and (e + C) = ex 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  INTEGRALS 289 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 290 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  n x n +1  = x ; ∫ x dx = n (i) + C , n ≠ –1 dx  n + 1  n +1 Particularly, we note that d dx ( x) =1 ; ∫ dx = x + C d (ii) dx (sin x) = cos x ; ∫ cos x dx = sin x + C d (iii) dx ( – cos x ) = sin x ; ∫ sin x dx = – cos x + C d ( tan x) = sec2 x ; ∫ sec 2 (iv) x dx = tan x + C dx d ( – cot x ) = cosec 2 x ; ∫ cosec 2 (v) x dx = – cot x + C dx d (vi) dx (sec x ) = sec x tan x ; ∫ sec x tan x dx = sec x + C d (vii) dx ( – cosec x) = cosec x cot x ; ∫ cosec x cot x dx = – cosec x + C d 1 dx ( –1 (viii) dx sin x = ) ; ∫ = sin – 1 x + C 1 – x2 1–x 2 d 1 dx ( –1 (ix) dx – cos x = ;) ∫ = – cos –1 x+ C 1 – x2 1–x 2 d 1 dx (x) dx ( ) tan – 1 x = 1 + x2 ; ∫ 1 + x2 = tan –1 x+ C d 1 dx (xi) dx ( – cot – 1 x =)1 + x2 ; ∫ 1 + x2 = – cot –1 x+ C  INTEGRALS 291 d 1 dx ( –1 (xii) dx sec x = ) ; ∫x = sec– 1 x + C x x2 – 1 2 x –1 d 1 dx ( –1 (xiii) dx – cosec x = ) ; ∫x = – cosec– 1 x + C x x2 – 1 2 x –1 d x ( e ) = ex ; ∫e x (xiv) dx = ex + C dx d 1 1 (xv) dx ( log | x |) = ; x ∫ x dx = log | x | +C d  ax  ax ∫ a dx = x x (xvi)  = a ; +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 Geometrical interpretation of indefinite integral Let f (x) = 2x. Then ∫ f (x ) dx = x + C. For different values of C, we get different 2 integrals. But these integrals are very similar geometrically. Thus, y = x2 + C, where C is arbitrary constant, represents a family of integrals. By assigning different values to C, we get different members of the family. These together constitute the indefinite integral. In this case, each integral represents a parabola with its axis along y-axis. Clearly, for C = 0, we obtain y = x2 , a parabola with its vertex on the origin. The curve y = x2 + 1 for C = 1 is obtained by shifting the parabola y = x2 one unit along y-axis in positive direction. For C = – 1, y = x2 – 1 is obtained by shifting the parabola y = x2 one unit along y-axis in the negative direction. Thus, for each positive value of C, each parabola of the family has its vertex on the positive side of the y-axis and for negative values of C, each has its vertex along the negative side of the y-axis. Some of these have been shown in the Fig 7.1. Let us consider the intersection of all these parabolas by a line x = a. In the Fig 7.1, we have taken a > 0. The same is true when a < 0. If the line x = a intersects the parabolas y = x2, y = x2 + 1, y = x2 + 2, y = x2 – 1, y = x2 – 2 at P0 , P1, P2, P–1 , P–2 etc., dy respectively, then at these points equals 2a. This indicates that the tangents to the dx curves at these points are parallel. Thus, ∫ 2 x dx = x + C = FC ( x) (say), implies that 2 292 MATHEMATICS Fig 7.1 the tangents to all the curves y = F C (x), C ∈ R, at the points of intersection of the curves by the line x = a, (a ∈ R), are parallel. Further, the following equation (statement) ∫ f ( x) dx = F (x ) + C = y (say) , represents a family of curves. The different values of C will correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself. This is the geometrical interpretation of indefinite integral. 7.2.2 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.  