Integration Strategies

Integration Strategies - INTEGRATION E STRATEGIES T he set...

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E INTEGRATION STRATEGIES T he set of exercises in this section consists of an assortment of indeFnite integrals that can be evaluated using the techniques developed in Chapter 7. Your task is to decide which techniques are best suited to the given integral and to carry out the computation. Let’s review the tools at our disposal. The two most powerful tools are substitution and integration by parts. In addition, we have special methods for treating speciFc types of integrals: (A) Trigonometric integrals: use trigonometric identities and reduction formulas (Sec- tion 7.3) (B) Square root expressions x 2 ± a 2 and a 2 x 2 : use trigonometric substitution (Section 7.4) or hyperbolic substitution (Section 7.5) (C) Rational functions: use the method of partial fractions (Section 7.6) When confronted with an integration problem, you may ask yourself the following ques- tions: 1. Can the integral be evaluated directly (perhaps after algebraic manipulation or a sub- stitution)? 2. Is the integral amenable to integration by parts (or perhaps substitution followed by integration by parts)? 3. Does the integral fall into one of the types (A)–(C) described above? Or can it be reduced to one of these types by a substitution? You may also refer to the Table of Integrals at the back of the book. This table contains a list of standard integration formulas, including reduction formulas for trigono- metric, exponential, and inverse trigonometric integrals. Sometimes it is necessary to transform your integral before applying one of the formulas listed in the table. In all cases, be on the lookout for the simplest method. EXAMPLE 1 Look for the Simplest Method Evaluate the integrals (a) Z x 7 dx x 4 1 and (b) Z x 2 x 4 1 . Solution At Frst glance, both integrals appear to be candidates for the method of partial fractions. However, the Frst integral may be treated much more simply using the substi- tution u = x 4 1. E1

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E2 APPENDIX E INTEGRATION STRATEGIES (a) Let u = x 4 1and du = 4 x 3 dx .Then x 4 = u + x 7 = x 4 x 3 = 1 4 ( u + 1 ) 1 Z x 7 x 4 1 = 1 4 Z ( u + 1 ) u = 1 4 Z µ 1 + 1 u ( simplify algebraically ) = 1 4 u + 1 4 ln | u |+ C = 1 4 ( x 4 1 ) + 1 4 ln | x 4 1 C We may absorb the term 1 4 into the constant C and write our result as Z x 7 x 4 1 = 1 4 x 4 + 1 4 ln | x 4 1 C (b) Direct substitution cannot be used to evaluate the second integral. We must Fnd the The substitution u = x 4 1 is not effective in integral (b) of Example 1 because x 2 = x 1 x 3 = 1 4 ( u + 1 ) 1 / 4 The substitution does not lead to a simpler integral: Z x 2 x 4 1 = 1 4 Z ( u + 1 ) 1 / 4 u partial fraction decomposition using the factorization x 4 1 = ( x 1 )( x + 1 )( x 2 + 1 ) : x 2 x 4 1 = 1 4 x 1 1 4 x + 1 + 1 2 x 2 + 1 Z x 2 x 4 1 = 1 4 Z x 1 1 4 Z x + 1 + 1 2 Z x 2 + 1 = 1 4 ln | x 1 |− 1 4 ln | x + 1 1 2 tan 1 x + C EXAMPLE 2 Find a Simplifying Substitution ±ind approaches to evaluate (a) Z e x 1 e 2 x and (b) Z e 2 x 1 e 2 x .
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This note was uploaded on 10/12/2009 for the course MATH calculus 3 taught by Professor Ningzhong during the Spring '08 term at University of Cincinnati.

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Integration Strategies - INTEGRATION E STRATEGIES T he set...

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