math made simple :)

1. Calculus Cheat Sheet Calculus Cheat Sheet
Limits Evaluation Techniques
Definitions Continuous Functions L’Hospital’s Rule
Precise Definition : We say lim f ( x ) = L if Limit at Infinity : We say lim f ( x ) = L if we If f ( x ) is continuous at a then lim f ( x ) = f ( a ) f ( x) 0 f ( x) ± ¥
x ®a x ®¥ x® a If lim = or lim = then,
x ®a g ( x ) 0 x® a g ( x ) ±¥
for every e > 0 there is a d > 0 such that can make f ( x ) as close to L as we want by
whenever 0 < x - a < d then f ( x ) - L < e . Continuous Functions and Composition f ( x) f ¢( x)
taking x large enough and positive. lim = lim a is a number, ¥ or -¥
f ( x ) is continuous at b and lim g ( x ) = b then x® a g ( x ) x ®a g ¢ ( x )
x® a
“Working” Definition : We say lim f ( x ) = L
x® a
There is a similar definition for lim f ( x ) = L
x ®- ¥
x ®a
( x ®a
)
lim f ( g ( x ) ) = f lim g ( x ) = f ( b ) Polynomials at Infinity
p ( x ) and q ( x ) are polynomials. To compute
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)
by taking x sufficiently close to a (on either side x 2 + 4 x - 12 ( x - 2 )( x + 6 ) lim factor largest power of x out of both
Infinite Limit : We say lim f ( x ) = ¥ if we lim = lim x ®±¥ q ( x)
of a) without letting x = a . x ®a x® 2 x2 - 2 x x® 2 x ( x - 2)
can make f ( x ) arbitrarily large (and positive) p ( x ) and q ( x ) and then compute limit.
x+6 8
Right hand limit : lim+ f ( x ) = L . This has
x® a
the same definition as the limit except it
by taking x sufficiently close to a (on either side
of a) without letting x = a .
= lim
x ®2 x
Rationalize Numerator/Denominator
= =4
2
lim
3x2 - 4
= lim 2 5
(
x 2 3 - 42
x ) 3 - 42
= lim 5 x = -
3
requires x > a . 3- x 3- x 3+ x
x ®-¥ 5 x - 2 x 2 x ®- ¥ x
x (
-2 ) x ®-¥
x -2 2
There is a similar definition for lim f ( x ) = -¥ lim 2
x ®9 x - 81
= lim 2
x ®9 x - 81 3 + Piecewise Function
x® a x
Left hand limit : lim- f ( x ) = L . This has the ì x 2 + 5 if x < -2
x ®a except we make f ( x ) arbitrarily large and 9- x -1 lim g ( x ) where g ( x ) = í
= lim 2 = lim
same definition as the limit except it requires
( )
( x - 81) 3 + x x®9 ( x + 9 ) 3 + x ( ) î1 - 3x if x ³ -2
x ®-2
negative. x ®9
x<a. Compute two one sided limits,
Relationship between the limit and one-sided limits -1 1 lim- g ( x ) = lim- x 2 + 5 = 9
= =-
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
x® a x ®a x ®a x ®a x ®a x® a
lim+ g ( x ) = lim+ 1 - 3x = 7
lim f ( x ) ¹ lim- f ( x ) Þ lim f ( x ) Does Not Exist Combine Rational Expressions x ®-2 x ®-2
x® a + x® a x ®a
1æ 1 1ö 1 æ x - ( x + h) ö One sided limits are different so lim g ( x )
lim ç - ÷ = lim ç ÷ x ®-2
Properties h ®0 h x + h
è x ø h ®0 h ç x ( x + h ) ÷
è ø doesn’t exist. If the two one sided limits had
Assume lim f ( x ) and lim g ( x ) both exist and c is any number then, been equal then lim g ( x ) would have existed
x ®a x ®a 1 æ -h ö -1 1 x ®-2
= lim ç ÷ = lim =- 2
h ®0 h ç x ( x + h ) ÷
1. lim é cf ( x )ù = c lim f ( x ) é f ( x ) ù lim f ( x ) è ø
h ®0 x ( x + h ) x and had the same value.
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 )
ë û n Some Continuous Functions
5. lim é f ( x ) ù = élim f ( x ) ù
n
x ®a x ®a 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.
