Math 408D
Exam 2
Name:
π/2 clearly, show all work. The exam is worth 100points, each question is equally weighted.
Directions: Write
2π/3
( −1 / 2, 3 / 2)
(0,1)
π/3
( 1 / 2, 3 / 2)
5π/6
2π/3
(− 3 / 2, 1 / 2)
( −1 / 2, 3 / 2)
π/2
(0,1)
sec θ =
π/6
π/3
( 1 / 2, 3 / 2) ( 3 / 2, 1 / 2)
π3π/4
(− 2 /(−1,0)
2, 2 / 2)
π/6
11π/6
( 3 / 2, −1 1/ 2)
/ 2)
π 5π/4
(− 2 / (−1,0)
2 , − 2 / 2)
7π/4 θ = 0 or 2 π
(1,0)
( 2 / 2 (x,y)
, − 2=/ 2)
5π/3
( 1 / 2, − 3 / 2)
11π/6
( 3 / 2, − 1 / 2)
x = cos θ
y = sin θ
5π/4
7π/4
(− 2 / 2 , − 2 / 2)
( 2 / 2 , − 2 / 2)
5π/3
4π/3
3π/2
2
1
1
cos2 θ
( −1 / 2, sin
− 3 /θ2)= 2 −
( 1 / 2, − 3 / 2)
(0,−1)
2 cos(2θ)
( −1 / 2, −
7π/6
(− 3 / 2, − 1 / 2)
4π/3
3 / 2)
3π/2
(0,−1)
1
φ)θ+
x =−cos
2 [sin(θ
sin θ cos φ =
sin(θ
y =+
sinφ)]
θ
sin θ sin φ =
1
cos θ
cot θ =
1
tan θ
sin2 θ + cos2 θ = 1
π/4 θ = 0 or 2 π
( 2 / 2,(x,y)
2 / =2) (1,0)
5π/6
7π/6
(− 3 / 2, 1 / 2)
(− 3 / 2, − 1 / 2)
sin θ
cos θ
1
csc θ =
sin θ
tan θ =
π/4
( 2 / 2, 2 / 2)
3π/4
(− 2 / 2, 2 / 2)
1 + tan2 sin
θ =θ sec2 θ
1 + cot2 θ = csc2 θ
tan θ =
cos θ
1 sin(A ± B) = sin A cos
1 B ± cos A sin B
1
sec θ =
csc θ =
cot θ =
cos θ
sin θ
tan θ
cos(A ± B) = cos A cos B ∓ sin A sin B
sin2 θ + cos2 θ = 1
1 + tan2 θ = sec2 θ
1 + cot2 θ = csc2 θ
√ .
√ .
.
2 = 1.414
3 = 1.732
π = 3.142
sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B
√
=
1
2
+
1
2 [cos(θ
1
2
√ .
3 = 1.732
.
2 = 1.414
cos(2θ)
.
π = 3.142
sin θ cos θ =
− φ) − cos(θ + φ)]
cos θ cos φ =
1
2
sin(2θ)
1
2 [cos(θ
− φ) + cos(θ + φ)]
1 −1
1
sinx2 θ = cos(cos
sin(sin−1 x) =
x) = x
2 − 2 cos(2θ)
2
1
1
−1
cos
cos(2θ)
tan(tan−1
x) θ==x 2 + 2csc(csc
x) = x
1
sin−1
θ cos
sec(sec
x) θ==x 2 sin(2θ)
cot(cot−1 x) = x
sin(θ + φ)]
sin θ cos
φ = 1 [sin(θ − φ) +−1
sin−1 (sin θ)2 = θ
cos (cos θ) = θ
sin θ−1
sin φ = 1 [cos(θ − φ) −
cos(θ + φ)]
tan (tan θ)2= θ
csc−1 (csc θ) = θ
cos θ cos φ = 1 [cos(θ − φ)
+ cos(θ + φ)]
sec−1 (sec θ) = 2θ
cot−1 (cot θ) = θ
−1
−1
−1
sin(sinln(pq)
x) =
x) =
x
x) = x
= xln(p) +cos(cos
ln(q)
ln(p/q)
=tan(tan
ln(p) − ln(q)
−1
csc(csc
x) = x
r
ln(p
) = r ln(p)
−1
sec(sec
ln(e)x)
==
1 x
−1
cot(cot
ln(1) = 0x) = x
sin−1 (sin
θ
ep eqθ)==ep+q
−1
p r (csc
(ecsc
) = erpθ) = θ
sec−1e(sec
1
= θ)
e =θ
−1
cot
e0 =(cot
1 θ) = θ
−1
cos−1 (cos θ) =
p−q (tan θ) = θ
epθ/eq = etan
ln(ep ) = p
ln(pq) = ln(p) + ln(q)
eln p = p
ln(p/q) = ln(p) − ln(q)
ax = ex ln a
ln(e) = 1
loga x = (ln x)/(ln a)
ln(pr ) = r ln(p)
p q
p+q
p q
′
e =
= x]
ep−q
[sin x]′ =e cos
x e [cos x]′ = − sin x e /e[tan
= sec2 x
p r
rp
′ ) =e
[csc x](e
= − csc x cot x
p
[c]′ = 0 ln(e ) = p [xp ]′ = pxp−1
x
x ln a
logcfa ′x = (ln x)/(ln a)[f ± g]′a=
=
[cf ]′ =
f ′ e± g ′
′
[sin x] = cos x
[c]′ = 0
[sin−1 x]′ =
′
[sin−1 x]Z
=b
a
Z
Z
eln p[e=x ]p′ = ex
1
= ex tan x
[sec x]′ e= sec
′
f
gf ′ − f g ′
p ′
p−1 ′
′
′
′
= pg
= fx]′ (g)
[cos x] = − sin x = [tan2x]′ = sec2 x [g ][csc
x]′ = −g csc x cot x[f (g)] [sec
= gsec x tan x
g
g
′
[xp ]′ = pxp−1
[ex ]′ = ex
[cf ]′ = cf ′
[f ± g]′ = f ′ ± g ′
−1
1
√ 1
[sec−1 x]′ = x√x12 −1
[cos−1 x]′ = √1−x
[tan−1 x]′ = 1+x
[csc−1 x]′ = x√−1
2
2
1−x2
x2 −1
′
f
gf ′ − f g ′
[g p ]′ = pg p−1 g′′
[f (g)]′ =xf′ ′ (g) g ′
=
[loga x] = (ln 1a) x
[a ] = (ln a) ax
[ln x]′ = x1g
g 2[ln |x|]′ = x1
−1
[cos−1 x]′ = √1−x
[tan−1 x]′ =
2
′
f (x) dx = F (b) − F (a), F (x) = f (x).
