Formulas
Upcoming SlideShare
Loading in...5
×
 

Like this? Share it with your network

Share

Formulas

on

  • 872 views

formulas

formulas

Statistics

Views

Total Views
872
Views on SlideShare
872
Embed Views
0

Actions

Likes
0
Downloads
61
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Formulas Document Transcript

  • 1. Derivative FormulasGeneral Rulesd d d [ f (x) + g(x)] = f (x) + g (x) [ f (x) − g(x)] = f (x) − g (x) [c f (x)] = c f (x)dx dx dxd d d f (x) f (x)g(x) − f (x)g (x) [ f (g(x))] = f (g(x))g (x) [ f (x)g(x)] = f (x)g(x) + f (x)g (x) =dx dx dx g(x) [g(x)]2Power Rulesd n d d d √ 1 (x ) = nx n−1 (c) = 0 (cx) = c ( x) = √dx dx dx dx 2 xExponentiald x d x d d [e ] = e x [a ] = a x ln a eu(x) = eu(x) u (x) er x = r er xdx dx dx dxTrigonometricd d d (sin x) = cos x (cos x) = −sin x (tan x) = sec2 xdx dx dxd d d (cot x) = −csc2 x (sec x) = sec x tan x (csc x) = −csc x cot xdx dx dxInverse Trigonometricd 1 d 1 d 1 (sin−1 x) = √ (cos−1 x) = − √ (tan−1 x) =dx 1 − x2 dx 1 − x2 dx 1 + x2d 1 d 1 d 1 (cot−1 x) = − (sec−1 x) = √ (csc−1 x) = − √dx 1 + x2 dx |x| x 2 − 1 dx |x| x 2 − 1Hyperbolicd d d (sinh x) = cosh x (cosh x) = sinh x (tanh x) = sech2 xdx dx dxd d d (coth x) = −csch2 x (sech x) = −sech x tanh x (csch x) = −csch x coth xdx dx dxInverse Hyperbolicd 1 d 1 d 1 (sinh−1 x) = √ (cosh−1 x) = √ (tanh−1 x) =dx 1 + x2 dx x2 − 1 dx 1 − x2d 1 d 1 d 1 (coth−1 x) = (sech−1 x) = − √ (csch−1 x) = − √dx 1 − x2 dx x 1 − x2 dx |x| x 2 + 1
  • 2. Table of IntegralsForms Involving a + bu 1 1 1. du = ln |a + bu| + c a + bu b u 1 2. du = 2 (a + bu − a ln |a + bu|) + c a + bu b u2 1 3. du = 3 [(a + bu)2 − 4a(a + bu) + 2a 2 ln |a + bu|] + c a + bu 2b 1 1 u 4. du = ln +c u(a + bu) a a + bu 1 b a + bu 1 5. du = 2 ln − +c u 2 (a + bu) a u auForms Involving (a + bu)2 1 −1 6. du = +c (a + bu)2 b(a + bu) u 1 a 7. du = 2 + ln |a + bu| + c (a + bu)2 b a + bu u2 1 a2 8. du = 3 a + bu − − 2a ln |a + bu| + c (a + bu)2 b a + bu 1 1 1 u 9. du = + 2 ln +c u(a + bu)2 a(a + bu) a a + bu 1 2b a + bu a + 2bu10. du = 3 ln − 2 +c u 2 (a + bu)2 a u a u(a + bu)Forms Involving a + bu √ 211. u a + bu du = (3bu − 2a)(a + bu)3/2 + c 15b2 √ 212. u 2 a + bu du = (15b2 u 2 − 12abu + 8a 2 )(a + bu)3/2 + c 105b3 √ 2 2na √13. u n a + bu du = u n (a + bu)3/2 − u n−1 a + bu du b(2n + 3) b(2n + 3) √ a + bu √ 114. du = 2 a + bu + a √ du u u a + bu √ √ a + bu −1 (a + bu)3/2 (2n − 5)b a + bu15. du = − du, n = 1 u n a(n − 1) u n−1 2a(n − 1) u n−1
