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- 1. Calculus Cheat Sheet Calculus Cheat Sheet Evaluation Techniques Limits Continuous Functions L’Hospital’s Rule Definitions f ( x) 0 f ( x) ± ¥ If f ( x ) is continuous at a then lim f ( x ) = f ( a ) Precise Definition : We say lim f ( x ) = L if Limit at Infinity : We say lim f ( x ) = L if we = or lim = x ®a If lim then, x® a x ®¥ x® a g ( x ) x® a g ( x ) ±¥ 0 can make f ( x ) as close to L as we want by for every e > 0 there is a d > 0 such that f ( x) f ¢ ( x) whenever 0 < x - a < d then f ( x ) - L < e . Continuous Functions and Composition a is a number, ¥ or -¥ = lim taking x large enough and positive. f ( x ) is continuous at b and lim g ( x ) = b then lim x ®a g ( x ) x ®a g ¢ ( x ) x® a ( ) There is a similar definition for lim f ( x ) = L “Working” Definition : We say lim f ( x ) = L lim f ( g ( x ) ) = f lim g ( x ) = f ( b ) Polynomials at Infinity x ®-¥ p ( x ) and q ( x ) are polynomials. To compute x® a x® a x® a if we can make f ( x ) as close to L as we want except we require x large and negative. Factor and Cancel p ( x) ( x - 2)( x + 6) x2 + 4 x - 12 lim factor largest power of x out of both by taking x sufficiently close to a (on either side Infinite Limit : We say lim f ( x ) = ¥ if we x®±¥ q ( x ) = lim lim of a) without letting x = a . x ( x - 2) x - 2x 2 x® a x ®2 x ®2 p ( x ) and q ( x ) and then compute limit. can make f ( x ) arbitrarily large (and positive) x+6 8 Right hand limit : lim+ f ( x ) = L . This has ) ( = lim = =4 x 2 3 - 42 by taking x sufficiently close to a (on either side 3 - 42 x® a x ®2 x 2 3x2 - 4 3 of a) without letting x = a . x = lim 2 5 = lim 5 x = - the same definition as the limit except it lim ( ) Rationalize Numerator/Denominator x®-¥ 5 x - 2 x -2 -2 2 x®- ¥ x x®-¥ requires x > a . 2 3- x 3- x 3+ x x x There is a similar definition for lim f ( x ) = -¥ = lim 2 lim 2 Piecewise Function x®9 x - 81 x®9 x - 81 3 + x ®a Left hand limit : lim- f ( x ) = L . This has the x ì x 2 + 5 if x < -2 except we make f ( x ) arbitrarily large and lim g ( x ) where g ( x ) = í 9- x -1 x® a = lim = lim ) ( ) î1 - 3 x if x ³ -2 ( x 2 - 81)(3 + x x®9 ( x + 9 ) 3 + x x®-2 same definition as the limit except it requires x ®9 negative. x<a. Compute two one sided limits, -1 lim g ( x ) = lim- x 2 + 5 = 9 Relationship between the limit and one-sided limits 1 = =- lim f ( x ) = L Þ lim+ f ( x ) = lim- f ( x ) = L lim+ f ( x ) = lim- f ( x ) = L Þ lim f ( x ) = L (18 )( 6 ) 108 x®-2- x®-2 lim g ( x ) = lim+ 1 - 3 x = 7 x® a x® a x® a x ®a x ®a x ®a lim+ f ( x ) ¹ lim f ( x ) Þ lim f ( x ) Does Not Exist x®-2+ Combine Rational Expressions x®-2 One sided limits are different so lim g ( x ) - x® a 1 æ x - ( x + h) ö x ®a x ®a 1æ 1 1ö - ÷ = lim ç ÷ x®-2 lim ç x ø h ®0 h ç x ( x + h ) ÷ h ®0 h x + h è è ø doesn’t exist. If the two one sided limits had Properties been equal then lim g ( x ) would have existed Assume lim f ( x ) and lim g ( x ) both exist and c is any number then, 1 æ -h ö -1 1 x®-2 x ®a x® a = lim ç ÷ = lim =- 2 h ®0 h ç x ( x + h ) ÷ é f ( x ) ù lim f ( x ) h®0 x ( x + h ) 1. lim écf ( x ) ù = c lim f ( x ) and had the same value. x è ø x ®a ë û provided lim g ( x ) ¹ 0 ú= x® a x ®a 4. lim ê x ®a g ( x ) û lim g ( x ) x® a ë x® a 2. lim é f ( x ) ± g ( x ) ù = lim f ( x ) ± lim g ( x ) Some Continuous Functions x ®a ë û x ®a n 5. lim é f ( x )ù = é lim f ( x )ù n x ®a x ®a ë û ë x ®a û Partial list of continuous functions and the values of x for which they are continuous. 