[ln |x|]′ = x1
[ln x]′ = x1
√ 1
1−x2
b
Z
1
1+x2
.
e = 2.718
ln(1) = 0
e0 =x]1′ = − csc2 x
[cot
.
e g]
=′ 2.718
[f
= f ′g + f g′
[cot x]′ = − csc2 x
[f g]′ = f ′ g + f g ′
−1
[cot−1 x]′ = 1+x
2
√
Z−1
x x2 −1
[sec−1 x]′ = x√x12 −1 [cot−1 x]′ =
f (x) dx = F (x) + C, F ′ (x) = f (x).
[loga x]′ = (ln 1a) x
[ax ]′ = (ln a) ax
[csc−1 x]′ =
Z
Z
sin x dx = − cos x + C
cos x dx = sin x + C
tan x dx = ln | sec x| + C
f (x) dx = F (b) − F (a), F ′ (x) = f (x).
f (x) dx = F (x) + C, F ′ (x) = f (x).
Za
Z
Z
csc x dx = ln | csc x − cot x| + C
sec x dx = ln | sec x + tan x| + C
cot x dx = ln | sin x| + C
Z
Z
sin x dx = − cos x + C
csc x dx = ln | csc x − cot x| + C
Z
cos x dx = sin x + C
(CONTINUED)
Z
sec x dx = ln | sec x + tan x| + C
Z
Z
tan x dx = ln | sec x| + C
cot x dx = ln | sin x| + C
−1
1+x2
[ln |x|]′ = x1
[loga x]′ = (ln 1a) x
[ax ]′ = (ln a) ax
[ln x]′ =x1 ′
f
gf ′ − f g ′
p ′
p−1r ′
′
[g d] =dypgln(p g) = r ln(p)[f (g)] = f ′ (g)
g′ = 1
=
dy
ln(pq) =
ln(e)
&ln(1) = 0& dx
dyln(p)dt+ ln(q) g ln(p/q)g 2= ln(p) −
d2ln(q)
y
dt ( dx )
Slope of C:
= dx .
= dx .
Concavity of C:
Net area between C and x-axis:
y dx = y dt dt.
2
Z
Z
dx
dx
0
dt
ep eq b= ep+q dt
ep /eq′ = ep−q
(ep )r = erp
e1 = e
e
=1
′
f (x)
dx = F (b)
f (x).′
'−1 F (x) = f−1(x).′
' =−1F (x)
'1 (
−1
1 F (x) = −1
−1
1 '
−1f (x) dx
′ + C,
−1 − ′F (a),
−1
′
′
(
√
√
√
√
= ' 2 [tan x] = 1+x2 '& [csc x] = x xdy2 −1 '[sec x] = x x2 −1 [cot . '&x] = 1+x
[sin x] a= 1−x
'& 2 [cosπ/2 x]
2'
dy 2π/3 '1−x
dx 2
2 dx
2(0,1)
2 dt'.
ln
p
x
x ln a
' 2πy
'2.718
'
)
+
(
)
dt
.
Area
of
S
:
)
+
(
)
Volume
of
S
:
πy
(
Lengthln(e
of Cp ) :='' p2π/3 ( dx
e
=
p
log
x
=
(ln
x)/(ln
a)
a
=
e
e
=
a dt
dt
dt
'
'
dt
'
dt dt .
( −1 / 2, 3 / 2)
1
1
1 ( 1 / 2, 3 / 2)
′
′
x ′ sin θ
x
′
[ln |x|]Z = x
[loga x] = (ln a) x
[a ] = (ln a) a
[ln x] = x
Z 3π/4
tan θ =Z
π/4
cos θ
Simple
parametric
curves
in standard
form.
(− 2 / 2, 2 / 2) x dx = − cos
Cx
= sin
C
tan x dx = ln | sec
x| x]
+′C= − csc2 x
( 2 / 2,x]′2 =
/ 2)sec2cos
[sin x]′ = cossin
x
[cos x]′ = x−+sin
[tan
x x dx[csc
x]′ x
=+
[sec x]′ =
[cot
1− csc x cot x
1 sec x tan x
1
ZZ b
Z
Z
sec
θ
=
csc
θ
=
cot
θ
=
Z
5π/6
Line:π/6 x = x0 + (x1 − x0 )t,cos
y=
y0 + (y1 − y0 )t.
θ
sin
θ
tan
θ
(− 3′ / 2, 1 / 2) csc x dx =
′ dx ′= ln
csc
x−
cot x| +
=]′ln=| sec
C [f
cotf+x
| sin[f
f (x) dx
F| (b)
−
F (a),
F
(x)
f (x).
f (x)
dx±=g]F′ (x)
C,g F ′ (x)
=x|
f′+
(x).
