Integrals of Functions of a Real Variable I. The Indefinite Integral x22 xx xe x) x (ln xdx) e x () x (ln ] x ln e ) x 1 ( ) e x ( 2 [ A Problem-Solving Approach with Computer Algebra Support A Learning Tool A Reference of Methods and Techniques A Source of Modern Algorithms (Hermite-Horowitz, Rothstein-Trager, Risch, Adamchik-Marichev) Solomon M. Antoniou SKEMSYS Scientific Knowledge Engineering and Management Systems 2 In memory of my grandfather Solomon Chattab In memory of my grandmother Sarah Chambar 3 Preface This book is an attempt to systematize the methods of integration of functions of a real variable. The book is mainly addressed to the student who wishes an in depth knowledge of integration techniques. We have tried to include the following: Good Taxonomy Explicit Calculations Easy to remember tricks Modern Information Technology Tools, like Computer Algebra Systems (Symbolic Languages) Modern Algorithms In a separate report, we are going to include some other issues concerning mainly Learning Lab Metacognitive Tools like Concept Maps and Vee Diagrams Questions based on Blooms Taxonomy and Gardners Multiple Intelligence The book can be used both as a learning tool and as a reference of methods, techniques and formulas. Solomon M. Antoniou Corinth, May 2012 solomon_antoniou@yahoo.com 4 Bibliography [1] T. Apostol Calculus Vol.I, Blaisdel Publishing Company 1962 [2] T. Apostol Mathematical Analysis Addison-Wesley 1957 [3] F. Ayres and E. Mendelson Theory and Problems of Differential and Integral Calculus Third Edition, Schaums Outline Series [4] R. Bartle and D. Sherbert Introduction to Real Analysis Wiley 1982 [5] G. N. Berman A Problem Book in Mathematical Analysis Mir Publishers 1980 [6] R. Ellis and D. Gulick Calculus with Analytic Geometry Saunders, Fifth Edition 1994 [7] H. Flanders Calculus W. H. Freeman, 1985 [8] I.S. Gradshteyn and I.M. Ryzhik: Table of Integrals, Series and Products Academic Press, 1996. Sixth Edition [9] G. H. Hardy: A Course of Pure Mathematics Cambridge University Press, 10th Edition, 1952 [10] G. H. Hardy: The Integration of Functions of a Single Variable Second Edition, Cambridge University Press, 1916 [11] A. Jeffrey and H-H. Dai: Handbook of Mathematical Formulas and Integrals Elsevier 2008. Fourth Edition 5 [12] K. Kuratowski: Introduction to Calculus Second Edition, Pergamon 1969 [13] N. Piskunov: Differential and Integral Calculus Mir Publishers, Moscow 1969 [14] W. Rudin Principles of Mathematical Analysis Second Edition, Mc-Graw Hill 1964 [15] V. I. Smirnov A Course of Higher Mathematics Vol. I, Vol. III, Pergamon 1964 [16] M. Spivac Calculus McGraw Hill 1968 [17] A.M. - . . 1988 [18] G. Thomas and R. Finney Calculus Ninth Edition, Addison-Wesley 1996 6 Web Sites of the Computer Algebra Systems AXIOM www.open-axiom.org MATLAB www.mathworks.com Mathematica www.wolfram.com Maple www.maplesoft.com Reduce reduce-algebra.com SAGE www.sagemath.org Macsyma www.symbolics-dks.com Maxima maxima.sourceforge.net Scientific WorkPlace www.mackichan.com 7 Volume I The Indefinite Integral Contents 1. Definition and Properties of the Indefinite Integral 8 2. Elementary Examples of Integration . 12 3. The Method of Substitution 47 4. Integration by Parts . 80 5. Integrals of Polynomial Functions 121 6. Integrals of Rational Functions .. 138 7. Integrals of Exponential Functions 389 8. Integrals of Logarithmic Functions 430 9. Integrals of Trigonometric Functions 474 10. Integrals of Hyperbolic Functions . 640 11. Integrals of Irrational Functions 653 12. Integrals of Inverse Trigonometric Functions . 875 13. Integrals of Inverse Hyperbolic Functions .. 891 14. Elliptic Integrals 900 15. The Risch Algorithm .. 918 16. The Adamchik-Marichev Algorithm . 935 1 Definition and Properties of the Indefinite Integral 1.1. The Primitive of a Real Function. Let R I : f be a function defined on an interval I. A function F differentiable in I such that I x ) x ( f ) x ( F e = ' is called a primitive of the function f in I. Example 1. The function 4 x 2 x 3 x ) x ( F3 4 + = is the primitive of the function 2 x 9 x 4 ) x ( f2 3 + = , since ) x ( f 2 x 9 x 4 ) x ( F2 3 = + = ' R xe Example 2. The function 6 x e 4 x 3 sin32) x ( F2 x + + = is the primitive of the function x 2 e 4 x 3 cos 2 ) x ( fx + = , since ) x ( f x 2 e 4 x 3 cos 2 ) x ( Fx = + = ' R xe It is obvious that if F is a primitive of the function f, i.e.. Chapter 1-Definition and properties of the indefinite integral 9 I x ) x ( f ) x ( F e = ' , then the function C ) x ( F + will also be a primitive of the function ) x ( f , since I x ) x ( f ) x ( F ) C ) x ( F ( e = ' = ' + Therefore if F is a primitive of f, then the set of primitives of f will be C F+ . We have the following basic Theorem: Theorem 1. - If ) x ( F is a primitive of the function ) x ( f on I then C ) x ( F + is a primitive of ) x ( f on I. - Inversely, every primitive of the function ) x ( f has the form C ) x ( F + Proof. - If ) x ( F is a primitive of the function ) x ( f on I then C ) x ( F + is also a primitive of the function ) x ( f R Ce , since I x ) x ( f ) x ( F ) C ) x ( F ( e = ' = ' + - If, other than the function ) x ( F there is another primitive ) x ( G of the function ) x ( f , then we would have I x ) x ( f ) x ( G and ) x ( f ) x ( F e = ' = ' from which it follows that ) x ( G ) x ( F ' = ' . This means that the functions ) x ( F and ) x ( G will differ only by a constant: C ) x ( G ) x ( F = from which we get that C ) x ( G ) x ( F + = 1.2. The Indefinite Integral. Indefinite integral of a function ) x ( f on an interval I is the set of all the primitive functions of f, denoted by }dx ) x ( f . We would therefore have C ) x ( F dx ) x ( f + =} The function ) x ( f appearing under the integral sign, is called the integrand. Chapter 1-Definition and properties of the indefinite integral 10 We also have the equivalence R C and I x ) x ( f ) x ( F C ) x ( F dx ) x ( f e e = ' + =} It is also obvious that R C and I x C ) x ( f dx ) x ( f e e + = '} 1.3. Properties of the Indefinite Integral The indefinite integral admits the following properties: [1] } } = dx ) x ( f dx ) x ( f , for every non-zero real constant . [2] } } } = dx ) x ( g dx ) x ( f dx )] x ( g ) x ( f [ The above two properties can be written as a single property = + + +}dx )] x ( f ) x ( f ) x ( f [n n 2 2 1 1 . } } } + + + = dx ) x ( f dx ) x ( f dx ) x ( f n n 2 2 1 1 for any non-zero real constants n 2 1 , , , . We list below a Table of Integrals, which the student should know (preferably to be remembered). These are the absolutely necessary integrals one must know. In most of the cases we shall examine below, every integral to be evaluated, using some appropriate techniques, can in principle be converted into a combination of integrals appearing in the Table. The proof of the relations appearing in the Table is based on the equivalence ) x ( f ) x ( F C ) x ( F dx ) x ( f = ' + =} For example C x sin dx x cos + =} since x cos ) C x (sin = ' + Chapter 1-Definition and properties of the indefinite integral 11 1.4 Table of Elementary Integrals 1. C x dx + =} 2. C1 nxdx x1 nn ++=} +, 1 n = 3. C | x | lnxdx+ =} 4. C e dx ex x + =} 5. C x cos dx x sin + =} 6. C x sin dx x cos + =} 7. } + = C x cotx sindx2 8. } + = C x tanx cosdx2 9. } + =+C x arctanx 1dx2 10. } + =C x arcsinx 1dx2 11. } + + =C a x x lna xdx2 22 2 2 Elementary Examples of Integration Learning Objectives In this Section the student will learn how to evaluate elementary integrals. In this case the integral transforms into a combination of integrals and then each one is evaluated using the Table. Trying to evaluate integrals we always use methods which convert them into one of the forms listed in the Table of Integrals. The procedure will become clearer through the following examples. The examples we consider make use of the Properties and the Table. More complicated cases require special methods and techniques, like substitution, integration by parts, partial fraction decomposition or even advanced algorithms. In next sections we introduce a taxonomy of cases and consider a number of specialized methods. Chapter 2-Elementary examples of integration 13 2.1 Case I. Integrals of the form }dx a . Example 1. Evaluate the integral }= dx 3 I Solution. C x 3 dx 3 dx 3 I + = = = } } General Case. The evaluation of integrals of the form }dx a may proceed as follows: C x a dx a dx a + = = } } 2.2 Case II. Integrals of the form }dx x an ( R a e , 1 n > ) Example 1. Evaluate the integral }= dx x 5 I3 Solution. In our case we recognize that 5 is a constant factor and therefore can be factored out of the integral, using the property } } = dx ) x ( f dx ) x ( f . We thus obtain } } = = dx x 5 dx x 5 I3 3 Next, evaluate the integral }dx x3, which is of the form }dx xn and appears as the second case in the Table of Integrals. We thus have C x45C4x5 C1 3x5 dx x 5 dx x 5 I44 1 33 3 + = + = ++= = = +} } In practice we usually write C x45C4x5 dx x 5 dx x 5 I443 3 + = + = = = } } General Case. The evaluation of integrals of the form }dx x an ( R a e , 1 n > ) may proceed as follows: Chapter 2-Elementary examples of integration 14 C x1 naC1 nxa dx x a dx x a1 n1 nn n ++= ++= = ++} }, e a , 1 n > 2.3 Case III. Integrals of the form dxxa} ( R a e ) Example 1. Evaluate the integral dxx2} Solution. C | x | ln 2 dxx12 dxx2+ = = } } General Case. The evaluation of integrals of the form dxxa} may proceed as follows: C | x | ln a dxx1a dxxa+ = = } } 2.4 Case IV. Integrals of Polynomial Functions These are integrals of the form }= dx ) x ( P I where ) x ( P is an n-th degree polynomial. In this case we convert the integral into a sum of integrals of the form dx x an} ( 0 n > ) which are evaluated according to 1 nxa dx x a1 nn+= +} ( 1 n > ) and x a dx a =} Example 1. Evaluate the integral } + = dx ) 2 x x 3 x 2 ( I2 3 Solution. In this case, in order to evaluate the integral, we have to write it in the form of sum of four integrals, according to the linearity property. We thus get } + = dx ) 2 x x 3 x 2 ( I2 3 } } } } + = dx 2 dx x dx x 3 dx x 22 3 Chapter 2-Elementary examples of integration 15 C x 22x3x34x22 3 4+ + = C x 2 x21x x212 3 4 + + = A remark is in order. Instead of using four constants for each one of the particular integrals, we use just a single constant C. Example 2. Evaluate the integral } = dx ) 2 x 3 ( I2 Solution. Since 4 x 12 x 9 ) 2 x 3 (2 2 + = , we get = + = = } }dx ) 4 x 12 x 9 ( dx ) 2 x 3 ( I2 2 = + = } } }dx 4 dx x 12 dx x 92 C x 4 x 6 x 3 C x 42x123x92 32 3+ + = + + = 2.5 Case V. Integrals of the form dxxan} ( R a e , 2 n > ) Example 3. Evaluate the integral }= dxx2I3 Solution. In this case we remark that the integral can be expressed as } = dx x 2 I3 making use of the known property nnxx1 = . Therefore using the second case from the Table of Integrals, we get = ++ = = = + } }C1 3x2 dx x 2 dx x 2 I1 33 3 Cx1C x C2x2222+ = + = += Example 1. Evaluate the integral dxx54} Chapter 2-Elementary examples of integration 16 Solution. = ++ = = = + } } }C1 4x5 dx x 5 dxx15 dxx51 444 4 Cx135C3x533+ = + = General Case. The evaluation of integrals of the form dxxan} ( R a e , 2 n > ) may proceed as follows: = ++ = = = + } } }C1 nxa dx x a dxx1a dxxa1 nnn n Cx11 naC x1 na1 n1 n + = + = + , 2 n > 2.6 Case VI. Integrals of Rational Functions Example 1. Evaluate the integral dxx3 x 2 x 5I2} + = Solution. On performing division of the numerator by the denominator, we have x32 x 5x3xx 2xx 5x3 x 2 x 52 2+ = + =+ Therefore the integral becomes } } |.|

