Common derivatives integrals_reduced
- 1. Common Derivatives and Integrals Common Derivatives and Integrals Derivatives Integrals Basic Properties/Formulas/Rules Basic Properties/Formulas/Rules d ( cf ( x ) ) = cf ¢ ( x ) , c is any constant. ( f ( x ) ± g ( x ) )¢ = f ¢ ( x ) ± g ¢ ( x ) ò cf ( x ) dx = c ò f ( x ) dx , c is a constant. ò f ( x ) ± g ( x ) dx = ò f ( x ) dx ± ò g ( x ) dx dx b b d n ( x ) = nxn-1 , n is any number. d ( c ) = 0 , c is any constant. ò a f ( x ) dx = F ( x ) a = F (b ) - F ( a ) where F ( x ) = ò f ( x ) dx dx dx b b b b b ò a cf ( x ) dx = c ò a f ( x ) dx , c is a constant. ò a f ( x ) ± g ( x ) dx = ò a f ( x ) dx ± òa g ( x ) dx æ f ö¢ f ¢ g - f g ¢ ( f g )¢ = f ¢ g + f g ¢ – (Product Rule) ç ÷ = – (Quotient Rule) a b a ègø g2 ò a f ( x ) dx = 0 ò a f ( x ) dx = -òb f ( x ) dx d ( ) f ( g ( x ) ) = f ¢ ( g ( x ) ) g ¢ ( x ) (Chain Rule) b c b b dx ò a f ( x ) dx = ò a f ( x ) dx + òc f ( x ) dx ò a c dx = c ( b - a ) ( ln g ( x )) = g (x ) g¢ x b dx e( ) d g ( x) = g ¢( x ) e ( ) g x d dx ( ) If f ( x ) ³ 0 on a £ x £ b then ò a f ( x ) dx ³ 0 b b If f ( x ) ³ g ( x ) on a £ x £ b then ò a f ( x ) dx ³ ò a g ( x ) dx Common Derivatives Polynomials Common Integrals d d d d n d dx (c) = 0 dx ( x) = 1 dx ( cx ) = c dx ( x ) = nxn-1 dx ( cxn ) = ncxn -1 Polynomials 1 ò dx = x + c ò k dx = k x + c ò x dx = n + 1 x n+1 n + c, n ¹ -1 Trig Functions ó 1 dx = ln x + c 1 òx òx -n d d d ô -1 dx = ln x + c dx = x - n +1 + c , n ¹ 1 ( sin x ) = cos x ( cos x ) = - sin x ( tan x ) = sec2 x õx -n + 1 dx dx dx p p p+q ó 1 dx = 1 ln ax + b + c 1 q +1 q d d d òx dx = x +c= +c q q ( sec x ) = sec x tan x ( csc x ) = - csc x cot x ( cot x ) = - csc2 x ô x dx dx dx õ ax + b a p q +1 p+ q Inverse Trig Functions Trig Functions d ( sin -1 x ) = 1 2 d ( cos-1 x ) = - 1 2 d ( tan -1 x ) = 1 +1x2 ò cos u du = sin u + c ò sin u du = - cos u + c ò sec u du = tan u + c 2 dx 1- x dx 1- x dx ò sec u tan u du = sec u + c ò csc u cot udu = - csc u + c ò csc u du = - cot u + c 2 d dx ( sec-1 x ) = 12 d dx ( csc-1 x ) = - 12 d dx ( cot -1 x ) = - 1 +1x2 ò tan u du = ln sec u + c ò cot u du = ln sin u + c x x -1 x x -1 1 ò sec u du = ln sec u + tan u + c ò sec u du = 2 ( sec u tan u + ln sec u + tan u ) + c 3 Exponential/Logarithm Functions d x d x 1 ( a ) = a x ln ( a ) (e ) = ex ò csc u du = ln csc u - cot u + c ò csc 3 u du = 2 ( - csc u cot u + ln csc u - cot u ) + c dx dx d ( ln ( x ) ) = 1 , x > 0 d ( ln x ) = 1 , x ¹ 0 d 1 dx x dx x dx ( log a ( x ) ) = x ln a , x > 0 Exponential/Logarithm Functions au ò e du = e + c ò a du = +c ò ln u du = u ln ( u ) - u + c u u u Hyperbolic Trig Functions ln a d d d e au ( sinh x ) = cosh x ( cosh x ) = sinh x ( tanh x ) = sech 2 x òe au sin ( bu ) du = ( a sin ( bu ) - b cos ( bu ) ) + c ò ue du = ( u - 1) e u u +c dx dx dx a + b2 2 d d d ( sech x ) = - sech x tanh x ( csch x ) = - csch x coth x ( coth x ) = - csch 2 x eau cos ( bu ) du = 2 e au ( a cos ( bu ) + b sin ( bu ) ) + c ó 1 du = ln ln u + c dx dx dx ò a + b2 ô õ u ln u Visit http://tutorial.math.lamar.edu for a complete set of Calculus I & II notes. © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus I & II notes. © 2005 Paul Dawkins