3=
/e
2,x
1 / 2)sec x dx[cf
[c] = 0
[xp=
]′ln=
pxp−1
[e′xC
]′( =
cf ′ x + tan x| +
=
±
g]
=Cf ′ g + f g ′
2
2
2
a
θ + cos
θ = 1a, b in x,
1+
θ=
csc2 θ
Circle w/radius
r: x = x0 + r cos(t), y = y0 + r sin(t). sinEllipse
w/radii
y :tan2xθ==xsec
a cos(t),1y+=cot
y02+
b sin(t).
0 +θ
π
θ = 0 or 2 π
′
′
′
f
gf −(x,y)
f g = (1,0)
(−1,0)
′
′
[g p ]′ = pg p−1 g ′ sin(A ±[fB)
(g)]=′ =
(g)
=
sinfAZ
cosg B ± cos A sin B
2
Z
Z Polar
g
g
coordinates.
(CONTINUED)
π/2
!
!
7π/6 sin x 2π/3
11π/6
dx = − cos
cos x dx = sin x + C
tan x dx = ln | sec x| + C
(0,1)x + Cπ/3
cos(A points
± B) =incos
A cos Bsatisfying
∓ sin A sin
B h(θ).
vgraph:
du
(− 3 / 2, − 1 (/ −1
2) / 2, 3 / 2)
( 3 / 2, − 1 / 2) u dv = uv −
(
1
/
2,
3
/
2)
Polar-Cartesian
relation:
x
=
r
cos
θ,
y
=
r
sin
θ.
Polar
xy-plane
r=
Z
Z
Z
−1
1
−1
−1
1
−1
1
−1sin′ θ
−1
′
−1
′
−1
′
′
√
√
√
√
[cot−1 x]′ = 1+x
[cos x] =
[tan x] = 1+x2 [csc x]√= .x x2 −1 tan[sec
[sin x] 3π/4
=
2
√
5π/4
7π/4
θ = . x]cot=x xdxx2=−1ln
2
π/4
&. |1sin
csc x1−x
dx2 = ln | csc x − cot 1−x
x| +
C
sec x dx = ln | sec x + 2tan
x|
+C
2 x| + C
cos
θ
=
1.414
3
=
1.732
π
=
3.142
eqs of polar graph:
Polar area of polar graph: 2 r dθ.
(−(− 22/ /22,, − 2 2/ 2)
/Parametric
2)
(( 22//22,,x−2=
/h(θ)
2) cos θ, y = h(θ) sin θ.
/22)
−1
! x=b
!
x ′θ=g1
4π/3 [ln 3π/2
1
' 1x]′!=x=b 1
' [a!
[ln |x|]′ = x1
] = (ln(b)
a) ax
x]′ u=g(b)
= x1 5π/3
(ln a) x
sec θ [log
='& a(
θ=
cotdθ,
θ = x = g(θ).
'csc
5π/6
π/6 u = g(x).
( −1 / 2, −g ′ (x)
3 / 2)dx =(0,−1)
( 1f/(u)
2, − du,
3 / 2)
f (g(x))
f
(x)
dx
=
f (g(θ))g ′ (θ)
dr
2 +(
2 dθ '.
'
cos
θ
sin
θ
tan
θ
Length
of
polar
graph:
)
r
(− 3x=a
/ 2, 1 / 2)
'
'
u=g(a)
x=a dθ
θ=g−1 (a)
( 3 / 2, 1 / 2)
(CONTINUED)
2
2
2
2
2
x = cos θ
y = sin θ
sin θ + cos θ = 1
1 + tan θ = sec θ
1 + cot θ = csc2 θ
Zπ b
Z
θ = 0 or 2 π
2 + x2
′
′
π/2 − F (a),
(x,y)
(1,0)a
a(−1,0)
dx x=IfF
= =flim
(x).
(x)
=AFcos
(x)
+±
C,lim
= x(possibly
f (x).
xB
Absolute
valuef (x)
rule:
lim
|anθπ/3
| = 0, F (x)
then
an = 0.x
Ifθ lim
|a
| dx
=sin
∞,
then
an(x)
±∞).
n=
x2π/3
2 dne
=(b)
a tan
x = a sin
= a sec
θ
sin(A
±fB)
cos2FA
sin
B
(0,1)
x − a
n→∞
n→∞θ
n→∞
n→∞
a
θ( −1 / 2, 3 2/ 2)
θ
2
1
1
1
( 1 / 2, 3 / 2)
sin
sin θ = 12 − 21 cos(2θ)
7π/6a
11π/6
sinθθcos
a2 cos
− x2 θ = 2 + 2 cos(2θ)
a θ = 2 sin(2θ)
Squeeze rule: If anπ/4≤( b3n/ ≤
cn1 / and
lim an = L andcos(A
lim c±n B)
= tan
L,
then
lim
bnsin
=AL.sin B
θ=
=
cos
A
cos
B
∓
3π/4
(− 3 / 2, − 1 / 2)
2,
−
2)
n→∞
n→∞
1
θ φ = 1 [cos(θ − φ) + cos(θ + φ)]
⎡2 / φ2,Z=2 /12)[sin(θ − φ) ⎤
Z φ = n→∞
cosZcos
θ cos
sin θ(−cos
( 2 / 2,sin
2 /θ2)sin
2) < deg(Q) + sin(θ + φ)]
2 [cos(θ − φ) − cos(θ + φ)]
2
deg(P5π/4
√
√
7π/4
π/2
1
1
.
.
.E
Ax
D
P
sin x 2π/3
dxMonotonic
= − cos x sequence
+C
= sin x and
+ C monotonic,
tan
dx =π ln
| sec
+ CE21
theorem: Ifcos
an xisdx
bounded
is x
convergent.
