Formulas
Geometry:
Circles:
Area =  r2
Circumference = 2 r
4
Spheres: Volume = 3  r3
Surface Area = 4 r2
Cylinders: Volume = (Area of base)  (Height)
1
Cones:
Volume = 3 (Area of base)  (Height)
Trigonometric Identities:
sin2x + cos2x = 1
tan2x + 1 = sec2x
ctn2x + 1 = csc2x
sin(x+y) = sin x cos y + cos x sin y
cos 2x = cos2x - sin2x
sin 2x = 2 sin x cos x
sin x sin y =
cos(x-y) - cos(x+y)
2
sin x cos y =
sin(x+y) + sin(x-y)
2

y = 

cos(x+y) = cos x cos y - sin x sin y
cos x cos y =
sin2x =


 is that angle y such that x = 


sin-1x
cos-1x
tan-1x
cot-1x
sec-1x
csc-1x
1 - cos(2x)
2
cos(x-y) + cos(x+y)
2
cos2x =
- 2  y  2
 0y
  - < y <
 and  0 < y < 
  0  y < or   y <
0 < y  or  < y 
sin y
cos y
tan y
cot y
sec y
csc y


2
2

2

2
sin-1(- x) = - sin-1 x
Hyperbolic Functions:
ex – e-x
sinh x =
2
cosh x
ex + e-x
coth x = sinh x = ex - e-x
cosh2x - sinh2x = 1
cos-1x = 2 - sin-1x
ex + e-x
2
1
2
sech x = cosh x = ex + e-x
cosh x =
tanh2x + sech2x = 1
3
2
3
2






cos-1(- x) =  - cos-1x
sinh(x+y) = sinh x cosh y + cosh x sinh y
1 + cos(2x)
2
sinh x
ex - e-x
tanh x = cosh x = ex + e-x
1
2
csch x = sinh x = ex - e-x
coth2x = 1 + csch2x
cosh(x+y) = cosh x cosh y + sinh x sinh y
cosh 2x = cosh2x + sinh2x
sinh 2x = 2 sinh x cosh x
sinh2x =

y = 

cosh(2x) - 1
2
cosh2x =

 
=
 

sinh-1x
cosh-1x
tanh-1x
coth-1x
sech-1x
csch-1x
cosh(2x) + 1
2



 is that y such that x = 



ln(x + x2 + 1)
ln(x + x2 -1)
1
ln( 11 +- xx )
2
1
2
ln( xx +1
)
-1
ln1 +
1 - x2
x

ln1 +
x2 + 1 
x




sinh y
cosh y and y  0
tanh y
coth y
sech y and y  0
csch y
Derivatives
dy
=
dx
f(x+h) - f(x)
h
h0
f’(x) = lim
Linear approximation (3 ways of writing the same thing)
f(x)  f(xo) + f’(xo)(x-xo)
f(x+x)  f(x) + f’(x) x
y 
dy
dx
x
Quadratic approximation
f”(xo)
f”(xo)
f(x+x)  f(x) + f’(x) x + 2 (x)2
f(x)  f(xo) + f’(xo)(x-xo) + 2 (x-xo)2
y 
dy
dx
x +
1 d2y
2 dx2
(x)2
Mean Value Theorem
(Slope of secant line = slope of tangent line at intermediate point. Same as
linear approximation but uncertainty is where derivative is evaluated.)
f(x) = f(xo) + f’(c)(x-xo)
f(x+x) = f(x) + f’(c) x
f(x)
lim g(x) =
L’Hospital’s Rule
xc
f '(x)
lim g'(x)
xc
y
x
dy
 dx (c)
provided f(x) and g(x) are either both 0 or both 
Summation Formulas
n(n+1)
1 + 2 + 3 + ... + n =
2
n2(n+1)2
13 + 23 + 33 + ... + n3 =
4
n(n+1)(2n+1)
12 + 22 + 32 + ... + n2 =
6
n+1
x -1
1 + x + x2 + ... + xn =
x-1
Integrals
n
b
 f(x) dx = lim

n
a
b-a
n ] n
 f[a + k(b-a)
k=1
b
Area = 
 L(x) dx
L(x) = Length of cross section through x.
a
2
b
Volume = 
 A(x) dx
A(x) = Area of cross section through x.
a
b
2
2
Volume = 
 [f(x) - g(x) ] dx
For the solid created by rotating a region about the x axis. The
a
region is bounded above by y = f(x), below by y = g(x), on the
left by x = a and on the right by x = b.
b
Volume = 2
 x[f(x) - g(x)] dx
For the solid created by rotating the same region about the y
a
axis.
b
Work
= 
 F(x) dx
F(x) = Force on object at position x.
a
Table of Derivatives and Integrals
d
2
dx tan x = sec x
d
2
dx cot x = - csc x
d
dx sec x = tan x sec x
d
dx csc x = - cot x csc x
d x
x
dx a = a ln(a)
d
1
-1
dx sin x = 1 - x2
2