\| + =+ = dxx32 x 5 dxx3 x 2 x 5I2 which can be further split into three particular integrals, each evaluated separately =|.|

\| + =+ = } }dxx32 x 5 dxx3 x 2 x 5I2 = + = } } }dxx13 dx 2 dx x 5 C | x | ln 3 x 22x52+ + = Chapter 2-Elementary examples of integration 17 C | x | ln 3 x 2 x252 + + = Example 2. Evaluate the integral dx1 x3 xI22} ++= Solution. We remark that the integrand can be written as 1 x211 x21 x1 x1 x2 ) 1 x (1 x3 x2 2 222222++ =++++=+ + +=++. Therefore the integral becomes =++ =||.|

\|++ =++= } } } }dx1 x12 dx dx1 x21 dx1 x3 xI2 2 22 C x arctan 2 x + + = The integrals in cases 2.7-2.16 and 2.20 can also be evaluated using the method of substitution, to be explained in next Sections. 2.7 Case VII. Integrals of the form dx x a} ( R a e ) In this case we use the known property 21x x = for converting a square root into power with a rational exponent. Therefore we have = ++= = = +} } }C121xa dx x a dx x a dx x a12121 C x x3a 2C x3a 2C23xa2323+ = + = + = Chapter 2-Elementary examples of integration 18 2.8 Case VIII. Integrals of the form: dx x x an} ( R a e , 1 n > ) The evaluation of integrals of the form dx x x an} ( R a e , 1 n > ) may proceed as follows: = = = } } }dx x x a dx x x a dx x x a21n n n = ++= ++ += = + + + ++}C23 n 2xa C121nxa dx x a211 n 121n21n C x x3 n 2a 2C23 n 2xa1 n211 n++= ++= ++ + Example. Evaluate the integral dx x x 22} Solution. = = = } } }dx x x 2 dx x x 2 dx x x 2212 2 2 = + = ++ += = + + ++}C27x2 C1212x2 dx x 2213 1212212 C x x743 + = 2.9 Case IX. Integrals of the form: dxxa} ( R a e ) The evaluation of integrals of the form dxxa} ( R a e ) may proceed as follows: = ++ = = = + } } }C121xa dx x a dxx1a dxxa12121 Chapter 2-Elementary examples of integration 19 C x a 2 C21xa21+ = + = 2.10 Case X. Integrals of the form dxxxan} ( R a e , 1 n > ) The evaluation of integrals of the form dxxxan} ( R a e , 1 n > ) may proceed as follows: = = = } } } dx x a dxxxa dxxxa21n21n n C x x1 n 2a 2C21 n 2xa C121nxan21n 121n++= ++= ++ = + + Example. Evaluate the integral dxxx) 3 (3} Solution. = = = } } } dx x 3 dxxx3 dxxx) 3 (213213 3 C x x76C27x3 C1213x33213 1213+ = + = ++ = + + 2.11 Case XI. Integrals of the form dxxxan} ( R a e , 1 n > ) The evaluation of integrals of the form dxxxan} ( R a e , 1 n > ) may proceed as follows: Chapter 2-Elementary examples of integration 20 = = = } } } dx x a dxxxa dxxxan21n21n Cxxn 2 3a 2C2n 2 3xa C1 n21xa1 n211 n 1 n21+ = += ++ = + + + Example. Evaluate the integral dxxx52} Solution. = = = } } } dx x 5 dxxx5 dxxx52212212 Cx10C21x5 C1 221x5211 221+ = += ++ = + 2.12 Case XII. Integrals of the form dx x a Inm}= Example 5. Evaluate the Integral dx x I32}= Solution. Using the known property 3232x x = , the integral takes the form dx x dx x I3232} } = = Therefore, using the second case from the Table of Integrals, we get = + = ++= = = +} }C35xC132xdx x dx x I351323232 C x x53C x533235+ = + = Chapter 2-Elementary examples of integration 21 General Case. The evaluation of integrals of the form dx x anm} may proceed as follows: Since nmnmx x = , we have = ++= = = +} }C1nmxa dx x a dx x a I1nmnmnm C x xn mn aCnn mxanm1nm++= ++= + 2.13 Case XIII. Integrals of the form dx x x a Inm p} = General Case. The evaluation of integrals of the form dx x x anm p} may proceed as follows: = = = } }dx x x a dx x x a Inmpnm p = ++ += ++ += = + + + ++}Cnn m pnxa C1nmpxa dx x anm1 p 1nmpnmp C x xn m n pn aC x xn m pnn a nm 1 pnm1 p + + += + + += + + Example. Evaluate the integral dx x x 5 I32 3} = Solution. Since 3232x x = , we have 31132332332 3x x x x x x = = = +. Therefore Chapter 2-Elementary examples of integration 22 = ++= = = +} }C1311x5 dx x 5 dx x x 5 I131131132 3 = + = + = + = +C x x1415C x1415C314x5324324 314 C x x1415 32 4 + = 2.14 Case XIV. Integrals of the form dxxxa Inmp}= General Case. The evaluation of integrals of the form dxxxa Inmp}= may proceed as follows: = = = } } dx x x a dxxxa Inmpnmp = + += + += = + +}Cnm n pnxa Cnm1 pxa dx x anm1 pnm1 pnmp Cxxm n pnn aC x xm n pnn anm1 pnm1 p + += + += ++ Example. Evaluate the integral dxxx) 3 ( I234} = Solution. We have Chapter 2-Elementary examples of integration 23 310324324234x xxxxx= = = Therefore = ++ = = = +} }C1310x3 dx x 3 dxxx) 3 ( I1310310234 = + = + = + = +C x x139C x139C313x3314314 313 C x x139 34 + = 2.15 Case XV. Integrals of the form dxxaInm}= General Case. The evaluation of integrals of the form dxxanm} may proceed as follows: = ++ = = = + } }C1nmxa dx x a dxxaI1nmnmnm = + + = ++ = + C x xn mn aCnn mxanm1nm Cxxn mn anm ++ = Example. Evaluate the Integral dxx5I32}= Chapter 2-Elementary examples of integration 24 Solution. Using the known properties 3232 32xx1x1 = = , the integral takes the form dx x 5 dxx5I3232 } } = = . Therefore, using the second case from the Table of Integrals, we get C132x5 dx x 5 dxx5I1323232 ++ = = = + } } C x 15 C31x5331+ = + = 2.16 Case XVI. Integrals of the form dxxxa Ipnm}= General Case. The evaluation of integrals of the form dxxxa Ipnm}= may proceed as follows: = = = } } dx x x a dxxxa Ipnmpnm = ++ = ++ = = + + }Cnn pn mxa C1 pnmxa dx x anmp 1 1 pnmpnm C x xn pn mn aC x xn pn mn a nm p 1nmp 1 + + = + + = Chapter 2-Elementary examples of integration 25 Example. Evaluate the integral dxxx7 I332}= Solution. We have 37332332332x xxxxx = = = Therefore = ++ = = = + } }C137x7 dx x 7 dxxx7 I13737332 = + = += ++ = + C x x421C34x7 C137x731134137 Cx x1421Cx x1421331 + = + = 2.17 Case XVII. Integrals of Trigonometric Functions Example 1. Evaluate the integral }= dx x tan I2 Solution. We can convert x tan2 in a form which can be integrated using the Table. We see that 1x cos1x cosx cosx cos1x cosx cos 1x cosx sinx tan2 222 22222 = == = Therefore = =||.|

\| = = } } } }dx dxx cos1dx 1x cos1dx x tan I2 22 C x x tan + = Chapter 2-Elementary examples of integration 26 Example 2. Evaluate the integral dxx sin 3x sin 2 x sin 5I2} = . Solution. Since 32x sin35x sin 3x sin 2x sin 3x sin 5x sin 3x sin 2 x sin 52 2 = = we have =|.|

\| == } }dx32x sin35dxx sin 3x sin 2 x sin 5I2 = + = = } }C x32) x cos (35dx32dx x sin35 C x32x cos35+ = Example 3. Evaluate the integral dxx cos 2x cos x 2 sinI2} += . Solution. Since =+=+x cos 2x cos x cos x sin 2x cos 2x cos x 2 sin2 2 x cos21x sinx cos 2x cosx cos 2x cos x sin 22+ = + = we have =|.|

\| + =+= } }dx x cos21x sin dxx cos 2x cos x 2 sinI2 C x sin21x cos dx x cos21dx x sin + + = + = } } Example 4. Evaluate the integral dxx cosx 2 cosI2}= . Solution. Since x cos12x cos1x cosx cos 2x cos1 x cos 2x cosx 2 cos2 2 22222 = == Chapter 2-Elementary examples of integration 27 the integral takes the form = = |.|

\| = = } } } }dxx cos1dx 2 dxx cos12 dxx cosx 2 cosI2 2 2 C x tan x 2 + = 2.18 Case XVIII. Integrals of Exponential Functions Example 1. Evaluate the integral dxe 4e 3 e 2Ixx x 2} = Solution. We can simplify the expression under the integral as 43e21e 4e 3e 4e 2e 4e 3 e 2xxxxx 2xx x 2 = = Therefore the integral becomes =|.|

\| == } }dx43e21dxe 4e 3 e 2Ixxx x 2 C x43e21dx43dx e21x x + = = } } 2.19 Case XIX. More Integrals. Example 1. Evaluate the integral dxx 1x 4 x 4 x 1I23 2} + = Solution. We have = + = + 22 223 2x 1) x 1 ( x 4 x 1x 1x 4 x 4 x 1 x 4x 11x 1) x 1 ( x 4x 1x 122222+=+= Therefore = +=||.|