- 2. Common Derivatives and Integrals Common Derivatives and Integrals Inverse Trig Functions ó 1 æu ö Trig Substitutions du = sin -1 ç ÷ + c ò sin -1 ô u du = u sin -1 u + 1 - u 2 + c If the integral contains the following root use the given substitution and formula. õ a -u2 2 èaø a ó 1 1 æuö 1 a 2 - b2 x2 Þ x = sin q and cos2 q = 1 - sin 2 q ô 2 du = tan -1 ç ÷ + c ò tan -1 u du = u tan -1 u - ln (1 + u 2 ) + c b õ a +u 2 a èaø 2 a b2x2 - a 2 Þ x = sec q and tan 2 q = sec 2 q -1 ó 1 1 æuö b du = sec-1 ç ÷ + c ò cos -1 ô u du = u cos - 1 u - 1 - u 2 + c õ u u2 - a2 a èaø a a2 + b2 x2 Þ x = tan q and sec 2 q = 1 + tan 2 q b Hyperbolic Trig Functions Partial Fractions ò sinh u du = cosh u + c ò cosh u du = sinh u + c ò sech u du = tanh u + c ó P ( x) 2 If integrating ô dx where the degree (largest exponent) of P ( x ) is smaller than the ò sech tanh u du = - sech u + c ò csch coth u du = - csch u + c ò csch 2 u du = - coth u + c õ Q (x) degree of Q ( x ) then factor the denominator as completely as possible and find the partial ò tanh u du = ln ( cosh u ) + c ò sech u du = tan sinh u + c -1 fraction decomposition of the rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in the denominator we get term(s) in the Miscellaneous decomposition according to the following table. ó 1 du = 1 ln u + a + c ó 1 du = 1 ln u - a + c ô 2 ô 2 õ a - u2 2a u - a õ u - a2 2a u + a Factor in Q ( x ) Term in P.F.D Factor in Q ( x ) Term in P.F.D u a2 ò a + u du = a 2 + u 2 + ln u + a 2 + u 2 + c A1 A2 Ak 2 2 A + +L + ( ax + b ) k 2 2 ax + b ax + b ( ax + b ) 2 ( ax + b ) k ax + b u 2 a2 ò u 2 - a 2 du = u - a 2 - ln u + u 2 - a 2 + c Ax + B A1 x + B1 +L + Ak x + Bk 2 2 ( ax + bx + c ) 2 k ax 2 + bx + c ax + bx + c ( ax2 + bx + c ) 2 k u 2 a2 æuö ax 2 + bx + c ò a - u du = a - u + sin -1 ç ÷ + c 2 2 2 2 2 èaø u-a Products and (some) Quotients of Trig Functions a2 æ a -u ö ò 2au - u 2 du = 2 2 au - u 2 + cos-1 ç 2 è a ø ÷+c ò sin x cos x dx n m 1. If n is odd. Strip one sine out and convert the remaining sines to cosines using Standard Integration Techniques sin 2 x = 1 - cos2 x , then use the substitution u = cos x Note that all but the first one of these tend to be taught in a Calculus II class. 2. If m is odd. Strip one cosine out and convert the remaining cosines to sines using cos 2 x = 1 - sin 2 x , then use the substitution u = sin x u Substitution 3. If n and m are both odd. Use either 1. or 2. ò a f ( g ( x ) ) g¢ ( x ) dx then the substitution u = g ( x ) will convert this into the b Given 4. If n and m are both even. Use double angle formula for sine and/or half angle formulas to reduce the integral into a form that can be integrated. g ( b) integral, ò f ( g ( x ) ) g ¢ ( x ) dx = ò ò b f ( u ) du . tan n x sec m x dx a g ( a) 1. If n is odd. Strip one tangent and one secant out and convert the remaining Integration by Parts tangents to secants using tan 2 x = sec 2 x - 1 , then use the substitution u = sec x The standard formulas for integration by parts are, 2. If m is even. Strip two secants out and convert the remaining secants to tangents b b b using sec 2 x = 1 + tan 2 x , then use the substitution u = tan x ò udv = uv - ò vdu òa udv = uv a - òa vdu 3. If n is odd and m is even. Use either 1. or 2. Choose u and dv and then compute du by differentiating u and compute v by using the 4. If n is even and m is odd. Each integral will be dealt with differently. Convert Example : cos 6 x = ( cos 2 x ) = (1 - sin 2 x ) 3 3 fact that v = ò dv . Visit http://tutorial.math.lamar.edu for a complete set of Calculus I & II notes. © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus I & II notes. © 2005 Paul Dawkins