7. cos ( x ) and sin ( x ) for all x. 1. Polynomials for all x. 6. lim é n f ( x ) ù = n lim f ( x ) 3. lim é f ( x ) g ( x ) ù = lim f ( x ) lim g ( x ) x ®a ë û x ®a x ®a ë û 2. Rational function, except for x’s that give 8. tan ( x ) and sec ( x ) provided x ®a x ®a division by zero. 3p p p 3p Basic Limit Evaluations at ± ¥ x ¹ L, - ,L 3. n x (n odd) for all x. ,- , , Note : sgn ( a ) = 1 if a > 0 and sgn ( a ) = - 1 if a < 0 . 2 222 4. n x (n even) for all x ³ 0 . 9. cot ( x ) and csc ( x ) provided 5. e x for all x. 1. lim e x = ¥ & lim e x = 0 5. n even : lim x n = ¥ x ¹ L , -2p , -p , 0, p , 2p ,L 6. ln x for x > 0 . x®- ¥ x ®± ¥ x®¥ lim ln ( x ) = ¥ lim- ln ( x ) = - ¥ 6. n odd : lim x n = ¥ & lim x n = -¥ 2. & x ®¥ x ®¥ x ®- ¥ x ®0 Intermediate Value Theorem 7. n even : lim a x n + L + b x + c = sgn ( a ) ¥ b Suppose that f ( x ) is continuous on [a, b] and let M be any number between f ( a ) and f ( b ) . 3. If r > 0 then lim =0 x ®± ¥ x ®¥ x r 8. n odd : lim a xn + L + b x + c = sgn ( a ) ¥ Then there exists a number c such that a < c < b and f ( c ) = M . 4. If r > 0 and x r is real for negative x x ®¥ 9. n odd : lim a xn + L + c x + d = - sgn ( a ) ¥ b then lim r = 0 x ®-¥ x ®-¥ x © 2005 Paul Dawkins © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
- 2. Calculus Cheat Sheet Calculus Cheat Sheet Chain Rule Variants Derivatives The chain rule applied to some specific functions. Definition and Notation ( ) f ( x + h) - f ( x) ( ) d d n-1 é f ( x )ù = n é f ( x ) ù f ¢ ( x ) cos é f ( x ) ù = - f ¢ ( x ) sin é f ( x ) ù n If y = f ( x ) then the derivative is defined to be f ¢ ( x ) = lim 1. 5. dx ë û ë û ë û ë û . dx h ®0 h ( ) ( ) d f ( x) d = f ¢( x)e tan é f ( x )ù = f ¢ ( x ) sec é f ( x ) ù f ( x) 2 2. e 6. ë û ë û If y = f ( x ) then all of the following are If y = f ( x ) all of the following are equivalent dx dx f ¢( x) ( sec [ f ( x)]) = f ¢( x ) sec [ f ( x)] tan [ f ( x )] ( ) d d ln é f ( x ) ù = notations for derivative evaluated at x = a . equivalent notations for the derivative. 7. 3. ë û f ( x) dx dx = ( f ( x) ) = Df ( x ) df dy d f ¢ ( x ) = y¢ = df dy f ¢ ( a ) = y ¢ x= a = = Df ( a ) = f ¢( x ) = ( ) d tan -1 é f ( x ) ù = ( ) d dx dx dx sin é f ( x ) ù = f ¢ ( x ) cos é f ( x ) ù dx x =a dx x =a 8. ë û 4. ë û ë û 1 + ë f ( x )û 2 é ù dx dx Interpretation of the Derivative If y = f ( x ) then, 2. f ¢ ( a ) is the instantaneous rate of Higher Order Derivatives The nth Derivative is denoted as The Second Derivative is denoted as 1. m = f ¢ ( a ) is the slope of the tangent change of f ( x ) at x = a . 