  • 3. √ √ 1 1 a + bu − a16a. √ du = √ ln √ √ + c, a > 0 u a + bu a a + bu + a 1 2 a + bu16b. √ du = √ tan−1 + c, a < 0 u a + bu −a −a √ 1 −1 a + bu (2n − 3)b 1 17. √ du = − √ du, n = 1 u n a + bu a(n − 1) u n−1 2a(n − 1) u n−1 a + bu u 2 √ 18. √ du = 2 (bu − 2a) a + bu + c a + bu 3b u2 2 √ 19. √ du = (3b2 u 2 − 4abu + 8a 2 ) a + bu + c a + bu 15b3 un 2 √ 2na u n−1 20. √ du = u n a + bu − √ du a + bu (2n + 1)b (2n + 1)b a + buForms Involving a 2 + u 2, a > 0 √ √ 21. a 2 + u 2 du = 1 u a 2 + u 2 + 1 a 2 ln u + a 2 + u 2 + c 2 2 √ √ 22. u 2 a 2 + u 2 du = 1 u(a 2 + 2u 2 ) a 2 + u 2 − 1 a 4 ln u + a 2 + u 2 + c 8 8 √ √ a2 + u2 a+ a2 + u2 23. du = a 2 + u 2 − a ln +c u u √ √ a2 + u2 a2 + u2 24. du = ln u + a2 + u2 − +c u2 u 1 25. √ du = ln u + a2 + u2 + c a2 + u2 u2 1 1 26. √ du = u a 2 + u 2 − a 2 ln u + a2 + u2 + c a2 + u2 2 2 1 1 u 27. √ du = ln √ +c u a 2 + u2 a a + a2 + u2 √ 1 a2 + u2 28. √ du = − +c u 2 a2 + u2 a2 uForms Involving a 2 ¯¯ u 2 , a > 0 √ u 29. a 2 − u 2 du = 1 u a 2 − u 2 + 1 a 2 sin−1 + c 2 2 a √ u 30. u 2 a 2 − u 2 du = 1 u(2u 2 − a 2 ) a 2 − u 2 + 1 a 4 sin−1 + c 8 8 a √ √ a2 − u2 a+ a2 − u2 31. du = a 2 − u 2 − a ln +c u u √ √ a2 − u2 a2 − u2 u 32. du = − − sin−1 + c u2 u a
  • 4. 1 u33. √ du = sin−1 + c a2 − u2 a √ 1 1 a+ a2 − u234. √ du = − ln +c u a2 − u2 a u u2 1 1 u35. √ du = − u a 2 − u 2 + a 2 sin−1 + c a2 − u2 2 2 a √ 1 a2 − u236. √ du = − +c u2 a2 − u2 a2 uForms Involving u 2 ¯¯ a 2 , a > 0 √ √37. u 2 − a 2 du = 1 u u 2 − a 2 − 1 a 2 ln u + u 2 − a 2 + c 2 2 √ √38. u 2 u 2 − a 2 du = 1 u(2u 2 − a 2 ) u 2 − a 2 − 1 a 4 ln u + u 2 − a 2 + c 8 8 √ u2 − a2 |u|39. du = u 2 − a 2 − a sec−1 +c u a √ √ u2 − a2 u2 − a240. du = ln u + u2 − a2 − +c u2 u 141. √ du = ln u + u2 − a2 + c u2 − a2 u2 1 142. √ du = u u 2 − a 2 + a 2 ln u + u2 − a2 + c u2 − a2 2 2 1 1 |u|43. √ du = sec−1 +c u u2 − a2 a a √ 1 u2 − a244. √ du = +c u2 u2 − a2 a2 uForms Involving 2au ¯¯ u 2 1 1 a−u45. 2au − u 2 du = (u − a) 2au − u 2 + a 2 cos−1 +c 2 2 a 1 1 a−u46. u 2au − u 2 du = (2u 2 − au − 3a 2 ) 2au − u 2 + a 3 cos−1 +c 6 2 a √ 2au − u 2 a−u47. du = 2au − u 2 + a cos−1 +c u a √ √ 2au − u 2 2 2au − u 2 a−u48. du = − − cos−1 +c u2 u a 1 a−u49. √ du = cos−1 +c 2au − u 2 a u a−u50. √ du = − 2au − u 2 + a cos−1 +c 2au − u 2 a
  • 5. u2 1 3 a−u51. √ du = − (u + 3a) 2au − u 2 + a 2 cos−1 +c 2au − u 2 2 2 a √ 1 2au − u 252. √ du = − +c u 2au − u2 auForms Involving sin u OR cos u53. sin u du = −cos u + c54. cos u du = sin u + c55. sin2 u du = 1 u − 2 1 2 sin u cos u + c56. cos2 u du = 1 u + 2 1 2 sin u cos u + c57. sin3 u du = − 2 cos u − 3 1 3 sin2 u cos u + c58. cos3 u du = 2 3 sin u + 1 3 sin u cos2 u + c 1 n−159. sinn u du = − sinn−1 u cos u + sinn−2 u du n n 1 n−160. cosn u du = cosn−1 u sin u + cosn−2 u du n n61. u sin u du = sin u − u cos u + c62. u cos u du = cos u + u sin u + c63. u n sin u du = −u n cos u + n u n−1 cos u du + c64. u n cos u du = u n sin u − n u n−1 sin u du + c 165. du = tan u − sec u + c 1 + sin u 166. du = tan u + sec u + c 1 − sin u 167. du = −cot u + csc u + c 1 + cos u 168. du = −cot u − csc u + c 1 − cos u sin(m − n)u sin(m + n)u69. sin(mu) sin(nu) du = − +c 2(m − n) 2(m + n) sin(m − n)u sin(m + n)u70. cos(mu) cos(nu) du = + +c 2(m − n) 2(m + n)
