u = tan (θ/2)

Now, if you happened to know a good number of trigonometric identities and understand the method of partial fraction decomposition, you could solve almost all integrals of rational functions in sine and cosine. Hence, the "fundamental substitution". Unfortunately, it is very complicated and difficult to invoke.

The purpose of this substitution is to rephrase an integral dealing with sines and cosines into the 'easier' language of rational functions of polynominals, for which we already have a method of integrating.

Invoking the name of the beast

This subtle substitution conceals three substitutions. First, we require the differential substitution (dθ in terms of du) that permits us to use standard u-substitution techniques:

u = tan (θ / 2)
θ = 2 tan-1 u
dθ = (2 du) / (1 + u2)

Now, it's also possible to obtain a substitution between u and cos θ:

u = tan (θ / 2)
u = (sin θ) / (1 + cos θ)
u = (1 - cos θ) / (sin θ)


These are both obscure trig identities that fall from the sine and cosine half-angle forumlas. Multiply the last two equations together.

u2 = (1 - cos θ) / (1 + cos θ)
cos θ = (1 + u2) / (1 - u2)

And, to complete the trio, we obtain a substitution between u and sin θ:

u = tan (θ / 2)
u = sin (θ) / (1 + cos θ)
u * [ 1 + (1 - u2) / (1 + u2) ] = sin θ
sin θ = 2u / (1 + u2)

Thus, we have our trio of substitutions.

An Example

As an example, consider ∫ dθ / (sin θ + cos θ). This integral is enough to scare the pants off of a high school calculus student, were it to appear on the final. None of the 'traditional' tricks will work on it, I guarantee.

Applying the substitution, however:

        2 du
      -------
      1 + u^2           2 du
----------------- = -------------
  2u      1 - u^2   -u^2 + 2u + 1
------- + -------
1 + u^2   1 + u^2

Ah, now this is managable. The denominator obviously factors:

-u^2 + 2u + 1 = (u + 1 - √2)(u + 1 + √2)

Which permits the use of partial fractions:

       2              A             B
------------- = ------------ + ------------
-u^2 + 2u + 1   (u + 1 - √2)   (u + 1 + √2)

2 = A(u + 1 - √2) + B(u + 1 + √2)
A + B = 0; B = - A
2 = A(1 - √2) - A(1 + √2)
2 = -2A√2
A = -1/√2; so B = 1/√2.

The integral of the two terms on the RHS is easy — they're only logarithms. So, after all that effort, we have the desired integral:

- ln (u + 1 - √2) + ln (u + 1 + √2)
-----------------------------------
                 √2

Remember, though: u = tan (θ/2), so the full solution looks more like:

1    |tan (θ/2) + 1 + √2|
-  ln|------------------| + C
√2   |tan (θ/2) + 1 - √2|

With this method, most integrals which are rational functions of sine and cosine can be solved. Unfortunately, as you can see, it requires a great deal of effort. Therefore, it is something of a last resort, which immediately got it culled from modern calculus textbooks. Nobody does integration by hand anymore — that is, no-one but high school students and over-ambitious math majors.