1. Polynomials for all x. 7. cos ( x ) and sin ( x ) for all x.
3. lim é f ( x ) g ( x )ù = lim f ( x ) lim g ( x ) 6. lim é n f ( x ) ù = n lim f ( 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.
Basic Limit Evaluations at ± ¥ 3. n
x (n odd) for all x. 3p p p 3p
x ¹ L, - , - , , ,L
2 2 2 2
Note : sgn ( a ) = 1 if a > 0 and sgn ( a ) = -1 if a < 0 . 4. n x (n even) for all x ³ 0 .
x 9. cot ( x ) and csc ( x ) provided
1. lim e x = ¥ & lim e x = 0 5. n even : lim x n = ¥ 5. e for all x.
x®¥ x®- ¥ x ®± ¥ 6. ln x for x > 0 . x ¹ L , -2p , -p , 0, p , 2p ,L
2. lim ln ( x ) = ¥ & lim- ln ( x ) = - ¥ 6. n odd : lim x n = ¥ & lim x n = -¥
x ®¥ x ®0 x ®¥ x ®- ¥
Intermediate Value Theorem
3. If r > 0 then lim
b
=0 7. n even : lim a x n + L + b x + c = sgn ( a ) ¥ Suppose that f ( x ) is continuous on [a, b] and let M be any number between f ( a ) and f ( b ) .
x ®± ¥
x ®¥ x r
4. If r > 0 and x r is real for negative x 8. n odd : lim a x n + L + b x + c = sgn ( a ) ¥ Then there exists a number c such that a < c < b and f ( c ) = M .
x ®¥
b
then lim r = 0 9. n odd : lim a x + L + c x + d = - sgn ( a ) ¥
n
x ®-¥
x ®- ¥ x
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

2. Calculus Cheat Sheet Calculus Cheat Sheet
Derivatives Chain Rule Variants
Definition and Notation The chain rule applied to some specific functions.
If y = f ( x ) then the derivative is defined to be f ¢ ( x ) = lim
h ®0
f ( x + h) - f ( x )
h
. 1.
d
dx ë
( û
n
)ë û
n-1
é f ( x )ù = n é f ( x ) ù f ¢ ( x ) 5.
d
dx ë û ( )
cos é f ( x ) ù = - f ¢ ( x ) sin é f ( x ) ù
ë û
If y = f ( x ) then all of the following are If y = f ( x ) all of the following are equivalent
2.
dx
e (
d f ( x)
)
= f ¢ ( x ) e f (x) 6.
d
dx ë û ( )
tan é f ( x ) ù = f ¢ ( x ) sec 2 é f ( x ) ù
ë û
f ¢ ( x) d
equivalent notations for the derivative. notations for derivative evaluated at x = a . 3.
d
(
ln é f ( x ) ù =
ë û ) 7. (sec [ f ( x )]) = f ¢( x ) sec [ f ( x)] tan [ f ( x )]
df dy d df dy dx f ( x) dx
f ¢ ( x ) = y¢ = = = ( f ( x ) ) = Df ( x ) f ¢ ( a ) = y ¢ x= a = = = Df ( a ) f ¢( x)
dx dx dx dx x =a dx x =a 4.
d
( )
sin é f ( x ) ù = f ¢ ( x ) cos é f ( x ) ù
ë û ë û
8.
d
ë û (
tan -1 é f ( x ) ù = )
1 + é f ( x )ù
2
dx dx
ë û
Interpretation of the Derivative
If y = f ( x ) then, 2. f ¢ ( a ) is the instantaneous rate of Higher Order Derivatives
The Second Derivative is denoted as The n th Derivative is denoted as
1. m = f ¢ ( a ) is the slope of the tangent change of f ( x ) at x = a . 2
d f dn f
f ¢¢ ( x ) = f ( ) ( x ) = 2 and is defined as f ( ) ( x ) = n and is defined as
2 n
line to y = f ( x ) at x = a and the 3. If f ( x ) is the position of an object at dx dx
equation of the tangent line at x = a is time x then f ¢ ( a ) is the velocity of
given by y = f ( a ) + f ¢ ( a )( x - a ) . the object at x = a .
f ¢¢ ( x ) = ( f ¢ ( x ) )¢ , i.e. the derivative of the ¢
( )
f ( x ) = f ( n -1) ( x ) , i.e. the derivative of
(n)
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
d Find y¢ if e 2 x -9 y + x 3 y 2 = sin ( y ) + 11x . Remember y = y ( x ) here, so products/quotients of x and y
1. ( c f )¢ = c f ¢ ( x ) 5. (c ) = 0 will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to
dx
( f ± g )¢ = f ¢ ( x ) ± g ¢ ( x ) d n differentiate as normal and every time you differentiate a y you tack on a y¢ (from the chain rule).
