Complex Integrals

Sometimes, integrals cannot be solved in terms of the elementary integral shortcuts.  When this occurs, techniques such as substitution, integration by parts, or partial fractions.

As an example, I will show how to integral f(x) = (4-x^2)^1.5.
First, x must be substituted for u, with the equations x = 2sin(u) and dx = 2cos(u) du.
Then, I(f(x)) = I((2cos(u))^3(2cos(u)) du) = 16 I(cos(u)^4 du)
This can be simplified using the identity cos(u)^2 = (1+cos(2u))/2
Using the identity, I(f(x)) = 16 I((1+cos(2u))(1+cos(2u))/4) du)
Distributing, I(f(x)) = 4 I(1 + 2cos(2u) + cos(2u)^2 du)
Using the identity cos(u)^2 = (1+cos(2u))/2 again, I(f(x)) = 4 I(1 + 2cos(2u) + (1+cos(4u))/2 du)
Distributing, I(f(x)) = 2 I(3 + 4cos(2u) + cos(4u) du)
This can be simplified using the equations a = 2u, da = 2du, b = 4u, db = 4du
Then, I(f(x)) = 6 I(du) + 4 I(cos(a) da) + 1/2 I(cos(b) db)
Evaluating, I(f(x)) = 4sin(a) + 1/2sin(b) + 6u
Substituting for u, I(f(x)) = 4sin(2u) + 1/2sin(4u) + 6u
Using the identity sin(2u) = 2sin(u)cos(u), I(f(x)) = 8sin(u)cos(u) + 2sin(2u)cos(2u) + 6u
Using the identity cos(2u) = 1 - 2sin(u)^2, I(f(x)) = 8sin(u)cos(u) + 4sin(u)cos(u)(1-2sin(u)^2) + 6u
Distributing, I(f(x)) = 8sin(u)cos(u) + 4sin(u)cos(u) - 8sin(u)^3cos(u) + 6u
Using the identity cos(u) = (1-sin(u))^.5, I(f(x)) = 8sin(u)(1-sin(u)^2)^.5 + 4sin(u)(1-sin(u)^2)^.5 - 8sin(u)^3(1-sin(u)^2)^.5 + 6u
Substituting for x, I(f(x)) = 4x(1-1/4x^2)^.5 + 2x(1-1/4x^2)^.5 - x^3(1-1/4x^2)^.5 + 6arcsin(1/2x)
Factoring, I(f(x)) = (6x-x^3)(1-1/4x^2)^.5 + 6arcsin(1/2x)
Finally, I(f(x)) =1/4x(6-x^2)(4-x^2)^.5 + 6arcsin(1/2x)

Integrals and Integral Shortcuts

An example of an integral.  I will be using "I" instead of the integral sign at the beginning.

If the integral of a function f(x) is f'(x), the integral of the function is f(x).  Thus, the integral is the "antiderivative" of f'(x).

The shortcuts for integrals are usually more helpful than those for derivatives, since integrals are more complex to calculate by hand.  Some shortcuts include:

I(u^n du) = u^(n+1)/(n+1) when n != -1
I(du/u) = ln(u)
I(sin(u) du) = -cos(u)
I(cos(u) du) = sin(u)
I(tan(u) du) = ln|sec(u)|
I(cot(u) du) = ln|sin(u)|
I(sec(u) du) = ln|sec(u) + tan(u)|
I(csc(u) du) = ln|csc(u) - cot(u)|
I(sec(u)^2 du) = tan(u)
I(csc(u)^2 du) = -cot(u)
I(sec(u)tan(u) du) = sec(u)
I(csc(u)cot(u) du) = -csc(u)
I(a^u du) = a^u/ln(a)
I(e^u du) = e^u
I(sinh(u) du) = cosh(u)
I(cosh(u) du) = sinh(u)
I(tanh(u) du) = ln(cosh(u))
I(coth(u) du) = ln|sinh(u)|
I(sech(u) du) = arctan(sinh(u))
I(csch(u) du) = -arccoth(cosh(u))
I(sech(u)^2 du) = tanh(u)
I(csch(u)^2 du) = -coth(u)
I(sech(u)tanh(u) du) = -csch(u)
I(du/(a^2-u^2)^.5 = arcsin(u/a)
I(du/(u^2+-a^2)^.5) = ln|u + (u^2 +- a^2)^.5|
I(du/(u^2+a^2)) = 1/a arctan(u/a)
I(du/(u^2-a^2)) = 1/(2a) ln|(u-a)/(u+a)|
I(du/(u(a^2+-u^2)^.5)) = 1/a(ln|u/(a+(a^2+-u^2)^.5)|
I(du/(u(u^2-a^2)) = 1/a arccos(a/u)
I((u^2+-a^2)^.5 du) = u/2(u^2+-a^2)^.5 +- a^2/2 ln|u + (u^2+-a^2)^.5|
I((a^2-u^2)^.5 du) = u/2(a^2-u^2)^.5 + a^2/2 arcsin(u/a)
I(e^(au) sin(bu) du) = e^(au)(a sin(bu) - b cos(bu))/(a^2+b^2)

