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Antiderivative

see math notation, function, derivative, calculus notation

note a function has an infinite number of antiderivatives

note a function continuous on an interval has a unique general family of antiderivatives on that interval of the form F x : c with RR c

in other words, if F x is one antiderivative of f x and f x is continuous on its function › domain, then any other antiderivative of f x can be rewritten into the form F x : c with RR c --- proof --- https://youtu.be/HfACrKJ_Y2w?t=38531

notation

dd F = f, where #todo inv

Antidifferentiation

procedure computing an antiderivative reciprocal derivative rules recursively

u substitution

integrating a function by substituting u for a function of x

useful when the derivative of u is in the integrand

$ f = $ (f -- dd u \\ `u) \\ u #todo inv --- me

proof #todo inv

$ f = $dd ($ f \\ `u) \\ u by identities of function composition

= $ (f \\ `u | dd `u) \\ u by the chain derivative rules

= $ (f \\ `u -- (dd u \\ `u)) \\ u by the inverse derivative rules

= $ (f -- dd u \\ `u) \\ u since f`h -- g`h = f-g \\ h

--- me

example

F = $ x. 2x"sin" x2

let u = x. x2 as its derivative is in the integrand and compute dd u = x. 2x

substituting, F = $ ((x. 2x"sin" x2) -- (x. 2x) \\ (x2. x)) \\ (x. x2)

computing the integral, F = $ (x. "sin" x) \\ (x. x2) = x. c.."cos" x2 with RR c

integration by parts

for the integral of products

rewrites a hard-to-compute integral as an easy-to-compute integral

$ v'dd u = u'v .. $ u'dd v

proof integrate both sides of the product derivative rules --- me and https://youtu.be/7gigNsz4Oe8?t=3908

example

F = $ x. x'[x]

let dd v = x. [x] as it is easier to integrate. then, u = x. x, dd u = x. 1 and v = $ x. [x] = x. [x]

substituting, F = (x. x)(x. [x]) .. $ (x. [x])(x. 1)

computing the integral, F = x. x'[x] .. [x] : c with RR c

using trigonometric identities

see trigonometric functions

procedure

with an integral of the form $ x. ["trig1" x]p_1 | ["trig2" x]p_2 with p_1 being odd and p_2 even, compute the integral as follows:

  1. separate out ["trig1" x]p_1 into ["trig1" x](p_1 : 1) | "trig1" x
  2. use u substitution with u = "trig1" x and use pythagorean trigonometric identities to rewrite ["trig2" x]p_2 as a function of "trig1" x
  3. compute the integral

procedure

with an integral of the form $ x. ["trig2" x]p_2 with p_2 being even, compute the integral as follows:

  1. rewrite ["trig2" x]p_2 as [["trig2" x]2](p_2 -- 2)
  2. use double-angle trigonometric identities to rewrite ["trig2" x]2 as a function of "trig1" 2x
  3. multiply out with the power p_2 -- 2 and compute the resulting integrals separately

note the technique above can also be used to compute integrals of the product of multiple trigonometric functions raised to even powers

example

$ "tan"2 = $ -"cos"2 .. 1 = x. "tan" x .. x : c

computing $ "cot"2 can be done similarly

example

$ -"cos" = /-"cos" : "tan"\

computing $ -"sin" can be done similarely

proof --- https://youtu.be/7gigNsz4Oe8?t=6366

procedure

with an integral with respect to x containing the form -- \a .. x2/ with a being a constant, compute the integral as follows:

  1. substitute x2 with [b2 | "trig" aa]2 such that the form -- \a .. x2/ can be rewritten as -- b \1 .. ["trig" aa]2/
  2. simplify the form to eliminate the square root using pythagorean trigonometric identities
  3. compute the integral

using partial fractions

for the integral of rational functions

procedure

compute the integral of a rational function as follows:

  1. factor out the denominator into its list of n : 1 prime factors f
  2. set the function equal to :A-f and solve for for the list of reals A
  3. compute the integral of the partial fraction, $ :A-f

note this method will only work if both

  1. the denominator factors into distinct linear factors (ensures no contradictory equations)
  2. the degree of the numerator is lower than the degree of the denominator (ensures the number of unknowns is equal to the number of coefficients in the numerator)