Lesson 10 techniques of integration
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The document discusses various techniques for integration including integration by parts, trigonometric substitution, algebraic substitution, reciprocal substitution, and partial fraction decomposition. Integration by parts allows one to integrate products of functions. Trigonometric substitution transforms integrals into ones involving trigonometric functions that can be evaluated using basic formulas. Algebraic substitution rationalizes irrational integrals. Partial fraction decomposition expresses rational functions as sums of simpler fractions to facilitate integration.
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Lesson 10 techniques of integration
- 2. OBJECTIVES At the end of the lesson, the student should be able to: • find an antiderivative using integration by parts. • use trigonometric substitution to solve an integral. • use algebraic substitution to solve an integral. • use reciprocal substitution to solve an integral. • evaluate an indefinite integral involving rational functions of sine and cosine. • use partial decomposition with linear factors and quadratic factors to integrate rational functions.
- 3. INTEGRATION BY PARTS • This technique can be applied to a wide variety of functions and is particularly useful for integrands involving products of algebraic and transcendental functions. • If u and v are functions of x and have continuous derivatives, then 𝒖𝒅𝒗 = 𝒖𝒗 − 𝒗𝒅𝒖
- 4. GUIDELINES FOR INTEGRATION BY PARTS 1. Try letting dv be the most complicated portion of the integrand that fits a basic integration rule. Then u will be the remaining factor(s) of the integrand. 2. Try letting u be the portion of the integrand whose derivative is a function simpler than u. Then dv will be the remaining factor(s) of the integrand. Note: dv always includes the dx of the original integrand.
- 5. SUMMARY OF COMMON INTEGRALS USING INTEGRATION BY PARTS 1. For integrals of the form 𝑥 𝑛 𝑒 𝑎𝑥 𝑑𝑥, 𝑥 𝑛 𝑠𝑖𝑛𝑎𝑥𝑑𝑥, 𝑜𝑟 𝑥 𝑛 𝑐𝑜𝑠𝑎𝑥𝑑𝑥 let 𝑢 = 𝑥 𝑛 𝑎𝑛𝑑 𝑙𝑒𝑡 𝑑𝑣 = 𝑒 𝑎𝑥 𝑑𝑥, 𝑠𝑖𝑛𝑎𝑥𝑑𝑥, 𝑜𝑟 𝑐𝑜𝑠𝑎𝑥𝑑𝑥. 2. For the integrals of the form 𝑥 𝑛 𝑙𝑛𝑥𝑑𝑥, 𝑥 𝑛 𝑎𝑟𝑐𝑠𝑖𝑛𝑎𝑥𝑑𝑥, 𝑜𝑟 𝑥 𝑛 𝑎𝑟𝑐𝑡𝑎𝑛𝑎𝑥𝑑𝑥 let 𝑢 = 𝑙𝑛𝑥, 𝑎𝑟𝑐𝑠𝑖𝑛𝑎𝑥, 𝑜𝑟 𝑎𝑟𝑐𝑡𝑎𝑛𝑎𝑥 𝑎𝑛𝑑 𝑙𝑒𝑡𝑑𝑣 = 𝑥 𝑛 𝑑𝑥.
- 6. 3. For integrals of the form 𝑒 𝑎𝑥 𝑠𝑖𝑛𝑏𝑥𝑑𝑥 𝑜𝑟 𝑒 𝑎𝑥 𝑐𝑜𝑠𝑏𝑥𝑑𝑥 let 𝑢 = 𝑠𝑖𝑛𝑏𝑥 𝑜𝑟 𝑐𝑜𝑠𝑏𝑥 𝑎𝑛𝑑 𝑙𝑒𝑡 𝑑𝑣 = 𝑒 𝑎𝑥 𝑑𝑥. TABULAR FORM In problems involving repeated applications of integration by parts, a tabular form can help to organize the work. This method works well for integrals of the form 𝑥 𝑛 𝑠𝑖𝑛𝑎𝑥𝑑𝑥, 𝑥 𝑛 𝑐𝑜𝑠𝑎𝑥𝑑𝑥, 𝑎𝑛𝑑 𝑥 𝑛 𝑒 𝑎𝑥 𝑑𝑥.
