Calculus: Early Transcendentals, by Anton, 7th Edition,ch08

# Calculus - Early Transcendentals

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317 CHAPTER 8 Principles of Integral Evaluation EXERCISE SET 8.1 1. u = 3 2 x, du = 2 dx, 1 2 u 3 du = 1 8 u 4 + C = 1 8 (3 2 x ) 4 + C 2. u = 4 + 9 x, du = 9 dx, 1 9 u 1 / 2 du = 2 3 · 9 u 3 / 2 + C = 2 27 (4 + 9 x ) 3 / 2 + C 3. u = x 2 , du = 2 xdx, 1 2 sec 2 u du = 1 2 tan u + C = 1 2 tan( x 2 ) + C 4. u = x 2 , du = 2 xdx, 2 tan u du = 2 ln | cos u | + C = 2 ln | cos( x 2 ) | + C 5. u = 2 + cos 3 x, du = 3 sin 3 xdx, 1 3 du u = 1 3 ln | u | + C = 1 3 ln(2 + cos 3 x ) + C 6. u = 3 x 2 , du = 3 2 dx, 2 3 du 4 + 4 u 2 = 1 6 du 1 + u 2 = 1 6 tan 1 u + C = 1 6 tan 1 (3 x/ 2) + C 7. u = e x , du = e x dx, sinh u du = cosh u + C = cosh e x + C 8. u = ln x, du = 1 x dx, sec u tan u du = sec u + C = sec(ln x ) + C 9. u = cot x, du = csc 2 xdx, e u du = e u + C = e cot x + C 10. u = x 2 , du = 2 xdx, 1 2 du 1 u 2 = 1 2 sin 1 u + C = 1 2 sin 1 ( x 2 ) + C 11. u = cos 7 x, du = 7 sin 7 xdx, 1 7 u 5 du = 1 42 u 6 + C = 1 42 cos 6 7 x + C 12. u = sin x, du = cos x dx, du u u 2 + 1 = ln 1 + 1 + u 2 u + C = ln 1 + 1 + sin 2 x sin x + C 13. u = e x , du = e x dx, du 4 + u 2 = ln u + u 2 + 4 + C = ln e x + e 2 x + 4 + C 14. u = tan 1 x, du = 1 1 + x 2 dx, e u du = e u + C = e tan 1 x + C 15. u = x 2 , du = 1 2 x 2 dx, 2 e u du = 2 e u + C = 2 e x 2 + C 16. u = 3 x 2 + 2 x, du = (6 x + 2) dx, 1 2 cot u du = 1 2 ln | sin u | + C = 1 2 ln sin | 3 x 2 + 2 x | + C 17. u = x, du = 1 2 x dx, 2 cosh u du = 2 sinh u + C = 2 sinh x + C

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318 Chapter 8 18. u = ln x, du = dx x , du u = ln | u | + C = ln | ln x | + C 19. u = x, du = 1 2 x dx, 2 du 3 u = 2 e u ln 3 du = 2 ln 3 e u ln 3 + C = 2 ln 3 3 x + C 20. u = sin θ, du = cos θdθ, sec u tan u du = sec u + C = sec(sin θ ) + C 21. u = 2 x , du = 2 x 2 dx, 1 2 csch 2 u du = 1 2 coth u + C = 1 2 coth 2 x + C 22. dx x 2 3 = ln x + x 2 3 + C 23. u = e x , du = e x dx, du 4 u 2 = 1 4 ln 2 + u 2 u + C = 1 4 ln 2 + e x 2 e x + C 24. u = ln x, du = 1 x dx, cos u du = sin u + C = sin(ln x ) + C 25. u = e x , du = e x dx, e x dx 1 e 2 x = du 1 u 2 = sin 1 u + C = sin 1 e x + C 26. u = x 1 / 2 , du = 1 2 x 3 / 2 dx, 2 sinh u du = 2 cosh u + C = 2 cosh( x 1 / 2 ) + C 27. u = x 2 , du = 2 xdx, 1 2 du sec u = 1 2 cos u du = 1 2 sin u + C = 1 2 sin( x 2 ) + C 28. 2 u = e x , 2 du = e x dx, 2 du 4 4 u 2 = sin 1 u + C = sin 1 ( e x / 2) + C 29. 4 x 2 = e x 2 ln 4 , u = x 2 ln 4 , du = 2 x ln 4 dx = x ln 16 dx, 1 ln 16 e u du = 1 ln 16 e u + C = 1 ln 16 e x 2 ln 4 + C = 1 ln 16 4 x 2 + C 30. 2 πx = e πx ln 2 , 2 πx dx = 1 π ln 2 e πx ln 2 + C = 1 π ln 2 2 πx + C EXERCISE SET 8.2 1. u = x , dv = e x dx , du = dx , v = e x ; xe x dx = xe x + e x dx = xe x e x + C 2. u = x , dv = e 3 x dx , du = dx , v = 1 3 e 3 x ; xe 3 x dx = 1 3 xe 3 x 1 3 e 3 x dx = 1 3 xe 3 x 1 9 e 3 x + C 3. u = x 2 , dv = e x dx , du = 2 x dx , v = e x ; x 2 e x dx = x 2 e x 2 xe x dx . For xe x dx use u = x , dv = e x dx , du = dx , v = e x to get xe x dx = xe x e x + C 1 so x 2 e x dx = x 2 e x 2 xe x + 2 e x + C
Exercise Set 8.2 319 4. u = x 2 , dv = e 2 x dx , du = 2 x dx , v = 1 2 e 2 x ; x 2 e 2 x dx = 1 2 x 2 e 2 x + xe 2 x dx For xe 2 x dx use u = x , dv = e 2 x dx to get xe 2 x dx = 1 2 xe 2 x + 1 2 e 2 x dx = 1 2 xe 2 x 1 4 e 2 x + C so x 2 e 2 x dx = 1 2 x 2 e 2 x 1 2 xe 2 x 1 4 e

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