Hi Larry,
It's a very troublesome integral.
Start out by setting tanx = u^2. This then makes the integrand
2u^2/(1+u^4) du.
This is just a rational function of u, so it can be integrated by using
partial fractions (if it's still called that).
Write 1+u^4 as a difference of squares by adding and subtracting 2u^2, then
factor the difference of squares as usual.
Use partial fractions to write the integrand as a sum of two terms of the
form u/(quadratic function of u).
Complete the square in the denominator, and shift the integration variable.
You will then get an answer which is a sum of logs and arctans.
The result is:
(1/(2sqrt(2)))*ln[{tanx + 1 - sqrt(2tanx)}/{tanx + 1 + sqrt(2tanx)}]
+
(1/sqrt(2))*arctan{sqrt(2tanx)/(1-sqrt(tanx))}
(where I used the addition formula for the arctan to pretty it up a bit).
As I said, it's sort of a mess....
If you have a lot of free time now that classes are finished, you might try
showing that the integral of ln(cosx) from 0 to pi/2 is (-)ln2...
Or find cos20cos40cos80 (all in degrees)...
mark snyder
fitchburg state college
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