MA110Formulas - Antiderivatives Z If Z dF(u f(u du = F(u C then = f(u du un du = un 1 C n 6= \u22121 n 1 Z 1 du = ln |u| C Z u eu du = eu C Z au u C 1 6= a

# MA110Formulas - Antiderivatives Z If Z dF(u f(u du = F(u C...

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Antiderivatives If Z f ( u ) du = F ( u ) + C , then dF ( u ) du = f ( u ). Z u n du = u n +1 n + 1 + C, n 6 = - 1 Z 1 u du = ln | u | + C Z e u du = e u + C Z a u du = a u ln a + C, 1 6 = a > 0 Z sin u du = - cos u + C Z cos u du = sin u + C Z sec 2 u du = tan u + C Z csc 2 u du = - cot u + C Z sec u tan u du = sec u + C Z csc u cot u du = - csc u + C Z tan u du = ln | sec u | + C Z cot u du = ln | sin u | + C Z sec u du = ln | sec u + tan u | + C Z csc u du = - ln | csc u + cot u | + C Z du a 2 - u 2 = sin - 1 u a + C, a > 0 Z du u 2 + a 2 = 1 a tan - 1 u a + C Trigonometric Identities sin 2 x + cos 2 x = 1 sec 2 x - tan 2 x = 1 csc 2 x - cot 2 x = 1 sin( x + y ) = sin x cos y + cos x sin y cos( x + y ) = cos x cos y - sin x sin y tan( x + y ) = tan x + tan y 1 - tan x tan y sin 2 x = 2 sin x cos x cos 2 x = 1 - 2 sin 2 x = 2 cos 2 xlevelsets - 1 sin 2 x = 1 2 (1 - cos 2 x ) cos 2 x = 1 2 (1 + cos 2 x ) tan 2 x = 2 tan x 1 - tan 2 x Derivatives of Inverse Trigonometric Functions d dx sin - 1 x = 1 1 - x 2 d dx cos - 1 x = - 1 1 - x 2 d dx tan - 1 x = 1 1 + x 2 d dx cot - 1 x = - 1 1 + x 2