Common Derivatives Integrals - tutorial.math.lamar.edu

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 ) cf ( x ) dx = c f ( x ) dx , c is a constant. f ( x ) g ( x ) dx = f ( x ) dx g ( x ) dx dx b b d n ( x ) = nx n-1 , n is any number. d ( c ) = 0 , c is any constant. a f ( x ) dx = F ( x ) a = F ( b) - F ( a ) where F ( x ) = f ( x ) dx dx dx b b b b b a cf ( x ) dx = c a f ( x ) dx , c is a constant. a f ( x ) g ( x ) dx = a f ( x ) dx a g ( x ) dx f f g - f g . ( f g ) = f g + f g (Product Rule) = (Quotient Rule) a b a g g2 a f ( x ) dx = 0 a f ( x ) dx = - b f ( x ) dx d ( ).

2 F ( g ( x ) ) = f ( g ( x ) ) g ( x ) (Chain Rule). b c b b dx a f ( x ) dx = a f ( x ) dx + c f ( x ) dx a c dx = c ( b - a ). g ( x) b dx e( ). d g (x). = g ( x) e ( ). g x d dx ( ln g ( x ) ) =. g ( x). If f ( x ) 0 on a x b then a f ( x ) dx 0. b b If f ( x ) g ( x ) on a x b then a f ( x ) dx 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 ) = nx n-1 dx ( cx n ) = ncx n -1 Polynomials 1. dx = x + c k dx = k x + c x dx = n + 1 x + c, n -1. n n +1. Trig Functions 1 dx = ln x + c 1. x dx = ln x + c x dx = x - n +1 + c, n 1. -1 -n d d d.

3 ( sin x ) = cos x ( cos x ) = - sin x ( tan x ) = sec 2 x x -n + 1. dx dx dx p p p+ q 1 dx = 1 ln ax + b + c 1 +1 q d d d x dx = xq +c = +c q q ( sec x ) = sec x tan x ( csc x ) = - csc x cot x ( cot x ) = - csc 2 x x dx dx dx ax + b a p q +1 p+q Inverse Trig Functions Trig Functions d 1 d 1 d 1. ( sin -1 x ) = ( cos -1 x ) = - ( tan -1 x ) = cos u du = sin u + c sin u du = - cos u + c sec u du = tan u + c 2. dx 1 - x2 dx 1 - x2 dx 1 + x2. sec u tan u du = sec u + c csc u cot udu = - csc u + c csc u du = - cot u + c 2. d ( sec -1 x ) = 12 d ( csc-1 x ) = - 12. d 1. dx dx dx ( cot -1 x ) = - 1 + x 2 tan u du = ln sec u + c cot u du = ln sin u + c x x -1 x x -1.

4 1. sec u du = ln sec u + tan u + c 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 csc u du = ln csc u - cot u + c csc 3. u du =. 2. ( - csc u cot u + ln csc u - cot u ) + c dx dx d 1 d 1 d 1. dx ( ln ( x ) ) = x , x > 0 dx ( ln x ) = x , x 0 dx ( log a ( x ) ) = x ln a , x > 0 Exponential/Logarithm Functions au e du = e + c a du = +c 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 e au sin ( bu ) du = ( a sin ( bu ) - b cos ( bu ) ) + c ue du = ( u - 1) e u u +c dx dx dx a + b2 2.

5 D d d ( sech x ) = - sech x tanh x ( csch x ) = - csch x coth x ( coth x ) = - csch 2 x ( ). e au ( a cos ( bu ) + b sin ( bu ) ) + c 1 du = ln ln u + c =. au dx dx dx e cos bu du . a 2 + b2 u ln u Visit for a complete set of Calculus I & II notes. 2005 Paul Dawkins Visit for a complete set of Calculus I & II notes. 2005 Paul Dawkins Common Derivatives and Integrals Common Derivatives and Integrals Inverse Trig Functions 1 u Trig Substitutions du = sin -1 + c sin u du = u sin -1 u + 1 - u 2 + c -1. If the integral contains the following root use the given substitution and formula. a -u2 2. a . a 1 1 u 1 a2 - b2 x2 x = sin q and cos 2 q = 1 - sin 2 q 2 du = tan -1 + c tan -1.

6 U du = u tan -1 u - ln (1 + u 2 ) + c b a +u 2. a a 2 a b2 x2 - a 2 x = sec q and tan 2 q = sec 2 q - 1. 1 1 u b du = sec -1 + c cos u du = u cos -1 u - 1 - u 2 + c -1.. u u2 - a2 a a a a 2 + b2 x 2 x = tan q and sec 2 q = 1 + tan 2 q b Hyperbolic Trig Functions Partial Fractions sinh u du = cosh u + c sech u tanh u du = - sech u + c sech u du = tanh u + c P ( x). 2. If integrating dx where the degree (largest exponent) of P ( x ) is smaller than the cosh u du = sinh u + c csch u coth u du = - csch u + c csch 2. u du = - coth u + c Q ( x). degree of Q ( x ) then factor the denominator as completely as possible and find the partial tanh u du = ln ( cosh u ) + c sech u du = tan sinh u + c -1.

7 Fraction decomposition of the rational expression. Integrate the partial fraction decomposition ( ). For each factor in the denominator we get term(s) in the Miscellaneous decomposition according to the following table. 1 du = 1 ln u + a + c 1 du = 1 ln u - a + c 2 2. a - u2 2a u - a u - a2 2a u + a Factor in Q ( x ) Term in Factor in Q ( x ) Term in u 2 a2. a + u du = a + u 2 + ln u + a 2 + u 2 + c A1 A2 Ak 2 2. A + +L +. ( ax + b ). k 2 2 ax + b 2 ax + b ax + b ( ax + b )2 ( ax + b ). k u a u 2 - a 2 du = u 2 - a 2 - ln u + u 2 - a 2 + c Ax + B A1 x + B1. +L +. Ak x + Bk 2 2. ( ax + bx + c ). k ax 2 + bx + c 2.

8 Ax + bx + c ( ax 2 + bx + c ). 2 k u 2 a2 u ax + bx + c 2. a - u du = a - u + sin - 1 + c 2 2 2. 2 2 a . u-a Products and (some) Quotients of Trig Functions a2 a -u . 2au - u 2 du =. 2. 2au - u 2 + cos -1 . 2 a . +c 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 - cos 2 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.

9 If n and m are both odd. Use either 1. or 2. 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, f ( g ( x ) ) g ( x ) dx = . b f ( u ) du . n tan 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.

10 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 udv = uv - vdu a udv = uv a - a 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 = dv . Visit for a complete set of Calculus I & II notes. 2005 Paul Dawkins Visit for a complete set of Calculus I & II notes. 2005 Paul Dawkins