\|+= } } }dx x 4 dxx 11dx x 4x 11I2 2 C x 2 x arcsin2 + + = Chapter 2-Elementary examples of integration 28 Example 2. Evaluate the integral } +=5 xdxI2 Solution. C 5 x x ln5 xdxI22 + + + =+=} Example 3. Evaluate the integral dx2 x3 2 xI22} + + += Solution. We have 1 x31 x11 x31 x1 x1 x3 1 x222 2222+++=++++=+ + + Therefore =+++=+ + += } } }dx1 x13 dx1 x1dx1 x3 1 xI2222 C x arctan 3 1 x x ln2 + + + + = 2.20 More Examples Example 1. Evaluate the integral dxx 57 x x x 4 x 5I322 3} + = Solution. We simplify the integrand first. We have = + = + 3232322323322 3x 57x 5x xx 5x 4x 5x 5x 57 x x x 4 x 5 = + =3232322323x157xx x51xx54xx 32653437x57x51x54x + = Therefore the integral becomes Chapter 2-Elementary examples of integration 29 = + = } } } } dx x57dx x51dx x54dx x I32653437 = ++ ++++= + + + +C132x57165x51134x54137x132165134137 = + + = C31x57611x5137x54310x3161137310 C x521x x556x x3512x x103 3653233 + + = Example 2. Evaluate the integrals (I) dx x x I21 } = (II) dx x x32I32 } = (III) dxxx 2I323 }= Solution. (I) We have 25212212 2x x x x x x = = = +. Therefore C125xdx x dx x x I1252521 ++= = = +} } C x x72C27x327+ = + = Chapter 2-Elementary examples of integration 30 (II) We have 34311313x x x x x x = = = +. Therefore = = = = } } }dx x32dx x x32dx x x32I343 32 = + = + = ++= +C x x72C37x32C134x3231237134 C x x72 32 + = (III) We have 3132132 32x 2 x 2xx 2xx 2= = = . Therefore = ++= = = +} }C131x2 dx x 2 dxxx 2I13131323 C x x23C x x46C34x233134+ = + = + = Chapter 2-Elementary examples of integration 31 Computer Algebra Systems We list below the most important computer algebra systems. 1. Maple. We list below three examples of integration using Maple. In the first example, we first define a function which we call f which is 3x 4 , using Maples format. > f:=4*x^3; := f 4 x3 We then use the command > int(f,x); x4 We may also use > Int(f,x)=int(f,x); = d(]((4 x3x x4 In the second example, we define a function g > g(x):=(1/2)*x^(-1/2); := ( ) g x12 x > int(g(x),x); x > Int(g(x),x)=int(g(x),x); = d(]((((12 x x x In the third example, we define a function 32x1) x ( h = by > h:=x->x^(-2/3); Chapter 2-Elementary examples of integration 32 := h x1x( ) / 2 3 and then we perform the integration using the command > int(h(x),x); 3 x( ) / 1 3 > Int(h(x),x)=int(h(x),x); = d(](((((1x( ) / 2 3 x 3 x( ) / 1 3 In all the previous examples we have first defined the function under integration using three different ways and then we used the command: int(expr, x). Using the command Int(expr, x), the program puts the known symbol of integration in front of the result of integration. It is obvious that there is not need to define the function first. We may go directly to the evaluation of the integral: > int((2*x+a),x); + x2a x > int((x^2+a)/(x^1+1),x); + + x22 x ( ) ln + x 1 a ( ) ln + x 1 In the above two examples, the function also contains a parameter. 2. Mathematica. In the first example, we define the function x 2 x 3 ) x ( f2 = by I n[ 1] : = f 3 x^2 2 xOut [ 1] = 2 x 3 x2 and then we use the integration command: ] x , f [ Integrate I n[ 2] : = Integrate f, xOut [ 2] = x2x3 Chapter 2-Elementary examples of integration 33 Mathematica contains also a nice interface, which can be used instead of writing code. The second example performs the integration of the function x 2 x 32 : I n[ 3] : = 3 x2 2 x xOut [ 3] = x2x3 The third example defines a function the Mathematical way and then use again the integration command: I n[ 4] : = g x_ : x ^ 2 3 xI n[ 5] : = Integrate g x , xOut [ 5] = 3 x22 x33 We should also mention that there is an online tool, called The Integrator by Wolfram Research, which can be used freely for evaluating integrals. The address of the site is integrals.wolfram.com Notes. It might sometimes appear that an integral we evaluate by hand would give different result compared to the answer we get using Computer Algebra Systems. This is due to the fact that many Systems use different integration algorithms. On the other hand, we will notice that we get different answers using different Systems. However we may establish that all the answers are equivalent each other. Chapter 2-Elementary examples of integration 34 2a Exercises on Elementary Integration Case I. Integrals of the form }dx a General Case. C x a dx a dx a + = = } } Exercise 1. Evaluate the integrals (1) }dx 4 (2) } dx ) 3 ( (3) } + dx ) a 1 ( (4) } + dx ) 6 ( Answers. (1) C x 4 + (2) C x 3 + (3) C x ) a 1 ( + + (4) C x ) 6 ( + + Case II. Integrals of the form }dx x an ( R a e , 1 n > ) General Case. C x1 naC1 nxa dx x a dx x a1 n1 nn n ++= ++= = ++} } Exercise 2. Evaluate the integrals (1) }dx x 32 (2) } dx x ) 2 (3 (3) }dx x 54 (4) } dx x ) 6 (3 Answers. (1) C x3+ (2) C x214+ (3) C x5+ (4) C x234+ Case III. Integrals of the form dxxa} ( R a e ) Chapter 2-Elementary examples of integration 35 General Case. C | x | ln a dxx1a dxxa+ = = } } Exercise 3. Evaluate the integrals (1) dxx2} (2) dxx4} (3) dxx6} (4) dxx13} Answers. (1) C | x | ln 2 + (2) C | x | ln 4 + (3) C | x | ln 6 + (4) C | x | ln 13 + Case IV. Integrals of Polynomial Functions These are integrals of the form }= dx ) x ( P I where ) x ( P is an n-th degree polynomial. In this case we convert the integral into a sum of integrals of the form dx x an} ( 0 n > ) which are evaluated according to the known formulas 1 nxa dx x a1 nn+= +} ( 1 n > ) and x a dx a =} Exercise 4. Evaluate the integrals (1) dx ) 3 x 2 (} (2) dx ) x 3 x (2} (3) dx ) 2 x x 3 (2} (4) dx ) 1 x 3 x 2 (3} + (1) Hint. } } } = dx 3 dx x 2 dx ) 3 x 2 ( Answer. C x 3 x2+ (2) Hint. dx x 3 dx x dx ) x 3 x (2 2} } } = Answer. C x233x23+ (3) Hint. } } } } = dx 2 dx x dx x 3 dx ) 2 x x 3 (2 2 Answer. C x 22xx23+ (4) Hint. } } } } + = + dx dx x 3 dx x 2 dx ) 1 x 3 x 2 (3 3 Answer. C x x23x212 4+ + Chapter 2-Elementary examples of integration 36 Case V. Integrals of the form dxxan} ( R a e , 2 n > ) General Case. = ++ = = = + } } }C1 nxa dx x a dxx1a dxxa1 nnn n Cx11 naC x1 na1 n1 n + = + = + , 2 n > Exercise 5. Evaluate the integrals (1) dxx33 } (2) dxx22 } (3) dxx54 } (4) dxx116 } (1) Hint. 33x 3x3 = Answer. Cx 23C x2322+ = + (2) Hint. 22x 2x2 = Answer. Cx2C x 21+ = + (3) Hint. 44x 5x5 = Answer. Cx 35C x3533+ = + (4) Hint. 66x 11x11 = Answer. Cx 511C x51155+ = + Case VI. Integrals of Rational Functions Exercise 6. Evaluate the integrals (1) dxx1 x 3} (2) dx1 x2 x22} + (3) dxx 24 x 2 x 322} + (4) dxx 32 x x 533} + (1) Hint. x13x1xx 3x1 x 3 = = Answer. C | x | ln x 3 + (2) Hint. 1 x311 x31 x) 1 x (1 x3 ) 1 x (1 x2 x2 2 222222+ =+++=+ +=+ Answer. C x arctan 3 x + Chapter 2-Elementary examples of integration 37 (3) Hint. 22 2 2222x 2x123x 24x 2x 2x 2x 3x 24 x 2 x 3 + = + =+ Answer. Cx2| x | ln x23+ (4) Hint. 3 23 3 3333x32x3135x 32x 3xx 3x 5x 32 x x 5 + = + = + Answer. Cx 31x 31x352 + + Case VII. Integrals of the form dx x a} ( R a e ) In this case we convert the square root into power with a rational exponent using the property 21x x = . Therefore we have the following general scheme: = ++= = = +} } }C121xa dx x a dx x a dx x a12121 C x x3a 2C x3a 2C23xa2323+ = + = + = Exercise 7. Evaluate the integrals (1) dx x 5} (2) dx x ) 2 (} (3) dx x ) 7 (} (4) dx x 6} Answers. (1) x x310 (2) x x34 (3) x x314 (4) x x 4 Case VIII. Integrals of the form dx x x an} ( R a e , 1 n > ) Integrals of this form are evaluated according to the general scheme: = = = } } }dx x x a dx x x a dx x x a21n n n Chapter 2-Elementary examples of integration 38 = ++= ++ += = + + + ++}C23 n 2xa C121nxa dx x a211 n 121n21n C x x3 n 2a 2C23 n 2xa1 n211 n++= ++= ++ + Exercise 8. Evaluate the integrals (1) dx x x 32} (2) dx x x ) 4 (} (3) dx x x 53} (4) dx x x ) 6 (4} (1) Hint. 25212212 2x 3 x 3 x x 3 x x 3 = = = + Answer. C x x76C x76327+ = + (2) Hint. 23x 4 x x ) 4 ( = Answer. C x x58C x58225+ = + (3) Hint. 273x 5 x x 5 = Answer. C x x910C x910429+ = + (4) Hint. 294x 6 x x ) 6 ( = Answer. C x x1112C x11125211+ = + Case IX. Integrals of the form dxxa} ( R a e ) Integrals of this form are evaluated according to the general scheme: = ++ = = = + } } }C121xa dx x a dxx1a dxxa12121 C x a 2 C21xa21+ = + = Chapter 2-Elementary examples of integration 39 Exercise 9. Evaluate the integrals (1) dxx1} (2) dxx3} (3) dxx5} (4) dxx7} Answers. (1) C x 2 + (2) C x 6 + (3) C x 10 + (4) C x 14 + Case X. Integrals of the form dxxxan} ( R a e , 1 n > ) Integrals of this form are evaluated according to the general scheme: = = = } } } dx x a dxxxa dxxxa21n21n n C x x1 n 2a 2C21 n 2xa C121nxan21n 121n++= ++= ++ = + + Exercise 10. Evaluate the integrals (1) dxxx33} (2) dxxx52} (3) dxxx) 8 (} (4) dxxx64} (1) Hint. 253x 3xx3 = Answer. C x x76C x76327+ = + (2) Hint. 232x 5xx5 = Answer. C x x 2 C x 2225+ = + (3) Hint. 21x 8xx) 8 ( = Answer. C x x316C x31623+ = + (4) Hint. 274x 6xx6 = Answer. C x x34C x34429+ = + Chapter 2-Elementary examples of integration 40 Case XI. Integrals of the form dxxxa In}= ( R a e , 1 n > ) Integrals of this form are evaluated according to the general scheme: = = = = } } } dx x a dxxxa dxxxa In21n21n Cxxn 2 3a 2C2n 2 3xa C1 n21xa1 n211 n 1 n21+= += ++ = + + + Exercise 11. Evaluate the integrals (1) dxxx6} (2) dxxx) 5 (4 } (3) dxxx45} (4) dxxx123} (1) Hint. 21x 6xx6 = Answer. C x 12 C x 1221+ = + (2) Hint. 274x 5xx) 5 ( = Answer. Cx x1710C x710225+ = + (3) Hint. 295x 4xx4 = Answer. Cx x178C x78327+ = + (4) Hint. 253x 12xx12 = Answer. Cx x8C x 823+ = + Case XII. Integrals of the form dx x anm} Integrals of this form are evaluated according to the general scheme: = ++= = +} }C1nmxa dx x a dx x a1nmnmnm Chapter 2-Elementary examples of integration 41 C x xn mn aCnn mxanm1nm++= ++= + Exercise 12. Evaluate the integrals (1) dx x 332} (2) dx x ) 5 (45} (3) dx x 1134} (4) dx x ) 14 (53} (1) Hint. 3232x 3 x 3 = Answer. C x x59C x59 3235+ = + (2) Hint. 4545x 5 x ) 5 ( = Answer. C x x920C x920 4249+ = + (3) Hint. 3434x 11 x 11 = Answer. C x x733C x733 3237+ = + (4) Hint. 5353x 14 x ) 14 ( = Answer. C x x435C x435 5358+ = + Case XIII. Integrals of the form dx x x anm p} General Case. = = } }dx x x a dx x x anmpnm p = ++ += ++ += = + + + ++}Cnn m n pxa C1nmpxa dx x anm1 p 1nmpnmp C x xn m n pn aC x xn m n pn anm 1 pnm1 p ++ += + + += + + Exercise 13. Evaluate the integrals (1) dx x x 532 2} (2) dx x x ) 12 (43 3} Chapter 2-Elementary examples of integration 42 (3) dx x x ) 15 (43 2} (4) dx x x52 3} (1) Hint. 3832 2x 5 x x 5 = Answer. C x x1115C x111532 3311+ = + (1) Hint. 41543 3x 12 x x ) 12 ( = Answer. C x x1948C x194843 4419+ = + (1) Hint. 41143 2x 15 x x ) 15 ( = Answer. C x x 4 C x 443 3415+ = + (1) Hint. 41543 3x x x = Answer. C x x194C x19443 4419+ = + Case XIV. Integrals of the form dxxxanmp} General Case. = = } } dx x x a dxxxanmpnmp = + += + += = + +}Cnm n pnxa Cnm1 pxa dx x anm1 pnm1 pnmp Cxxm n pnn aC x xm n pnn anm1 pnm1 p + += + += ++ Exercise 14. Evaluate the integrals (1) dxxx) 4 (323} (2) dxxx5435} (3) dxxx12544} (4) dxxx) 15 (752} (1) Hint. 37323x 4xx) 4 ( = Answer. C x x56C x56 33310+ = + Chapter 2-Elementary examples of integration 43 (2) Hint. 417435x 5xx5 = Answer. 45421x x2120C x2120 = + (3) Hint. 516544x 12xx12 = Answer. C x x720C x720 54521+ = + (4) Hint. 79752x 15xx) 15 ( = Answer. C x x16105C x1610572 2716+ = + Case XV. Integrals of the form dxxanm} General Case. = ++ = = + } }C1nmxa dx x a dxxa1nmnmnm = + + = ++ = + C x xn mn aCnn mxanm1nm Cxxn mn anm + + = Exercise 15. Evaluate the integrals (1) dxx143} (2) dxx232} (3) dxx1243} (4) dxx1653} (1) Hint. 4343xx1 = Answer. C x 4 C x 4441+ = + (2) Hint. 3232x 2x2 = Answer. C x 6 C x 6331+ = + Chapter 2-Elementary examples of integration 44 (3) Hint. 4343x 12x12 = Answer. C x 48 C x 48441+ = + (4) Hint. 5353x 16x16 = Answer. C x 40 C x 405252+ = + Case XVI. Integrals of the form dxxxapnm} General Case. = = } } dx x x a dxxxapnmpnm = ++ = ++ = = + + }Cnn pn mxa C1 pnmxa dx x anmp 1 1 pnmpnm C x xn n p mn aC x xn n p mn anm p 1nmp 1 + + = + + = Exercise 16. Evaluate the integrals (1) dxxx5432} (2) dxxx) 6 (243} (3) dxxx13253} (4) dxxx) 7 (352} (1) Hint. 310432432x 5 x 5xx5 = = . Answer. Cx x1715C x7153237+ = + (2) Hint. 45243243x 6 x 6xx6 = = . Answer. Cx24C x 24441+ = + (3) Hint. 57253253x 13 x 13xx13 = = Answer. Cx1265C x2655252+ = + Chapter 2-Elementary examples of integration 45 (4) Hint. 513352352x 7 x 7xx) 7 ( = = Answer. 535858x x1835x1835x835 = = Case XVII. Integrals of Trigonometric Functions Exercise 17. Evaluate the integrals (1) dxx sin3 x cos 222} (2) dxx cos2 x sin 322} + (1) Hint. = = =x sin1 x sin 2x sin3 ) x sin 1 ( 2x sin3 x cos 2222222 x sin12x sin1x sinx sin 22 2 22 = = Answer. C x cot x 2 + + (2) Hint. =+ =+ =+x cos5 x cos 3x cos2 ) x cos 1 ( 3x cos2 x sin 3222222 x cos53x cos5x cosx cos 32 2 22+ = += Answer. C x tan 5 x 3 + + Case XVIII. Integrals of Exponential Functions Exercise 18. Evaluate the integrals (1) dxe 3e 8 e 5xx x 2} + (2) dxee 3 e 2x 2x 2 x 3} (1) Hint. 38e35e 3e 8e 3e 5e 3e 8 e 5xxxxx 2xx x 2+ = + =+ Answer. C x38e35x + + (2) Hint. 3 e 2ee 3ee 2ee 3 e 2xx 2x 2x 2x 3x 2x 2 x 3 = = Answer. C x 3 e 2x+ Chapter 2-Elementary examples of integration 46 Case XIX. Miscellaneous Cases Exercise 19. Evaluate the integrals (1) dxx 1x 4 x 4 x 123 2} + (2) } + 5 xdx2 (3) dx1 x3 1 x22} + + + (4) dxx 57 x x x 4 x 5322 3} + (1) Hint. = + += + 232223 2x 1x 4 x 4x 1x 1x 1x 4 x 4 x 1 x 4x 11x 1) x 1 ( x 4x 112222 +=+= Answer. C x 2 x arcsin2+ + (2) Answer. C | 5 x x | ln2+ + + (3) Hint. 1 x31 x11 x31 x1 x1 x3 1 x222 2222+++=++++=+ + + Answer. C x arctan 3 | 1 x x | ln2+ + + + (4) Hint. = + = + 3232322323322 3x 57x 5x xx 5x 4x 5x 5x 57 x x x 4 x 5 32653437323223322323x57x51x54x x57x51x54x + = + = Answer. = + + C x521x556x3512x1033161137310 C x521x x556x x3512x x103 3653233 + + = 3 The Method of Substitution Learning Objectives In this Section the student will learn how to evaluate integrals using the method of substitution. Many cases of evaluation of indefinite integrals }dx ) x ( f the function ) x ( f under integration (i.e. the integrand) is a rather complicated expression. The integral either does not belong to any of the known forms listed in the Table of Integrals, or cannot be reduced to any of those forms, using simple algebraic manipulations. Here is where some methods come in. The most method is the method of substitution. This method converts the integral into another one, simpler, contained often to the Table, and thus can easily be evaluated. In this section we shall introduce the method in its simple, obvious version. More complicated, special cases of substitution (like trigonometric substitutions, hyperbolic substitutions, Euler substitutions, Chebyshev substitutions) will be considered in the next sections. Chapter 3-The method of substitution 48 The two most common cases are the following: First Case. Let the integrand be a function Y X : f . We have to evaluate the integral }dx ) x ( f We substitute the independent variable x by an expression ) u ( : ) u ( x = where X U : is an 1 1 function, having an inverse U X :1 . Considering the composition )) u ( ( f ) u )( f ( = , since du ) u ( dx ' = , we have du ) u ( )) u ( ( f dx ) x ( f ' = Therefore the integral transforms into } } } = ' = du ) u ( g du ) u ( )) u ( ( f dx ) x ( f where the new transformed integral }du ) u ( g can be evaluated using the Table of Elementary Integrals. It might also happen to have be converted into a more simpler integral. At the end of the calculation, after evaluating the integral }du ) u ( g , we have to substitute u by ) x (1 : ) x ( u ) x ( u1 1du ) u ( g du ) u ( )) u ( ( f dx ) x ( f = = } } } = ' = Example. Evaluate the integral } 2x 1dx Solution. Under the substitution u sin x = ( u sin ) u ( = ) we have du u cos dx = , u cos u sin 1 x 12 2 = = and x arcsin u = Therefore = = = == } } }x arcsin ux arcsin u2du duu cosu cosx 1dx Chapter 3-The method of substitution 49 C x arcsin C ux arcsin u + = + = = This formula is already known from the Table of Integrals. Second Case. In this case we substitute an expression contained in the function ) x ( f under integration by a new variable u: ) x ( u = We then convert the integral into another one, containing the variable u only, which can in turn be evaluated. Therefore we suppose that there exists a differentiable function ) x ( and a function ) u ( g such that the integrand dx ) x ( f can be written in the form dx ) x ( )) x ( ( g dx ) x ( f ' = Introducing the transformation (usually called substitution) ) x ( u = , dx ) x ( du ' = we can convert the original integral into ) x ( udu ) u ( g dx ) x ( )) x ( ( g dx ) x ( f| =} } } = ' = Example. Evaluate the integral dx x 1 x 22} + Solution. We consider the substitution 2x 1 u + = (2x 1 ) x ( + = ). We then have dx x 2 du = and u x 12 = + Therefore the integral transforms into = + = = ++ =+ =} }C u u32du u dx x 1 x 222x 1 ux 1 u2 C x 1 ) x 1 (322 2 + + + = Note 1. We should stress the fact that there is not any unique substitution transformation which can convert an integral into a simpler, easy-to-evaluate integral. In this section we only consider some obvious forms of substitution. In Chapter 3-The method of substitution 50 next sections we are going to consider special forms of substitution, depending on the form of the integrand. Note 2. We note that there are - Special methods of substitution, like trigonometric substitutions, hyperbolic substitutions, Euler substitutions, Chebyshev substitutions - Special techniques like integration by parts and partial fraction decomposition - Special algorithms, like the Ostrogradsky algorithm, the Hermite-Horowitz algorithm, the Rothstein-Trager algorithm, the Risch algorithm and the Adamchik-Marichev algorithm. We below list some examples of using the method of substitution, classified according to the type of function under integration. Case I. Integrals of Polynomial and Rational Functions Example 1. Evaluate the integral } + + + = dx ) 7 x 5 x )( 5 x 2 ( I2 Solution. We see that under the substitution 7 x 5 x u2 + + = , we have dx ) 5 x 2 ( du + = Therefore the integral becomes C2udu u2+ =} Going back to the original variable, we have the value of the integral C ) 7 x 5 x (21dx ) 7 x 5 x )( 5 x 2 ( I2 2 2 + + + = + + + = } Example 2. Evaluate the integral } + + = dx ) 1 x x 5 x 4 ( ) 1 x 10 x 12 ( I3 2 3 2 Solution. We see that under the substitution 1 x x 5 x 4 u2 3 + = , we have dx ) 1 x 10 x 12 ( du2 + = Therefore the integral becomes Chapter 3-The method of substitution 51 C4udu u43 + =} Going back to the original variable, we have the value of the integral = + + = }dx ) 1 x x 5 x 4 ( ) 1 x 10 x 12 ( I3 2 3 2 C ) 1 x x 5 x 4 (414 2 3 + + = Example 3. Evaluate the integral dx5 x 3 x3 x 2I2} + = Solution. We see that under the substitution of the denominator by a new variable u, 5 x 3 x u2 + = we have dx ) 3 x 2 ( du = . The integral then becomes }= duu1I ,which belongs to one of the known integrals of the Table. Therefore C | 5 x 3 x | ln C | u | ln duu1dx5 x 3 x3 x 2I22 + + = + = =+ = } }. Example 4. Evaluate the integral dxx 1xI3} += Solution. Under the substitution x 1 u + = , we have dx du = . Therefore the integral transforms into = + =} }duu1 u 3 u 3 uduu) 1 u (2 3 3 =|.|