 sec x dx = tan x
d
-1
dx cos x =
 1 dx = - cos-1x
 1 - x2

 1 dx = tan-1x
 1 + x2

-1
1 - x2
d
1
-1x =
tan
dx
1 + x2
d
-1
-1
dx cot x = 1 + x2
2

 csc x dx = - cot x
 tan x sec x dx = sec x


 cot x csc x dx = - csc x
ax
x

a
dx
=

ln(a)
 1 dx = sin-1x
 1 - x2

d
1
-1x =
sec
dx
x x2 - 1
 1 dx = - cot-1x
 1 + x2

1

 x x2 - 1 dx = sec-1x

d
-1
-1
dx csc x = x x2 - 1
1

 x x2 - 1 dx = - csc-1x

d
dx sinh x = cosh x

 cosh x dx = sinh x
3
d
dx cosh x
d
dx tanh x
d
dx coth x
d
dx sech x
d
dx csch x
= sinh x
 sinh x dx = cosh x

= sech2 x
2

 sech x dx = tanh x
= - csch2 x
2

 csch x dx = - coth x
= - tanh x sech x
 tanh x sech x dx = - sech x

= - coth x csch x

 coth x csch x dx = - csch x
d
-1
dx sinh x =
1
1 + x2
d
-1
dx cosh x =
1
x2 - 1
1

 1 + x2 dx = sinh-1x

 1 dx = cosh-1x
 2
 x -1
d
1
-1
dx tanh x = 1 - x2
d
-1
-1x =
sech
dx
x 1 - x2
 1 dx = tanh-1x
 1 - x2

 1 dx = coth-1x
 1 - x2

1

 x 1 - x2 dx = - sech-1x

d
-1
-1x =
csch
dx
x x2 + 1
1

dx = - csch-1x

 x x2 + 1
d
1
-1
dx coth x = 1 - x2
More Integrals

 u dv = uv - 
 v du

 sec x dx = ln( tan x + sec x)

 csc x dx = ln( csc x - cot x)
1
n
n-1

 sin x dx = - n sin x cos x
n

 cos x dx =
n

 tan x dx =
 cotnx dx =

n

 sec x dx =
n

 csc x dx =
1
cosn-1x sin x
n
1
tann-1x
n-1
1
- n-1 cotn-1x
1
tan x secn-2x
n-1
1
- n-1 cot x cscn-2x
+
+
n-1
 sinn-2x dx
n 
n-1
 cosn-2x dx
n 
-
n-2

 tan x dx
-
 cotn-2x dx

n-2
n-2
+ n-1 
 sec x dx
n-2
n-2
+ n-1 
 csc x dx
n
2m+1
n
2 m
2
2

dx = 
 sin x cos
 u (1-u ) du after substituting u = sin x and using cos x = 1 – sin x
Similarly when sin is to an odd power
4
2n
2m

 sin x cos dx Use sin2x =
1 - cos(2x)
2
and cos2x =
1 + cos(2x)
2
n
2m
n
2 m-1
2
2

 tan x sec dx = 
 u (1+u ) du after substituting u = tan x and using sec x = 1 + tan x
2n+1
m
2
n m-1
2
2

 tan x sec dx = 
 (u -1) u du after substituting u = sec x and using tan x = sec x - 1
2n
2m+1
2
n
2m+1

dx = 
dx after tan2x = sec2x – 1. Now expand out and use
 tan x sec
 (sec x - 1) sec
n
reduction formula for 
 sec x dx
Integral involving a2 – x2 Let x = a sin u
Integral involving a2 + x2 Let x = a tan u
Integral involving x2 – a2 Let x = a sec u
b
4ac - b2
a(x + )2 +
.
2a
4a
Integral involving ax2 +bx + c Complete the square: ax2 +bx + c =
b
Then substitute u = x + 2a which reduces the integral to one of the three previous
 anxn + an-1xn-1 +  + a1x + a0
Integral of a rational function 
dx. First, divide denominator into
m
m-1
 bmx + bm-1x +  + b1x + b0
numerator to get a polynomial plus a rational function where the numerator has degree less than the
denominator. Integrate the polynomial as usual. For the new rational function, factor the
denominator into the product of linear and quadratic factors and split it up using partial fractions:
anxn + an-1xn-1 +  + a1x + a0
A1
A2
B
C
Dx + E
= x-r + x-r +  + x-s + (x-s)2 + x2+bx+c. To find the numerators,
2 2