2 dn f df f ¢¢ ( x ) = f ( 2) ( x ) = 2 and is defined as f ( n) ( x ) = n and is defined as line to y = f ( x ) at x = a and the 3. If f ( x ) is the position of an object at dx dx time x then f ¢ ( a ) is the velocity of equation of the tangent line at x = a is ( ) ¢ , i.e. the derivative of the ¢ f ¢¢ ( x ) = ( f ¢ ( x ) ) f ( x ) = f ( ) ( x ) , i.e. the derivative of ( n) n-1 given by y = f ( a ) + f ¢ ( a )( x - a ) . the object at x = a . first derivative, f ¢ ( x ) . () the (n-1)st derivative, f ( n -1) x . Basic Properties and Formulas If f ( x ) and g ( x ) are differentiable functions (the derivative exists), c and n are any real numbers, Implicit Differentiation Find y¢ if e2 x -9 y + x3 y 2 = sin ( y ) + 11x . Remember y = y ( x ) here, so products/quotients of x and y ( c f )¢ = c f ¢ ( x ) d (c) = 0 1. 5. will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to dx ± g )¢ = f ¢ ( x ) ± g ¢ ( x ) (f differentiate as normal and every time you differentiate a y you tack on a y¢ (from the chain rule). ( x ) = n x n-1 – Power Rule dn 2. 6. After differentiating solve for y¢ . dx ( f g )¢ = f ¢ g + f g ¢ – Product Rule 3. ( ) f ( g ( x) ) = f ¢ ( g ( x) ) g¢ ( x) d e2 x-9 y ( 2 - 9 y¢ ) + 3x 2 y 2 + 2 x 3 y y¢ = cos ( y ) y¢ + 11 7. æ f ö¢ f ¢ g - f g ¢ dx 11 - 2e 2 x -9 y - 3 x 2 y 2 4. ç ÷ = 2e 2 x -9 y - 9 y¢e 2x - 9 y + 3x 2 y 2 + 2 x 3 y y¢ = cos ( y ) y¢ + 11 – Quotient Rule y¢ = Þ This is the Chain Rule g2 2 x3 y - 9e2 x -9 y - cos ( y ) ègø ( 2 x y - 9e x - cos ( y ) ) y ¢ = 11 - 2e2 x - 9 y - 3 x 2 y 2 2 -9 y 3 Common Derivatives ( a ) = a x ln ( a ) d d dx ( x) = 1 ( csc x ) = - csc x cot x Increasing/Decreasing – Concave Up/Concave Down dx dx dx Critical Points ( e ) = ex x = c is a critical point of f ( x ) provided either d d dx ( sin x ) = cos x ( cot x ) = - csc2 x Concave Up/Concave Down 1. If f ¢¢ ( x ) > 0 for all x in an interval I then dx dx dx 1. f ¢ ( c ) = 0 or 2. f ¢ ( c ) doesn’t exist. ( sin -1 x ) = 1 2 ( ln ( x ) ) = 1 , x > 0 d d d ( cos x ) = - sin x f ( x ) is concave up on the interval I. 1- x dx dx dx x 2. If f ¢¢ ( x ) < 0 for all x in an interval I then Increasing/Decreasing ( ln x ) = 1 , x ¹ 0 d d ( tan x ) = sec 2 x ( cos x ) = - 1 2 d 1. If f ¢ ( x ) > 0 for all x in an interval I then -1 f ( x ) is concave down on the interval I. dx dx x 1- x dx f ( x ) is increasing on the interval I. ( log a ( x )) = x ln a , x > 0 d d 1 ( sec x ) = sec x tan x ( tan -1 x ) = 1 +1x2 d 2. If f ¢ ( x ) < 0 for all x in an interval I then dx dx Inflection Points dx x = c is a inflection point of f ( x ) if the f ( x ) is decreasing on the interval I. concavity changes at x = c . 3. If f ¢ ( x ) = 0 for all x in an interval I then f ( x ) is constant on the interval I. © 2005 Paul Dawkins © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
