Top 20 Integral Shortcuts, Calculus Chapter 6 Combined, Calc 2 Flashcards, Derivatives of Inverse Trig, Exponentials, and Log functions, Hyperbolic Trig Identities and Basic Derivatives and Integrals, Integrals / Derivatives / Identities for Trig / H...

d/dx[cos u]

-(sin u)u'

d/dx [ cos(x) ]

-sin(x)

∫1/xsqrt(x^2-a^2) dx

1/asec^-1|x/a|+c

∫1/(a^2+x^2) dx

1/atan^-1x/a+c

d/dx [ arcsinh(x) ]

1/sqrt(x^2 + 1)

∫a dx

ax+c

∫csch(x)

ln|tanh(x/2) | + C

∫1/x dx

ln|x|+c

d/dx [ sec(x) ]

sec(x)tan(x)

d/dx [ tan(x) ]

sec^2(x)

d/dx [ tanh(x) ]

sech^2(x)

∫secxtanx dx

secx+c

sec^2(x) (pythageorean identity)

tan^2(x) + 1

d/dx[arctan u]

u'/(1+u²)

∫arccsc(x)dx

x * arccos x - sqrt(1-x2) + C

∫arctan(x)dx

x * arctan x - (1/2) ln(1+x^2) + C

∫lnx dx

xlnx-x+c

cos^2(x) (power reduction formula)

(1+cos(2x))/2

sin^2(x) (power reduction formula)

(1-cos(2x))/2

cos(ax)cos(bx) (product to sum)

(1/2)(cos(a + b) + cos(a - b))

sin(ax)sin(bx) (product to sum)

(1/2)(cos(a - b) - cos(a + b))

sin(ax)cos(bx) (product to sum)

(1/2)(sin(a + b) + sin(a - b))

cos(ax)sin(bx) (product to sum)

(1/2)(sin(a + b) - sin(a - b))

(def) arctanh(x)

(1/2)ln((1 + x)/(1 - x))

(def) arccoth(x)

(1/2)ln((x + 1)/(x - 1))

∫du//(u√(u²+a²))

(1/a)arcsec(|u|/a) + C

∫du/(a^2 + u^2)

(1/a)arctan(u/a) + C

∫du/(a²+u²)

(1/a)arctan(u/a) + C

∫(1/u)(1/sqrt(u^2-a^2))

(1/a)sec^-1|u/a|+C, where u^2 > a^2

∫1/(a^2 + u^2)

(1/a)tan^-1(u/a) + C

∫a^x dx

(a^x/lna)+c

d/dx[sin u]

(cos u)u'

cosh²x (type 2)

(cosh2x+1)/2

sinh²x (type 2)

(cosh2x-1)/2

(def) coth(x)

(e^x + e^(-x))/(e^x - e^(-x))

(def) cosh(x)

(e^x + e^(-x))/2

(def) tanh(x)

(e^x - e^(-x))/(e^x + e^(-x))

(def) sinh(x)

(e^x - e^(-x))/2

d/dx[sec u]

(sec u tan u)u'

d/dx[tan u]

(sec²u)u'

∫x^n dx

(x^n+1/n+1)+c

d/dx[csc u]

-(csc u cot u)u'

d/dx[cot u]

-(csc²u)u'

d/dx [ arcsech(x) ]

-1/(x * sqrt(1 - x^2))

d/dx [ arccsch(x) ]

-1/(|x| * sqrt(1 + x^2))

cos(ax) - cos(bx) (sum to product)

-2sin((a+b)/2)sin((a-b)/2)

∫sin u du

-cos u + C

∫sin(x)dx

-cos(x) + C

∫sinx dx

-cosx+c

∫csc^2(x)dx

-cot(x) + C

1 - csc^2(x) (pythagorean identity)

-cot^2(x)

∫csc^2x dx

-cotx+c

∫csc(x)cot(x)dx

-csc(x) + C

d/dx [ csc(x) ]

-csc(x)cot(x)

csc

-csc*cot

cot

-csc^2

d/dx [ cot(x) ]

-csc^2(x)

d/dx [ csch(x) ]

-csch(x)coth(x)

d/dx [ coth(x) ]

-csch^2(x)

(cschx)'

-cschxcothx

(cothx)'

-csch²x

∫cscxcotx dx

-cscx+c

d/dx [ arccot(u) ]

-du/(1+u^2)

d/dx [ arccsc(u) ]