\| + = }duu13 u 3 u2 = + = } } } }duu1du 3 du u 3 du u2 = + + = C | u | ln u 32u33u2 3 C | u | ln u 3 u23u312 3 + + = Chapter 3-The method of substitution 52 Going back to the original variable, we obtain the value of the integral C | x 1 | ln ) x 1 ( 3 ) x 1 (23) x 1 (31dxx 1xI2 33+ + + + + + =+= } Example 5. Evaluate the integral dx) x 1 (xI3} += Solution. Under the substitution x 1 u + = , we have dx du = . Therefore the integral transforms into = = =} } } } }duu1duu1duu1duuuduu1 u3 2 3 3 3 Cu 21u1du u du u23 2 + + = = } } Going back to the original variable, we obtain the value of the integral C) x 1 ( 21x 11dx) x 1 (xI2 3 ++++ =+= } Example 6. Evaluate the integral dxx 1x 2I4} += Solution. Under the substitution 2x u = , we have dx x 2 du = . Therefore C ) x arctan( C u arctanu 1dudxx 1x 2I22 4 + = + =+=+= } } Example 7. Evaluate the integral } +=2 2x adxI Solution. Under the substitution t a x = , we have dt a dx = . Therefore, since ) t 1 ( a t a a x a2 2 2 2 2 2 2 + = + = + , the integral transforms to C t arctana1t 1dxaa) t 1 ( adx a2 2 2 2 + =+=+ } } Chapter 3-The method of substitution 53 Going back to the original variable, we obtain Caxarctana1x adxI2 2 +|.|

\|=+= } Case II. Integrals of Irrational Functions Example 1. Evaluate the integral dx 3 x x I3 2} + = Solution. We see that under the substitution 3 x u3 + = we get dx x 3 du2= which implies dx x du312= . Therefore the integral becomes du u31I }= which can be evaluated using the Table. We have = + = ++= = = +} }C23u31C121u31du u31du u31I2312121 C u u92+ = Therefore = + = = + = } }C u u92du u31dx 3 x x I3 2 C 3 x ) 3 x (923 3 + + + = Example 2. Evaluate the integral } =2 2x adxI ( 0 a > ) Solution. Under the substitution t a x = , we have dt a dx = . Therefore, since ) t 1 ( a t a a x a2 2 2 2 2 2 2 = = , the integral transforms into C t arcsint 1dtt 1 adt a) t 1 ( adt a2 2 2 2 + === } } } Chapter 3-The method of substitution 54 Going back to the original variable, since axt = , we obtain Caxarcsinx adxI2 2 +|.|

\|== } Example 3. Evaluate the integral dxx 1x 8I83} = Solution. Under the substitution 4x u = , since dx x 4 du3= , the integral becomes C u arcsin 2 duu 122 + =} Going back to the original variable, we obtain the value of the integral C ) x arcsin( 2 dxx 1x 8I483+ == } Example 4. Evaluate the integral dx1 x4 x 3I2} ++= Solution. We split the integral into a sum of two integrals =+++=++= } } }dx1 x4dx1 xx 3dx1 x4 x 3I2 2 2 2 12 2I I 3 dx1 x14 dx1 xx3 + =+++= } } The integral dx1 xxI21 } += can be evaluated by the substitution 1 x u2 + = . Since dx x 2 du = , the integral transforms into uudu21=}. Therefore we obtain 1 x dx1 xxI221 + =+= } The integral dx1 x1I22 } += is known from the Table with value Chapter 3-The method of substitution 55 1 x x ln dx1 x1I222 + + =+= } Therefore we obtain the value of the integral C 1 x x ln 4 1 x 3 dx1 x4 x 3I2 22 + + + + + =++= } Case III. Integrals of Logarithmic Functions Example 1. Evaluate the integral }=x ln xdxI Solution. Under the substitution x ln u = , since dxx1du = , the integral transforms into C | u | lnudu+ =} Going back to the original variable, we obtain as the value of the integral C | x ln | lnx ln xdxI + = = } Example 2. Evaluate the integral dxxx lnI3}= Solution. Under the substitution x ln u = , since dxx1du = , we get C x ln41C4udu u dxxx lnI4433+ = + = = = } } Example 3. Evaluate the integral dx) x ln 1 ( x1I2} += Solution. Under the substitution x ln u = , we have dxx1du = . Therefore C ) x arctan(ln C u arctanu 1dudx) x ln 1 ( x1I2 2 + = + =+=+= } } Case IV. Integrals of Exponential Functions Chapter 3-The method of substitution 56 Example 1. Evaluate the integral dxe 1eIx 2x} += Solution. Under the substitution xe u = , we have dx e dux= . Therefore C ) e arctan( C u arctanu 1dudxe 1eIx2 x 2x+ = + =+=+= } } Example 2. Evaluate the integral dxe1 eIxx 2} = Solution. We first split the integral into a sum of two integrals: = = == } } } } } dx e dx e dxe1dxeedxe1 eIx xx xx 2xx 2 dx e ex x} = Under the substitution x u = , we have dx du = . Therefore x u u xe e du e dx e = = = } } The value of the integral thus becomes C e e dxe1 eIx xxx 2+ + == } Example 3. Evaluate the integral dx a Ix}= , 0 a > Solution. Using the identity a ln x xe a = and the substitution a ln x u = , dx ) a (ln du = , we have C ea ln1du ea ln1dx e dx a Iu u a ln x x + = = = = } } } C aa ln1C ea ln1x a ln x + = + = Example 4. Evaluate the integral dx23 7 2 5Ixx x} + = Solution. Chapter 3-The method of substitution 57 =+= + = } } }dx23 7dx22 5dx23 7 2 5Ixxxxxx x dx237 dx 5 dx237 dx 5x x} } } } |.|