- 3. Calculus Cheat Sheet Calculus Cheat Sheet Extrema Related Rates Absolute Extrema Relative (local) Extrema Sketch picture and identify known/unknown quantities. Write down equation relating quantities 1. x = c is an absolute maximum of f ( x ) 1. x = c is a relative (or local) maximum of and differentiate with respect to t using implicit differentiation (i.e. add on a derivative every time f ( x ) if f ( c ) ³ f ( x ) for all x near c. you differentiate a function of t). Plug in known quantities and solve for the unknown quantity. if f ( c ) ³ f ( x ) for all x in the domain. Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one 2. x = c is a relative (or local) minimum of 2. x = c is an absolute minimum of f ( x ) The bottom is initially 10 ft away and is being starts walking north. The angleq changes at f ( x ) if f ( c ) £ f ( x ) for all x near c. if f ( c ) £ f ( x ) for all x in the domain. pushed towards the wall at 1 ft/sec. How fast 0.01 rad/min. At what rate is the distance 4 between them changing when q = 0.5 rad? is the top moving after 12 sec? 1st Derivative Test If x = c is a critical point of f ( x ) then x = c is Fermat’s Theorem If f ( x ) has a relative (or local) extrema at 1. a rel. max. of f ( x ) if f ¢ ( x ) > 0 to the left x = c , then x = c is a critical point of f ( x ) . of x = c and f ¢ ( x ) < 0 to the right of x = c . We have q ¢ = 0.01 rad/min. and want to find 2. a rel. min. of f ( x ) if f ¢ ( x ) < 0 to the left x¢ is negative because x is decreasing. Using Extreme Value Theorem x¢ . We can use various trig fcns but easiest is, If f ( x ) is continuous on the closed interval of x = c and f ¢ ( x ) > 0 to the right of x = c . Pythagorean Theorem and differentiating, x¢ x Þ sec q tan q q ¢ = sec q = x 2 + y 2 = 15 2 Þ 2 x x¢ + 2 y y¢ = 0 [a, b] then there exist numbers c and d so that, 3. not a relative extrema of f ( x ) if f ¢ ( x ) is 50 50 After 12 sec we have x = 10 - 12 ( 1 ) = 7 and We know q = 0.05 so plug in q ¢ and solve. 1. a £ c, d £ b , 2. f ( c ) is the abs. max. in the same sign on both sides of x = c . 4 x¢ sec ( 0.5 ) tan ( 0.5 )( 0.01) = so y = 152 - 7 2 = 176 . Plug in and solve [a, b] , 3. f ( d ) is the abs. min. in [a, b] . 50 2nd Derivative Test for y¢ . x¢ = 0.3112 ft/sec If x = c is a critical point of f ( x ) such that 7 7 ( - 1 ) + 176 y¢ = 0 Þ y¢ = ft/sec Finding Absolute Extrema Remember to have calculator in radians! f ¢ ( c ) = 0 then x = c 4 4 176 To find the absolute extrema of the continuous function f ( x ) on the interval [ a , b ] use the 1. is a relative maximum of f ( x ) if f ¢¢ ( c ) < 0 . Optimization 2. is a relative minimum of f ( x ) if f ¢¢ ( c ) > 0 . following process. Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for 1. Find all critical points of f ( x ) in [ a, b ] . one of the two variables and plug into first equation. Find critical points of equation in range of 3. may be a relative maximum, relative 2. Evaluate f ( x ) at all points found in Step 1. minimum, or neither if f ¢¢ ( c ) = 0 . variables and verify that they are min/max as needed. Ex. Determine point(s) on y = x 2 + 1 that are 3. Evaluate f ( a ) and f ( b ) . Ex. We’re enclosing a rectangular field with 500 ft of fence material and one side of the closest to (0,2). Finding Relative Extrema and/or 4. Identify the abs. max. (largest function field is a building. Determine dimensions that Classify Critical Points value) and the abs. min.