2.
6.
dx
( x ) = n x n-1 – Power Rule After differentiating solve for y¢ .
3. ( f g )¢ = f ¢ g + f g ¢ – Product Rule
7.
d
( )
f ( g ( x )) = f ¢ ( g ( x ) ) g ¢ ( x ) e 2 x -9 y ( 2 - 9 y ¢ ) + 3 x 2 y 2 + 2 x3 y y ¢ = cos ( y ) y ¢ + 11
æ f ö¢ f ¢ g - f g ¢ dx
11 - 2e 2 x -9 y - 3 x 2 y 2
4. ç ÷ = – Quotient Rule This is the Chain Rule 2e 2 x -9 y - 9 y¢e 2 x -9 y + 3 x 2 y 2 + 2 x3 y y¢ = cos ( y ) y¢ + 11 Þ y¢ =
ègø g2 2 x 3 y - 9e 2 x -9 y - cos ( y )
( 2x y - 9e x
3 2 -9 y
- cos ( y ) ) y ¢ = 11 - 2e2 x-9 y - 3x 2 y 2
Common Derivatives
d d d x
dx
( x) = 1
dx
( csc x ) = - csc x cot x
dx
( a ) = a x ln ( a ) Increasing/Decreasing – Concave Up/Concave Down
Critical Points
d d d x
dx
( sin x ) = cos x
dx
( cot x ) = - csc2 x
dx
(e ) = e x x = c is a critical point of f ( x ) provided either Concave Up/Concave Down
1. If f ¢¢ ( x ) > 0 for all x in an interval I then
1. f ¢ ( c ) = 0 or 2. f ¢ ( c ) doesn’t exist.
d d
dx
( cos x ) = - sin x
dx
( sin -1 x ) = 1 2 d
dx
( ln ( x ) ) = 1 , x > 0
x
f ( x ) is concave up on the interval I.
1- x
d d 1 Increasing/Decreasing 2. If f ¢¢ ( x ) < 0 for all x in an interval I then
dx
( tan x ) = sec2 x d
( cos x ) = - 1 2
-1
dx
( ln x ) = x , x ¹ 0 1. If f ¢ ( x ) > 0 for all x in an interval I then
f ( x ) is concave down on the interval I.
dx 1- x
d d 1 f ( x ) is increasing on the interval I.
( sec x ) = sec x tan x d 1
( tan x ) = 1 + x2
-1 ( log a ( x )) = x ln a , x > 0
dx
dx
dx 2. If f ¢ ( x ) < 0 for all x in an interval I then Inflection Points
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.
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

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
if f ( c ) ³ f ( x ) for all x in the domain. 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.
2. x = c is a relative (or local) minimum of Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one
2. x = c is an absolute minimum of f ( x ) The bottom is initially 10 ft away and is being starts walking north. The angle q 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
4
0.01 rad/min. At what rate is the distance
is the top moving after 12 sec? between them changing when q = 0.5 rad?
1st Derivative Test
Fermat’s Theorem If x = c is a critical point of f ( x ) then x = c is
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
Extreme Value Theorem 2. a rel. min. of f ( x ) if f ¢ ( x ) < 0 to the left x¢ is negative because x is decreasing. Using
x¢ . We can use various trig fcns but easiest is,
If f ( x ) is continuous on the closed interval Pythagorean Theorem and differentiating,
of x = c and f ¢ ( x ) > 0 to the right of x = c . x x¢
x 2 + y 2 = 152 Þ 2 x x ¢ + 2 y y ¢ = 0 sec q = Þ sec q tan q q ¢ =
[ a, b ] then there exist numbers c and d so that, 3. not a relative extrema of f ( x ) if f ¢ ( x ) is
After 12 sec we have x = 10 - 12 ( 1 ) = 7 and
50 50
We know q = 0.05 so plug in q ¢ and solve.