- 7. TABULAR FORM Cxxxxxxxdx cos2sin2cossin 22 differential integral 2x -cosx 2 -sinx 0 cosx xx sin2 2 x xsin + - +
- 8. TABULAR FORM xexexexxde xxxx cossincoscos differential integral -sinx -cosx xex cos xcos + - x e x e x e Cxexexxdexe xxxx sincoscoscos Cxexexxde xxx sincoscos2 Cxexexxde xxx sin 2 1 cos 2 1 cos
- 9. TABULAR FORM xdex x 3 xxdx 5cossin
- 10. EXAMPLE Find the integral. 1. 𝑐𝑠𝑐3 𝑥𝑑𝑥 6. 𝑒−2𝑥 𝑥3 𝑑𝑥 2. 𝑥𝑒2𝑥 (2𝑥+1)2 𝑑𝑥 7. 𝑐𝑜𝑠 𝑥𝑑𝑥 3. 𝑎𝑟𝑐𝑡𝑎𝑛𝑥𝑑𝑥 8. 𝑙𝑛 𝑥2 + 1 𝑑𝑥 4. 𝑥3 𝑥2+1 𝑑𝑥 9. 𝑥4 𝑙𝑛𝑥𝑑𝑥 5. 𝑒3𝑥 𝑐𝑜𝑠4𝑥𝑑𝑥 10. 𝑥 5+4𝑥 𝑑𝑥
- 11. TRIGONOMETRIC SUBSTITUTION • A change of variable in which a trigonometric function is substituted for the variable of integration is called a trigonometric substitution. In many cases, this type of substitution is used, like algebraic substitution to rationalize certain irrational integrands. However, It can also be used in some cases to simplify the integrand even if no radicals are present.
- 12. • If the integrand contains the combination : 𝒂 𝟐 − 𝒖 𝟐 let u= a sin𝜽 𝒂 𝟐 + 𝒖 𝟐 let u= a tan𝜽 𝒖 𝟐 − 𝒂 𝟐 let u=a sec𝜽 • In all cases, a is a constant. It is easy to show that each of the above substitutions will reduced the corresponding combination to a perfect square. Thus, 𝒂 𝟐 − 𝒖 𝟐 becomes 𝒂 𝟐 𝒄𝒐𝒔 𝟐 𝜽 𝒂 𝟐 + 𝒖 𝟐 becomes 𝒂 𝟐 𝒔𝒆𝒄 𝟐 𝜽 𝒖 𝟐 − 𝒂 𝟐 becomes 𝒂 𝟐 𝒕𝒂𝒏 𝟐 𝜽 • If the integrand involves only the square root of any of the combinations, it is automatically rationalized by thesubstitution prescribed.
- 13. EXAMPLE Find the indefinite integral. 1. 𝑑𝑥 𝑥2 𝑎2−𝑥2 2. 4+𝑥2 𝑥4 𝑑𝑥 3. 𝑥2 𝑑𝑥 (9𝑥2+1) 3 2 4. 1 4+4𝑥2+𝑥2 𝑑𝑥 Evaluate the definite integral. 1. 1 2 𝑥3 𝑥2 − 1𝑑𝑥 2. 0 2𝑎 𝑑𝑥 𝑥 2𝑎𝑥−𝑥2
- 14. INTEGRATION BY ALGEBRAIC SUSTITUTION • If the substitution involves only algebraic terms, it is called an algebraic substitution. Generally, the purpose of algebraic substitution is to rationalize irrational integrands. Thus, this type of substitution usually involves replacement of radical expression by a new variable. • If a definite integral is to be evaluated by using a substitution, it is usually preferable to change the limits so as to correspond with the change in variable. In this manner, there will be no need to return to the original variable of integration.