1cotx|
2 =P 1.414
3B=an1.732
cscthen
θ+=
2 / 5π/6
2 , b−2 2−/ 2)
( 2bx
/ 2π/6
,+
− c)(x
2 / 2) + d)(x
,(− ⎣
4ac < 0 ⎦(0,1)
=
+
+ = 3.142
+θ = tan θ2 + · · ·
, Q =π/3
(ax2 +
+sec
e)2θ·=
· · cos=⇒
2
θ
sin
θ
(5π/3
1 / 2, 3 / 2)
/ 2)
Q 3 / 2, 1 / 2)
ax + bx +
Z( −1 d/ 2,̸= 3e4π/3
Z
Z c x + d x + e (x + e)
(−
⎧ Q
3π/2
( 3 / 2, 1 / 2)
sin
θ
0,
if −1 <θ r=<
1 −1
−1
2= =
2 x) = x2
( −13π/4
/ 2,=
−xxdx
3 / 2)
( 1=
/ 2,
3C
/ 2) tan(tan−1
sin(sin−1
x)
cos(cos
x=2 ln
x)
x)dx
cot(cot
csc
= ln
|(0,−1)
csc x−1
−x)
cot
x|x−+
secx)x =
dxsin
| csc(csc
sec
tan
x
x|cot
+C
⎨−1
π/4
θ+
cosx2 +
θ⎪
=
1 x|=+xCtan
1 + sec(sec
tan2cos
θ cot
=θ sec
θ x ln | sin
1+
θ = csc θ
1,
if r = 1
n
θ
=
0
or
2
π
(− 2 / 2, π2 / 2)
( 2 / 2,sequence
2 / 2)
Exponential
rule:
lim
r
=
.
(−1,0)
(x,y) =curve
(1,0) C:
(x,
y) = (f
g(t)), aif ≤
t ≤1 b. −1 1
x = cos θ −1 y = sin θ Parametric
1 (t),
1
n→∞
r>
⎪ ∞,
⎩
sin(A
± θB)csc
= sin
B −1
sin−1 (sin
cos (cos θ) = θ
tan−1 (tan θ) = sec
θ θ = csc−1 (csc
θ) =
(secBθ)±=cos
θ A sincot
(cot θ) = θ
θsec
=A cos
θ=
5π/6 θ) = θ
π/6
dne,
if
r
≤
−1
(Let S be surface obtained by rotating C
x-axis assumingsin
y≥
cosabout
θ
θ 0).
tan θ
(− 3 / 2, 1 7π/6
/ 2)
(11π/6
3 / 2, 1 / 2)
(CONTINUED)
2
2
2
2
2
cos(A
±
B)
=
cos
A
cos
B
∓
sin
A
sin
B
(− 3 / 2, − 1 / 2)
( 3 / 2, − 1 / 2) 2 sin1 θ +1 cos θ = 1
1 + tan θ = sec θ 1
1 + cot θ = csc2 θ
1
− 21 cos(2θ)
=
+ 2 cos(2θ)
sin θ cos θ = 2 sin(2θ)
sin2 θ =
∞
π
θ = 0 or 2 π cos θr.
dy+2 ln(q)
d 2dy ln(p
2
ln(pq)
=
ln(p)
ln(p/q)
=
ln(p)
−
ln(q)
)
=
r
ln(p)
ln(e)
=
1
ln(1) = 0
c/(1. − r), if |r|
( dxn−1
) = √
√< .1 .
7π/4 (x,y) = (1,0)
d y
(−1,0) 5π/4dy
. & y dx = & y dx dt.
dt
Geometric
series
rule:
1 dt cr
1 πA=sin
1
Slope
of
C:
=
=
.
Concavity
of
C:
.
Net
area
between
C
and
x-axis:
2
=
1.414
3
=
1.732
3.142
sin(A
±
B)
=
sin
A
cos
B
±
cos
B
[sin(θ
−
φ)
+
sin(θ
+
φ)]
sin
θ
sin
φ
=
[cos(θ
−
φ)
−
cos(θ
+
φ)]
cos
θ
cos
φ
=
[cos(θ
−
φ) + cos(θ dt
+ φ)]
sin
θ
cos
φ
=
divergent,
if
|r|
≥
1
(− 2 / 2 , − 22/ 2)dx
( 2 / 2 , − 2 / 2) dx2 2
dx
dx
2
n=1dt
p q
5π/3ep /eq =11π/6
4π/3 dt
e7π/6
e = ep+q
ep−q
(ep )r = erp
e1 = e
e0 = 1
3π/2
∞
'
' / 2)
( 1 / '2, − 3 / 2)
( cos(A ± B) =' cos
(0,−1)
A cos B ∓ sin A sin B
(− 3 / 2, − 1( /−1
2) / 2, − 3(
( 3 / 2, − 1 / 2)
'
'&
'
'&
Series divergence
test: If lim an ̸=''&0 or lim dxan2dne,dy 2then''
an diverges.
x ln a
n→∞
n→∞
)2 + ( dy
)2 dte''ln
. p=p
Area
of S : ' 2πy
)(ln+x)/(ln
( dt ) a)
dt'. n=1 ax−1
Volume
of Se =
:. '2.718
πy 2 dx
dt'.
( dt
Lengthln(e
of Cp ) :='' p ( dx
log
x
=
=
e
−1
−1
−1
−1
−1
dt
dt
dt
a
√
√
5π/4
7π/4
θ
y = sin
sin(sin x) = x x = cos cos(cos
x) θ= x
tan(tan x)∞
=x
csc(csc
x)
sec(sec
x) = x . cot(cot x) = x
. =x
.
