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Calculus 2 Reviews
Calculus 2 Unit 7 ( Test 2 )
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Trig formulas
Calculus 2 Reviews > Calculus 2 Unit 7 ( Test 2 )
12 cards
Derivatives Review
Calculus 2 Reviews > Calculus 2 Unit 7 ( Test 2 )
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Calculus 2 Reviews > Calculus 2 Unit 7 ( Test 2 )
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Cards (86)
Integration By Parts Formula
int
(
udv
) =
uv
-
int
(
vdu
)
ILATE (Whichever function comes first in the following list should be u)
I :
Inverse Trig
L :
Logarithmic Functions
A :
Algebraic Functions
(
Polynomials
)
T :
Trigonometric Functions
E :
Exponential Functions
General rule for integration by parts
Choose "u" so that the integral of "
v
" is possible AND so that the
derivative
of "
u
" is "
better
" or
easier to work with
Make sure you pick dv to easily
integrate
!
Always check if you can do
u-substitution
before doing integration by parts.
Also, if you are doing integration by parts more than
two
times in a problem, make sure not to forget to
add
all the integration to the
final
answer!
When you integrate ln(x), you will need to implement the
dx value
!
Reduction formula for
∫
sin
n
(
x
)
d
x
,
n
≥
2
\int_{ }^{ }\sin^{n}\left(x\right)dx\ ,\ n\ \ge\ 2
∫
sin
n
(
x
)
d
x
,
n
≥
2
-1/n sin^(n-1) x cos(x) + (n-1)/n int ( sin^(n-2) x dx )
sin^2 x =
2sin
x
cos
x
sin^2 (x) + cos^2 (x) =
1
1 + tan^2 (x) =
sec^2 (x)
1 + cot^2 (x) =
csc^2 (x)
sin^2 (x) = (
1/2
[
1
-
cos(2x)
])
cos^2 (x) = (
1/2
[
1
+
cos(2x)
])
If the power of "sin" is
odd
, keep one factor of "sin" and use
Pythagorean
identity to change sin^2 (x) =
1
-
cos^2
(x)
If the power of "cos" is
odd
, do the same for "cos"
If both are odd, pick one that will give power "
2
" ( if possible )
If both powers are even, use the
half-angle
identity
General idea is that with trig to the odd power,
break it into one even
and
one odd
in order to use the
trig identities.
Int ( tan^n(x) sec^m(x)) dx
Int ( cot^n(x) csc^m(x)) dx
If power of "tan" is ODD, keep 1 factor of sec(x)tan(x) and use
tan^2(x) = sec^2(x) -1
If power of "sec" is EVEN,
strip off "sec^2(x)
" and use sec^2(x) = tan^2(x) + 1
Same format for int( cot^n(x) csc^m(x))dx
For integrals where sine and cosine have different angles:
sin(mx)sin(nx) =
1/2
[ cos(mx - nx) -
cos(mx + nx)
]
sin(mx)cos(nx) = 1/2 [ sin(mx - nx) + sin(
mx
+ nx) ]
cos(mx)cos(nx) = 1/2 [ cos(mx - nx) + cos(mx + nx) ]
Trigonometric Substitution, sqrt (
a^2 - x^2
)
Side, sin(theta) =
x/a
-> x =
a sin(theta)
Use: 1 - sin^2(theta) = cos^2(theta)
Trigonometric Substitution, sqrt (a^2 +
x^2
)
Hypotenuse, tan(theta) =
x/a
->
x= a tan(theta)
Use: 1 + tan^2(theta) = sec^2(theta)
Trigonometric Substitution, sqrt (
x^2 - a^2
)
Side, sec(theta) = x/a -> x= a sec(theta)
Use:
sec^2(theta) - 1 = tan^2(theta)
In trigonometric substitution, you have to put sides in a relationship such that we get "
x/a
" for "
tan
", "
sin
" or "
sec
"
a^2 + b^2 = c^2
sqrt ( a^2 + b^2 ) = c
a = sqrt ( b^2 - c^2 )
sin =
opposite
/
hypotenuse
tan =
opposite
/
adjacent
sec =
hypotenuse / adjacent
cos =
adjacent
/
hypotenuse
soh
cah toa
In
trigonometric
substitution, don't forget to convert back from theta to terms of
x
!
∫
(
ln
(
x
)
)
n
d
x
=
\int_{ }^{ }(\ln(x))^{n}dx=
∫
(
ln
(
x
)
)
n
d
x
=
x
(
ln
(
x
)
)
n
−
n
∫
(
ln
(
x
)
)
n
−
1
d
x
x(\ln(x))^{n}-n\int_{ }^{ }(\ln(x))^{n-1}dx
x
(
ln
(
x
)
)
n
−
n
∫
(
ln
(
x
)
)
n
−
1
d
x
Reduction formula with u= ((ln(x))^n
cos^2(x) =
1/sec^2(x)
cos(2x)dx
= cos^2(x) - sin^2(x)
sin(2x) =
2 sin(x)cos(x)
cho
(
csc
) ,
sha
(
sec
),
cao
(
cot
)
ln(1) =
0
cos(pi/2) =
0
sin(pi/2) =
1
cos(0) =
1
sin(0) =
0
Trapezoidal Rule
(b - a)/2n
pattern = 1, 2, 2, 2, ... 2, 1
delta x = (
b -a
) /
n
Midpoint
Rule
Involves
finding
the
average
x̄ = 1/2 (
x2
-
x1
)
(
b
-
a
)/n
See all 86 cards