INTEGRALS 293 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 dx ∫ f (x) dx = dx ( F ( x) + C ) 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. 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. 294 MATHEMATICS 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 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 f2 (x) +... + kn f n (x)] dx = k1 ∫ f1 ( x) dx + k2 ∫ 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.  INTEGRALS 295 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 issin 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. (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 ∫ ∫ (x 2 + 2 e – x ) dx x (i) x2 dx (ii) ∫ (x 3 + 1) dx (iii) Solution (i) We have x3 – 1 ∫ x 2 dx = ∫ x dx – ∫ x– 2 dx (by Property V) 296 MATHEMATICS  x1 + 1   x– 2 + 1  =  + C1 –  + C 2  ; C , C are constants of integration  1 +1   – 2 +1  1 2 x2 x– 1 2 = + C1 – – C 2 = x + 1 + C1 – C2 2 –1 2 x x2 1 = + + C , where C = C 1 – 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 3 = 2 + x + C = x3 + x + C +1 5 3 3 3 1 1 ∫ (x 2 + 2 e – ) dx = ∫ x 2 dx + ∫ 2 e x dx – ∫ x dx x (iii) We have 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) ∫ cos 2 x dx Solution (i) We have ∫ (sin x + cos x ) dx = ∫ sin x dx + ∫ cos x dx = – cos x + sin x + C  INTEGRALS 297 (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 ∫ cos x2 dx = ∫ 2 dx – ∫ cos x 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. 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 ∫e – x2 dx by inspection since we can not find a function whose derivative is e– x 298 MATHEMATICS (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 = 4 + C = y5 + C 4+1 5 7.2.3 Comparison between differentiation and integration 1. Both are operations on functions. 2. Both satisfy the property of linearity, i.e., d d d (i) [k1 f1 (x) + k2 f 2 (x )] = k1 f1 (x) + k2 f 2 (x ) dx dx dx (ii) ∫[ k1 f1 ( x) + k 2 f2 ( x) ] dx = k1 ∫ f1 (x) dx + k2 ∫ f2 (x) dx Here k1 and k2 are constants. 3. We have already seen that all functions are not differentiable. Similarly, all functions are not integrable. We will learn more about nondifferentiable functions and nonintegrable functions in higher classes. 4. The derivative of a function, when it exists, is a unique function. The integral of a function is not so. However, they are unique upto an additive constant, i.e., any two integrals of a function differ by a constant. 5. When a polynomial function P is differentiated, the result is a polynomial whose degree is 1 less than the degree of P. When a polynomial function P is integrated, the result is a polynomial whose degree is 1 more than that of P. 6. We can speak of the derivative at a point. We never speak of the integral at a point, we speak of the integral of a function over an interval on which the integral is defined as will be seen in Section 7.7. 7. The derivative of a function has a geometrical meaning, namely, the slope of the tangent to the corresponding curve at a point. Similarly, the indefinite integral of a function represents geometrically, a family of curves placed parallel to each other having parallel tangents at the points of intersection of the curves of the family with the lines orthogonal (perpendicular) to the axis representing the variable of integration. 