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Theoretical Computer Science Cheat SheetπCalculusWallis’ identity:π =2· 2 · 2 · 4 · 4 · 6 · 6 ···Derivatives:1 · 3 · 3 · 5 · 5 · 7 ···1. d(cu)dx= cdud(u + v), 2. = dudx dx dx + dvdx ,Brouncker’s continued fraction expansion:π4 =1+ 1 24. d(un ) du= nun−132+ 2dx dx , d(u/v)5. = v( ) (dudx − u dv)dxdxv 2 , 6. d(ecu )dx2+ 522+2+···727. d(cu ) du= (ln c)cudxdxGregrory’s series:π4 =1− 1 3 + 1 5 − 1 7 + 1 9 −···Newton’s series:π6 = 1 2 + 12 · 3 · 2 3 + 1 · 32 · 4 · 5 · 2 5 + ···Sharp’s series:π6 = √ 1 (1 − 13 3 1 · 3 + 13 2 · 5 − 1 ···)3 3 · 7 +Euler’s series:π 26 = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 + ···π 28 = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + 1 9 2 + ···π 212 = 1 1 2 − 1 2 2 + 1 3 2 − 1 4 2 + 1 5 2 −···Partial FractionsLet N(x) and D(x) be polynomial functionsof x. We can break downN(x)/D(x) using partial fraction expansion.First, if the degree of N is greaterthan or equal to the degree of D, divideN by D, obtainingN(x)D(x) = (x)Q(x)+N′ D(x) ,where the degree of N ′ is less than that ofD. Second, factor D(x). Use the followingrules: For a non-repeated factor:N(x)(x − a)D(x) = Ax − a + N ′ (x)D(x) ,where [ ] N(x)A =.D(x)x=aFor a repeated factor:m−1N(x)(x − a) m D(x) = ∑ A k (x)(x − a) m−k +N′ D(x) ,whereA k = 1 k!k=0[ ( )]dk N(x)dx k .D(x)x=aThe reasonable man adapts himself to theworld; the unreasonable persists in tryingto adapt the world to himself. Thereforeall progress depends on the unreasonable.– George Bernard Shaw9.11.13.15.17.19.21.23.25.27.29.d(sin u)dxd(tan u)dxd(sec u)dxd(arcsin u)dxd(arctan u)dxd(arcsec u)dxd(sinh u)dxd(tanh u)dxd(sech u)dxd(arcsinh u)dxd(arctanh u)dxd(uv)3.dx= u dvdx + v dudx ,, 8.d(ln u)dxcu du= cedx ,= 1 duu dx ,= cos u dud(cos u), 10. = − sin u dudx dxdx ,= sec 2 u dud(cot u), 12. = csc 2 u dudx dxdx ,= tan u sec u dud(csc u), 14. = − cot u csc u dudx dxdx ,=d(arcsech u)31.dxIntegrals:∫ ∫1. cu dx = c3.6.8.10.12.14.∫∫∫1 dud(arccos u)√ , 16. = −1 du√1 − u2 dx dx 1 − u2 dx ,= 1 du1+u 2 dx1=u √ du1 − u 2 dx, 18.d(arccot u)dx, 20.d(arccsc u)dx== −11+u 2 dudx ,−1u √ 1 − u 2 dudx ,= cosh u dud(cosh u), 22. = sinh u dudx dxdx ,= sech 2 u dud(coth u), 24. = − csch 2 u dudx dxdx ,= − sech u tanh u dud(csch u), 26. = − csch u coth u dudx dxdx ,1 dud(arccosh u) 1 du= √ , 28. = √1+u2 dx dx u2 − 1 dx ,= 1 dud(arccoth u)1 − u 2 , 30. = 1 dudx dx u 2 − 1 dx ,−1=u √ du1 − u 2 dx, 32.d(arccsch u)dx=−1|u| √ 1+u 2 dudx .∫∫ ∫u dx, 2. (u + v) dx = udx+ v dx,x n dx = 1∫ ∫1n +1 xn+1 , n ≠ −1, 4.x dx =lnx, 5. e x dx = e x ,∫dx1+x 2 = arctan x, 7. u dv ∫dx dx = uv − v dudx dx,∫∫∫sin xdx= − cos x, 9.tan xdx= − ln | cos x|, 11.sec xdx=ln| sec x + tan x|, 13.arcsin x a dx = arcsin x a + √ a 2 − x 2 , a > 0,∫∫∫cos xdx= sin x,cot xdx=ln| cos x|,csc xdx=ln| csc x + cot x|,