1. a £ c, d £ b , 2. f ( c ) is the abs. max. in
4
the same sign on both sides of x = c .
x¢
so y = 152 - 7 2 = 176 . Plug in and solve sec ( 0.5 ) tan ( 0.5 )( 0.01) =
[ a, b ] , 3. f ( d ) is the abs. min. in [ a , b ] . nd
2 Derivative Test for y¢ . 50
If x = c is a critical point of f ( x ) such that 7 x¢ = 0.3112 ft/sec
Finding Absolute Extrema 7 ( - 1 ) + 176 y ¢ = 0 Þ y ¢ = ft/sec Remember to have calculator in radians!
f ¢ ( c ) = 0 then x = c
4
To find the absolute extrema of the continuous 4 176
function f ( x ) on the interval [ a , b ] use the 1. is a relative maximum of f ( x ) if f ¢¢ ( c ) < 0 .
Optimization
following process. 2. is a relative minimum of f ( x ) if f ¢¢ ( c ) > 0 . 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 ] . 3. may be a relative maximum, relative one of the two variables and plug into first equation. Find critical points of equation in range of
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. We’re enclosing a rectangular field with Ex. Determine point(s) on y = x 2 + 1 that are
3. Evaluate f ( a ) and f ( b ) .
Finding Relative Extrema and/or 500 ft of fence material and one side of the closest to (0,2).
4. Identify the abs. max. (largest function field is a building. Determine dimensions that
value) and the abs. min.(smallest function Classify Critical Points
will maximize the enclosed area.
value) from the evaluations in Steps 2 & 3. 1. Find all critical points of f ( x ) .
2. Use the 1st derivative test or the 2 nd
derivative test on each critical point.
Minimize f = d 2 = ( x - 0 ) + ( y - 2 ) and the
2 2
Mean Value Theorem Maximize A = xy subject to constraint of constraint is y = x 2 + 1 . Solve constraint for
If f ( x ) is continuous on the closed interval [ a, b ] and differentiable on the open interval ( a, b ) x + 2 y = 500 . Solve constraint for x and plug
x 2 and plug into the function.
into area.
f (b) - f ( a) x2 = y -1 Þ f = x 2 + ( y - 2 )
2
then there is a number a < c < b such that f ¢ ( c ) = . A = y ( 500 - 2 y )
b-a x = 500 - 2 y Þ
= y -1 + ( y - 2 ) = y2 - 3 y + 3
2
= 500 y - 2 y 2
Newton’s Method Differentiate and find critical point(s). Differentiate and find critical point(s).
f ( xn ) A¢ = 500 - 4 y Þ y = 125 f ¢ = 2y -3 Þ y=3 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
By 2nd deriv. test this is a rel. max. and so is
nd
By the 2 derivative test this is a rel. min. and
f ¢ ( xn )
the answer we’re after. Finally, find x. so all we need to do is find x value(s).
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. The 2 points are then ( 1
2 2 )
, 3 and - ( 1
2 2)
,3 .
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

4. Calculus Cheat Sheet Calculus Cheat Sheet
Integrals Standard Integration Techniques
Definitions Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class.
Definite Integral: Suppose f ( x ) is continuous Anti-Derivative : An anti-derivative of f ( x )
g (b )
ò a f ( g ( x ) ) g ¢ ( x ) dx = ò g(a ) f (u ) du
b
on [ a, b ] . Divide [ a , b ] into n subintervals of is a function, F ( x ) , such that F ¢ ( x ) = f ( x ) . u Substitution : The substitution u = g ( x ) will convert using
width D x and choose x from each interval. Indefinite Integral : ò f ( x ) dx = F ( x ) + c du = g ¢ ( x ) dx . For indefinite integrals drop the limits of integration.
*
i
¥
where F ( x ) is an anti-derivative of f ( x ) .