8. The derivative is used for finding some physical quantities like the velocity of a moving particle, when the distance traversed at any time t is known. Similarly, the integral is used in calculating the distance traversed when the velocity at time t is known. 9. Differentiation is a process involving limits. So is integration, as will be seen in Section 7.7.  INTEGRALS 299 10. The process of differentiation and integration are inverses of each other as discussed in Section 7.2.2 (i). 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 6. ∫ (4 e + 1) dx 7. ∫ x (1 – 2 ) dx 8. ∫ ( ax + bx + c ) dx 3x 2 2 x 2  1  x 3 + 5 x2 – 4 9. ∫ (2 x + e ) dx 2 x 10. ∫  x–   dx 11. x ∫ x2 dx x3 + 3 x + 4 x3 − x2 + x – 1 12. ∫ x dx 13. ∫ x –1 dx 14. ∫ (1 – x) x dx ∫ x ( 3 x + 2 x + 3) dx ∫ (2 x – 3cos x + e ) dx 2 x 15. 16. ∫ (2 x – 3sin x + 5 x ) dx ∫ sec x (sec x + tan x) dx 2 17. 18. sec2 x 2 – 3sin x 19. ∫ 2 dx 20. ∫ dx. cosec x cos 2 x 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 + 2 x2 + C (B) x + x +C 3 3 2 3 1 3 1 2 2 3 2 1 2 (C) x + 2 x2 + 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 4 1 129 3 1 129 (A) x + − (B) x + + x3 8 x4 8 4 1 129 3 1 129 (C) x + 3 + (D) x + 4 − x 8 x 8 300 MATHEMATICS 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). 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  INTEGRALS 301 (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 2t tan 4 t sec2 t dt x Thus, ∫ x t dx = ∫ = 2 ∫ tan t sec t dt 4 2 Again, we make another substitution tan t = u so that sec2 t dt = du u5 Therefore, 2 ∫ tan 4 t sec2 t dt = 2 ∫ u 4 du = 2 +C 5 2 tan 5 t + C (since u = tan t) = 5 2 5 = tan x + C (since t = x ) 5 tan 4 x sec 2 x 2 Hence, ∫ x dx = tan5 x + C 5 Alternatively, make the substitution tan x = t 1 (iv) Derivative of tan– 1x =. Thus, we use the substitution 1+ x 2 dx tan–1 x = t so that = dt. 1 + x2 sin (tan – 1 x) Therefore , ∫ dx = ∫ sin t dt = – cos t + C = – cos (tan –1x) + C 1 + x2 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 302 MATHEMATICS 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 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 cosec2 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)  INTEGRALS 303 Solution (i) We have ∫ sin x cos 2 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 ∫ sin x cos 2 x (sin x) dx = − ∫ (1 – t 2 ) t 2 dt 2 Therefore,  t3 t5  = – ∫ ( t – t ) dt = –  –  + C 2 4 3 5 1 1 = – cos3 x + cos5 x + C 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 304 MATHEMATICS 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) 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 + C 2 = log cos x + sin x + C 2 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 x2 1 x 9. (4 x + 2) x 2 + x + 1 10. 11. , x> 0 x– x x+ 4 3 1 5 x2 1 12. (x – 1) 3 x 13. 14. , x > 0, m ≠ 1 (2 + 3 x3 ) 3 x (log x) m x x 15. 16. e2 x + 3 17. 2 9 – 4x2 ex  INTEGRALS 305 –1 etan x e2 x – 1 e2 x – e– 2 x 18. 19. 20. 1 + x2 e 2x + 1 e2 x + e– 2 x sin – 1 x 21. tan2 (2x – 3) 22. sec2 (7 – 4x) 23. 1 – x2 2cos x – 3sin x 1 cos x 24. 25. 26. 6cos x + 4sin x 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. 1 + cos x 31. (1 + cos x ) 2 32. 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. ( x3 sin tan – 1 x4 ) 8 x 1+ x Choose the correct answer in Exercises 38 and 39. 10 x 9 + 10 x log e10 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. ∫ sin2 x cos 2 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 306 MATHEMATICS Solution (i) Recall the identity cos 2x = 2 cos2 x – 1, which gives 1 + cos 2x cos2 x = 2 1 1 1 ∫ cos ∫ (1 + cos 2x) dx = ∫ dx + ∫ cos 2 x dx 2 Therefore, x dx = 2 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 1 sin 5 x dx ∫ sin x dx ∫ sin 2 x cos 3 x dx 2 ∫ Then =  1  1  =  – 5 cos 5 x + cos x  + C 2   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 3 x sin3 x = 4 3 1 ∫ sin ∫ sin x dx – ∫ sin 3 x 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 2 x) sin x dx 3 Alternatively, Put cos x = t so that – sin x dx = dt t3 ∫ sin x dx = − ∫ 1 – t 2 dt = – ∫ dt + ∫ t 2 dt = – t + ( ) 3 Therefore, +C 3 1 cos 3x + C = – cos x + 3 Remark It can be shown using trigonometric identities that both answers are equivalent.  INTEGRALS 307 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 2x – cos 2α cos x – sin x 13. 14. 15. tan3 2x sec 2x cos x – cos α 1 + sin 2 x sin 3 x + cos3 x cos 2 x + 2sin 2 x 16. tan4 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 cos 3 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 − cos2 x 23. ∫ sin 2 x cos2 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. ∫ cos 2 (ex 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 2a x +a 308 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 ∫ 2 2 (4) = log x + x – a + C 2 2 x –a dx x (5) ∫ a –x2 2 = 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 22 = x –a ( 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 = 2 a  ∫ 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  =   = + 2a  a − x a + x   2 2 a –x 2 a  ( a + x) ( a − x)   INTEGRALS 309 dx 1  dx dx  Therefore, ∫ a 2 – 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 a sec2 θ d θ Therefore, ∫ x2 + a 2 = ∫ a 2 tan 2 θ + a 2 1 1 1 –1 x = ∫ dθ = θ + C = tan +C a a a a (4) Let x = a sec θ. Then dx = a sec θ tan θ d θ. dx a secθ tanθ d θ Therefore, ∫ x2 − a 2 = ∫ a 2 sec2θ − a 2 = ∫ secθ dθ = log secθ + tanθ + C1 x x2 = log + – 1 + C1 a a2 2 2 = log x + x – a − log a + C1 2 2 = log x + x – a + C , where C = C1 – log |a| (5) Let x = a sinθ. Then dx = a cosθ dθ. dx a cosθ d θ Therefore, ∫ a 2 − x2 = ∫ a 2 – a 2 sin 2θ x = ∫ d θ = θ + C = sin –1 +C a (6) Let x = a tan θ. Then dx = a sec2 θ dθ. dx a sec2 θ dθ Therefore, ∫ x2 + a 2 = ∫ a 2 tan 2 θ + a 2 = ∫ secθ dθ = log (secθ + tan θ) + C1 310 MATHEMATICS x x2 = log + + 1 + C1 a a2 2 2 = log x + x + a − log |a | + C1 2 2 = log x + x + a + C , 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. dx (7) To find the integral ∫ ax2 + bx + c , we write  2 b c  b   c b2 2  ax + bx + c = a  x + x +  = a  x + 2  + –   a a  2 a   a 4a 2   b c b2 2 Now, put x + = t so that dx = dt and writing – 2 = ± k. We find the 2a a 4a 1 dt  c b2  integral reduced to the form ∫ a t ± k2 2 depending upon the sign of  – 2  a 4a  and hence can be evaluated. dx (8) To find the integral of the type ∫ 2 ax + bx + c , proceeding as in (7), we obtain the integral using the standard formulae. px + q (9) To find the integral of the type ∫ ax2 + 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 (ax2 + 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.  