∞
2 = 1.414
3 = 1.732
π = 3.142
(− 2 / 2 , − 2 / 2)
( 2 / 2 , − 2 / 2)
rule:
an =
am+p−1 where m−1= n − p + 1.
5π/3substitution −1
4π/3 Series
3π/2
−1
−1 index
−1
in standard
sin (sin−1θ)/ 2,=− θ 3 / 2) cos
(cos θ)
= θ 3 /Simple
tan parametric
(tan θ) = θcurves
csc
(csc θ) =form.
θ
sec ′ (sec θ) = θ
cot−1 (cot
θ) = θ
′
′
(x1 / 2, − [tan
2)x]′ = sec2 x n=p [csc x]m=1
[sin x]′ = (cos
x 2 [cos x]
=1− sin
=
−
csc
x
cot
x
[sec
x]
=
sec
x
tan
x
[cot x]′ = − csc2 x
2
1
1
1
1 (0,−1)
Integral test: If f (x) is
continuous,
positive,
decreasing on [1,sin
∞)θ cos θ = sin(2θ)
cos
θ
=
+
cos(2θ)
sin θ = 2 − 2 cos(2θ)
2
2
∞
∞ x0 + (x1 −2 x0
Line: Z
x=
)t,
y = y0 + (y1 − y0 )t.
X
x = cos θ
y = sin θ
′
p ′
p−1
xboth
′
x
′n converge or diverge. ′
′
′1
′
1 and′
1
then
f
(x)
dx
a
[c] θ=cos
0φ
[x
[e ] =
=ln(p)
eθ sin−φ ln(q)
[cf ] −
=
cfr−) cos(θ
[f ±cos
g] ln(e)
f φ=
±=1g 2 [cos(θ −
[f
g]
==cos(θ
f0′ g ++f gφ)]
− ]φ)
+px
sin(θ + φ)]
sin
= 2 [cos(θ
φ)
+ φ)]
θ=
cos
φ)′ +
sin
= 2 [sin(θ
ln(pq)
= ln(p)
+=ln(q)
ln(p/q)
ln(p
= r ln(p)
ln(1)
n=1 w/radii a, b in x, y :
Circle w/radius r: x = x0 + r cos(t), y = y0 +1r sin(t).
Ellipse
x = x0 + a cos(t), y = y0 + b sin(t).
′
(1 p r
f ep /e
gfq ′=−efp−q
g′
2∞ p ′ 1
2
p−1
′ erp
′
′ sin
ep eq = sin
ep+q
) g=
e1 θg=′cos
e θ = 21 sin(2θ)
e0 = 1
cosX
cos(2θ)
θ = 12 − 21 cos(2θ)
[gθ =
] 1=2 +
pg(e
(g)]
=
p[f>
1 = f (g)
2 convergent
2
−1
−1g
−1
−1
−1
−1
g tan(tan
rule:Polar
=
sin(sin x)1 = x
cos(cos x) = x p-series
x)1= xcoordinates.
csc(csc
x)
=
x
sec(sec
x)
=
x
cot(cot
x) = x
1
np − φ)
divergent
p≤1
.
sin
θ
sin
φ
=
[cos(θ
−
cos(θ
+
φ)]
cos
θ
cos
φ
=
[cos(θ
−
φ)
+
cos(θ + φ)]
sin θ cos φ
=
p2 [sin(θ − φ) + sin(θ + φ)]ln p
x
x
ln
a
n=1
2
ln(e ) = p
e
=p
loga x = (ln x)/(ln a)
a =e 2
e = 2.718
−1y = r sin θ.
−1
relation:
Polar
points
insec
xy-plane
satisfying
rcot
= −1
h(θ).
sin−1 (sinPolar-Cartesian
θ) = θ1
cos−1
(cos θ) =−1
θx = r cos
tanθ,
(tan θ) =1 θ
csc−1graph:
(csc θ) =
θ
(sec θ)
=θ
(cot θ) =
−1θ
[sec−1 x]′ = x√x12 −1& [cot−1 x]′ = 1+x
[cos−1 x]′ = √1−x
[tan−1 x]′ = 1+x
[csc−1 x]′ = x√−1
[sin−1 x]′ = √1−x
2
2
2
2
x2 −1
1 2
polar graph: x′ = h(θ)
θ, y = h(θ)
sin θ.−1 Polar area of ′polar
r dθ. ′−1
−1
−1 graph:
′ −1 x)Parametric
′eqs of−1
2 cos
′ csc(csc
2
sin(sin
=
x
cos(cos
x)
=
x
tan(tan
x)
=
x
x)
=
x
sec(sec
x)
=
x
cot(cot
x)
=
x
[sin x] = cos x
[cos x] =′ − sin
x
[tan x] =
sec
x
[csc
x] 'a=
−
x cot
x] = sec x tan x
[cot x] = − csc2 x
1 x
′
Comparison
tests:
b ′csc
positive
' [sec
n , x]
[ln |x|]
= x1
[log
[ax ]′ = (ln a) ax
[ln x] = 1
r n =P
(ln
a) x
'& a(
'
P = ln(p) − ln(q)
ln(pq) = ln(p) + ln(q) x ln(p/q)
ln(p
)
=
r
ln(p)
ln(e)
=
1
ln(1) = 0
2 + ( dr
' n , then
polar
dθ''. sec−1′ (sec θ)
If xLength
bn converges
an)2converges
−1
′ −1 (sin θ) = θ
p−1 θ) = θ
x of
′ r (csc
′ = ′θ
′ (cot ′θ) = θ ′
dθ
tane−1
(tan
θ)and
=graph:
θan ]′≤='bcsc
θ)
=
θ
cot
[c]sin
=0
[xp ]′cos
= −1
px(cos
[e P
]′ =
[cf
cf −1
[f
±
g]
=
f
±
g
[f
g]
= f g + fg
P
p q
p+q
p Ifq
p−q
p
r
rp
1
0
b
diverges
and
a
≥
b
,
then
a
diverges
n
n
e Ze = e
e /e = en
(en ) = e
e
=
e
e
=
1
Z
′
b
P
P
a′n
′
′ −=
′
f
gf
f
g
If
lim
c
>
0
then
both
a
,
b
converge
or
diverge.
n
p ′
p−1
′ n If lim f|a
′= ∞,
′ + C,lim Fa (x)
dx =IfF (b)
(x).an =[g0.