- 15. EXAMPLE Evaluate by algebraic substitution. 1. 𝑑𝑥 1+ 3 𝑥+1 2. 𝑥2 −4 𝑥 dx 3. 𝑑𝑥 𝑥 +4 𝑥 4. 𝑑𝑥 𝑥+ 2𝑥+1 5. sin 𝑥 𝑑𝑥 6. 𝑑𝑥 1+ 𝑥 7. 0 𝑎 𝑥3 𝑎2 − 𝑥2 𝑑𝑥 8. 0 4 𝑙𝑛 𝑧 + 2 𝑑𝑧
- 16. RECIPROCAL SUBSTITUTION Another substitution which is quite useful is 𝒙 = 𝟏 𝒛 , 𝒅𝒙 = − 𝒅𝒛 𝒛 𝟐 which is called reciprocal substitution. This method unlike the previous substitution will not convert an irrational integrand to a rational one. However, when it is indicated, this substitution will transform the integral so that generally the integration formulas can be applied.
- 17. EXAMPLE • Evaluate the following integrals. 1. 4−𝑥2 𝑥4 𝑑𝑥4. 5 4 5 3 𝑑𝑥 𝑥2 𝑥2−1 2. 𝑑𝑥 𝑥 𝑥2+4𝑥−4 5. 1 6 3+𝑥2 𝑥4 𝑑𝑥 3. 1 4 𝑑𝑥 𝑥 𝑥2+𝑥−2
- 18. HALF ANGLE SUBSTITUTION • An integral which is a rational functions of the trigonometric function of an angle u can be transformed by means of the substitution 𝑧 = 𝑡𝑎𝑛 1 2 𝑢, which is equivalent to the relations: • 𝑠𝑖𝑛𝑢 = 2𝑧 1+𝑧2 , 𝑐𝑜𝑠𝑢 = 1−𝑧2 1+𝑧2, 𝑑𝑢 = 2𝑑𝑧 1+𝑧2
- 19. EXAMPLE Find or evaluate the following integrals 1. 𝑑𝑥 5−3 cos 𝑥 2. 0 𝜋 2 𝑑𝑥 1+sin 𝑥+cos 𝑥 3. 2𝜋 3 𝜋 2 𝑑𝑥 tan 𝑥+sin 𝑥 4. 0 𝜋 6 𝑑𝑥 3+5 sin 3𝑥 5. 𝑑𝑥 4 sec 𝑥+3 6. 4 𝑐𝑠𝑐𝜃−𝑐𝑜𝑡𝜃 𝑑𝜃
- 20. INTEGRATION OF RATIONAL FUNCTION BY PARTIAL FRACTION DEFINITION • A rational function is a function which can be expressed as the quotient of two polynomial functions. That is, a function H is a rational function if 𝑯 𝒙 = 𝒇(𝒙) 𝒈(𝒙) ,where both f(x) and g(x)are polynomials. In general, we shall be concerned in integrating expressions of the form: 𝒇(𝒙) 𝒈(𝒙) 𝒅𝒙
- 21. The method of partial fractions is an algebraic procedure of expressing a given rational function as a sum of simpler fractions which is called the partial fraction decomposition of the original rational function. The rational function must be in its proper fraction form to use the partial fraction method.
- 22. • Four cases shall be considered.. Case 1. Distinct linear factor in the denominator. Case 2. Repeated linear factor in the denominator. Case 3. Distinct quadratic factor in the denominator. Case 4. Repeated quadratic factor in the denominator.