INTEGRALS 311 ( px + q) dx (10) For the evaluation of the integral of the type ax2 + 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) ∫ x2 − 16 (ii) ∫ 2 x − x2 Solution dx dx 1 x–4 (i) We have ∫ x2 − 16 = ∫ x2 – 4 2 = 8 log x+ 4 + C [by 7.4 (1)] dx dx (ii) ∫ 2 x − x2 =∫ 2 1 – ( x – 1) Put x – 1 = t. Then dx = dt. dx dt Therefore, ∫ 2x − x 2 = ∫ 1– t 2 –1 = sin (t ) + C [by 7.4 (5)] –1 = sin (x – 1) + C Example 9 Find the following integrals : dx dx dx (i) ∫ x2 − 6 x + 13 (ii) ∫ 3 x2 + 13 x − 10 (iii) ∫ 5x 2 − 2x 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 + 2 2 = 2 tan –1 Therefore, +C [by 7.4 (3)] 2 1 x–3 = tan– 1 +C 2 2 312 MATHEMATICS (ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand,  2 13 x 10  3x 2 + 13 x – 10 = 3  x + –   3 3  13   17   2 2 = 3   x +     –   6   6   (completing the square) dx 1 dx Thus = ∫ ∫ 3 x2 + 13 x − 10 3  13  2  17  2 x+  −   6 6  13 Put x + = t. Then dx = dt. 6 dx 1 dt Therefore, ∫ 3 x2 + 13 x − 10 = 3 ∫ 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  INTEGRALS 313 dx dx (iii) We have ∫ 5x 2 − 2x =∫  2x  5 x 2 –   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, ∫ 2 = 5 ∫ 2 5x − 2x  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 − 4 x + x2 dx Solution (i) Using the formula 7.4 (9), we express d x+2= A 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, ∫ 2 x 2 + 6 x + 5 = 4 ∫ 2 x2 + 6 x + 5 dx + 2 ∫ 2 x2 + 6 x + 5 1 1 = I1 + I2 (say)... (1) 4 2 314 MATHEMATICS In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt dt Therefore, I1 = ∫ t = log t + C1 2 = log | 2 x + 6 x + 5 | + C1... (2) dx 1 dx and I2 = ∫ 2x 2 + 6x + 5 = 2 ∫ 5 x2 + 3 x + 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 2 t + C 2 [by 7.4 (3)] 2 t +  2×  2 2 –1  3 –1 = tan 2  x +  + C 2 = tan ( 2 x + 3 ) + C 2... (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  INTEGRALS 315 x+3 1 (– 4 – 2 x) dx + dx Therefore, ∫ 2 dx = – 2 ∫ 5 − 4x − x 2 ∫ 5 − 4 x − x2 5 − 4x − x 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 – 4x – 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 ∫ –1 Therefore, I2 = = sin + C2 [by 7.4 (5)] 2 3 −t 2 3 –1 x+ 2 = sin + C2... (3) 3 Substituting (2) and (3) in (1), we obtain x+3 x+ 2 C ∫ 5 – 4x – x 2 = – 5 – 4x – x2 + sin – 1 3 + C , where C = C2 – 1 2 EXERCISE 7.4 Integrate the functions in Exercises 1 to 23. 3 x2 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 + 2 x4 1 − x6 x –1 x2 sec2 x 7. 8. 9. x2 – 1 x6 + a 6 tan 2 x + 4 316 MATHEMATICS 1 1 1 10. 11. 2 12. x2 + 2 x + 2 9 x + 6x + 5 7 – 6 x – x2 1 1 1 13. 14. 15. ( x – 1)( x – 2 ) 8 + 3x – x2 ( x – a )( x – b ) 4 x +1 x+2 5x − 2 16. 17. 18. 2 2x + x – 3 2 x –1 1 + 2 x + 3 x2 6x + 7 x+2 x+2 19. 20. 21. ( x – 5)( x – 4) 4x – x 2 2 x + 2x + 3 x+3 5x + 3 22. 2 23.. x – 2x − 5 x2 + 4 x + 10 Choose the correct answer in Exercises 24 and 25. dx 24. ∫ x2 + 2 x + 2 equals (A) x tan–1 (x + 1) + C (B) tan–1 (x + 1) + C (C) (x + 1) tan–1 x + C (D) tan–1 x + C dx 25. ∫ 9 x − 4 x2 equals 1  9x − 8  1  8x − 9  (A) sin –1  +C (B) sin –1  +C 9  8  2  9  1  9x − 8  1  9x − 8  sin –1  sin –1  +C (C)  +C (D) 2  9  3  8  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  INTEGRALS 317 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 ( x – a)2 (x – a ) 2 px2 + 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 px2 + qx + r A Bx + C 5. + 2 , (x – a ) ( x2 + bx + c) x – a x + bx + c where x2 + bx + c cannot be factorised furthe