=′ F (x)
= (possibly
f. (x).
Absolute value
rule:
lim−|aFn(a),
| =n→∞
0, F b(x)
then= flim
dne
±∞).
n | dx
] = pgln(p
[f (x)
(g)]
g=
p f (x)
xthen
np2
ln(e
e=ln p =
logarxg) = (ln
x)/(ln
a) = f (g)
a
exn→∞
e = 2.718
ln(pq)
ln(p) + ln(q)
ln(p/q)
r ln(p)
ln(e)
=
1ln a n ln(1)
=0
n→∞ g
n→∞− ln(q)
n→∞
a)= p
g = ln(p)
Squeeze rule: Ifp anq ≤ bnp−q
≤ cn and lim an = pL rand rp
lim c = L, then1 lim bn = L.
ep eq = ep+q
e /e = e
(e ) = en→∞ n
e n→∞
=e
e0 = 1
n→∞
Z′
Z 2′
Z′ ′
1
−1
1
−1
1
−1
−1
′
−1
−1
′
P
′
′
′
′ ′ = −1 2
√ x
√
[sec
x]
=x
[cot
[cos
[tan=
x]n bx=
[csc= −x]csc=and
=x√1−x2[cosAlternating
[sin x] =x]cos
x] = −x]sin=Series
x √ [tan
x]
sec
[csc
x cot
= sec
tan−1
x
[cot−1x]x]
= − csc2 x
2 is x]
2
2 −1 bn[sec
1+x
test:
(−1)
bbounded
decreasing
=
0x]then
x xlim
xdxx2=
n =
Monotonic
Ifcos
ann xifis
then
anxtan
isconverges.
convergent.
. + C 1+x
= − cos x sequence
+ C 1−x
dx
sin
x and
+ C monotonic,
x
psin x dx
ln ptheorem:
x ln a ln | sec x|
n→∞
ln(e ) = p
e
=p
loga x = (ln x)/(ln a)
a =e
e = 2.718
1
′
x ′ Z′
x Z
′ ′⎧
′ ax ′
= x1
[a
= 1=
(lnfa)
= x1
[c]′ = 0 Z
[xp ]′[ln
= x]
px′ p−1
[ex ]′ [ln
= e|x|]
[cf ]′ [log
= cf
±g
[f g]′ = f ′ g + f g ′
a x] = 0,
<±r] g]
<
(ln a) x if −1[f
⎪
⎨tan x| + C
csc x dx = ln | csc x − cot x| + C
sec x dx = ln | sec x +
cot x dx = ln | sin x| + C
1,
if r = 1
Exponential
′
rule: [csclim
r=n −
=csc x cot x
′
′ x]
′ sec2 x
[sin x]′ = cos x
[cos x]′ = − sin
[tan
x]′p−1
= .sec
[cot x]′ = − csc2 x
fx
gf
− sequence
f′g=
∞,
if
r >′[sec
1 fx]′ (g)
p ′ n→∞
′⎪
′ x tan x
⎩
[g
]
=
pg
g
[f
(g)]
=
g
=
Z b
Z
g
g2
dne, if r ≤ −1
′ ′
f (x) dx
F (b)
− F (a), F
f (x) [f
dx±=g]F′ (x)
C,g ′ F ′ (x) [f
=g]
f′(x).
[c]′ = 0
[xp=
]′ =
pxp−1
[ex(x)
] ==exf (x).
[cf ]′ = cf ′
= f+′ ±
= f ′g + f g′
(CONTINUED)
a
.
−1 ′
−1
1∞Page 2 −1 c/(1
1
−11 ′
′ −1 x]′ = √−1
√1
r), if [sec
|r| <
x] =
[cot−1 x]′ = 1+x
[cos−1 x]′ f= ′√1−x
[tan
[csc
x]′ = x−
[sin−1 x]′ = √1−x
n−1
2
2
gf
−
f
g
2
2
1+x
p
′
p−1
′
x2 −1 ′
Geometric
series rule: [g ] =
cr pg = g ′
. g ′ x x2 −1
[f
(g)]
=
f
(g)
=
Z
Z
divergent, if |r| ≥Z1
2
g
g
n=1
1
′
x ′
x
[ln |x|]′ =cos
[logxa+x]C
= (ln 1a) x
[a] = (ln
[lncos
x]′x=+x1C
∞
sin x dx = −
tana)
x adx
= ln | sec x| + C
x x dx = sin
Series divergence test: IfZ lim an ̸= 0 or lim an dne, then
Z
Zan diverges.
′
[sin−1
−1
′
√ 1
−1
′
√ −1
−1 n→∞
′
1
n→∞
−1
′
√−1
−1
′
√1
−1
′
−1
cos θ = 2 + 2 cos(2θ)
7π/6
11π/6
( −1 / 2, sin
− 3 /θ2)= 2 −
( 1 / 2, − 3 / 2)
2 cos(2θ)
(0,−1)
cos(A ± B)
(− 3 / 2, − 1 / 2)
( 3 / 2, − 1 / 2)
φ)
+ sin(θ
sinln(p)
θ sin−φ ln(q)
= 21 [cos(θ −ln(p
φ) r−) cos(θ
+ φ)]
sin θ cos φ
= 12 [sin(θ
ln(pq)
= ln(p)
+θln(q)
= r ln(p)
x =−cos
y =+
sinφ)]
θln(p/q) =
√ .
