Calculus 2: Chapters 7.1-7.5 And 7.8

Front 1

Integration by Parts Formula

Back 1

∫ u dv = uv - ∫ v du

Front 2

Integration by Parts, Definite Integral Formula

Back 2

∫(a to b) u dv = uv|(a to b) - ∫(a to b) v du

Front 3

For ∫ sin^n(x)cos^m(x) dx and n is odd we:

Back 3

take a sine out and convert the rest to cosines using sin^2(x) = 1 - cos^2(x), then use the substitution u = cos(x)

Front 4

For ∫ sin^n(x)cos^m(x) dx and m is odd we:

Back 4

take a cosine out and convert the rest to sines using cos^2(x) = 1 - sin^2(x), then use the substitution u = sin(x)

Front 5

For ∫ sin^n(x)cos^m(x) dx and n and m are both odd we:

Back 5

take out either a cosine or a sine

Front 6

For ∫ sin^n(x)cos^m(x) dx and n and m are both even we:

Back 6

use double angle and/or half angle formulas to reduce the integral into a form that can be integrated

Front 7

For ∫ tan^n(x)sec^m(x) dx and n is odd we:

Back 7

take a tangent out and a secant out and convert the rest to secants using tan^2(x) = sec^2(x) - 1, then use the substitution u = sec(x)

Front 8

For ∫ tan^n(x)sec^m(x) dx and m is even we:

Back 8

take 2 secants out and convert the rest to tangents using sec^2(x) = 1 + tan^2(x), then use the substitution u = tan(x)

Front 9

For ∫ tan^n(x)sec^m(x) dx and n is odd and m is even we:

Back 9

we take out either a tan(x)sec(x) or sec^2(x)

Front 10

For ∫ tan^n(x)sec^m(x) dx and n is even and m is odd we:

Back 10

deal with each integral sepreatley

Front 11

If integral contains √ (a^2 - x^2), what is the substitution, identity, and quadrant we use?

Back 11

u = a sin(ϴ)

cosϴ = √ (1 - sin^2(ϴ))

In quadrant 1 and 4

Front 12

If integral contains √ (a^2 + x^2), what is the substitution, identity, and quadrant we use?

Back 12

u = a tan(ϴ)

secϴ = √ (1 + tan^2(ϴ))

In quadrant 1 and 4

Front 13

If integral contains √ (x^2 - a^2), what is the substitution, identity, and quadrant we use?

Back 13

u = a sec(ϴ)

tanϴ = √ (sec^2(ϴ) - 1)

In quadrant 1 and 3

Front 14

Partial Fractions: ∫ P(x)/Q(x)

If P(x) has a smaller degree than Q(x) then we:

Back 14

factor the denominator and find the partial fraction decomposition

Front 15

Partial Fractions: ∫ P(x)/Q(x)

If P(x) has a larger degree than Q(x) then we:

Back 15

use long division to make the degree of P(x) smaller or equal to the degree of Q(x) and then factor the denominator and find the partial fraction decomposition

Front 16

if Q(x) is: ax+b then the PFD is

Back 16

A / ax+b

Front 17

if Q(x) is: (ax+b)^n then the PFD is

Back 17

A / ax+b + B / (ax+b)^2 + ... + Z / (ax+b)^n

Front 18

if Q(x) is: ax^2+bx+c then the PFD is

Back 18

(Ax + B) / (ax^2+bx+c)

Front 19

if Q(x) is: (ax^2+bx+c)^n then the PFD is

Back 19

(Ax + B)/(ax^2+bx+c) +...+ (Cx + D)/(ax^2+bx+c)^n

Front 20

convergent integral

Back 20

limit exists and is a finite number

Front 21

divergent integral

Back 21

limit either doesn’t exist or is (plus or minus) infinity.

Front 22

d/dx (sinx)

Back 22

cosx

Front 23

d/dx (cosx)

Back 23

-sinx

Front 24

d/dx (tanx)

Back 24

sec^2(x)

Front 25

d/dx (secx)

Back 25

secxtanx

Front 26

d/dx (cscx)

Back 26

-cscxcotx

Front 27

d/dx (cotx)

Back 27

-csc^2(x)

Front 28

d/dx (arcsinx)

Back 28

1/√ (1-x^2)

Front 29

d/dx (arccosx)

Back 29

-1/√ (1-x^2)

Front 30

d/dx (arctanx)

Back 30

1/(1+x^2)

Front 31

d/dx (arccotx)

Back 31

-1/(1+x^2)

Front 32

d/dx (arcsecx)

Back 32

1/(|x|√(x^2 - 1))

Front 33

d/dx (arccscx)

Back 33

-1/(|x|√(x^2 - 1))

Front 34

d/dx (a^x)

Back 34

a^x ln(a)

Front 35

d/dx (ln(x))

Back 35

1/|x|

Front 36

d/dx (log_a (x)

Back 36

1/(xlna)

Front 37

∫ 1/ (ax+b) dx

Back 37

(1/a) ln|ax+b| + C

Front 38

∫ (1/x) dx

Back 38

ln|x| + C

Front 39

∫ cos x dx

Back 39

sin x + C

Front 40

∫ sin x dx

Back 40

-cos x + C

Front 41

∫ sec^2 x dx

Back 41

tan x +C

Front 42

∫ secxtanx dx

Back 42

sec x + C

Front 43

∫ cscxcotx dx

Back 43

-cscx + C

Front 44

∫csc^2 x dx

Back 44

-cotx + C

Front 45

∫ tan x dx

Back 45

ln|secx| + C

Front 46

∫cot x dx

Back 46

ln|sin x| + C

Front 47

∫ sec x dx

Back 47

ln|sec x +tan x| + C

Front 48

∫ csc x dx

Back 48

ln|cscx-cotx| + C