1
2
(x-r1)(x-r2) (x-s) (x +bx+c)
cross multiply to get anxn + an-1xn-1 +  + a1x + a0 = A1(x-r2)(x-s)2(x2+bx+c) +
A2(x-r1)(x-r3)(x-s)2(x2+bx+c) +  + B(x-r1)(x-r2)(x-s)(x2+bx+c) + C(x-r1)(x-r2)(x2+bx+c) +
(Dx+E)(x-r1)(x-r2)(x-s)2. You will need to plug in additional values to find B, D and E.
1
A1 A2
Example: (x-4)(x-5) = x-4 + x-5  1 = A1(x-5) + A2(x-4). x = 4  A1 = -1. The resulting
fractions with linear denominators are easy to integrate. Complete the square, if necessary, to
integrate the ones with quadratic denominators.

b
b

b

lim 
 f(x) dx = b
 f(x) dx

lim 
 f(x) dx = a-
 f(x) dx

lim 
 f(x) dx = ca+
 f(x) dx
if f(x) is unbounded as x  a
a
b
a
b
a
a
-
0


 f(x) dx = 
 f(x) dx + 
 f(x) dx
-
-
0
c
Numerical Integration
Midpoint method
b

 f(x) dx  2(x) [ f(x1) + f(x3) + f(x5) +
+ f(xn-1) ]
where x =
b-a
and xj = a + j(x) and n is even
n
a
Simpson’s method
b
x

 f(x) dx  3 [ f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) +
a
5
+ 2f(xn-2) + 4f(xn-1) + f(xn) ]
with x, xj as above, n is even
Arc Length
b

L = length of curve y = f(x) where a  x  b = 

2
dy
1 + dx dx
 
a
Center of Mass
_ _
(x, y) = center of mass of thin plate of uniform density occupying the region bounded above by y = f(x)
and below y = g(x) and between x = a and x = b. Let A = area. Then
_
_
1 b
1 b
2
2

x = A
x
[f(x)
g(x)]
dx
y
=

2A  {[f(x)] - [g(x)] } dx
a
a
Parametric Equations of Curves: x = f(t) and y = g(t)
To write as y = F(x): Solve x = f(t) for t in terms of x and substitute in y = g(t). Similarly to write as
x = G(y)
Slopes:
dy
dx
=
dy
dt
dx
dt
Second derivatives:
b

a

d2y
d2x
=
d dy
( )
dt dx
dx
dt
Areas: 
 y dx = 
 g(t) f ′(t) dt where a = f() and b = f()


Arc Lengths: L = 

2
2
dx dy
+
 dt   dt  dt = length of the part corresponding to   t  
   

Polar Coordinates r and 
x2 + y2
y
tan  = x
Curve defined by a polar equation r = f(): x = f() cos  and y = f() sin 
r =
x = r cos 
y = r sin 
dr
sin  + r cos 
dy
d
Slopes: dx = dr
cos  - r sin 
d


Arc Lengths: L = 

1

2
 dr 
r2 +   d = length of the part corresponding to   t  
d

2
2
Areas: A = 2 
 {[f()] - [g()] } d = area of region where g()  r  f() and     

6
Conic Sections
y
(x - u)
a2
Ellipse:
2
+
(y - v)
b2
2
b
= 1
a  b  foci at (u  c, v)
c2 = a2 – b2
a
x
a
y
y
b
Hyperbola:
(x - u)2 (y - v)2
- b2
a2
foci at (u  c, v)
= 1
bx a
a
x
a
c2 = a2 + b2
y
bx a
Sequences and Series

Geometric series:
1
 xn = 1 - x
if -1 < x < 1
n=0

nth term test: lim an  0   an diverges
n
n=1


n=1
n=1


1
n=1
 | an | <    an converges
Absolute convergence:
Integral test: f(x) positive & decreasing  
 f(x) dx <    f(n) < 
Limit comparison test: Assume an > 0 & bn > 0 for all n


n=1

n=1

an
n bn
<  &  bn <    an < 
an
lim
n bn
> 0 &  bn =    an = 
lim
n=1
n=1
Ratio test:
an+1
n an
If an > 0 then
| an+1 |
lim
n |an |
lim

< 1   an < 
&
n=1

< 1   an converges
an+1
n an
lim

> 1   an = 
n=1
| an+1 |
lim
n |an |
&
n=1

> 1   an diverges
n=1
Root test:

n
If an > 0 then
n
lim
n
n
an < 1   an < 
lim
n
&
lim
n
n=1

an > 1   an = 
n=1

n
| an | < 1   an converges
&
lim
n
n=1

| an | > 1   an diverges
n=1

Alternating series test: an decreasing with n & lim an = 0   (-1)nan converges
n
n=1

Taylor series: f(x) =
f(n)(a) (x - a)n

n!
n=0
 f(x) - n f(j)(a) (x - a)j  < M |x - a|n+1


j!
(n+1)!


j=0
where M = maximum of | f(n+1)(t) | for t between a and x
7