(smallest function will maximize the enclosed area. 1. Find all critical points of f ( x ) . value) from the evaluations in Steps 2 & 3. 2. Use the 1st derivative test or the 2nd Minimize f = d 2 = ( x - 0 ) + ( y - 2 ) and the derivative test on each critical point. 2 2 Maximize A = xy subject to constraint of constraint is y = x 2 + 1 . Solve constraint for Mean Value Theorem x + 2 y = 500 . Solve constraint for x and plug If f ( x ) is continuous on the closed interval [ a , b ] and differentiable on the open interval ( a , b ) x 2 and plug into the function. into area. x2 = y -1 Þ f = x2 + ( y - 2 ) f (b) - f ( a) 2 A = y ( 500 - 2 y ) then there is a number a < c < b such that f ¢ ( c ) = . x = 500 - 2 y Þ b-a = y - 1+ ( y - 2) = y2 - 3 y + 3 2 = 500 y - 2 y 2 Differentiate and find critical point(s). Differentiate and find critical point(s). Newton’s Method f ¢ = 2y - 3 Þ y= 3 A¢ = 500 - 4 y Þ y = 125 f ( xn ) 2 If xn is the n guess for the root/solution of f ( x ) = 0 then (n+1)st guess is xn+1 = xn - th nd By 2nd deriv. test this is a rel. max. and so is f ¢ ( xn ) By the 2 derivative test this is a rel. min. and so all we need to do is find x value(s). the answer where after. Finally, find x. provided f ¢ ( xn ) exists. x = 500 - 2 (125 ) = 250 x 2 = 3 - 1 = 1 Þ x = ± 12 2 2 ( ) ( ) The dimensions are then 250 x 125. , 3 and - 1 1 ,3 . The 2 points are then 2 2 2 2 © 2005 Paul Dawkins © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
- 4. Calculus Cheat Sheet Calculus Cheat Sheet Standard Integration Techniques Integrals Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. Definitions Definite Integral: Suppose f ( x ) is continuous Anti-Derivative : An anti-derivative of f ( x ) () ò a f ( g ( x )) g ¢ ( x ) dx = ò g (a ) f ( u ) du u Substitution : The substitution u = g ( x ) will convert b gb on [ a , b ] . Divide [ a , b ] into n subintervals of is a function, F ( x ) , such that F ¢ ( x ) = f ( x ) . using Indefinite Integral : ò f ( x ) dx = F ( x ) + c du = g ¢ ( x ) dx . For indefinite integrals drop the limits of integration. width D x and choose x from each interval. * i ¥ where F ( x ) is an anti-derivative of f ( x ) . ò a f ( x ) dx = n å f ( x ) D x . cos ( x3 ) dx cos ( x3 ) dx = ò cos ( u ) dx 2 8 b 2 ò 1 5x ò 1 5x 5 2 * 2 Then lim Ex. 13 i ®¥ =1 i = 5 sin ( u ) 1 = 5 ( sin (8 ) - sin (1) ) u= x Þ du = 3 x dx Þ x dx = du 3 2 2 8 1 3 3 3 Fundamental Theorem of Calculus x = 1 Þ u = 13 = 1 :: x = 2 Þ u = 2 3 = 8 Part I : If f ( x ) is continuous on [ a, b ] then Variants of Part I : d u(x) f ( t ) dt = u ¢ ( x ) f éu ( x ) ù dx ò a g ( x ) = ò f ( t ) dt is also continuous on [ a , b ] x b b Integration by Parts : ò u dv = uv - ò v du and ò a u dv = uv - ò v du . Choose u and dv from ë û b a a a db integral and compute du by differentiating u and compute v using v = ò dv . f ( t ) dt = - v¢ ( x ) f é v ( x )ù dx ò v( x) d and g ¢ ( x ) = f ( t ) dt = f ( x ) . x dx ò a ë û ò xe -x 5 ò3 ln x dx Part II : f ( x ) is continuous