\|+ =|.|

\| + = C2323ln7x 5x+|.|

\||.|

\|+ = Example 5. Evaluate the integral dx e e 1 Ix x} = Solution. Under the substitution xe 1 u = , we have dx e dux = . Therefore the integral becomes C u u32du u + = } Going back to the original variable, we obtain the value of the integral C e 1 ) e 1 (32dx e e 1 Ix x x x + = = } Example 6. Evaluate the integral dxxeIx}= Solution. Under the substitution x u = , since dxx 21du = , the integral transforms into C e 2 du e 2u u + =} Going back to the original variable, we obtain the value of the integral C e 2 dxxeIxx+ = = } Example 7. Evaluate the integral dxe 1eIx 2x} = Chapter 3-The method of substitution 58 Solution. Under the substitution xe u = , we have dx e dux= . Therefore C ) e arcsin( C u arcsinu 1dudxe 1eIx2 x 2x+ = + === } } Example 8. Evaluate the integral dx4 12Ixx} = Solution. Under the substitution x2 u = , we have 2 xu 1 4 1 = and dx 2 ) 2 (ln dux= . Therefore the integral becomes C u arcsin2 ln1u 1du2 ln12 + =} Going back to the original variable, the integral has the value C ) 2 arcsin(2 ln1dx4 12Ixxx+ == } Case V. Integrals of Trigonometric Functions Example 1. Evaluate the integral dx x tan I }= Solution. Since x cosx sinx tan = , we have dxx cosx sinI }= . Under the substitution x cos u = , dx x sin du = , the integral becomes C | u | ln duu1+ = }. Going back to the original variable, we get C | x cos | ln dx x tan I + = = } Example 2. Evaluate the integral dx x cot I }= Solution. Since x sinx cosx cot = , we have dxx sinx cosI }= . Under the substitution x sin u = , dx x cos du = the integral becomes Chapter 3-The method of substitution 59 C | u | ln duu1+ =}. Going back to the original variable, we get C | x sin | ln dx x cot I + = = } Example 3. Evaluate the integral dx x cos e Ix sin}= Solution. Under the substitution x sin u = , since dx x cos du = , the integral transforms into C e du eu u + =} Going back to the original variable, we obtain C e dx x cos e Ix sin x sin + = = } Example 4. Evaluate the integral dxex sinIx cos}= Solution. Under the substitution x cos u = , since dx x sin du = , the integral transforms into } } = du eeduuu Under a second substitution u t = , since du dt = , the above integral takes the form C e dt et t + =} Going back to the original variable, since u t = and x cos u = , the value of the integral is given by C e dxex sinIx cosx cos + = = } Example 5. Evaluate the integral dxx) x ( cosI }= Chapter 3-The method of substitution 60 Solution. Under the substitution x u = , since dxx 21du = , the integral transforms into C u sin 2 du u cos 2 + =} Going back to the original variable, we obtain the value of the integral C ) x ( sin 2 dxx) x ( cosI + = = } Example 6. Evaluate the integral dxx sin x cosx 2 sinI4 4} = Solution. We simplify the denominator first. We have = + = ) x sin x (cos ) x sin x (cos x sin x cos2 2 2 2 4 4 x 2 cos ) x sin x (cos 12 2 = = Therefore the integral becomes dxx 2 cosx 2 sinI }= Under the substitution x 2 cos u = , dx x 2 sin 2 du = , we have C | u | ln21udu21I + = = } Going back to the original variable, we obtain C | x 2 cos | ln21I + = Example 7. Evaluate the integral } +=x tan 1 x cosdxI2 Solution. Using the substitution x tan 1 u + = , we have dxx cos1du2= Therefore the integral becomes Chapter 3-The method of substitution 61 C u 2udu+ =} Going back to the original variable, we find the value of the integral C x tan 1 2x tan 1 x cosdxI2 + + =+= } Case VI. Integrals of Inverse Trigonometric Functions Example 1. Evaluate the integral dxx 1x arctanI2} += Solution. Under the substitution x arctan u = , we have dxx 11du2+= . Therefore C ) u (arctan21C2udu u dxx 1x arctanI222 + = + = =+= } } C )] x tan arctan(arc [212 + = Case VII. Integrals with two successive substitutions In evaluating integrals, there might appear the need for more than one substitution. Let us list some Examples of this kind. Example 1. Evaluate the integral dx5 e 3eIx 2x} += Solution. Using the substitution xe u = , we get dx e dux= and the integral transforms into } }+=+35udu315 u 3du22 Under a new substitution t35u = , since dt35du = and Chapter 3-The method of substitution 62 ) 1 t (3535t3535u222 + = +||.|

\|= + , the last integral becomes C t arctan151dt1 t1151dt) 1 t (35353122 + =+=+ } } Since xe53u53t = = , the value of the integral is C e53arctan151Ix+||.|

\|= Example 2. Evaluate the integral dx9 xx 4I62} += Solution. We use the substitution 3x u = first. We then have dx x 3 du2= and thus du31dx x2 = . Therefore the integral becomes } + 9 udu342 (1) Under the further transformation t 3 u = , since ) 1 t ( 9 9 ) t 3 ( 9 u2 2 2 + = + = + and dt 3 du = the integral in (1) transforms into C t arctan941 tdt94) 1 t ( 9dt 3342 2 + =+=+ } } We now have to go back to the original variable in reverse order. Since 3ut = and 3x u = , we find the value of the integral to be C3xarctan94dx9 xx 4I362+||.|

\|=+= } Chapter 3-The method of substitution 63 Example 3. Evaluate the integral dx) x ln(sinx cotI }= Solution. We use the substitution x sin u = first. We then have dx x cos du = and thus u ln ududx) x ln(sin ) x (sinx cosdx) x ln(sinx cot= = . Therefore the integral becomes }u ln udu (1) Under a second substitution u ln t = , since ududt = , the integral in (1) becomes C | t | lntdt+ =} We now have to go back to the original variable in reverse order. Since u ln t = and x sin u = , we find the value of the integral to be C | ) x ln(sin | ln dx) x ln(sinx cotI + = = } Example 4. Evaluate the integral dx) x ln 3 1 ( xx lnI2} += Solution. Under the substitution x ln u = , we have dxx1du = . Therefore the integral transforms into duu 3 1u2} + Under a second substitution 2u 3 1 t + = , since du u 6 dt = , the previous integral transforms into C | t | lntdt61+ =} Going back to the original variable, since 2u 3 1 t + = and x ln u = , we obtain the value of the integral Chapter 3-The method of substitution 64 C ) x ln 3 1 ( ln61dx) x ln 3 1 ( xx lnI22 + + =+= } Example 5. Evaluate the integral dxx axI4 2} = ( 0 a > ) Solution. Under the substitution 2x u = , we have dx x 2 du = . Therefore, since 2 2 4 2u a x a = , the integral transforms to } 2 2u adu21 Under a further substitution t a u = , since dt a du = and ) t 1 ( a u a2 2 2 2 = , the last integral transforms into C t arctan21t 1dt21) t 1 ( adt a212 2 2 + == } } Going back to the original variable, since aut = and 2x u = , we obtain Caxarcsin21dxx axI24 2 +||.|

\|== } Case VIII. Integrals with two different substitutions Example 1. Evaluate the integral dxx 1x arcsin x 2I2} = Solution. The integral can be split into 2 12 2 2I I dxx 1x arcsindxx 1x 2dxx 1x arcsin x 2I === } } } Chapter 3-The method of substitution 65 For the integral dxx 1x 2I21 } = we make the substitution 2x 1 u = . Therefore, since dx x 2 du = , the integral becomes u 2udu = }. Going back to the original variable, we find 221x 1 2 dxx 1x 2I == } For the integral dxx 1x arcsinI22 } = we make the substitution x arcsin t = . Therefore, since dxx 11dt2= , the integral becomes t t32dt t =} Going back to the original variable, we find that x arcsin ) x (arcsin32dxx 1x arcsinI22 == } We thus find that the original integral has the value C x arcsin ) x (arcsin32x 1 2 dxx 1x arcsin x 2I22 + + == } Case VIII. Advanced Examples Example 1. Evaluate the integral dx) 1 x (xI2 47} += Solution. We use the substitution 1 x u4 + = . We then have 1 u x4 = , dx x 4 du3= and thus du41) 1 u ( ) dx x ( x dx x3 4 7 = = . Therefore the integral transforms into = =} } }duu141duuu41duu1 u412 2 2 Chapter 3-The method of substitution 66 Cu141| u | ln41duu141udu412 + + = = } } Going back to the original variable, we obtain the value of the integral: C) 1 x ( 41| 1 x | ln41dx) 1 x (xI442 47+++ + =+= } Example 2. Evaluate the integral dxx 2 1xI23} += Solution. Under the substitution 2x 2 1 u + = , we find 21 ux2 = and then du41dx x = . The term dx x3 thus becomes du81 udu4121 u) dx x ( x2 =|.|

\| = . Therefore the integral transforms into = =} } }duu181duuu81duu1 u81 C u41u u121duu181du u81+ = = } } Going back to the original variable, we find the following value of the integral: C x 2 141x 2 1 ) x 2 1 (121dxx 2 1xI2 2 223+ + + + =+= } Example 3. Evaluate the integral dx2 x1 xI23} ++= Solution. We split the integral into a sum of two integrals: 2 12 2323I I dx2 x1dx2 xxdx2 x1 xI + =+++=++= } } } For the integral 1I , under the substitution 2 x u2 + = , we find 2 u x2 = and then du21dx x = . The term dx x3 thus becomes du22 udu21) 2 u ( ) dx x ( x2 = = . Chapter 3-The method of substitution 67 Therefore the integral 1I transforms into = =} } }duu1duuu21duu2 u21 u 2 u u31duu1du u21 = = } } Going back to the original variable, we find the following value of the integral: 2 x 2 2 x ) 2 x (31dx2 xxI2 2 2231 + + + =+= } The integral 1I is elementary, giving the value |.|

\| + + =+= }2 x x ln dx2 x1I222 Therefore we find for the original integral = + =++= } 2 123I I dx2 x1 xI C 2 x x ln 2 x 2 2 x ) 2 x (312 2 2 2 +|.|

\| + + + + + + = Example 4. Evaluate the integral dx e 1 Ix} = Solution. We try to get rid of the square root. Therefore we consider the substitution 1 e ux = Squaring both members of the previous relation and solving with respect to xe , we find 2 xu 1 e + = Inverting the above equation, we find ) u 1 ln( x2+ = Therefore Chapter 3-The method of substitution 68 duu 1u 2dx2+= The integral thus transforms into =+ +=+=+ } } }duu 11 ) u 1 (2 duu 1u2 duu 1u 2u22222 =+ =||.|

