CALCULUSIICH12

# CALCULUSIICH12 - CHAPTER 1 FURTHER TECHNIQUES IN...

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Unformatted text preview: CHAPTER 1 FURTHER TECHNIQUES IN CONSTRUCTING ANTIDERIVATIVES CONTENTS 1.1. Review of Definite and Indefinite Integrals 1.2. Trigonometric Integrals 1.3. Partial Fractions 1.4. Integration Using Tables and Computer Algebra Systems 1.5. Improper Integrals. 1.2. TRIGONOMETRIC INTEGRALS 1.2.1. Trigonometric Integrals 1.2.2. Trigonometric Substitution 1.2.1 Trigonometric Integrals Example. Evaluate C x x C u u du u xdx x xdx x xdx +- = +- =- =- = = 3 3 1 3 3 1 2 2 2 3 sin sin ) 1 ( cos ) sin 1 ( cos cos cos Solution. The direct application of the Substitution Rule would not work: we need one cosine factor and the rest is converted into sine since du= cos x dx if u= sin x . In fact cos 3 x = cos 2 x cos x = (1 - sin 2 x ) cos x Hence we have xdx 3 cos However this method will not work when the integrand contains only even powers of sine and cosine. In this case we take advantage of the half-angle identities: Of course we have to make the substitution u =2 x ) 2 cos 1 ( cos and ) 2 cos 1 ( sin 2 1 2 2 1 2 x x x x + =- = Example. Find Solution. We can use the half-angle formula 2 sin xdx Note. We can use the recursive formula proved by the integration by parts method [ ] 2 ) sin ( ) 2 sin ( ) 2 sin ( ) 2 cos 1 ( sin 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 =--- =- =- = x x dx x xdx 1.2.2 Trigonometric Substitution1....
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## CALCULUSIICH12 - CHAPTER 1 FURTHER TECHNIQUES IN...

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