What is the Integral of (tan x)^1/2 ?

Date: 03/07/99 at 17:09:46
From: Neal
Subject: Integration of (tan x)^1/2

What is the integral of (tan x)^1/2 ?

My first substitution was u^2 = tan x and I got the following result:
2 times the Integral of u^2 du /(u^4 +1)

Date: 03/08/99 at 00:08:42
From: Doctor Pete
Subject: Re: Integration of (tan x)^1/2

You are on the right track. All you need to do now is use partial 
fraction decomposition to express the integrand as a sum of fractions. 
In particular, you are probably thinking this is impossible, since the 
polynomial u^4 + 1 has no real roots. But in fact, u^4 + 1 is the 
product of two irreducible quadratics. To see how this happens, 
suppose for the moment that a factorization exists; i.e.

     u^4 + 1 = (u^2 + Pu + 1)(u^2 + Qu + 1)

where P, Q are constants to be determined. Note that I have assumed 
the coefficients of the u^2 and constant terms to be 1 (why?). 
Multiplying the right-hand side out gives

     u^4 + 1 = u^4 + (P+Q)u^3 + (PQ+2)u^2 + (P+Q)u + 1

whereupon equating the coefficients on both sides gives the following 
system of equations for P, Q:

     P + Q = 0
     PQ + 2 = 0

Therefore, P = -Q, and there are two solutions which are symmetric:

     {(P,Q)} = {(Sqrt[2], -Sqrt[2]), (-Sqrt[2], Sqrt[2])}

This makes sense because the factorization we assumed is identical 
for both solutions. So we have found that

     u^4 + 1 = (u^2 + Sqrt[2]u + 1)(u^2 - Sqrt[2]u + 1)

So, it now remains to write the fraction u^2 / (u^4 + 1) in the form

     (Au + B)/(u^2 + Sqrt[2]u + 1) + (Cu + D)/(u^2 - Sqrt[2]u + 1)

where A, B, C, D are constants to be determined. This is accomplished 
by summing the two fractions by cross-multiplying, and equating the 
numerator of the result with the numerator u^2. This again gives a 
system of equations for the unknowns  A, B, C, D.  Finally, each of 
these two fractions is integrated by completing the square in the 
denominator and using the appropriate trigonometric substitution. 
These steps I leave to you.

- Doctor Pete, The Math Forum
  http://mathforum.org/dr.math/