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# Class Note for MATH 250A with Professor Lega at UA

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Date Created: 02/06/15

Calculus and Differential Equations l MATH 250 A Methods of integration ll Methods o tegration Calculus and Differential Equations The method of partial fractions 0 The purpose of the method of partial fractions is to find antiderivatives of rational functions ie functions of the form PX f X where P and Q are ol nomials QM p y o The method involves three steps 0 If d P 2 d Q first use long division and re write f as d H lt d Q7 where N and H are polynomials Then apply the method to the rational function HXQX 9 If d P lt d Q find the partial fraction decomposition of PXQX e Integrate each of the terms appearing in the partial fraction decomposition of f to obtain an antiderivative of f Methods of integration Calculus and Differential Equations The method of artial fractions continued a To do this we need to be able to perform each of the steps separately They are 0 Long division of polynomials 9 Partial fraction decomposition of PXQX where d P lt d Q 9 Integration of terms that typically appear in a decomposition into partial fractions Such terms are of the form A and BxC Xia 24rbltcquot7 where n 2 1 and X2 bx c is irreducible 0 Example for step 1 Divide X3 by X2 3X 2 tegration Calculus and Differential Equations Xles of aplication 7 i 0 We have already used partial fractions when solving the logistic equation 9 Solve the following differential equation y1y272y3 dX y25 39 9 Solve the differential equation dX y4 with the following initial conditions 0 y0 2 9 y0 1 Methods of integration Calculus and Differential Equations Trigonometric substitutions Trigonometric substitutions take advantage of known algebraic relationship between a Sines cosines and tangents cos20 l sin20 1 d cos0 7 sin0 d 2 tanW 7 1tan 0 7 d E sin0 7 cos0 1 cos20 o Hyperbolic sines cosines and tangents cosh20 7 sinh20 1 i cosh0 sinh0 i sinh0 cosh0 d0 d0 d 2 1 tanhW 7 17tanh 0 7 W Trigonometric substitutions continued 0 For integrands that involve 32 7 X2 3 gt 0 note that le g a and try the substitution X asin0 9 Since the integrand will involve cos20 and the dX will be given by dX acos0 d0 one can expect to be able to simplify the integral after such a substitution 0 Example Show that dX arcsin C dX m 0 Similarly for integrands that involve X2 7 32 a gt 0 one can change variables so that X gt 0 and then try X acosh0 since X2 2 32 9 Examples Show that X2 7 32 dX can be written as a2 sinh20 d0 after a substitution Methods of Integration Calculus and Differential Equations Calculus and Differential Equations Methods of integration Halfangle substitutions o Half angle substitutions are useful to find antiderivatives of products andor ratios of sines and cosines 0 Indeed let ttan02 Then 1722 1t27 2t cos0 1 t2 sin0 7 dt 51 d0 0 A product or ratio of sines and cosines will thus be transformed into a rational function of t which we know how to integrate using partial fractions 0 t C 3 2gtl Calculus and Differential Equations 0 Example Show that ln i sin0 Methods of integration Partial fraction decomposition P X X into partial fractions proceed as follows To decompose the rational function where d P lt d Q 0 Factor the denominator QX into terms of the form X 7 a and X2 bxc where n 21 and X2 bxc is irreducible 9 For each factor of the form X7 a the partial fraction decomposition of PXQX will include terms of the form A1 A2 X7a7 X7a27 A1 7 Xiay39v39 An 7 Xia 39 9 To find An multiply by X 7 a and set X a into the resulting equation 0 To find the Aj s j y n multiply by X7 a and substitute in appropriate values of X Methods of integration Calculus and Differential Equations Partial fraction decomposition continued 6 For each factor of the form X2 bX c th e partial fraction decomposition of PXQX will include terms of the form 81X C1 BjXCj Xzbxc7 7 X2 l bxcj7 H an C 7 X2bxc 39 9 To find the Bj s and Cj s multiply by X2 I bx c expand and equate the coefficients of the various powers of X in both sides of the resulting equation Example Find the partial fraction decomposition of X25 XI x1X272x339 Mamas o tegration Calculus and Differential Equations Integration of a partial fraction decomposition Typical terms in a partial fraction decomposition are of the form A and BxC Xia X2 bxc 39 0 Terms of the form n gt1 X i a 9 I7 1 then X a dx lnlxial C o lfngt 1 then Methods of integration Calculus and Differential Equations Integration of a partial fraction decomposit B C 1 n 2 1 X2 bX c 0 Compare the numerator to the derivative of X2 9 Terms of the form ion continued bxc BxC dx 2xbiCdX X2bxc X2bxc 5 D dx T 2 U L lt2bltc 7 bB whereuX2bxcand DCiT 9 Thus we can integrate provided we know how to find an antiderivative of lX2 bX c 9 Note that since X2 bx c is irreducible one can write X2bXCltX tegration b2 d2 where a 2 c 7 4 Calculus and Differential Equations Integration of a partial fraction decomposition continued 1 b n etu Then 9 To integrate x 2d2 d M dX i 1 du lltx 92 NZ T d2 2 1 0 lfn1then dX fl du fl 1 i C Xg2d2Tdu21Tdarcan d 2d du 1u2 u2Llln W c052 20 d0 Alternatively integrate by parts and find a recursive formula o If n gt 1 let 9 arctanu Then d0 and Methods of integration Calculus and Differential Equations

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