-du/(|u|sqrt(u^2 - 1))

d/dx [ arccos(u) ]

-du/sqrt(1 - u^2)

∫tan(x)dx

-ln(cos(x))

∫csc(x)dx

-ln(csc(x) + cot(x)) + C

∫tan u du

-ln|cos u| + C

∫csc u du

-ln|csc u + cot u| + C

∫cscx dx

-ln|cscx+cotx|+c

d/dx [ sech(x) ]

-sech(x)tanh(x)

(sechx)'

-sechxtanhx

cos

-sin

1 - sec^2(x) (pythagorean identity)

-tan^2(x)

d/dx[arccot u]

-u'/(1+u²)

d/dx[arccsc u]

-u'/(|u|√(u²-1))

d/dx[arccos u]

-u'/√(1-u²)

ln(0+) or 1/0−

-∞

1/∞ or e^-∞

0

cosh^2(x) - sinh^2(x) (pyth. identity)

1

coth^2(x) - csch^2(x) (pyth. identity)

1

sin^2(x) + cos^2(x)

1

tanh^2(x) + sech^2(x) (pyth. identity)

1

ln(x)

1 ---- x

logx

1 ----- x*ln(a)

csc^2(x) (pythagorean identity)

1 + cot^2(x)

coth^2(x) (pyth. identity)

1 + csch^2(x)

cosh^2(x) (pyth. identity)

1 + sinh^2(x)

sin^2(x) (pythageorean identity)

1 - cos^2(x)

-cot^2(x)

1 - csc^2(x)

-tan^2(x) (pythagorean identity)

1 - sec^2(x)

tanh^2(x) (pyth. identity)

1 - sech^2(x)

cos^2(x) (pythageorean identity)

1 - sin^2(x)

sech^2(x) (pyth. identity)

1 - tanh^2(x)

cosh²x (type 1)

1+sinh²x

tanh²x

1-sech²x

sech²x

1-tanh²x

U-substitution/Variable Substitution 5 Key Steps

1. Declare a variable u and set it equal to an algebraic expression that appears in the integral, and then substitute u for this expression in the integral. The expression you should use is the inner function you would identify if using the chain rule. 2. Differentiate u to find du/dx, and then isolate all x variables on one side of the equal sign. 3. Make another substitution to change dx and all other occurrences of x in the integral to an expression that includes du. 4. Integrate using u as your new variable of integration. 5. Express this answer in terms of x.

How to solve this indeterminate form with L'Hopital's Rule: The limit of f(x)^g(x) When f(x) = 0 and g(x) = 0 When f(x) = ∞ and g(x) = 0 When f(x) = 1 and g()x) = ∞ Ex: Lim x^x x->0

1. Set the Limit Equal to y. lim x^x = y x->0 2. Take the natural log of both sides, and then do some log rolling: ln lim x^x = ln y x->0 Roll the log inside the limit = lim ln x^x x->0 Roll the exponent over the log = lim x ln x x->0 3. Evaluate this limit as you would have with 0 * ∞. = lim ln x/ (1/x) Apply L'Hopital's rule 4. Solve for y ln y = 0 y= 1

Key steps when Integrating Polynomials

1. Use sum rule to break polynomial into its terms. Integrate each separately. 2. Use the constant multiple rule to move the coefficient of each term outside its respective integral. 3. Use the Power Rule to evaluate each integral. (You only need to add a single C to the end of the resulting expression)

d/dx [ arccoth(x) ]

1/(1 - x^2)

d/dx [ arctanh(x) ]

1/(1 - x^2)

d/dx [ arccosh(x) ]

1/sqrt(x^2 - 1)

(def) sech(x)

2/(e^x + e^(-x))

(def) csch(x)

2/(e^x - e^(-x))

cos(ax) + cos(bx) (sum to product)

2cos((a+b)/2)cos((a-b)/2)

sin(ax) - sin(bx) (sum to product)

2cos((a+b)/2)sin((a-b)/2)

sin(ax) + sin(bx) (sum to product)

2sin((a+b)/2)cos((a-b)/2)

sin(2x) (double angle)

2sin(x)cos(x)

sinh(2x) (double angle)

2sinh(x)cosh(x)

tan(2x) (double angle)

2tan(x)/(1-tan^2(x))

tanh(2x) (double angle)

2tanh(x)/(1+tanh^2(x))