ò a f ( x ) dx = nlim å f ( x ) D x . cos ( x3 ) dx cos ( x3 ) dx = ò
b 2 2 8
Then
®¥
*
i Ex. ò 1 5x
2
ò 1 5x
2
1 3
5
cos ( u ) du
i =1
u = x 3 Þ du = 3 x 2dx Þ x 2dx = 1 du (sin (8 ) - sin (1) )
8
3 = 5 sin ( u ) 1 =
3
5
3
Fundamental Theorem of Calculus x = 1 Þ u = 1 = 1 :: x = 2 Þ u = 2 = 8
3 3
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
x
g ( x ) = ò f ( t ) dt is also continuous on [ a, b ]
b b b
ë û Integration by Parts : ò u dv = uv - ò v du and ò a u dv = uv a
- ò v du . Choose u and dv from
a a
d x d b
f ( t ) dt = -v¢ ( x ) f év ( x ) ù integral and compute du by differentiating u and compute v using v = ò dv .
dx ò v( x)
and g ¢ ( x ) = f ( t ) dt = f ( x ) .
dx ò a
ë û
ò xe
-x 5
Part II : f ( x ) is continuous on [ a , b ] , F ( x ) is d u( x )
f ( t ) dt = u¢ ( x ) f [ u ( x ) ] - v¢ ( x ) f [ v ( x ) ]
Ex. dx Ex. ò3 ln x dx
dx ò v( x) u=x dv = e- x Þ du = dx v = -e- x
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 + òe
-x -x -x
dx = - xe dx = - xe- x - e- x + c
dx = ( x ln ( x ) - x )
5 5 5 5
ò3 ln x dx = x ln x 3 - ò3
b
then ò f ( x ) dx = F ( b ) - F ( a ) . 3
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
b b b b b For ò sin n x cos m x dx we have the following : For ò tan n x secm x dx we have the following :
òa f ( x ) ± g ( x ) dx = òa f ( x ) dx ± òa g ( x ) dx ò a cf ( x ) dx = c ò a f ( x ) dx , c is a constant 1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and
a b b
òa f ( x ) dx = 0 òa f ( x ) dx = ò f ( t ) dt cosines using sin 2 x = 1 - cos 2 x , then use convert the rest to secants using
a
b a
the substitution u = cos x . tan 2 x = sec 2 x - 1 , then use the substitution
òa f ( x ) dx = -òb f ( x ) dx
b b
ò f ( x ) dx £ ò
a a
f ( x ) dx 2. m odd. Strip 1 cosine out and convert rest
2.
u = sec x .
m even. Strip 2 secants out and convert rest
b a
to sines using cos 2 x = 1 - sin 2 x , then use
If f ( x ) ³ g ( x ) on a £ x £ b then ò f ( x ) dx ³ ò g ( x ) dx the substitution u = sin x . to tangents using sec 2 x = 1 + tan 2 x , then
a b
b
3. n and m both odd. Use either 1. or 2. use the substitution u = tan x .
If f ( x ) ³ 0 on a £ x £ b then ò f ( x ) dx ³ 0 4. n and m both even. Use double angle 3. n odd and m even. Use either 1. or 2.
a
b
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 ) integral into a form that can be integrated. dealt with differently.
2 (1 + cos ( 2 x ) ) , sin ( x ) = 2 (1 - cos ( 2 x ) )
a
Trig Formulas : sin ( 2 x ) = 2 sin ( x ) cos ( x ) , cos 2 ( x ) = 1 2 1
Common Integrals
ò k dx = k x + c ò ò tan u du = ln sec u + c ò tan sin 5 x
ò cos x dx
3
cos u du = sin u + c Ex. x sec5 x dx Ex. 3
ò x dx = xn +1 + 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
n 1 3 5 2 4 5 4 2 2
(sin x ) sin x
ò cos x dx = ò cos x dx = ò cos x dx
sin x sin x sin x
n +1 3 3 3
ò x dx = ò x dx = ln x + c ò sec u du = tan u + c ò a + u du = a tan ( a ) + c = ò ( sec 2 x - 1) sec 4 x tan x sec xdx
-1 1 2 1 u
1 -1
(1-cos x ) sin x 2 2
2 2
=ò dx ( u = cos x )
3
ò a x + b dx = a ln ax + b + c ò sec u tan u du = sec u + c
cos x
ò a - u du = sin ( a ) + c = ò ( u 2 - 1) u 4du
1 1 1 u -1
2 2 ( u = sec x ) = -ò
(1-u ) 2 2
du = -ò 1-2u +u du
2 4
ò ln u du = u ln ( u ) - u + c ò csc u cot udu = - csc u + c u 3 u 3
= sec x - sec x + c
1 7 1 5
7 5
= 1 sec 2 x + 2 ln cos x - 1 cos 2 x + c
ò e du = e + c ò csc u du = - cot u + c
u u 2 2 2
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins