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Set u = sec x tan x then we find du = (sec x tan x sec 2 x) dx substitute du = (sec x tan x sec 2 x) dx, u = sec x tan x.
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Title POTEMindd Author christopherdinardo Created Date 11/19/ PM. 0 U" X ^ ³ _ ± ± 3 x = 0Q` § a ³ x = h 0 i Q x = " 0 # ° a4 § ° b Q§ ^ n = 1 Q n = 2 x 0 1 0 A » ± D 2 4 6 1 3 Q 2 ´² Q 0 7 5 6 40 1 3 ²´ ¡§ 0 0 7 c 74 6 ³± ³ 5 x. E a d b ^ d _ ^ c a b a ` ^ _ ^ \ Z Y X h g f i i k k k k k k k k k k k k j l l n n n n n n n j k k k k k k k k k k k k k m o o o o } ~ ~ ~ ~ } { u z s r u y x t x w v t u t s r q p.
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= n = 1 r = n = 2x;Z³§ N(A) = f0g > » Ax = 2 0 1 3 1 ¨ Q± » Q§x \U 8 ³ ¦ » § » 3 Q±³» 6 ³ Q§V 4 8 0 5 7 Ax = 0;. X n =0 (1) n (2 n 1) s < s> 0) (6) is the Diric hlet series asso ciated with the quadratic c haracter mo dulo 4 The functions 2 (2 s)and 4 1) app ear as the Mellin transforms of 2 S 2 and C 1 resp ectiv ely,and 4 the Mellin transforms of S 1, C 2 T and are also simply related to These results are presen ted in T able 1, where 2 R, s C. Search the world's information, including webpages, images, videos and more Google has many special features to help you find exactly what you're looking for.
U = x 2 2x3 Then the derivative of u is Now "pretend" that the differentiation notation is an arithmetic fraction, and multiply both sides of the previous equation by dx getting or du = (2x2) dx Make substitutions into the original problem, removing all forms of x, resulting in = e u C = e x. Â Ê < 2 § ó ¼ æ Ã ¡ c Ý å Ù ´ ) ¦ Æ ¸ ¡ ¦ Ý ² ç Ù ¶ î ² ¦ ² ¸ î Å Æ v Ê $ r Ù ^ É ø ¦ ä ® É ö í ì Ö ¦ Ô Ù ´ > > J $ r Ù ^ ø r Æ r × ã å J > * Æ Ê ² Á ï Æ Ê Å ä Å ¦ ¾ º Ã ì ¾ Á. U P V W X O P Y L Z N P M N W H P Z N \ \ O Q H L W K N X W M R ^ O L M H N P _ ` S a L M H N P L W S ^ ^ N L Q J , C ?.
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