\|+ = } } }duu 112 du 2 duu 111 22 2 C u arctan 2 u 2 + = Going back to the original variable, we find the following value of the integral: C e 1 arctan 2 e 1 2 dx e 1 Ix x x + = =} Chapter 3-The method of substitution 69 3a Exercises on the Method of Substitution I) Integrals of Polynomial and Rational Functions Exercise 1. Evaluate the integrals (1) } + dx ) 1 x 3 x 5 )( 3 x 10 (2 (2) } + + dx ) 4 x 5 x 2 )( 10 x 8 (2 (3) } + + dx ) 8 x 7 x 3 )( 7 x 6 (3 2 (4) } + + + dx ) 5 x 2 x 4 x 2 ( ) 2 x 8 x 6 (4 2 3 2 (1) Hint. Let 1 x 3 x 5 u2 + = . Then dx ) 3 x 10 ( du = . The integral transforms into C2udu u2+ =}. Answer. C ) 1 x 3 x 5 (212 2 + + (2) Hint. Let 4 x 5 x 2 u2 + = . Then dx ) 5 x 4 ( du + = . Therefore dx ) 10 x 8 ( du 2 + = . The integral becomes C u du u 22 + =} Answer. C ) 4 x 5 x 2 (2 2 + + (3) Hint. Let 8 x 7 x 3 u2 + + = . Then dx ) 7 x 6 ( du + = and thus dx ) 7 x 6 ( du = . The integral becomes C4udu u43 + = }. Chapter 3-The method of substitution 70 Answer. C ) 8 x 7 x 3 (414 2 + + + (4) Hint. Let 5 x 2 x 4 x 2 u2 3 + + = . Then dx ) 2 x 8 x 6 ( du2 + = . Therefore dx ) 2 x 8 x 6 ( du2 + = . The integral takes the form C5udu u54 + = }. Answer. C ) 5 x 2 x 4 x 2 (515 2 3 + + + Exercise 2. Evaluate the integrals (1) dx3 x 5 x 25 x 42 } + (2) dxx 1x3} (3) dx) x 1 (x 22 } (4) dxx 1x 362} + (1) Hint. Let 3 x 5 x 2 u2 + = . Then dx ) 5 x 4 ( du = . The integral becomes C | u | lnudu+ =}. Answer. C | 3 x 5 x 2 | ln2 + + (2) Hint. Let x 1 u = . Then u 1 x = . Therefore du dx = and 3 3) u 1 ( x = . The integral becomes = + = } }duuu u 3 u 3 1duu) u 1 (3 2 3 C u31u23u 3 | u | ln du u du u 3 du 3udu3 2 2 + + + = + + = } } } }. Answer. C ) x 1 (31) x 1 (23) x 1 ( 3 | x 1 | ln3 2 + + + (3) Hint. Let x 1 u = . Then u 1 x = . Therefore du dx = . The integral becomes C | u | ln 2u2udu2udu2 duu) u 1 ( 22 2 + + = + = } } }. Answer. C | x 1 | ln 2x 12+ + (4) Hint. Let 3x u = . Then dx x 3 du2= . The integral becomes Chapter 3-The method of substitution 71 C u arctanu 1du2 + =+}. Answer. C ) x arctan(3 + II) Integrals of Irrational Functions Exercise 3. Evaluate the integrals (1) dx 4 x x 53 2} + (2) dx x 2 x 5 ) 1 x 10 (4 3} (3) } =2x 9dxI (4) dxx 1x 5I62} = (5) dx3 x5 x 4I2} = (6) dx5 x7 x 8I2} ++= (1) Hint. Let 4 x u3 + = . Then dx x 3 du2= . The integral takes the form C u u3235du u35+ =}. Answer. C 4 x ) 4 x (9103 3 + + + (2) Hint. Let x 2 x 5 u4 = . Then dx ) 1 x 10 ( 2 du3 = . The integral becomes C u u3221du u21+ =}. Answer. C x 2 x 5 ) x 2 x 5 (314 4 + (3) Hint. Let u 3 x = . Then du 3 dx = and ) u 1 ( 9 ) u 3 ( 9 x 92 2 2 = = . The integral becomes C u arcsinu 1du) u 1 ( 9du 32 2 + == } }. Answer. C3xarcsin +|.|

\| (4) Hint. Let 3x u = . Then dx x 3 du2= . The integral becomes C u arcsin35u 1du352 + =}. Answer. C ) x arcsin(353 + (5) Hint. We have 2 12 2 2I I dx3 x5dx3 xx 4dx3 x5 x 4 ==} } } Chapter 3-The method of substitution 72 For the integral 1I we use the substitution 3 x u2 = . We then have dx x 2 du = . The integral becomes u 4udu2 =}. The integral 2I is elementary (it is the number 11 appearing in the Table). Answer. C | 3 x x | ln 5 3 x 42 2 + + (6) Hint. We have 2 12 2 2I I dx5 x7dx5 xx 8dx5 x7 x 8+ =+++=++} } }. For the integral 1I we use the substitution 5 x u2 + = . We then have dx x 2 du = . The integral transforms into u 8udu4 =}. The integral 2I is elementary. Answer. C | 5 x x | ln 7 5 x 82 2 + + + + + . III) Integrals of Logarithmic Functions Exercise 4. Evaluate the integrals (1) }=2) x (ln xdxI (2) dxx) x (lnI4}= (3) dxx ln 1 x1I2} += (4) dxx ln 1 x3I2} = (1) Hint. We use the substitution x ln u = . Since dxx1du = , the integral transforms into Cu1udu2 + =}. Answer. Cx ln1+ (2) Hint. We use the substitution x ln u = . Since dxx1du = , the integral transforms into C5udu u54 + =}. Answer. C ) x (ln515 + . Chapter 3-The method of substitution 73 (3) Hint. We use the substitution x ln u = . Since dxx1du = , the integral transforms into C | u 1 u | lnu 1duI22 + + + =+= } Answer. C | ) x (ln 1 x ln | ln2 + + + (4) Hint. We use the substitution x ln u = . Since dxx1du = , the integral transforms into C u arcsin 3u 1du32 + =}. Answer. C ) x arcsin(ln 3 + IV) Integrals of Exponential Functions Exercise 5. Evaluate the integrals (1) dxe 1eIx 2x} += (2) dxe3 e 2 eIxx x 2} += (3) dx 5 Ix}= (4) dx23 4 2 3Ixx x} = (1) Hint. We use the substitution xe u = . Since dx e dux= and 2 x 2u 1 e 1 + = + , the integral transforms into C | u 1 u | lnu 1du22 + + + =+}. Answer. C | e 1 e | lnx 2 x + + + . (2) Hint. Since = |.|

\| + = +} }dxe32 e dxe3 e 2 exxxx x 2 dx e 3 dx 2 dx ex x} } } + = . The last integral is evaluated using x u = . Answer. C e 3 x 2 ex x + + + (3) Hint. Since 5 ln x xe 5 = , we use the substitution 5 ln x u = . Therefore dx ) 5 (ln du = . The integral then becomes C5 lnedu e5 ln1uu + =}. Chapter 3-The method of substitution 74 Answer. C5 ln5x+ . In general Ca lnadx axx + =}, 0 a > . (4) Hint. We have dx234 dx 3 dx234 3 dx23 4 2 3xxxxx x} } } } |.|

\| =||.|

\| = . Answer. C23) 2 / 3 ln(4x 3x+|.|

\| Exercise 6. Evaluate the integrals (1) dx e e 2 3 Ix x} = (2) dxx 2eIx} = (3) dxe 4eIx 2x} = (4) dx9 13Ixx} += (1) Hint. We use the substitution xe 2 3 u = . Since dx e 2 dux = , the integral transforms into C u u3221du u21+ = }. Answer. C e 2 3 ) e 2 3 (31x x + (2) Hint. We use the substitution x u = . Since x 2dxdu = , the integral transforms into C e du eu u + = }. Answer. C ex + . (3) Hint. We use the substitution xe u = . Since dx e dux= and 2 x 2u 4 e 4 = the integral transforms into } 2u 4du. Using a second substitution t 2 u = , the last integral transforms into C t arcsin) t 1 ( 4dt 22 + =}. Going back to the original variable, we get the following Answer. C2earcsinx+||.|

\|. (4) Hint. We use the substitution x3 u = . Since dx 3 ) 3 (ln dux= and Chapter 3-The method of substitution 75 2 xu 1 9 1 + = + , the integral transforms into C3 ln| u 1 u | lnu 1du3 ln122 ++ +=+}. Answer. C3 ln| 9 1 3 | lnx x++ +. V) Integrals of Trigonometric Functions Exercise 7. Evaluate the integrals (1) dx x sin ex cos} (2) dxex cosx sin } (3) dxx 3) x 2 ( sin} (4) dxx cos 5x sin2} + (5) } x cot 3 2 x sindx2 (6) dx x sin x cos5} (1) Hint. We use the substitution x cos u = . Since dx x sin du = , the integral transforms into C e du eu u + = }. Answer. C ex cos + (2) Hint. We use the substitution x sin u = . Since dx x cos du = , the integral transforms into C e du eu u + =}. Answer. C ex sin + (3) Hint. We use the substitution x 2 u = . Since dxx1du = , the integral transforms into C u cos31du u sin31+ =}. Answer. C ) x 2 ( cos31+ (4) Hint. We use the substitution x cos u = . Since dx x sin du = , the integral transforms into C | u 5 u | lnu 5du22 + + + =+}. Answer. C | x cos 5 x cos | ln2 + + + (5) Hint. We use the substitution x cot 3 2 u = . Since dxx sin3du2= , the integral transform into C u32udu31+ =}. Answer. C x cot 3 232+ Chapter 3-The method of substitution 76 (6) Hint. We use the substitution x sin u = . Since dx x cos du = , the integral transforms into C6udu u65 + =}. Answer. C6x sin6+ VI) Integrals of Inverse Trigonometric Functions Exercise 8. Evaluate the integrals (1) dxx 1x arcsinI2} = (2) dxx 1) x (arctan23} + (1) Hint. We use the substitution x arcsin u = . Since dxx 11du2= , the integral transforms into C2udu u I2+ = = }. Answer. C2) x (arcsin2+ (2) Hint. We use the substitution x arctan u = . Since dxx 11du2+= , the integral transforms into C4udu u I43 + = = }. Answer. C4) x (arcsin4+ VII) Integrals with two successive substitutions Exercise 9. Evaluate the integrals (1) dx4 eeIx 2x} += (2) dx16 xx 5I62} += (3) dx) x ln(cosx tanI }= (4) dx) x ln 3 2 ( xx ln 5I2 } += (1) Hint. We use the substitution xe u = . Since dx e dux= , the integral transforms into } + 4 udu2. Under a second substitution t 2 u = , the integral transforms into C t arctan21) 1 t ( 4dt 22 + =+}. Going back to the original variable, we get the following Answer. C2earctan21x+||.|

\|. Chapter 3-The method of substitution 77 (2) Hint. We use the substitution 3x u = . Since dx x 3 du2= , the integral transforms into } +16 udu352. Under a second transformation t 4 u = , the previous integral transforms into C t arctan125) 1 t ( 16dt 4352 + =+}. Answer. C4xarctan1253+||.|

\|. (3) Hint. We use the substitution ) x ln(cos u = . Since dx x tan du = , the integral transforms into C | u | lnudu+ = }. Answer. C | ) x ln(cos | ln + . (4) Hint. We use the substitution x ln u = . Since xdxdu = , the integral transforms into duu 3 2u 52} +. Under a second transformation 2u 3 2 t + = , since du u 6 dt = , the last integral transforms into C | t | ln65tdt65+ =}. Answer. C ) x ln 3 2 ( ln652 + + . VIII) Integrals with two different substitutions Exercise 10. Evaluate the integrals (1) dxx 1x arcsin 2 x 32} (2) dxx 1x arctan 6 x 52 } ++ (1) Hint. We have 2 12 2 2I I dxx 1x arcsin 2dxx 1x 3dxx 1x arcsin 2 x 3 ==} } } For the integral 1I we use the substitution 2x 1 u = . Since dx x 2 du = , this integral transforms into u 3udu23 = }. For the integral 2I we use the substitution x arcsin t = . Since dxx 11dt2= , this integral transforms into Chapter 3-The method of substitution 78 2t dt t 2 = }. Going back to the original variables, we get the following Answer. C ) x (arcsin x 1 32 2 + (2) Hint. We have 2 12 2 2I I dxx 1x arctan 6dxx 1x 5dxx 1x arctan 6 x 5+ =+++=++} } } For the integral 1I we use the substitution 2x 1 u + = . Since dx x 2 du = , this integral transforms into | u | ln25udu25=}. For the integral 2I we use the substitution x arctan t = . Since dxx 11dt2+= , this integral transforms into t t 4 dt t 6 =}. Going back to the original variable, we obtain the following Answer. C x arctan x arctan 4 ) x 1 ln(252 + + + VIII) Supplementary Examples Exercise 11. Evaluate the integrals (1) dx) 1 x (x4 59} + (2) dxx 4 1x 223} + (1) Hint. We use the substitution 1 x u5 + = . Since dx x 5 du4= , we have the following transformation du51) 1 u ( ) dx x ( x dx x4 5 9 = . Therefore the integral becomes Cu1151u1101udu51duudu51duu) 1 u (513 2 4 3 4 + + = =} } } Answer. C) 1 x (1151) 1 x (11013 5 2 5 ++ ++ (2) Hint. We use the substitution 2x 4 1 u + = . We then get dx x 8 du = and 41 ux2 = . Since we have the transformation Chapter 3-The method of substitution 79 du ) 1 u (161du8141 u2 ) dx x ( x 2 dx x 22 3 =|.|

