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)