Partial Derivatives

Partial differentiation is the same as regular differentiation, but it keeps all variables in a multivariable function the same except for one variable.  The partial derivative d/dx(f(a,b,c,...x,y,z)) can be defined through the limit definition: d/dx(f(a,b,c,...x,y,z)) = lim(h->0) (f(a,b,c...,x + h,y,z) - f(a,b,c,...,x,y,z))/h.  For example, the derivative of f(x, y)=xy is lim(h->0) (f(x + h,y) - f(x,y))/h) = lim(h->0) (((x + h)y - xy)/h) = lim(h->0) ((xy + hy - xy)/h) = lim(h->0) (hy/h) = y.

Shortcuts for Differentiation

While the limit definition of the derivative can allow differentiation, it is often tedious and the shortcuts for differentiation can save time.  Some shortcuts include:

d/dx(C)=0
d/dx(u^n) = nu^(n-1) du/dx
d/dx(sin(u)) = cos(u) du/dx
d/dx(cos(u)) = -sin(u) du/dx
d/dx(tan(u)) = sec(u)^2 du/dx
d/dx(cot(u)) = -csc(u)^2 du/dx
d/dx(sec(u)) = sec(u)tan(u) du/dx
d/dx(csc(u)) = -csc(u)cot(u) du/dx
d/dx(logau) = (logeu)/u du/dx
d/dx(ln(u)) = 1/u du/dx
d/dx(a^u) = a^u ln(a) du/dx
d/dx(e^u) = e^u du/dx
d/dx(arcsin(u)) = 1/(1-u^2)^.5 du/dx
d/dx(arccos(u)) = -1/(1-u^2)^.5 du/dx
d/dx(arctan(u)) = 1/(1+u^2) du/dx
d/dx(arccot(u)) = -1/(1+u^2) du/dx
d/dx(arcsec(u)) = +-1/(u(u^2-1)^.5) du/dx (+ if u > 1 or - if u < -1)
d/dx(arccsc(u)) = +-1/(u(u^2-1)^.5) du/dx (- if u > 1 or + if u < -1)
d/dx(sinh(u)) = cosh(u) du/dx
d/dx(cosh(u)) = sinh(u) du/dx
d/dx(tanh(u)) = sech(u)^2 du/dx
d/dx(coth(u)) = -csch(u)^2 du/dx
d/dx(sech(u)) = -sech(u)tanh(u) du/dx
d/dx(csch(u)) = -csch(u)coth(u) du/dx
d/dx(arcsinh(u)) = 1/(1+u^2)^.5 du/dx
d/dx(arccosh(u)) = 1/(u^2-1)^.5 du/dx
d/dx(arctanh(u)) = 1/(1-u^2) du/dx
d/dx(arccoth(u)) = 1/(1-u^2) du/dx
d/dx(arcsech(u)) = 1/(u(1-u^2)^.5) du/dx
d/dx(arccsch(u)) = -1/(u(u^2+1)) du/dx

Limit Definition of the Derivative

The slope of the line tangent to a curve can be represented by the equation of the derivative.

The derivative is one of the fundamentals of calculus; it shows how a function changes as its input changes. It can be thought of as the slope of the tangent line on a graph on the xy plane, or as the velocity of an object if the position of the object has a relationship to the time, t. The derivative (f'(x)) can be defined as the slope of a secant as the difference in the x values of the endpoints of the secant approaches 0 (i.e. f'(x) = lim(h->0) (f(x + h) - f(x))/h).  For example, the derivative of f(x)=x is lim(h->0) (f(x + h) - f(x))/h) = lim(h->0) ((x + h - x)/h) = lim(h->0) (h/h) = 1.