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 Z u 3 du = 1 8 u 4 + C = 1 8 (3 2 x ) 4 + C 2. u =4+9 =9 dx, 1 9 Z 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 Z sec 2 udu = 1 2 tan u + C = 1 2 tan( x 2 )+ C 4. u = x 2 xdx, 2 Z tan = 2ln | cos u | + C = | cos( x 2 ) | + C 5. u = 2 + cos 3 = 3 sin 3 xdx, 1 3 Z du u = 1 3 ln | u | + C = 1 3 ln(2 + cos 3 x C 6. u = 3 x 2 = 3 2 dx, 2 3 Z du 4+4 u 2 = 1 6 Z du 1+ u 2 = 1 6 tan 1 u + C = 1 6 tan 1 (3 x/ 2) + C 7. u = e x = e x dx, Z sinh = cosh u + C = cosh e x + C 8. u =ln = 1 x dx, Z sec u tan = sec u + C = sec(ln x C 9. u = cot = csc 2 xdx, Z e u du = e u + C = e cot x + C 10. u = x 2 xdx, 1 2 Z du 1 u 2 = 1 2 sin 1 u + C = 1 2 sin 1 ( x 2 C 11. u = cos 7 = 7 sin 7 xdx, 1 7 Z u 5 du = 1 42 u 6 + C = 1 42 cos 6 7 x + C 12. u = sin = cos xdx, Z du u u 2 +1 = ln ¯ ¯ ¯ ¯ ¯ u 2 u ¯ ¯ ¯ ¯ ¯ + C = ln ¯ ¯ ¯ ¯ ¯ p 1 + sin 2 x sin x ¯ ¯ ¯ ¯ ¯ + C 13. u = e x = e x dx, Z du 4+ u 2 ³ u + p u 2 +4 ´ + C ³ e x + p e 2 x ´ + C 14. u = tan 1 = 1 x 2 dx, Z e u du = e u + C = e tan 1 x + C 15. u = x 2 = 1 2 x 2 dx, 2 Z e u du e u + C e x 2 + C 16. u x 2 +2 =(6 x +2) dx, 1 2 Z cot = 1 2 ln | sin u | + C = 1 2 ln sin | 3 x 2 x | + C 17. u = = 1 2 x dx, Z 2 cosh = 2 sinh u + C = 2 sinh x + C
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318 Chapter 8 18. u =ln x,du = dx x , Z du u | u | + C | ln x | + C 19. u = = 1 2 x dx, Z 2 du 3 u =2 Z 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, Z sec u tan udu = sec u + C = sec(sin θ )+ C 21. u = 2 x ,du = 2 x 2 dx, 1 2 Z csch 2 = 1 2 coth u + C = 1 2 coth 2 x + C 22. Z dx x 2 3 ¯ ¯ ¯ x + p x 2 3 ¯ ¯ ¯ + C 23. u = e x = e x dx, Z du 4 u 2 = 1 4 ln ¯ ¯ ¯ ¯ 2+ u 2 u ¯ ¯ ¯ ¯ + C = 1 4 ln ¯ ¯ ¯ ¯ e x 2 e x ¯ ¯ ¯ ¯ + C 24. u = 1 x dx, Z cos = sin u + C = sin(ln x C 25. u = e x = e x dx, Z e x dx 1 e 2 x = Z du 1 u 2 = sin 1 u + C = sin 1 e x + C 26. u = x 1 / 2 = 1 2 x 3 / 2 dx, Z 2 sinh = 2 cosh u + C = 2 cosh( x 1 / 2 C 27. u = x 2 xdx, 1 2 Z du sec u = 1 2 Z cos = 1 2 sin u + C = 1 2 sin( x 2 C 28. 2 u = e x , 2 du = e x dx, Z 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 = 2 x ln 4 dx = x ln 16 dx, 1 ln 16 Z 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 ln 2 , Z 2 dx = 1 π ln 2 e ln 2 + C = 1 π ln 2 2 + C EXERCISE SET 8.2 1. u = x , dv = e x dx , du = dx , v = e x ; Z xe x dx = xe x + Z e x dx = xe x e x + C 2. u = x , dv = e 3 x dx , du = dx , v = 1 3 e 3 x ; Z xe 3 x dx = 1 3 xe 3 x 1 3 Z e 3 x dx = 1 3 xe 3 x 1 9 e 3 x + C 3. u = x 2 , dv = e x dx , du xdx , v = e x ; Z x 2 e x dx = x 2 e x 2 Z xe x dx . For Z xe x dx use u = x , dv = e x dx , du = dx , v = e x to get Z xe x dx = xe x e x + C 1 so Z x 2 e x dx = x 2 e x 2 xe x +2 e x + C
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Exercise Set 8.2 319 4. u = x 2 , dv = e 2 x dx , du =2 xdx , v = 1 2 e 2 x ; Z x 2 e 2 x dx = 1 2 x 2 e 2 x + Z xe 2 x dx For Z xe 2 x dx use u = x , dv = e 2 x dx to get Z xe 2 x dx = 1 2 xe 2 x + 1 2 Z e 2 x dx = 1 2 xe 2 x 1 4 e 2 x + C so Z x 2 e 2 x dx = 1 2 x 2 e
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Calculus: Early Transcendentals, by Anton, 7th Edition,ch08...

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