Continuous interest

A = P × e^rt

Compound interest

A = P(1 + r/n)^nt

How to solve this indeterminate form with L'Hopital's Rule: When f(x) = ∞ and g(x) = ∞, the limit of f(x) - g(x) is in the indeterminate form ∞ - ∞. For example: Lim [(cot x - csc x)] x->0+

A little tweaking with the basic trig identities does the trick lim(cos x/sinx - 1/sinx) x->0+

Antiderivative of Trig functions F': cos x = F': sin x = F': sec²x = F': csc²x = F': secx tanx= F': cscx cotx =

Antiderivative of Trig functions F': cos x = sin x + C F': sin x = -cos x + C F': sec²x =tan x + C F': csc²x = -cotx + C F': secx tanx= sec x + C F': cscx cotx = -csc x +C

L'Hopital's Rule

C can be any real number as well as ∞ or -∞. Sometimes you may need to apply L'Hopital's Rule more than once.

Derivative Constant Multiple Rule: d/dx af(x)=

Derivative Constant Multiple Rule: d/dx af(x)= a [d/dx f(x)] Move the constant outside of a derivative before you differentiate

Derivative Sum Rule: d/dx [f(x) + g(x)] =

Derivative Sum Rule: d/dx f(x) + d/dx (g(x) The derivative of the sum of functions equals the sum of the derivatives of those functions

Derivatives of the Trig Functions d: sin x = d: cos x = d: tan x = d: cot x = d: sec x = d: csc x =

Derivatives of the Trig Functions d: sin x = cos x d: cos x = -sin x d: tan x = sec²x d: cot x = - csc²x d: sec x = secx tanx d: csc x = -cscx cotx

Find Integral (Anti-Derivative) F': 0 = F': 1 = F': e^x = F': 1/x = F': n^x =

Find Integral (Anti-Derivative) F': 0 = C F': 1 = x + C F': e^x = e^x + C F': 1/x = ln x +C F': n^x = n^x/ln n + C

Find the Derivative d: n = d: x = d: e^x = d: n^x = d: ln x = d: 1/x =

Find the Derivative d: a = 0 d: x = 1 d: e^x = e^x d: n^x = n^x ln n d: ln x = 1/x d: 1/x = ln|x|

Fundamental Theorem Of Calculus (How to solve definite integrals)

Find the derivative of the integral. Plug in the numbers subtracting the top from the bottom.

Simpson's Rule

First, divide the area that you want to approximate into an even number of regions by drawing a number of vertical lines. The width is normally represented by (b-a/n) but in Simpson's Rule it is adjusted to be (b-a/3n) Then that width is multiplied by the corresponding f(x) with coefficients for the height starting and ending at 1 and alternating between 4 and 2 in between.

Indeterminate Forms of Limits

If the limit approaches either 0/0 or +-∞/+-∞

Half Life

P(t)= P₀ × e^kt

Population growth

P(t)=P₀ × e^kt

How to solve this indeterminate form with L'Hopital's Rule: When f(x) = 0 and g(x) = ∞, the limit of f(x) * g(x) is in the indeterminate form 0 * ∞. Rewrite as...

Rewrite f(x) * g(x) to f(x) / [1/g(x)] The limit of this new function is the indeterminate form 0/0, which allows you to use L'Hopital's Rule.

How to find the derivative of x^x

Set it equal to y y = x^x Natural log both sides ln y = ln x^x Get the exponent on the other side ln y = x ln x Solve

Three Key Rules for Integration

Sum Rule Constant Multiple Rule Power Rule

Law of cooling

T(t)= (T₀ − Ts) × e^kt + Ts

The Power Rule: d/dx x^n

The Power Rule: d/dx x^n = n x^n-1 The power rule tells you that to find the derivative of x raised to any power, bring down the exponent as the coefficient of x, and then subtract 1 from the exponent and use this as your new exponent.

The Product Rule:

The Product Rule: The derivative of the first function times the second plus the derivative of the second times the first

The Quotient Rule

The derivative of the top function times the bottom minus the derivative of the bottom times the top, over the bottom squared.

How to remember the second two square identities

Use the first trig identity Divide every term in the first square identity by cos²x to produce one new equation and by sin²x to produce the other.

The Chain Rule

Write down the derivative of the outer function, then write everything found inside of it and finally multiply this result by the derivative of the next function inward.