  • 6. cos(n − m)u cos(m + n)u71. sin(mu) cos(nu) du = − +c 2(n − m) 2(m + n) sinm−1 u cosn+1 u m−172. sinm u cosn u du = − + sinm−2 u cosn u du m+n m+nForms Involving Other Trigonometric Functions73. tan u du = −ln |cos u| + c = ln |sec u| + c74. cot u du = ln |sin u| + c75. sec u du = ln |sec u + tan u| + c76. csc u du = ln |csc u − cot u| + c77. tan2 u du = tan u − u + c78. cot2 u du = −cot u − u + c79. sec2 u du = tan u + c80. csc2 u du = −cot u + c81. tan3 u du = 1 2 tan2 u + ln |cos u| + c82. cot3 u du = − 1 cot2 u − ln |sin u| + c 283. sec3 u du = 1 2 sec u tan u + 1 2 ln |sec u + tan u| + c84. csc3 u du = − 1 csc u cot u + 2 1 2 ln |csc u − cot u| + c 185. tann u du = tann−1 u − tann−2 u du, n = 1 n−1 186. cotn u du = − cotn−1 u − cotn−2 u du, n = 1 n−1 1 n−287. secn u du = secn−2 u tan u + secn−2 u du, n = 1 n−1 n−1 1 n−288. cscn u du = − cscn−2 u cot u + cscn−2 u du, n = 1 n−1 n−1 189. du = 1 u ± ln |cos u ± sin u| + c 1 ± tan u 2 190. du = 1 u ∓ ln |sin u ± cos u| + c 1 ± cot u 2
  • 7. 1 91. du = u + cot u ∓ csc u + c 1 ± sec u 1 92. du = u − tan u ± sec u + c 1 ± csc uForms Involving Inverse Trigonometric Functions 93. sin−1 u du = u sin−1 u + 1 − u2 + c 94. cos−1 u du = u cos−1 u − 1 − u2 + c 95. tan−1 u du = u tan−1 u − ln 1 + u 2 + c 96. cot−1 u du = u cot−1 u + ln 1 + u 2 + c 97. sec−1 u du = u sec−1 u − ln |u + u 2 − 1| + c 98. csc−1 u du = u csc−1 u + ln |u + u 2 − 1| + c √ 99. u sin−1 u du = 1 (2u 2 − 1) sin−1 u + 1 u 1 − u 2 + c 4 4 √100. u cos−1 u du = 1 (2u 2 − 1) cos−1 u − 1 u 1 − u 2 + c 4 4Forms Involving eu 1 au101. eau du = e +c a 1 1102. ueau du = u− 2 eau + c a a 1 2 2 2103. u 2 eau du = u − 2u+ 3 eau + c a a a 1 n au n104. u n eau du = u e − u n−1 eau du a a 1105. eau sin bu du = (a sin bu − b cos bu)eau + c a 2 + b2 1106. eau cos bu du = (a cos bu + b sin bu)eau + c a 2 + b2Forms Involving ln u107. ln u du = u ln u − u + c108. u ln u du = 1 u 2 ln u − 1 u 2 + c 2 4
  • 8. 1 1109. u n ln u du = u n+1 ln u − u n+1 + c n+1 (n + 1)2 1110. du = ln |ln u| + c u ln u111. (ln u)2 du = u(ln u)2 − 2u ln u + 2u + c112. (ln u)n du = u(ln u)n − n (ln u)n−1 duForms Involving Hyperbolic Functions113. sinh u du = cosh u + c114. cosh u du = sinh u + c115. tanh u du = ln (cosh u) + c116. coth u du = ln |sinh u| + c117. sech u du = tan−1 |sinh u| + c118. csch u du = ln |tanh 1 u| + c 2119. sech2 u du = tanh u + c120. csch2 u du = −coth u + c121. sech u tanh u du = −sech u + c122. csch u coth u du = −csch u + c 1123. √ da = sinh−1 a + c a2 + 1 1124. √ da = cosh−1 a + c a2 − 1 1125. da = tanh−1 a + c 1 − a2 1126. √ da = −csch−1 a + c |a| a 2 + 1 1127. √ da = −sech−1 a + c a 1 − a2