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## LinearAnalysis

by: Leopoldo Rutherford

39

0

2

# LinearAnalysis MATH287

Leopoldo Rutherford

GPA 3.8

RobertSchwennicke

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RobertSchwennicke
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## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Leopoldo Rutherford on Sunday October 11, 2015. The Class Notes belongs to MATH287 at Cuesta College taught by RobertSchwennicke in Fall. Since its upload, it has received 39 views. For similar materials see /class/221241/math287-cuesta-college in Mathematics (M) at Cuesta College.

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Date Created: 10/11/15
Math 287 Partial Fraction Decomposition X In order to integrate an integrand of the form it is often necessary to decompose this fraction into a sum of simpler fractions The process used to accomplish this is called partial fraction decomposition STEP 1 Divide if improper If N x Dx is an improper fraction that is if the degree of the numerator is greater than or equal to the degree of the denominator then divide then rewrite the fraction as a polynomial plus the fraction N1 x Dx where the polynomial is the quotient of N x Dx and N1 x is the remainder of N x Dx Then apply the following steps to the proper fraction N1 x Dx STEP 2 Factor the denominator Completely factor the denominator Dx into factors of the form px q and CDC2 bx C where ax2 bx C is irreducible STEP 3 Linear factors For each factor of the form px q the partial fraction decomposition must include the following sum A 1 22 mm px q px q px q of m fractions STEP 4 Quadratic factors For each factor of the form axz bx C the partial fraction decomposition must include the leC1 BzxC2 anCn ax2 bxc ax2 bxc2 ax2 bxc following sum of 71 fractions Practice Evaluate the following integrals J x3 2 Jx212x12dx x24x5 I x34x 4 m4 2x 5 2s1nx 3 I dx 439 i z 1x 4x x 0 c0sxc0s x1 Math 287 Trigonometric Substitution Your text contains a table on integrals in the inside front cover Formulasl 17 42 7 45 can be generated using a technique called trigonometric substitution These formulas apply to integrands which contain expressions of the form V612 x2 V a2 x2 or sz a2 The idea behind trigonometric substitution is to write the expression under the radical as a square and thereby remove the radical We will use two identities from trigonometry to accomplish this cos2 9 Sin2 9 l and l tanz t9 sec2 9 1 Make the substitution x a sin 9 with 9 restricted to the values S 9 S Then Val x2 xa2 a2 sin2 9 Jaz sin2 9 wa2 cos2 9 alcos 6 acost9 and dx a COS 9 The triangle to the right can help you seethat whiz x2 ac0st9 a x Ma et e sust1tut10n x atanH with 9 restricted to the values S 9 S Then Val x2 ch2 a2 tan2 9 aZ1tan2 t9 xa2 sec2 9 alsect9 asect9 and dx a sec2 9 The triangle to the right can help you seethat Val x2 asect9 3 Make the substitution x asecQ with 9 restricted to the values 0 S 9 S or 7239 S 9 S37 Then lxz a2 ch2 sec2 6 a2 czzsec2 9 1 1a2 tan2 9 altant9 atant9 and dx a sec 0 tan 0 The triangle to the right can help you seethat xxz a2 atant9 x a Practice Use the techniques above to compute the integrals l 2 IV4x2dx

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