\| = the integral transforms into C u41u u121udu81du u81duu1 u81+ = =} } } Answer. C x 4 141x 4 1 ) x 4 1 (1212 2 2 + + + + 4 Integration by Parts Learning Objectives In this Section the student will learn how to evaluate integrals using the integration by parts method. In applying the method of integration by parts, we shall use the following Theorem: Theorem 4.1. If ) x ( f and ) x ( g are integrable functions, then } } ' = ' dx ) x ( g ) x ( f ) x ( g ) x ( f dx ) x ( g ) x ( f (1) Proof. We start from the formula (Leibnitz rule) ) x ( g ) x ( f ) x ( g ) x ( f ] ) x ( g ) x ( f [ ' + ' = ' Integrating the above formula, we get } } } ' + ' = ' dx ) x ( g ) x ( f dx ) x ( g ) x ( f dx ] ) x ( g ) x ( f [ which is equivalent to } } ' + ' = dx ) x ( g ) x ( f dx ) x ( g ) x ( f ) x ( g ) x ( f from which we get relation (1). Note. Formula (1) can also be written as } } = ) x ( dg ) x ( f ) x ( g ) x ( f ) x ( df ) x ( g (2) Chapter 4-Integration by parts 81 4.1. Examples Example 1. Evaluate the Integral dx e x Ix}= Solution. In applying formula (1), it is instructive to form the table ) x ( f ' ) x ( g ) x ( f ) x ( g' xe x xe 1 Using formula (1) and the above table, we get C e e x dx e e x dx e x Ix x x x x + = = = } } Maple support: > f:=x*exp(x); := f x ex > int(f,x); ( ) + 1 x ex > Int(f,x)=int(f,x); = d(]((x exx ( ) + 1 x ex Example 2. Evaluate the Integral dx x ln x I3}= Solution. In applying formula (1), it is instructive to form the table ) x ( f ' ) x ( g ) x ( f ) x ( g' 3x x ln 4x4 x1 Using formula (1) and the above table, we get = = = = } } }dx x41x ln4xdxx14xx ln4xdx x ln x I34 4 43 C x161x ln4xC4x41x ln4x44 4 4+ = + = Maple support: Chapter 4-Integration by parts 82 > f:=x^3*log(x); := f x3( ) ln x > int(f,x); 14 x4( ) ln x x416 > Int(f,x)=int(f,x); = d(]((x3( ) ln x x 14 x4( ) ln x x416 Example 3. Evaluate the Integral dx x ln I }= Solution. In applying formula (1), it is instructive to form the table ) x ( f ' ) x ( g ) x ( f ) x ( g' 1 x ln x x1 Using formula (1) and the above table, we get C x x ln x dx x ln x dxx1x x ln x dx x ln I + = = = = } } } Maple support: > f:=log(x); := f ( ) ln x > int(f,x); x ( ) ln x x > Int(f,x)=int(f,x); = d(](( ) ln x x x ( ) ln x x Example 4. Evaluate the Integral dx x arctan I }= Solution. In applying formula (1), it is instructive to form the table ) x ( f ' ) x ( g ) x ( f ) x ( g' 1 x arctan x 2x 11+ Using formula (1) and the above table, we have Chapter 4-Integration by parts 83 =+ = = } }dxx 11x x arctan x dx x arctan I2 C ) x 1 ln(21x arctan x dxx 1x 221x arctan x22 + + =+ = } Note. The integral dxx 11x2} + was written in the form dxx 1x 2212} + and then was evaluated using the substitution 2x 1 u + = , converting it to C ) x 1 ln(21C | u | ln21udu212+ + = + =}. Maple support: > f:=arctan(x); := f ( ) arctan x > int(f,x); x ( ) arctan x12( ) ln + 1 x2 > Int(f,x)=int(f,x); = d(](( ) arctan x x x ( ) arctan x12( ) ln + 1 x2 Example 5. Evaluate the integral dx x arcsin I }= Solution. In applying formula (1), it is instructive to form the table ) x ( f ' ) x ( g ) x ( f ) x ( g' 1 x arcsin x 2x 11 Using formula (1) and the above table, we have = = = } }dxx 11x x arcsin x dx x arcsin I2 C x 1 x arcsin x dxx 1x 221x arcsin x22 + + =+ = } Chapter 4-Integration by parts 84 Note. The integral dxx 11x2} was written in the form dxx 1x 2212} and then was evaluated using the substitution 2x 1 u = , converting it into 2x 1 C uudu21 = + =}. Maple support: > f:=arcsin(x); := f ( ) arcsin x > int(f,x); + x ( ) arcsin x 1 x2 > Int(f,x)=int(f,x); = d(](( ) arcsin x x + x ( ) arcsin x 1 x2 Example 6. Evaluate the integrals dx x sin x I }= and dx x cos x J }= Solution. Using formula (1) and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' x sin x x cos 1 we have dx x cos x cos x dx x sin x I } } + = = C x sin x cos x + + = . Using formula (1) and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' x cos x x sin 1 we have dx x sin x sin x dx x cos x J } } + = = C x cos x sin x + = Maple support: Chapter 4-Integration by parts 85 > f:=x*sin(x); := f x ( ) sin x > int(f,x); ( ) sin x x ( ) cos x > Int(f,x)=int(f,x); = d(](x ( ) sin x x ( ) sin x x ( ) cos x > g:=x*cos(x); := g x ( ) cos x > int(g,x); + ( ) cos x x ( ) sin x > Int(g,x)=int(g,x); = d(](x ( ) cos x x + ( ) cos x x ( ) sin x Example 7. Evaluate the integrals dx x sin e Ix}= and dx x cos e Jx}= Solution. Using formula (1) and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' xe x sin xe x cos we have J x sin e dx x cos e x sin e dx x sin e Ix x x x = = = } } or x sin e J Ix= + Using formula (1) and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' xe x cos xe x sin we have I x cos e dx x sin e x cos e dx x cos e Jx x x x + = + = = } } or x cos e J Ix= + Chapter 4-Integration by parts 86 Solving the system x sin e J Ix= + and x cos e J Ix= + we obtain ) x cos e x sin e (21Ix x = and ) x cos e x sin e (21Jx x + = A generalization of this example is given in 9.21. Maple support: > f:=exp(x)*sin(x); := f ex( ) sin x > int(f,x); + 12ex( ) cos x12ex( ) sin x > Int(f,x)=int(f,x); = d(]((ex( ) sin x x + 12ex( ) cos x12ex( ) sin x > g:=exp(x)*cos(x); := g ex( ) cos x > int(g,x); + 12ex( ) cos x12ex( ) sin x > Int(g,x)=int(g,x); = d(]((ex( ) cos x x + 12ex( ) cos x12ex( ) sin x Example 8. Evaluate the integrals dxx sinxI2}= and dxx cosxJ2}= Solution. Using formula (1) and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' x sin12 x x cot 1 we obtain Chapter 4-Integration by parts 87 = + = + = = } } }dxx sinx cosx cot x dx x cot x cot x dxx sinxI2 C | x sin | ln x cot x dxx sin) x (sinx cot x + + ='+ = } Using formula (1) and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' x cos12 x x tan 1 we obtain = + = = = } } }dxx cosx sinx tan x dx x tan x tan x dxx cosxJ2 C | x cos | ln x tan x dxx cos) x (cosx tan x + =' = } Maple support: > f:=x/sin(x)^2; := f x( ) sin x2 > int(f,x); + x ( ) cot x ( ) ln ( ) sin x > Int(f,x)=int(f,x); = d(](((( x( ) sin x2 x + x ( ) cot x ( ) ln ( ) sin x > g:=x/cos(x)^2; := g x( ) cos x2 > int(g,x); + x ( ) tan x ( ) ln ( ) cos x > Int(g,x)=int(g,x); = d(](((( x( ) cos x2 x + x ( ) tan x ( ) ln ( ) cos x Chapter 4-Integration by parts 88 Example 9. Evaluate the integrals dx x tan x I2}= and dx x cot x J2}= Solution. We have the formula: 1x cos1x tan22 = . Therefore =|.|

\| = = } }dx 1x cos1x dx x tan x I22 C2x| x cos | ln x tan x dx x dxx cosx22 + = = } } where we have used the value of the integral dxx cosx2 } we have found in the previous Example. We have the formula: 1x sin1x cot22 = . Therefore =|.|

\| = = } }dx 1x sin1x dx x cot x J22 C2x| x sin | ln x cot x dx x dxx sinx22 + + = = } } where we have used the value of the integral dxx sinx2 } we have found in the previous Example. Maple support: > f:=x*tan(x)^2; := f x ( ) tan x2 > int(f,x); x ( ) tan x x2212( ) ln + 1 ( ) tan x2 > Int(f,x)=int(f,x); = d(]((x ( ) tan x2x x ( ) tan x x2212( ) ln + 1 ( ) tan x2 > g:=x*cot(x)^2; := g x ( ) cot x2 Chapter 4-Integration by parts 89 > int(g,x); + x12 x2( ) tan x( ) tan x( ) ln ( ) tan x12( ) ln + 1 ( ) tan x2 > Int(g,x)=int(g,x); = d(]((x ( ) cot x2x + x12 x2( ) tan x( ) tan x( ) ln ( ) tan x12( ) ln + 1 ( ) tan x2 Example 10. Evaluate the integral dx) x 1 (xI2 22} += Solution. Using formula (1) and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' 2 2) x 1 (x+ x ) x 1 ( 212+ 1 we obtain =+++ =+= } }dxx 1121) x 1 ( 2xdx) x 1 (xI2 2 2 22 C x arctan21) x 1 ( 2x2 + ++ = Maple support: > f:=x^2/(1+x^2)^2; := f x2( ) + 1 x22 > int(f,x); 12( ) arctan x x2 ( ) + 1 x2 > Int(f,x)=int(f,x); = d(](((((( x2( ) + 1 x22 x 12( ) arctan x x2 ( ) + 1 x2 Chapter 4-Integration by parts 90 Example 11. Evaluate the integral dx ) x (arctan x I2}= Solution. Using formula (1) and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' x 2) x (arctan 2x2 ) x (arctanx 122+ we obtain =+ = = } }dx ) x (arctanx 122x) x (arctan2xdx ) x (arctan x I22222 J ) x (arctan2xdx ) x (arctanx 1x) x (arctan2x222222 =+ = } The integral dx ) x (arctanx 1xJ22} += can be written as = |.|

\|+ = }dx ) x (arctanx 111 J2 L K dx ) x (arctanx 11dx ) x (arctan2 =+ = } } The integral dx ) x (arctan K }= was evaluated above in Example 4: ) x 1 ln(21x arctan x dx ) x (arctan K2+ = = } The integral dx ) x (arctanx 11L2 } += can be evaluated using the substitution x arctan u = . Since dxx 11du2+= , we find Chapter 4-Integration by parts 91 222) x (arctan212udu u dx ) x (arctanx 11L = = =+= } } Collecting everything together, we obtain = = = ) L K ( ) x (arctan2xJ ) x (arctan2xI2222 C ) x (arctan21) x 1 ln(21x arctan x ) x (arctan2x2 2 22+ + + + = Maple support: > f:=x*arctan(x)^2; := f x ( ) arctan x2 > int(f,x); + + 12( ) arctan x212( ) arctan x2x2x ( ) arctan x12( ) ln + 1 x2 > Int(f,x)=int(f,x); = d(]((x ( ) arctan x2x + + 12( ) arctan x212( ) arctan x2x2x ( ) arctan x12( ) ln + 1 x2 Example 12. Evaluate the integral dx ) x (arctan x I2}= Solution. Using formula (1) and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' 2x x arctan 3x3 2x 11+ we obtain =+ = = } }dxx 1x31) x (arctan3xdx ) x (arctan x I23 32 J31) x (arctan3x3 = The integral dxx 1xJ23} += can be evaluated as follows Chapter 4-Integration by parts 92 =+ =|.|

\|+ =+= } } } }dxx 1xdx x dxx 1xx dxx 1xJ2 2 23 ) x 1 ( ln212xdxx 1x 2212x2222+ =+ = } Therefore we obtain the value of the integral = = =}J31) x (arctan3xdx ) x (arctan x I32 C ) x 1 ( ln212x31) x (arctan3x22 3+||.|

