# Calculus Cheat Sheet - Pauls Online Math Notes

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Calculus Cheat Sheet - Pauls Online Math Notes

Definite Integral: Suppose f ( )on [ ab , ] . Divide [ , ]widthD x and chooseCalculus Cheat SheetIntegralsDefinitionsx is continuousab into n subintervals ofb*Then ( ) limå( i )a¥n®¥ i = 1*xifrom each interval.ò f x dx = f x D x .Anti-Derivative : An anti-derivative of f ( x )is a function, F( x ) , such that F¢ ( x) = f ( x).Indefinite Integral : ò f ( x) dx= F( x)+ cwhere F( x ) is an anti-derivative of f ( x ) .Fundamental Theorem of Calculusf x is continuous on [ ab , ] then Variants of Part I :xd ux ( )g( x) = ò f () t dt is also continuous on [ ab , ]f () t dt u ( x) f u( x)adxò = ¢ éaë ùûd xd bg¢ x = f t dt = f xdxò .f () t dt v ( x) f v( x)adxò =- ¢ évx ( ) ë ùûf x is continuous on[ ab , ] , F( x ) isd ux ( )f () t dt u x f ux v x fdxò = ¢ - ¢vx ( )F x = ò f x dx )Part I : If ( )and ( ) () ( )Part II : ( )an anti-derivative of f ( x ) (i.e. ( ) ( )bò f x dx F b F a .athen ( ) = ( ) - ( )( ) ± ( ) = ( ) ± ( )ò ò òò ± = ò ± òòf x g x dx f x dx g x dxb b b( ) ( ) ( ) ( )f x g x dx f x dx g x dxa a aa( ) 0a f x dx =ba( ) =- ( )ò òf x dx f x dxbabaIf f ( x) ³ g( x)on a£ x£ bthen ( ) ³ ( )ò òbIf f ( x ) ³ 0 on a£ x£ b then ( ) 0òaPropertiescf x dx=c f x dxb( ) ( )( ) [ ( )] ( ) [ vx ( )]ò ò , c is a constantbbò cf ( x) dx=cò f ( x)dx , c is a constantaba( ) = ()ò òòf x dx g x dxa f x dx ³baf x dx f t dtb( ) £ ( )baòf x dx f x dxbIf m£ f ( x)£ M on a£ x£ b then mb ( - a) £ f ( x) dx£ M ( b-a)òaaacos sinCommon Integralsò kdx= kx+cò udu = u+cu uò e du = 2+ cò csc udu =- cot u+cx n1 n 1, 1n 1+ò = ++cn ¹- ò sinudu =- cosu+c-1 1 2òx dx= ò x dx= ln x + c ò sec udu = tan u+c1ò dx= 1a ln ax+ b + cax+bò lnudu = uln ( u)- u+cò secutanudu = secu+cò cscucotudu =- cscu+cVisit http://tutorial.math.lamar.edu for a complete set of Calculus notes.òòòòtanudu = ln secu + csecudu = ln secu+ tan u + cu( )1 du 1 -1= tana2 u2 a a + c+1u( )du -1= sina2 u2a+ c-© 2005 Paul Dawkins

Calculus Cheat SheetStandard Integration TechniquesNote that at many schools all but the Substitution Rule tend to be taught in a Calculus II class.u Substitution : The substitution u = g( x)will convert ( )( )du = g¢x dx . For indefinite integrals drop the limits of integration.22 3Ex. ò ( )5x cos x dx1u = x Þ du = 3xdx Þ xdx=du3 2 2 13131 1 :: 232 8x= Þ u = = x= Þ u = =Integration by Parts : òudv= uv-ò vdu andòbgbò f ( g x ) g¢ ( x) dx=ò f ( u)du usinga( )( )ga( ) = ò ( )585= sin( u) = ( sin( 8) -sin()1)2 82 351 1 3integral and compute du by differentiating u and compute v using vEx.ò-xò x dx- x - xu = x dv= e Þ du = dx v=-eò- x - x - x - x - xxe dx=- xe + e dx=-xe - e + c5x cos x dx cos u du3 1 3bb bò udv = uv - vduaa ò . Choose u and dv fromaEx.ò53ln xdx= ò dv .u = ln x dv= dx Þ du = dx v=xò1x5 5 5 5( ) ( )( ( ) )lnxdx= xlnx - dx= xlnx -x3 3 33= 5ln 5 -3ln 3 -2Products and (some) Quotients of Trig Functionsn mn mFor ò sin xcosxdx we have the following : Forò tan xsecxdx we have the following :1. n odd. Strip 1 sine out and convert rest to2 2cosines using sin x= 1- cos x, then usethe substitution u = cos x.2. m odd. Strip 1 cosine out and convert rest2 2to sines using cos x= 1- sin x, then usethe substitution u = sin x.3. n and m both odd. Use either 1. or 2.4. n and m both even. Use double angleand/or half angle formulas to reduce theintegral into a form that can be integrated.1. n odd. Strip 1 tangent and 1 secant out andconvert the rest to secants using2 2tan x= sec x- 1, then use the substitutionu = sec x.2. m even. Strip 2 secants out and convert rest2 2to tangents using sec x= 1+ tan x, thenuse the substitution u = tan x.3. n odd and m even. Use either 1. or 2.4. n even and m odd. Each integral will bedealt with differently.sin 2x 2sin x cos x2cos x11 cos 2x2sin x1= 1-cos 2x( )Trig Formulas : ( ) = ( ) ( ) , ( ) = + ( ) , ( ) ( )2ò2( )Ex.òòtanxsec3 5xdx= ò2 4ò( sec x 1sec ) xtanxsecxdx2 4ò( u 1) udu ( u sec x)3 5 2 4tan sec tan sec tan secx xdx x x x xdx= -= - == sec x- sec x+c1 7 1 57 5Ex.òsin5xdx cos3x5 42 2sin x sin xsinx (sin x)sin x= =cos3x cos3x cos3x(1-cos 2x)2sinx= ò=cos3x(1-u2) 21 2u2 u4=- du- +ò =-u3 ò u3ò ò òdx dx dx( cos )dx u xdu= sec x+ 2ln cosx - cos x+c1 2 1 22 2Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.© 2005 Paul Dawkins