- 23. CASE I. DISTINCT LINEAR FACTOR IN THE DENOMINATOR For each linear factor 𝒂𝒙𝒊 + 𝒃𝒊 of the denominator, there corresponds a partial fraction having that factor as the denominator and a constant numerator; that is 𝒇(𝒙) 𝒈(𝒙) = 𝑨 𝒂 𝟏 𝒙 + 𝒃 𝟏 + 𝑩 𝒂 𝟐 𝒙 + 𝒃 𝟐 +∙∙∙ + 𝑵 𝒂 𝒏 𝒙 + 𝒃 𝒏 where A, B,....N are constants to be determined.Thus, 𝒇(𝒙) 𝒈(𝒙) = 𝑨 𝒂 𝟏 𝒙+𝒃 𝟏 𝒅𝒙+ 𝑩 𝒂 𝟐 𝒙+𝒃 𝟐 𝒅𝒙 + ⋯ 𝑵 𝒂 𝒏 𝒙+𝒃 𝒏 𝒅𝒙
- 24. CASE II: REPEATED LINEAR FACTORS If the linear factor 𝒂𝒙 + 𝒃 𝒏 appears as the denominator of the rational function for each repeated linear factor of the denominator, there corresponds a series of partial fractions, where A, B, C, …, N are constants to be determined The degree n of the repeated linear factor gives the number of partial fractions in a series. Thus, n32 bax N ... bax C bax B bax A dx bax N ...dx bax C dx bax B dx bax A dx )x(g )x(f n32
- 25. CASE III: QUADRATIC FACTORS • For each non-repeated irreducible quadratic factor 𝒂𝒙 𝟐 + 𝒃𝒙 + 𝒄 of the denominator there corresponds a partial fraction of the form where A, B, …..N are constants to be determined. Thus, nnn nn cxbxa MbxaN cxbxa DbxaC cxbxa BbxaA xg xf 2 22 2 2 22 11 2 1 111 )2( ... )2()2( )( )( nnn nn cxbxa MbxaN cxbxa DbxaC cxbxa BbxaA dx xg xf 2 22 2 2 22 11 2 1 111 )2( ... )2()2( )( )(
- 26. CASE IV: REPEATED QUADRATIC FACTORS For each repeated irreducible quadratic factor 𝒂𝒙 𝟐 + 𝒃𝒙 + 𝒄 𝒏 of the denominator there corresponds a partial fraction of the form where A, B, …..N are constants to be determined. Thus, n cbxax MbaxN cbxax DbaxC cbxax BbaxA xg xf )( )2( ... )( )2()2( )( )( 2222 n cbxax MbaxN cbxax DbaxC cbxax BbaxA xg xf )( )2( ... )( )2()2( )( )( 2222
- 27. GUIDELINES FOR SOLVING THE BASIC EQUATION • LINEAR FACTORS 1. Substitute the roots of the distinct linear factors in the basic equation. 2. For repeated linear factors, use the coefficients determined in guideline 1 to rewrite the basic equation. Then substitute other convenient values of x and solve for the remaining coefficients.
- 28. • QUADRATIC FACTORS 1. Expand the basic equation 2. Collect terms according to power of x. 3. Equate the coefficients of like powers to obtain a system of linear equations involving A, B, C, and so on. 4. Solve the system of linear equations.
- 29. EXAMPLE I Use partial fraction to find the integral. 1. 𝑥2+12𝑥+12 𝑥3−4𝑥 𝑑𝑥 6. 𝑥2−4𝑥+7 𝑥3−𝑥2+𝑥+3 𝑑𝑥 2. 𝑥2+3𝑥−4 𝑥3−4𝑥2+4𝑥 𝑑𝑥 7. 𝑥2+𝑥+3 𝑥4+6𝑥2+9 𝑑𝑥 3. 𝑥2 𝑥4−2𝑥2−8 𝑑𝑥 8. 𝑥3−𝑥+3 𝑥2+𝑥−2 𝑑𝑥 4. 𝑥 16𝑥4−1 𝑑𝑥 9. 6𝑥 𝑥3−8 𝑑𝑥 5. 𝑥2 𝑥2+9 2 𝑑𝑥 10. 𝑥 2𝑥−9 𝑥3−6𝑥2+12𝑥−8 𝑑𝑥
- 30. • II Use substitution to find integral. 1. 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 𝑐𝑜𝑠𝑥−1 𝑑𝑥 2. 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥+𝑠𝑖𝑛2 𝑥 𝑑𝑥 3. 𝑠𝑒𝑐2 𝑥 𝑡𝑎𝑛2 𝑥+5𝑡𝑎𝑛𝑥+6 𝑑𝑥 4. 𝑒 𝑥 𝑒 𝑥−1 𝑒 𝑥+4 𝑑𝑥 5. 𝑒 𝑥 𝑒2𝑥+1 𝑒 𝑥−1 𝑑𝑥