Cheat Sheet – Exam 2
Derivatives
1.
2.
3.
4.
5.
6.
7.
8.
9.
d
(tan x)  sec2 x
dx
d
(cot x)   csc2 x
dx
d
(sec x)  sec x tan x
dx
d
(csc x)   csc x cot x
dx
d
1
(sin 1 x) 
dx
1  x2
d
1
(cos 1 x)  
dx
1  x2
d
1
(tan 1 x) 
dx
1  x2
d
1
(cot 1 x)  
dx
1  x2
d
1
(sec1 x) 
dx
x x2 1
d
1
10.
(csc1 x)  
dx
x x2 1
d
11.
(sinh x)  cosh x
dx
d
12.
(cosh x)  sinh x
dx
Integrals
1  x 
 a2  x2 dx  sin  a   C
1
1
 x
14.  2
dx  tan 1    C
2
a x
a
a
1
13.
15.
16.
17.
18.
19.
x
1
x2  a2
dx 
1 1  x 
sec    C
a
a
 ln x dx  x ln x  x  C
 tan x dx  ln | sec x | C
 sec x dx  ln | sec x  tan x | C
 cot x dx   ln | csc x | C
20.
 csc x dx  ln | csc x  cot x | C
21.
 sec
3
x dx 
sec x tan x ln | sec x  tan x |

C
2
2
Trig Identities
22. sin 2 x  cos2 x  1
23. 1  tan 2 x  sec2 x
24. 1  cot 2 x  csc2 x
25. sin 2 x  2sin x cos x
26. cos 2 x  cos2 x  sin 2 x
27. sin A cos B  12 [sin( A  B)  sin( A  B)]
28. cos A cos B  12 [cos( A  B)  cos( A  B)]
29. sin A sin B  12 [cos( A  B)  cos( A  B)]
Right Angle Trigonometry
30. sin  
opp
hyp
hyp
csc  opp
31. cos  
adj
hyp
sec  hyp
adj
32. tan  
opp
adj
adj
cot   opp
Half-Angle Formulas
33. sin 2 x  12 (1  cos 2 x)
34. cos2 x  12 (1  cos 2 x)
0
0
π/6
sin x
cos x
1
3
tan x
0
1
2
2
π/4
π/3
2
3
2
1
1
3
2
1
2
π/2
1
2
2
3
0
Undef.
7.1 Integration by Parts
7.4 Partial Fractions

Case 2:
p ( x)
A
B
C



2
( x  2)( x  4)
x  2 x  4 ( x  4) 2
Case 3:
p( x)
A
Bx  C

 2
2
( x  2)( x  4) x  2 x  4
udv  uv   vdu
L – Logs
I – Inverse Trig
A – Algebraic
T – Trig
E – Exponential
7.2

Higher on the list = u
Lower on the list = dv
ba
n
x
Tn  2 [ f ( x0 )  2 f ( x1 )  ...  2 f ( xn1 )  f ( xn )]
7.7 x 
sin n x cos m x dx
a) If sin x has an odd power, pull a sin x aside and
convert the rest to cos x. Substitute u = cos x
b) If cos x has an odd power, pull a cos x aside and
convert the rest to sin x. Substitute u = sin x
c) If both have even powers, use the half-angle
formulas.

x
S n  3 [ f ( x0 )  4 f ( x1 )  2 f ( x 2 )  ...  2 f ( x n2 )  4 f ( x n 1 )  f ( xn )]
Pattern for Simpson’s Rule Coefficients:
1,4,2,4,2,4,…,4,2,4,2,4,1
Error Bounds:
| ET |
K (b  a)3
where | f ''( x) |  K
12n 2
| EM |
K (b  a)3
where | f ''( x) |  K
24n 2
| ES |
K (b  a)5
where | f (4) ( x) |  K
4
180n
secn x tan m x dx
2
a) If sec x has an even power, pull a sec x aside
and convert the rest to tan x. Substitute u =
tan x
b) If tan x has an odd power, pull a
tan x sec x aside and convert the rest to
sec x. Substitute u = sec x
c) If sec x has an odd power and tan x has an even
power, convert all tan x into sec x.

M n  x[ f ( x1 )  f ( x2 )  ...  f ( xn1 )  f ( xn )]
Type I:

cscn x cot m x dx
2
a) If csc x has an even power, pull a csc x aside and
convert the rest to cot x. Substitute u = cot x
b) If cot x has an odd power, pull a
cot x csc x aside and convert the rest to
csc x. Substitute u = csc x
c) If csc x has an odd power and cot x has an even
power, convert all cot x into csc x.

b



a
t
f ( x)dx  lim f ( x)dx
t  a
b
f ( x)dx  lim  f ( x)dx


t  t
f ( x)dx  lim
x 

a
t
t
f ( x)dx  lim  f ( x)dx
t  a
Type II: If f(x) is discontinuous at a.

b
a
b
f ( x)dx  lim  f ( x)dx
t a
t
If f(x) is discontinuous at b.

b
a
t
f ( x)dx  lim  f ( x)dx
t b
a
If f(x) is discontinuous at c for some a  c  b
7.3 Trig Substitution
a2  x2
x  a sin 
1  sin 2   cos2 
a x
2
x  a tan 
1  tan   sec 
x2  a2
x  a sec
sec2   1  tan 2 
2
7.8 Improper Integrals
2

b
a
t
b
f ( x)dx  lim  f ( x)dx  lim  f ( x)dx
x c
a
2
Don’t Forget +C
t c
t
Parametric Equations x  f (t ), y  g (t )
Eliminate the parameter: Solve for t and
substitute.
Slope of the tangent line: m 
d
dt
d2y
Concavity:

dx 2
dy

dx
dy
dt
dx
Vertical when
dt

Let S be the solid generated by rotating the region
bounded by y  f ( x), x  a, x  b, y  0 about
the y-axis. Then the volume of S is given by:
b
V   2 xf ( x) dx
ydx   g (t ) f '(t )dt
a

      dt
dx
2
2
dy
dt
dt
b
x  r cos 
r x y
y  r sin 
tan  
2
2
dy

dx
dr
d
dr
d
2
y
x
sin   r cos 
cos   r sin 
Equation of the tangent line: y  y1  m( x  x1 )
dr
d
Horizontal when
Vertical when
dr
d
sin   r cos  0
cos  r sin   0
Undefined when both are 0.

Area inside the curve:

Area between curves:
  r
Arc Length: L 

1
2

   
dr
d
r 2 d
2
1
1
2
2
 r22  d
 r 2 d
Function y  f ( x)
Arc Length: L 

b
a
1
  dx
dy 2
dx
Let S be the solid generated by rotating the region
bounded by x  f ( y), y  a, y  b, x  0 about
the x-axis. Then the volume of S is given by:
Polar Equations r  f ( )
Slope of tangent line:
2
a


the y-axis. Then the volume of S is given by:
b
dx
0
dt
Arc Length: L 
a
V    [ f ( y)] dy
dy
0
dt
Undefined when both are 0.
Area under the curve (left to right):

2
V    [ f ( x)] dx
Let S be the solid generated by rotating the region
bounded by x  f ( y), y  a, y  b, x  0 about
Equation of the tangent line: y  y1  m( x  x1 )
Horizontal when
the x-axis. Then the volume of S is given by:
b
dt
 dy dt 
 dx 
 dt 
dx
Volume of a Solid
Let S be the solid generated by rotating the region
bounded by y  f ( x), x  a, x  b, y  0 about
V   2 yf ( y) dy
a