8dx

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8dx

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Arvind P Vidhyarthi. p cos x + qsin x + r dx (integral of the form) (2) (a) I=3cosx+5sinx+72cosx+3sinx+8dx= a cos x + b sin x + c using pcosx+qsinx+racosx+bsinx+c Divide both of sides by (acos x + b sinx + c) and integrating, Solution: − I=3cosx+5sinx+72cosx+3sinx+8 dx 3 cos x + 5 sinx + 7 = l(2 cos x + 3 sinx + 8) + m(d. cof 2 cos x + 3 sin x + 8) + n ind l, m and n by comparing the coeficient of sinx, cos x and constant. 3 cos x + 5 sinx + 7 = l(2 cos x + 3 sinx + 8) + m(3 cos x − 2 sinx) + n . ANDERSDN IN 46016 R R 1 8DX 66 FARMLAND IN 47340 8IG 8DNE CHURCH RD R R 1 8DX 332 UNIDN KY 41091 30012 PEM8ERLY CTS W LAFAYETTE IN 47906 951 SHERW00D LK DR Ш2А SCHERERVILLE IN 46375 7447 GRANDVIEW INDIANAPDLIS IN 46260 R R 1 CAMP8ELLS8URG IN

47108.1316 1/2 W 4TH ST MARIDN IN 46952 R R 4 8DX 240 PLYMDUTH IN 46563 30012 PEM8ERLY CTS W LAFAYETTE IN 47906 R R 1 HARTSVILLE IN 47244 R R 3 8DX 25  + 2)dx, (d) J(3x)8dx, (e) J^i—dx. Solution ion: (a) [3x4 dx 3x5 + C (b) f(x34x + 7)dx x4 4x2 _ _ = + 7x + C 4 2 x4 = — 2x2+7x + C 4 (c) J(x5)(x + 2)dx = ((x23x10)dx •lOx + C x3 3x2 3 2 (d) f(3x)8dx = ^^ + C J* ; 9—1 (3x)9 _ (x3)9 _ = — +C or i — + C Consequently, when integrating the original function, the integration process has to include a constant: y = 6x3 − 4x2 +8x + C. The value of C is not always required, but it can be determined if we are given some extra information, such as y = 10 when x = 0, then C = 10. The notation for integration employs a curly

“S”.symbol ∫ , which may seem strange, but is short for sum and will be explained later. So, starting with dydx = 18x2−8x +8 we rewrite this as dy = ( 18x2−8x +8 ) dx and For example, By the substitution method with # (x 1 4)8 dx 5 1 9 (x 1 4)9 1 C u 5 x 1 4 (omitting the details) Since (x14)8 is more complicated than (x 2 2), we take dv5 (x 1 4)8 dx. #(x22)(x14)8dx 5 (x22)19(x14)92 #19(x14)9dx u dv u v v du £ u5x22 dv5(x14)8dx du5dx v5e(x14)8dx 519(x14)9 § 5 1 9(x 2 2)(x14)9 2 1 9 # (x14)9dx Taking out the 19 5 19(x2 2)(x 1 4)9 2 19110(x14)101C Integrating by the substitution method 5 19(x2 2)(x14)9 2 190(x14)101C Again we could check this For example, By the substitution method with # (x 1 4)8 dx 5 1 9 (x 1 4)9 1 C

u.5 x 1 4 (omitting the details) Since (x14)8 is more complicated than (x 2 2), we take dv5 (x 1 4)8 dx. #(x22)(x14)8dx 5 (x22)19(x14)92 #19(x14)9dx u dv u v v du £ u5x22 dv5(x14)8dx du5dx v5e(x14)8dx 519(x14)9 § 5 1 9(x 2 2)(x14)9 2 1 9 # (x14)9dx Taking out the 19 5 19(x2 2)(x 1 4)9 2 19110(x14)101C Integrating by the substitution method 5 19(x2 2)(x14)9 2 190(x14)101C Again we could check this Sedgwick Ltd Sedgwick Centre, London El 8DX, England. Tel (0171) 377 3456. Tx 882131. Fax (0171) 377 3199. Sedgwick Aviation Ltd Sedgwick Centre, London El 8DX, England. Tel (0171) 377 3456. Tx 882131. Fax (0171) 377 3199. Sedgwick Bankrisk Ltd Sedgwick House, The Sedgwick Centre, London El 8DX. England. Tel (0171) 377 3456. Tx 882131. Fax

(0171).377 3199. Sedgwick Energy Ltd Sedgwick Centre, London El 8DX, England. Tel (0171) 377 3456. Tx 882131.although brackets are not always used: y= ∫ 18x2−8x +8dx. This equation reads: “y is the integral of 18x2 − 8x +8 deex.” The dx reminds us that x is the independent variable. In this case we can write the answer: dy = 18x2 − 8x +8dx y= ∫ 18x2−8x +8dx = 6x3 − 4x2 +8x + C where C is some constant. Another example: dy = 6x 2 + 10xdx y= ∫ 6x2+ 10xdx = 2x3 +5x2 + C. Finally, dy = dx y= ∫ dx = x + C. The antiderivatives for the sine and cosine functions are written: ∫ sinxdx = −cosx Philippe Guyenne, David Nicholls, Catherine Sulem. Substituting this into the first equation yields the modified NLS equation 2

iut.= #(ow + 8Dx) – o(0), –– (*) _203(k0) '*) |ul°u o(2k) g +8°k u Go"(eIDx)0%|ul”, With Hamiltonian 203(k0) 1 4 #–'sia)" Co F 1 1 H = | |:(*noow)": (*)—co L8 2 1 £o oboi" |x. as derived from (41). Here again, these modified DS and NLS equations can be solved numerically by a pseudospectral method which is suitable for handling the Fourier multipliers o and