VM266 0847

integral of (u dv) = uv - integral of (v du)

x = a sinb if rad(a^2 - x^2) is in the integrandx = a tanb if rad(a^2 + x^2) is in the integrandx = a secb if rad(x^2 - a^2) is in the integrand

1. Look for a quadratic equation in the integrand. 2. Complete the square.3. a. set u = the perfect square from the completed square b. solve for x using u.4. plug in u accordingly and and x as well5. evaluate integral

1 - 1 functions are functions that never takes the same value twice.More formally:1. Let f be a function that maps from D -> R2. For every x in D, there exist a unique R. - Let x1, x2 be in D - f(x1) is not = f(x2)NOTE: A) #2 implies the converse: for every y in R, there exist a unique x in DB) Graphically, a function is 1 - 1 if a horizontal line never intersects the function's curve twice.

The inverse of a function that maps from D -> R is the function that maps from R -> D.Let y = f(x) be a 1 - 1 function that maps from D - > R. A function g is the inverse of f where the R(f) = D(g) and D(f) = R(g). Therefore, y = f(x) <=> x = g(y) The inverse function g is denoted as f^-1

If g is the inverse of f, then... i) g(f(x)) = x for every x in D(f) ii) f(g(x)) = x for every x in R(f)-------------------Ver1 proof is set as hyperlinkVer2 proof URL: http://i1321.photobucket.com/albums/u556/radicaldreamer18/Calculus%202%20Chapter%207/IMAG0299_zps9a60cf31.jpg

The natural logarithmic function is a logarithm with base e. In addition, it is also found that...ln(x) = integral {(1/t) dt} from 1 -> x----------------------If x > 1, ln(x) = integral {(1/t) dt} from 1 -> xIf 0 < x < 1, -ln(x) = integral {(1/t) dt} from x -> 1

Shortcut proof:1. By defn of ln's, ln(x) = integral {(1/t) dt} from 1 -> x.2. ln(x) must be differentiable because it is the antiderivative of something (1/t)3. (d/dx) ln(x) = (d/dx) integral {(1/t) dt} from 1 -> x4. (d/dx) ln(x) = 1/x, the derivative of an integral of a function is the function itself. -----------------------------(d/dx) ln(u) = (1/u)duproof:1. let y = f(x) = ln(u)2. (dy/dx) = (dy/du)(du/dx) 3. f'(x) = (d/du) ln(u) * du

integral of (1/u) du = u'/u; given that u is a function. Proven from chain rule of (d/dx) ln(x)(d/dx) ln(u) = (d/du) ln(u) * (du/dx) = (1/u)du

Derived using integration by parts!!!

1. log base x of x => ln(e) = 12. log(a) + log(b) = log(ab)3. log(a) - log(b) = log(a/b)4. a log(b) = log(b^a)----------------More basic stuff: log base a of y = x, then y = a^x

Given a differentiable equation, do the following to differentiate logarithmically:1. Take ln of both sides2. Differentiate both sides...IMPLICIT DIFFERENTIATION MAY BE NEEDED ALONG THE WAY3. Multily by the initial equation.

e is a positive number where ln(e) = 1 <=> e is the base of ln(x)

1. (e^p)(e^q) = e^(p+q)2. (e^p)/(e^q) = e^(p-q)3. (e^p)^q = e^pq

integral of 1/x = ln(x) + C

integral of a^x = (1/ln(a)) * a^x + Cintegral of a^u = (1/ln(a)) * a^u + C

If y = arc sin(x), then sin(y) = x.domain of arc sin [-1, 1]range of arc sin [-pi/2, pi/2]

i) sin(arc sin(x)) = x; if x is defined on [-1,1]ii) arc sin(sin(x)) = x; if x is defined on [-pi/2, pi/2]

if arc cos(x) = y, then cos(y) = x.x is defined on [-1, 1]y is defined on [0, pi]

i) cos(arc cos(x)) = x; if x is defined on [-1. 1]ii) arc cos(cos(x)) = x; if x is defined on [0, pi]

If arc tan(x) = y, then tan(y) = x.x is defined on (-inf, inf)y is defined on (-pi/2, pi/2)

i) tan(arc tan(x)) = x; if x is defined on (-inf, inf)ii) arc tan(tan(x)) = x; if x is defined on (-pi/2, pi/2)

If arc cot(x) = y, then cot(y) = x.x is defined on (-inf, inf)y is defined on [0, pi]

i) cot(arc cot(x)) = x; if x is defined on (-inf, inf)ii) arc cot(cot(x)) = x; if x is defined on (0, pi)

If arc sec(x) = y, then sec(y) = x.x is defined on (-inf, -1] U [1, inf)y is defined on [0, pi/2) U [pi, 3pi/2)

i) sec(arc sec(x)) = x; if x is defined on (-inf, -1] U [1, inf)ii) arc sec(sec(x)) = x; if x is defined on [0, pi/2} U [pi, 3pi/2)

If arc csc(x) = y, then csc(y) = xx must be defined on (-inf, -1] U [0, inf)y must be defined on [-pi/2, pi/2], hole at 0

i) csc(arc csc(x)) = x; if x is defined on (-inf, -1] U [1, inf)ii) arc csc(csc(x)) = x; if x is defined on [-pi/2, pi/2], hole at 0

sinh x = [(e^x) - (e^-x)] / 2

integral of cosh x = sinh x + C

cosh x = [(e^x) + (e^-x)] / 2

Integral of sinh x = cosh x + C

tanh (x) = sinh(x) / cosh(x) = (e^x - e^(-x)) / (e^x + e^(-x))

integral of sech^2 x = tanh x + C

coth (x) = cosh(x) / sinh (x) = (e^x + e^(-x)) / (e^x - e^(-x))

Integral of csch^2 x = -coth x + C

sech(x) = 1 / cosh(x) = 2 / (e^x + e^(-x))

Integral of sech(x)tanh(x) = -sech x + C

csch(x) = 1 / sinh(x) = 2 / (e^x - e^(-x))

Integral of csch(x)coth(x) = -csch x + C