15.17.19.21.23.25.26.29.33.36.∫∫∫∫∫∫∫∫∫∫Theoretical Computer Science Cheat SheetCalculus Cont.arccos x a dx = arccos x a − √ ∫a 2 − x 2 , a > 0, 16. arctan x a dx = x arctan x a − a 2 ln(a2 + x 2 ), a > 0,∫sin 2 ( )(ax)dx = 12a ax − sin(ax) cos(ax) , 18. cos 2 )(ax)dx =2a( 1 ax + sin(ax) cos(ax) ,∫sec 2 xdx= tan x, 20. csc 2 xdx= − cot x,sin n xdx= − sinn−1 x cos x+ n − 1 ∫∫sin n−2 x dx, 22. cos n xdx= cosn−1 x sin x+ n − 1 ∫cos n−2 x dx,n nn n∫∫∫tan n xdx= tann−1 x− tan n−2 x dx, n ≠1, 24. cot n xdx= − cotn−1 x− cot n−2 x dx, n ≠1,n − 1n − 1sec n xdx= tan x secn−1 xn − 1+ n − 2 ∫n − 1sec n−2 x dx, n ≠1,csc n xdx= − cot x cscn−1 x+ n − 2 ∫∫∫csc n−2 x dx, n ≠1, 27. sinh xdx= cosh x, 28. cosh xdx= sinh x,n − 1 n − 1∫∫∫tanh xdx=ln| cosh x|, 30. coth xdx=ln| sinh x|, 31. sech xdx= arctan sinh x, 32. csch xdx=ln ∣ tanhx∣ ,sinh 2 xdx= 1 4 sinh(2x) − 1 2 x,34. ∫cosh 2 xdx= 1 4 sinh(2x)+ 1 2 x,arcsinh x a dx = x arcsinh x a − √ x 2 + a 2 , a > 0, 37.∫35. ∫sech 2 xdx= tanh x,arctanh x a dx = x arctanh x a + a 2 ln |a2 − x 2 |,238.39.40.42.43.46.48.50.52.54.56.58.60.⎧∫⎨ x arccosh xarccosh x a dx = a − √ x 2 + a 2 ,⎩x arccosh x a + √ x 2 + a 2 ,∫dx(√a2 + x =ln x + √ )a 2 + x 2 , a > 0,2∫∫if arccosh x a> 0 and a>0,if arccosh x a< 0 and a>0,∫dx√a2a 2 + x 2 = 1 a arctan x a , a > 0, 41. √− x 2 dx = x 2 a2 − x 2 + a22 arcsin x a , a > 0,(a 2 − x 2 ) 3/2 dx = x 8 (5a2 − 2x 2 ) √ a 2 − x 2 + 3a48 arcsin x a , a > 0,∫∫dx√a2 − x = arcsin x 2 a , a > 0, 44. dxa 2 − x 2 = 1 ∣ ∣∣∣2a ln a + xa − x∣ ,∫ √a2 √ ∣± x 2 dx = x ∣∣x2 a2 ± x 2 ± a22 ln √ ∣ ∫∣∣+ a2 ± x 2 , 47.∫dxax 2 + bx = 1 ∣ ∣∣∣a ln xa + bx∣ ,∫45. dx(a 2 − x 2 ) = x3/2 a 2√ a 2 − x , 2dx∣ ∣∣x√x2 − a =ln √ ∣ ∣∣ + x2 − a 2 , a > 0,2∫49. x √ 2(3bx − 2a)(a + bx)3/2a + bx dx =15b 2 ,∫ √a + bxdx =2 √ ∫∫1a + bx + axx √ a + bx dx,51. x√ dx = 1√ √ ∣ √ lna + bx − a∣∣∣a + bx 2∣√ √ , a > 0,a + bx + a∫ √a2 − x 2dx = √ ax2 − x 2 a + √ a− a ln2 − x 2∫∣ x ∣ ,53. x √ a 2 − x 2 dx = − 1 3 (a2 − x 2 ) 3/2 ,∫x 2√ a 2 − x 2 dx = x 8 (2x2 − a 2 ) √ ∫∣a 2 − x 2 + a48 arcsin x a , a > 0, 55. dx∣∣∣∣√a2 − x = − 1 2 a ln a + √ a 2 − x 2x ∣ ,∫∫xdx√a2 − x = −√ a 2 − x 2 x 2 dx, 57. √ 2 a2 − x = − √ x2 2 a2 − x 2 + a22 arcsin x a,a>0,∫ √a2 + x 2dx = √ ax2 + x 2 a + √ a− a ln2 + x 2∫ √ ∣ x ∣ , 59. x2 − a 2dx = √ xx2 − a 2 − a arccos a|x| , a > 0,∫x √ ∫∣ ∣x 2 ± a 2 dx = 1 3 (x2 ± a 2 ) 3/2 dx∣∣∣, 61.x √ x 2 + a = 1 2 a ln x ∣∣∣a + √ ,a 2 + x 2
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