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SalasSV_08_01

# SalasSV_08_01 - CHAPTER CHAPTER 8 4 1 2 3 5 6 8 10 11 12 14...

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CHAPTER 4 THE MEAN-VALUE THEOREM AND APPLICATIONS 8 TECHNIQUES OF INTEGRATION 8.1 INTEGRAL TABLES AND REVIEW We begin by listing the more important integrals with which you are already familiar. 1. k dx = kx + C , k constant. 2. x r dx = 1 r + 1 x r + 1 + C , r = − 1. 3. dx x = ln | x | + C . 4. e x dx = e x + C . 5. p x dx = p x ln p + C , p > 0 constant, p = 1. 6. sin x dx = − cos x + C . 7. cos x dx = sin x + C . 8. tan x dx = ln | sec x | + C . 9. cot x dx = ln | sin x | + C . 10. sec x dx = ln | sec x + tan x | + C . 11. csc x dx = ln | csc x cot x | + C . 12. sec x tan x dx = sec x + C . 13. csc x cot x dx = − csc x + C . 14. sec 2 x dx = tan x + C . 15. csc 2 x dx = − cot x + C . 16. dx a 2 x 2 = sin 1 x a + C , a > 0 constant. 17. dx a 2 + x 2 = 1 a tan 1 x a + C , a > 0 constant. 446

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8.1 INTEGRAL TABLES AND REVIEW 447 18. dx x x 2 a 2 = 1 a sec 1 | x | a + C , a > 0 constant. 19. sinh x dx = cosh x + C . 20. cosh x dx = sinh x + C . For review we work out a few integrals involving u -substitutions. Example 1 Find x tan x 2 dx . SOLUTION Set u = x 2 , du = 2 x dx . Then x tan x 2 dx = 1 2 tan u du = 1 2 ln | sec u | + C = 1 2 | sec x 2 | + C . Formula 8 Example 2 Calculate 1 0 e x e x + 2 dx . SOLUTION Set u = e x + 2, du = e x dx . At x = 0, u = 3; at x = 1, u = e + 2. Thus 1 0 e x e x + 2 dx = e + 2 3 du u = ln | u | e + 2 3 Formula 3 = ln ( e + 2) ln 3 = ln 1 3 ( e + 2) = 0. 45. Example 3 Find cos 2 x (2 + sin 2 x ) 1 / 3 dx . SOLUTION Set u = 2 + sin 2 x , du = 2 cos 2 x dx . Then cos 2 x (2 + sin 2 x ) 1 / 3 dx = 1 2 1 u 1 / 3 du = 1 2 u 1 / 3 du = 1 2 ( 3 2 ) u 2 / 3 + C Formula 2 = 3 4 (2 + sin 2 x ) 2 / 3 + C . The final example involves in little algebra. Example 4 Find P = dx x 2 + 2 x + 5 . SOLUTION First we complete the square in the denominator: P = dx x 2 + 2 x + 5 = dx ( x 2 + 2 x + 1) + 4 = dx ( x + 1) 2 + 2 2 .
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