\|+ = or C ) x 1 ln(61x61) x (arctan3xdx ) x (arctan x I2 232+ + + = =} Maple support: > f:=x^2*arctan(x); := f ( ) arctan x x2 > int(f,x); + 13 x3( ) arctan x x2616( ) ln + 1 x2 > Int(f,x)=int(f,x); = d(]((( ) arctan x x2x + 13 x3( ) arctan x x2616( ) ln + 1 x2 4.2. Reduction Formulas. Example 1. Evaluate the integrals (1) dx e x Ix1 }= (2) dx e x Ix 22 }= (3) dx e x Ix 33 }= Solution. (1) This integral was evaluated previously in 4.1, Example 1: C e e x dx e x Ix x x1 + = =} (1) (2) Using the formula Chapter 4-Integration by parts 93 } } ' = ' dx ) x ( g ) x ( f ) x ( g ) x ( f dx ) x ( g ) x ( f and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' xe 2x xe x 2 we have 1x 2 x x 2 x 22I 2 e x dx e x 2 e x dx e x I = = = } } (2) where 1I is given by (1). (3) Using the formula } } ' = ' dx ) x ( g ) x ( f ) x ( g ) x ( f dx ) x ( g ) x ( f and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' xe 3x xe 2x 3 we have 2x 3 x 2 x 3 x 33I 3 e x dx e x 3 e x dx e x I = = = } } (3) where 2I is given by (2). We see from the above Examples that we need to repeat integration so many times as long as the exponent of x becomes higher and higher. It is further observed that each integral is being expressed in terms of the integral having exponent of x less by one: 0x x1I e x dx e x I = =} 1x 2 x 22I 2 e x dx e x I = =} 2x 3 x 33I 3 e x dx e x I = = } Therefore considering the integral dx e x Ix nn }= , one expects that 1 nx n x nnI n e x dx e x I = = } ( 1 n > ) (4) Chapter 4-Integration by parts 94 This is indeed the case by considering the following table ) x ( f ' ) x ( g ) x ( f ) x ( g' xe nx xe 1 nx n Using the table above we establish the formula 1 nx n x 1 n x n x nnI n e x dx e x n e x dx e x I = = = } } Therefore 1 nx nnI n e x I = , 1 n > The above formula is a reduction formula for the integral dx e x Ix nn }= . It is sometimes called iteration formula or recurrence relation. It means that the evaluation of dx e x Ix nn }= is reduced to the evaluation of 1 nI , the evaluation of 1 nI is reduced to the evaluation of 2 nI , etc using successively formula (3). Using the above formula we can evaluate for example dx e xx 4} without using four times integration by parts. For 4 n = we have from (4): 3x 44I 4 e x I = For 3 n = we have from (4): 2x 33I 3 e x I = For 2 n = we have from (4): 1x 22I 2 e x I = For 1 n = we have from (4): 0x1I e x I = where it is obvious that x x0e dx e I = = } Going from the last equation to the first one, we get - x x0x1e e x I e x I = = - x x x 2 x x x 21x 22e 2 e x 2 e x ) e e x ( 2 e x I 2 e x I + = = = Chapter 4-Integration by parts 95 - = + = = ) e 2 e x 2 e x ( 3 e x I 3 e x Ix x x 2 x 32x 33 x x x 2 x 3e 6 e x 6 e x 3 e x + = - = + = = ) e 6 e x 6 e x 3 e x ( 4 e x I 4 e x Ix x x 2 x 3 x 43x 44 x x x 2 x 3 x 4e 24 e x 24 e x 12 e x 4 e x + + = (an arbitrary constant has been omitted). Some remarks are now in order: - No all reduction formulas can be established using the integration by parts method. - There are reduction formulas depending on two (or more) indices. See for example reduction formulas for integration of trigonometric functions ( - Computer Algebra Systems evaluate immediately integrals which are evaluated by hand using reduction formulas. Maple support: > f:=x^4*exp(x); := f x4ex > int(f,x); ( ) + + 24 24 x 12 x24 x3x4ex > Int(f,x)=int(f,x); = d(]((x4exx ( ) + + 24 24 x 12 x24 x3x4ex 4.3. Integrals in which integration by parts follows a substitution. In many cases we have to make a substitution first and then integrate by parts. Example 1. Evaluate the integral }= dxx) x ln(lnI Chapter 4-Integration by parts 96 Solution. Using the substitution x ln u = , since xdxdu = , the integral becomes }= du u ln I which was evaluated before (Example 3, 4.1): C u u ln u du u ln + =} Going back to the original variable, we obtain the value of the integral: C ) x (ln ) x ln(ln ) x (ln dxx) x ln(lnI + = = } Maple support: > f:=log(log(x))/x; := f( ) ln ( ) ln xx > int(f,x); ( ) ln ( ) ln x ( ) ln x ( ) ln x > Int(f,x)=int(f,x); = d(]((((( ) ln ( ) ln xx x ( ) ln ( ) ln x ( ) ln x ( ) ln x Example 2. Evaluate the integral dx e x I2x 3} = Solution. Using the substitution 2x u = , since dx x 2 du = and du u21du21) u ( dx x x dx x2 3=|.|

\| = = the integral transforms into du e u21u}. The last integral was evaluated in 4.1, Example 1: u u ue e u du e u =} Going back to the original variable, we find Chapter 4-Integration by parts 97 C e21e ) x (21dx e x I2 2 2x x 2 x 3+ = = } Maple support: > f:=x^3*exp(-x^2); := f x3e( ) x2 > int(f,x); 12( ) + 1 x2e( ) x2 > Int(f,x)=int(f,x); = d(](((x3e( ) x2x 12( ) + 1 x2e( ) x2 Example 3. Evaluate the integral dx e Ix}= Solution. Using the substitution u x = , 2u x = and du u 2 dx = , the integral transforms into du e u 2u}. The last integral was evaluated in 4.1, Example 1: u u ue e u du e u =} Going back to the original variable, we find C e 2 e x 2 dx e Ix x x + = = } Maple support: > f:=exp(sqrt(x)); := f e( ) x > int(f,x); 2 e( ) xx 2 e( ) x > Int(f,x)=int(f,x); = d(](((e( ) xx 2 e( ) xx 2 e( ) x Example 4. Evaluate the integrals dx ) x cos(ln I }= and dx ) x sin(ln J }= Chapter 4-Integration by parts 98 Solution. Under the substitution x ln u = , since ue x = and du e dxu= , the integral I transforms into du e ) u (cosu}, which was evaluated before (in 4.1, Example 7): ) u cos e u sin e (21du e ) u (cosu u u + =} Therefore going back to the original variable, we find C )] x cos(ln x ) x sin(ln x [21dx ) x cos(ln I + + = = } Similarly, under the substitution x ln u = , since ue x = and du e dxu= , the integral J transforms into du e ) u (sinu}, which was evaluated before (in 4.1, Example 7): ) u cos e u sin e (21du e ) u (sinu u u =} Therefore going back to the original variable, we find C )] x cos(ln x ) x sin(ln x [21dx ) x sin(ln I + = = } Maple support: > f:=cos(log(x)); := f ( ) cos ( ) ln x > int(f,x); + 12( ) cos ( ) ln x x12( ) sin ( ) ln x x > Int(f,x)=int(f,x); = d(](( ) cos ( ) ln x x + 12( ) cos ( ) ln x x12( ) sin ( ) ln x x > g:=sin(log(x)); := g ( ) sin ( ) ln x > int(g,x); + 12( ) cos ( ) ln x x12( ) sin ( ) ln x x > Int(g,x)=int(g,x); Chapter 4-Integration by parts 99 = d(](( ) sin ( ) ln x x + 12( ) cos ( ) ln x x12( ) sin ( ) ln x x Example 5. Evaluate the integral dx x sin I }= Solution. We make the substitution x sin u = (1) Inverting the sine function, we find u arcsin x = Squaring both members of the previous relation, we obtain 2) u (arcsin x = (2) from which we calculate the differential: du ) u (arcsinu 12dx2= The integral thus becomes } du ) u (arcsinu 12u2 (3) Using the formula } } ' = ' du ) u ( g ) u ( f ) u ( g ) u ( f du ) u ( g ) u ( f and the table ) u ( f ' ) u ( g ) u ( f ) u ( g' 2u 1u u arcsin 2u 1 2u 11 we get for the integral in (3): duu 11) u 1 ( 2 ) u (arcsin u 1 222 2 } or C u 2 ) u (arcsin u 1 2 du 2 ) u (arcsin u 1 22 2 + + = + } Going back to the original variable, we get the following value of the integral: Chapter 4-Integration by parts 100 C x sin 2 x ) x sin ( 1 2 dx x sin I2 + + = =} C x sin 2 x ) x (cos 2 + + = Maple support: > f:=sin(sqrt(x)); := f ( ) sin x > int(f,x); 2 ( ) sin x 2 x ( ) cos x > Int(f,x)=int(f,x); = d(]((( ) sin x x 2 ( ) sin x 2 x ( ) cos x 4.4. Supplementary Exercises. Exercise 1. Evaluate the integral }= dxxx lnI32 Solution. Using the formula } } ' = ' dx ) x ( g ) x ( f ) x ( g ) x ( f dx ) x ( g ) x ( f and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' 32x1 x ln 3x 3 x1 we have = = = } }dxxx3 ) x (ln x 3 dxxx lnI3332 C x 9 ) x (ln x 3 dx x 3 ) x (ln x 33 3323 + = = } Maple support: > f:=log(x)/x^(2/3); := f( ) ln xx( ) / 2 3 > int(f,x); Chapter 4-Integration by parts 101 3 ( ) ln x x( ) / 1 39 x( ) / 1 3 > Int(f,x)=int(f,x); = d(](((((( ) ln xx( ) / 2 3 x 3 ( ) ln x x( ) / 1 39 x( ) / 1 3 Exercise 2. Evaluate the integral }= dxx) x (lnI22 Solution. Using the formula } } ' = ' dx ) x ( g ) x ( f ) x ( g ) x ( f dx ) x ( g ) x ( f and the table ) x ( f ' ) x ( g ) x ( f ) x ( g' 2x1 2) x (ln x1 x1) x (ln 2 we have J 2 ) x (lnx1dxxx ln2 ) x (lnx1dxx) x (lnI22222+ = + = = } } For the integral }= dxxx lnJ2, using the table ) x ( f ' ) x ( g ) x ( f ) x ( g' 2x1 x ln x1 x1 we have x1xx lndxx1xx lndxxx lnJ2 2 = + = = } } Collecting everything together, we obtain Cx) x ln 1 ( 2) x (lnx1dxx) x (lnI222++ = = } Maple support: > f:=log(x)^2/x^2; Chapter 4-Integration by parts 102 := f( ) ln x2x2 > int(f,x); ( ) ln x2x2 ( ) ln xx2x > Int(f,x)=int(f,x); = d(](((((( ) ln x2x2 x ( ) ln x2x2 ( ) ln xx2x Exercise 3. Evaluate the integral }= dx x cos x sin x I Solution. Since x 2 sin21) x cos x sin 2 (21x cos x sin = = , the integral takes the form } } = = dx x 2 sin x21dx x cos x sin x I Using the table ) x ( f ' ) x ( g ) x ( f ) x ( g' x 2 sin x 2x 2 cos 1 we have } } = = dx x 2 sin x21dx x cos x sin x I C2x 2 sin412x 2 cos xdx x 2 cos412x 2 cos x+ + = + = } C x 2 sin81) x 2 cos x (21+ + = Maple support: > f:=x*sin(x)*cos(x); := f x ( ) sin x ( ) cos x > int(f,x); + + 12 x ( ) cos x214( ) sin x ( ) cos x x4 Chapter 4-Integration by parts 103 > Int(f,x)=int(f,x); = d(](x ( ) sin x ( ) cos x x + + 12 x ( ) cos x214( ) sin x ( ) cos x x4 Exercise 4. Evaluate the integral }= dxx sinx cos xI3 Solution. Using the table ) x ( f ' ) x ( g ) x ( f ) x ( g' x sinx cos3 x x sin 212 1 and the integration by parts formula, we find C x cot21x sin 2xx sindx21x sin 2xdxx sinx cos xI2 2 2 3 + = + = = } } Maple support: > f:=x*cos(x)/sin(x)^3; := f x ( ) cos x( ) sin x3 > int(f,x); 12 x( ) sin x212( ) cos x( ) sin x > Int(f,x)=int(f,x); = d(]((((x ( ) cos x( ) sin x3 x 12 x( ) sin x212( ) cos x( ) sin x Exercise 5. Evaluate the integral }= dxe) 2 / x ( cosIx2 Solution. We have the formula 2x cos 1) 2 / x ( cos2 += Therefore the integral becomes = + =+= } } }dxex cos21dxe121dxex cos 121Ix x x Chapter 4-Integration by parts 104 J21e21dx x cos e21dx e21x x x + = + = } } Using the table ) x ( f ' ) x ( g ) x ( f ) x ( g' xe x cos xe x sin we have dx x sin e x cos e dx x cos e Jx x x} } = = We now have to calculate the integral dx x sin ex} . Using the table ) x ( f ' ) x ( g ) x ( f ) x ( g' xe x sin xe x cos we have J x sin e dx x cos e x sin e dx x sin ex x x x + = + = } } Therefore ) J x sin e ( x cos e dx x cos e Jx x x + = = } from which we obtain ) x cos x (sin e21Jx = Collecting everything together, we find C ) x cos x (sin e41e21Ix x + + = Maple support: > f:=cos(x/2)^2/exp(x); := f |\

|.||cos x22ex > int(f,x); Chapter 4-Integration by parts 105 +