5π/4
7π/4
2
=
1.414
(− 2 / 2 ,p− q2 / 2) p+q
2 / 2)
p ( q2 / 2 , −p−q
e e = e 4π/3
(ep )r = erp
5π/3e /e = e
3π/2
1 −1 ( 1 / 2, − 3 / 2)
1
−1
( −1x)
/ 2,=
− x23 /θ2)= cos(cos
(0,−1)
sin(sin−1
x) = x
tan(tan−1
x)2 θ==x 1 + 1csc(csc
x) = x
−
cos(2θ)
cos
cos(2θ)
sin
sin θ cos θ = 2 sin(2θ)
= cos A cos B ∓ sin
A sin B
cos ln(e)
θ cos φ==1 12 [cos(θ −
φ) +=cos(θ
ln(1)
0 + φ)]
√ .
.
3 = 1.732
π = 3.142
e1 = e
e0 = 1
−1
sec(sec
x) θ==x 21 sin(2θ)
cot(cot
x) = x
sin−1
θ cos
.
2
2
2
2
ln(ep ) = p
eln p = p
loga x = (ln x)/(ln a)
ax = ex ln a
e = 2.718
1
1
1
y =+
sinφ)]
θ
sin(θ
sin θ−1
sin φ = [cos(θ − φ) −
cos(θ + φ)]
cos θ cos φ = [cos(θ − φ)
+ cos(θ + φ)]
sin θ cos
φ = [sin(θx =−cosφ)θ+−1
sin−1 (sin θ)2 = θ
cos (cos θ) = θ
tan (tan θ)2= θ
csc−1 (csc θ) = θ
sec−1 (sec θ) = 2θ
cot−1 (cot θ) = θ
(
P− csc x cot x
2
[sin x]′ = cos x 2 [cos x]′ = − sin x
[tan x]′ = asec
x2 L
[csc
x]′ =
x]′ = sec x tan x
[cot x]′ = − csc2 x
<
an is absolutely [sec
convergent
n+1 1
1 1 1
1
1
−1
−
cos(2θ)
cos
θ
=
+
cos(2θ)
sin
θ
cos
θ
=
sin(2θ)
sin
θ
=
Ratio
Test:
lim
=
P
−1
−1
−1
−1
−1
r
2
2
2
2
2
sin(sinln(pq)
x) =
x) =
x
x) =Lx > 1 ln(p
csc(csc
x) = x
sec(sec
cot(cot
n→∞
= xln(p) +cos(cos
ln(q)
ln(p/q)
=tan(tan
ln(p)an− ln(q)
ln(p)
ln(e)x)
==
1 x
ln(1) = 0x) = x
a) =isr divergent
′
p ′
p−1
x ′
x
′
′ n
′
′
′
′
′
[c]
=
0
[x
]
=
px
[e
]
=
e
[f
±
g]
=
f
±
g
[f
g]
=
f ′ g ++f gφ)]
1 [cf ] = cf
1
1
sin θ sin φ = 2 [cos(θ − φ) − cos(θ + φ)]
cos θ cos φ = 2 [cos(θ − φ) + cos(θ
sin θ cos φ = 2 [sin(θ − φ) + sin(θ + φ)]
(
−1 p q
−1
−1
−1
−1
−1
p−q (tan θ) = θ
p r (csc
rpθ) = θ
1
P
sin (sin
θ
cos (cos
csc
sec e(sec
cot
e e θ)==ep+q
epθ/eq = etan
= θ)
e =θ
e0 =(cot
1 θ) = θ
p
θ)′=
L < 1 (e ) an= ise absolutely convergent
f
gf ′ −lim
f g ′ n |an | = p ′
Root
Test:
P
n→∞
[g L] >
= 1pg p−1 agn′ is divergent
[f (g)]′ = f ′ (g)xg ′ x ln a
=
.
p
g eln p =gp2
log
(lnx)x)/(ln
a −1
=x)
e =x
e = 2.718
−1 ) = p
−1
a x =−1
sin(sinln(e
x) = x
cos(cos−1 x) =
x
tan(tan−1 x) = x
csc(csc
=
x a) sec(sec
cot(cot
x) = x
r
ln(pq) = ln(p) + ln(q)
ln(p/q) = ln(p) − ln(q)
ln(p ) = r ln(p)
ln(e) = 1
ln(1) = 0
−1
−1 ′
−1 ′
−1(sin
−1θ
−1(cot
′ θ) =
−1(cos
−1(tan
−1
sin−1
θ) =
θ1
cos−1
θ)
=−1
tan
=1 θ2 [csc
csc
=θ
sec−1
(sec
θ) √
=1θ
cot0−1
√∞
p ′ q=
p+q
pθ 2q
r x](csc
rpθ)√−1
1 =
[sec
x]
[cot
[cos
x]′ X
=
[tan
x]2′ x=θ)1+x
[sin
∞√
∞
∞
′
′ p )Z
′e
2 −1
e =
e1−x
/e[tan
= x]
ep−q
(e
=csc
e=
=X
ex
e =x]x]
1′ ==−1+x
X
1−x2[cos x]′ = − sin
x xx
x tan
xc2n−1
[sin
x]′ =ex]cos
x
=
sec
[csc
x]
=
−
x
cot
[sec
x]
=
sec
x
[cot
csc22 x
dx eX
n+1
n
n−1
n
(x − a)
cn (x − a) =
ncn (x − a)
cn (x − a) dx = C +
dx n=0
nx + 1
1
1
1
′
′
x ′
′
.