on [ a , b ] , F ( x ) is d u( x) dx Ex. f ( t ) dt = u ¢ ( x ) f [ u ( x) ] - v¢ ( x ) f [ v( x ) ] dx ò v( x) Ex. -x -x u=x dv = e Þ du = dx v = -e an anti-derivative of f ( x ) (i.e. F ( x ) = ò f ( x ) dx ) u = ln x dv = dx Þ du = 1 dx v = x x ò xe dx = - xe + ò e dx = - xe - e + c -x -x -x -x -x ò3 ln x dx = x ln x 3 - ò3 dx = ( x ln ( x ) - x ) 3 5 5 5 5 then ò f ( x ) dx = F ( b ) - F ( a ) . b a = 5ln ( 5 ) - 3ln ( 3) - 2 Properties ò f ( x) ± g ( x) dx = ò f ( x) dx ± ò g ( x) dx ò cf ( x) dx = c ò f ( x) dx , c is a constant Products and (some) Quotients of Trig Functions For ò sin n x cos m x dx we have the following : For ò tan n x sec m x dx we have the following : ò a f ( x ) ± g ( x ) dx = ò a f ( x ) dx ± ò a g ( x ) dx ò a cf ( x ) dx = c ò a f ( x ) dx , c is a constant b b b b b 1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and ò a f ( x ) dx = 0 ò a f ( x ) dx = ò a f ( t ) dt a b b cosines using sin 2 x = 1 - cos 2 x , then use convert the rest to secants using tan 2 x = sec 2 x - 1 , then use the substitution the substitution u = cos x . ò a f ( x) dx = - òb f ( x) dx b a ò f ( x ) dx £ ò f ( x ) dx b b u = sec x . 2. m odd. Strip 1 cosine out and convert rest a a to sines using cos2 x = 1 - sin 2 x , then use 2. m even. Strip 2 secants out and convert rest ò f ( x) dx ³ ò g ( x ) dx If f ( x ) ³ g ( x ) on a £ x £ b then b a to tangents using sec 2 x = 1 + tan 2 x , then the substitution u = sin x . a b use the substitution u = tan x . 3. n and m both odd. Use either 1. or 2. If f ( x ) ³ 0 on a £ x £ b then f ( x ) dx ³ 0 ò b 4. n and m both even. Use double angle 3. n odd and m even. Use either 1. or 2. a and/or half angle formulas to reduce the 4. n even and m odd. Each integral will be If m £ f ( x ) £ M on a £ x £ b then m ( b - a ) £ ò f ( x ) dx £ M ( b - a ) b integral into a form that can be integrated. dealt with differently. 2 (1 + cos ( 2 x ) ) , sin ( x ) = 2 (1 - cos ( 2 x ) ) Trig Formulas : sin ( 2 x ) = 2sin ( x ) cos ( x ) , cos 2 ( x ) = a 2 1 1 Common Integrals ò tan ò k dx = k x + c ò cos u du = sin u + c ò tan u du = ln sec u + c ò cos x dx sin 5 x 3 x sec5 x dx Ex. Ex. 3 ò x dx = n+1 x + c, n ¹ -1 ò sin u du = - cos u + c ò sec u du = ln sec u + tan u + c ò tan x sec xdx = ò tan x sec x tan x sec xdx +1 n n 1 3 5 2 4 ò cos x dx = ò cos x dx = ò cos x dx 2 2 5 4 (sin x) sin x sin x sin x sin x 3 3 3 ò a + u du = a tan ( a ) + c = ò ( sec2 x - 1) sec 4 x tan x sec xdx ò x dx = ò x dx = ln x + c ò sec u du = tan u + c -1 -1 u 2 1 1 1 (1- cos x ) sin x ( u = cos x ) =ò 2 2 2 2 dx ò a x + b dx = a ln ax + b + c ò sec u tan u du = secu + c 3 ò a - u du = sin ( a ) + c = ò ( u 2 - 1) u 4 du cos x -1 ( u = sec x ) 1 u 1 1 (1-u ) = -ò du = - ò 22 1- 2u + u du 2 4 2 2 ò csc u cot udu = - csc u + c ò ln u du = u ln ( u ) - u + c 3 3 u u = 1 sec 7 x - 1 sec5 x + c = 1 sec 2 x + 2 ln cos x - 1 cos2 x + c 7 5 ò csc u du = - cot u + c ò e du = e + c 2 2 2 u u © 2005 Paul Dawkins © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.

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