posted by integral on .
Find the indefinite integral of 1-tanx/1+tanx
Dont know how to really approach this question. Should i use identities, or is there a power series i can use?
There are standard formulae for integrals of rational functions of trigonometric formulae. In this case, you can simplify things as follows.
Let's use the abbreviation:
t = tan(x)
s = sin(x)
c = cos(x)
We can write:
(c - s)^2/(c^2 - s^2) =
(c^2 + s^2 - 2cs)/(c^2 - s^2)
Then use the trigonometric identities:
c^2 + s^2 = 1
2 cs = sin(2x)
c^2 - s^2 = cos(2x)
1/(cos(2x)) - tan(2x)
Integrating tan(2x) is trivial. You can integrate 1/cos(2x) e.g. by putting
x = (pi/4 - u), so that cos(2x) =
[cos^2(u) + sin^2(u)]/[2 sin(u)cos(u)] =
1/2 [cot(u) + tan(u)]
which is trivial to integrate.
how did you get from (c-s)/(c+s) to
(c - s)^2/(c^2 - s^2)?
Multiply numerator and denominator by (c-s).