p
x n=0
n=1 eln p =
=
x]
=
[a
]
=
a)
=
′
p ′[ln x] p−1
xp
′ [ln |x|]
x
′ [log
′x n=0
′a(ln
′ exaln′a
a
ln(e
)
=
p
log
=
(ln
x)/(ln
a)
=
e g]
=′ 2.718
x
x
(ln
a)
x
[c] = 0 ln(pq) = ln(p)
[x ]+=ln(q)
px
[e ] ==ln(p)
e
cfar ) = r ln(p)
[f ± g] ln(e)
=f =
± 1g
[f
==f0′ g + f g ′
ln(p/q)
− ln(q) [cf ] =
ln(p
ln(1)
′
f ep /e
gfq ′=−efp−q
g′
p q
p r ′
1 ′
p∞′
p−1
′
e b= ep+q
(e
) = erp Z
e0 = 1
′ e Z
′
′
2
′ e g= e
][csc
=
pg
= fx]′ (g)
=
f (n)
[sin x] = cos x
[cos x] = − sin
x
[tan
x [gX
x]′(a)
= −g csc xncot x[f (g)] [sec
= sec x tan x ′
[cot x]′ = − csc2 x
′ g 2x] = sec
g
f (x) =
(x − a)
f (x) dx = F (b) − F (a), F (x) = f (x).
f (x) dx = F (x) + C, F (x) = f. (x).
n!
n=0
ln(eap ) = p p ′
eln p =x p′
ax = ex ln a
e = 2.718
′
p−1
x
′ loga ′x = (ln x)/(ln a)
[c] = 0
[x ] = px
[e ] = e
[cf ] = cf
[f ± g]′ = f ′ ± g ′
[f g]′ = f ′ g + f g ′
−1
1
−1
−1
1
−1
1
−1
′
−1
′
−1
′
−1
′
′
[sin x] = √1−x2 [cos x] = √1−x2 [tan x] = 1+x2 [csc x] = x√x2 −1 [sec x] = x√x2 −1 [cot−1 x]′ = 1+x
2
Z
Z
′
′
′ Z
f
gf − f′g
′
′
2
′
′
′
2
p [csc
′ x tan x
[sin x] = cossin
x x dx[cos
x] =′x−+sin
x = [tan x] = sec
x1 x[gdx
x] p−1
= −g′′csc x cot
x[f (g)]′[sec
=tan
[cot
]′ = sin
pg
= fx]′ (g)
gsec
′ cos
x = ln | sec
=−
x adx
x| x]
+ C= − csc x
[logxa+x]C
= (ln 1a) x
[ax ]′ = (ln
a)
[lncos
x] = x1C
g
g 2[ln |x|] = x
Z
Z
Z
[c]′ = 0
[xp ]′ = pxp−1
[ex ]′ = ex
[cf ]′ = cf ′
[f ± g]′ = f ′ ± g ′
[f g]′ = f ′ g + f g ′
csc x dx = ln | csc x − cot x| + C
sec x dx = ln | sec x + tan x| + C
cot x dx = ln | sin x| + C
−1
−1
1
1
′
[sec−1 x]′ = x√x12 −1 [cot−1 x]′ = 1+x
[cos−1 x]′ =′√1−x
[tan′ −1 x]′ = 1+x
[csc−1 x]′ = x√Z−1
[sin−1 x]Z
=b √1−x
2
2
2
2 ′
x2 −1
f
gf
′ − fg
′
p
′
p−1
′
′
′
′
f (x) dx = F (b) − F (a),= F (x) = f (x).
f (x)
dx
=
F (x)
+ C, F (x) = f (x).
[g
]
=
pg
g
[f
(g)]
=
f
(g)
g
a
g 2[ln |x|]′ = 1
[loga x]′ = (ln 1a) x
[ax ]′ = (ln a) ax
[ln x]′ = x1g
x
(CONTINUED)
Z′
Z
Z
−1
1
1
[sec−1 x]′ = x√x12 −1 [cot−1 x]′ =
[cos−1 x]′ = √1−x
[tan−1 x]′ = 1+x
[csc−1 x]′ = x√Z−1
[sin−1 x]Z
= √1−x
2
2
x2 −1
b x dx2 = − cos x + C
sin
cos
x
dx
=
sin
x
+
C
tan x dx = ln | sec x| + C
F (b)′ − F1 (a), F ′ (x) = f (x).
f (x) dxx=′ Z
F (x) + C,x F ′ (x) = f (x).
1
1
′
′
Z a f (x) dx =[ln
Z
[ln |x|] = x
[loga x] = (ln a) x
[a ] = (ln a) a
x] = x
csc x dx = ln | csc x − cot x| + C
sec x dx = ln | sec x + tan x| + C
cot x dx = ln | sin x| + C
ZZ
Z
Z
cos
x
dx
=
sin
x
+
C
tan+x C,
dx =Fln
′ | sec x| + C
F (x) = f (x).
f (x) dx = F (x)
(x) = f (x).
(CONTINUED)
Za
Z
Z
csc x dx = ln | csc x − cot x| + C
sec x dx = ln | sec x + tan x| + C
cot x dx = ln | sin x| + C
Z
Z
Z
sin x dx = − cos x + C
cos x dx = sin x + C
tan x dx = ln | sec x| + C
Z
Z
Z
(CONTINUED)
csc x dx = ln | csc x − cot x| + C
sec x dx = ln | sec x + tan x| + C
cot x dx = ln | sin x| + C
b
sin
x dxdx==−Fcos
f (x)
(b)x−+FC(a),
′
Z
(CONTINUED)
Page 3
−1
1+x2