∫-du/sqrt(1-u^2)

arccos(u) + C

∫1/√(x²-1)dx

arccoshx+c

∫du/√(a²+u²)

arcsin (u/a) + C

∫du/sqrt(1-u^2)

arcsin(u) + C

∫1/√(1+x²)dx

arcsinhx+c

∫arcsech(x)

arctan(sinh(x)) + C

∫sech(x)

arctan(sinh(x)) + C

∫1/(1-x²)dx

arctanhx+c

aⁿ

aⁿ × ln(a)

sin

cos

d/dx [ sin(x) ]

cos(x)

cos(2x) (double angle)

cos^2(x) - sin^2(x)

d/dx [ sinh(x) ]

cosh(x)

∫sinh(x)

cosh(x) + C

cosh(2x) (double angle)

cosh^2(x) + sinh^2(x)

sinh^2(x) (pyth. identity)

cosh^2(x) - 1

(sinhx)'

coshx

sinh²x (type 1)

cosh²x-1

∫-csc^2(x)

cot(x) + C

csc^2(x) - 1 (pythagorean identity)

cot^2(x)

csch^2(x) (pyth. identity)

coth^(x) - 1

∫-csc(x)cot(x)dx

csc(x) + C

1 + cot^2(x) (pythagorean identity)

csc^2(x)

cot^2(x) (pythageorean identity)

csc^2(x) - 1

d/dx [ arctan(u) ]

du/(1+u^2)

d/dx [ arcsec(u) ]

du/(|u|sqrt(u^2 - 1))

d/dx [ arcsin(u) ]

du/sqrt(1 - u^2)

∫e^x dx

e^x+c

eⁿ

eⁿ

Definite Integral

has limits a & b, find antiderivative, F(b) - F(a)

(def) arcsech(x)

ln(1+- sqrt(1-x^2))/x)

(def) arccsch(x)

ln(1+sqrt(1+x^2))/x)

∫arctanh(x)

ln(cosh(x)) + C

∫sec(x)dx

ln(sec(x) + tan(x)) + C

∫cot(x)dx

ln(sin(x))

∫arccoth(x)

ln(sinh(x)) + C

(def) arcsinh(x)

ln(x + sqrt(x^2 + 1))

(def) arccosh(x)

ln(x +- sqrt(x^2 - 1))

∫cotx dx

lnsinx+c

∫tanh(x)

ln|cosh(x)| + C

∫sec u du

ln|sec u + tan u| + C

∫sec x

ln|secx+tanx|+c

∫tanx dx

ln|secx|+c

∫cot u du

ln|sin u| + C

∫coth(x)

ln|sinh(x)| + C

Indefinite Integral

no limits, find antiderivative + C, use inital value to find C

∫sec(x)tan(x)dx

sec(x) + C

tan

sec^2

tan^2(x) + 1 (pythagorean identity)

sec^2(x)

tan^2(x) (pythageorean identity)

sec^2(x) - 1

(tanhx)'

sech²x

∫cos u du

sin u + C

∫cos(x)dx

sin(x) + C

∫1/sqrt(a^2 - u^2)

sin^-1(u/a) + C, where u^2 < a^2

∫1/sqrt(a^2-x^2) dx

sin^-1x/a+c

d/dx [ cosh(x) ]

sinh(x)

∫cosh(x)

sinh(x) + C

(coshx)'

sinhx

∫cosx dx

sinx+c

∫sec^2(x)dx

tan(x) + C

sec

tan*sec

sec^2(x) - 1 (pythagorean identity)

tan^2(x)

∫sec^2x dx

tanx+c

∫tan^2x

tanx-x+c

d/dx[arcsec u]

u'/(|u|√(u²-1))

d/dx[arcsin u]

u'/√(1-u²)

∫arccos(x)dx

x * arccos(x) - sqrt(1 - x^2) + C

∫arccosh(x)

x * arccosh(x) - sqrt(x-1)sqrt(x+1)

∫arccot(x)dx

x * arccot(x) + (1/2)ln(1 + x^2) + C

∫arccsch(x)

x * arccsch(x) + x * sqrt((1/x^2 + sinh(x))/(x^2 + 1)))

∫arcsec(x)dx

x * arcsec(x) - sgn(x) * ln |x + √(x² - 1)| + C

∫arcsin(x)dx

x * arcsin(x) + sqrt(1 - x^2) + C

e^∞ or ln(∞) or 1/0+

∞