# trigsub - ± u √ a 2 u 2 ² C 8 Z √ u 2-a 2 du = u 2...

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1. Use Ex. 50 in Exercises 7.1 to show that Z sec 3 udu = 1 2 sec u tan u + 1 2 ln | sec u + tan u | + C. 2. Use the formula in problem 1 to show Z tan 2 u sec udu = 1 2 sec u tan u - 1 2 ln | sec u + tan u | + C. In # 3–#8, derive each formula. 3. Z du a 2 + u 2 = ln ± u + a 2 + u 2 ² + C 4. Z du u 2 - a 2 = ln ³ ³ ³ u + u 2 - a 2 ³ ³ ³ + C 5. Z du ( a 2 + u 2 ) 2 = 1 2 a 3 ´ Arctan u a + au a 2 + u 2 µ + C 6. Z a 2 - u 2 du = a 2 2 Arcsin u a + u 2 a 2 - u 2 + C 7. Z a 2 + u 2 du = u 2 a 2 + u 2 + a 2 2 ln
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Unformatted text preview: ± u + √ a 2 + u 2 ² + C 8. Z √ u 2-a 2 du = u 2 √ u 2-a 2-a 2 2 ln ³ ³ ³ u + √ u 2-a 2 ³ ³ ³ + C 9. Use trigonometric substitutions to derive Basic Integral Formulas [14]–[16]....
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## This note was uploaded on 05/17/2011 for the course MAC 2312 taught by Professor Bonner during the Spring '08 term at University of Florida.

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