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November 28, 2014

November 28, 2014

Posted by **integral** on Saturday, July 30, 2011 at 1:46pm.

Dont know how to really approach this question. Should i use identities, or is there a power series i can use?

- calculus -
**Count Iblis**, Saturday, July 30, 2011 at 4:10pmThere 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:

(1-t)/(1+t) =

(c-s)/(c+s) =

(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)

to obtain:

(1-t)/(1+t) =

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) =

sin(2u. Then

1/sin(2u) =

[cos^2(u) + sin^2(u)]/[2 sin(u)cos(u)] =

1/2 [cot(u) + tan(u)]

which is trivial to integrate.

- calculus -
**integral**, Saturday, July 30, 2011 at 5:21pmhow did you get from (c-s)/(c+s) to

(c - s)^2/(c^2 - s^2)?

- calculus -
**Count Iblis**, Saturday, July 30, 2011 at 5:33pmMultiply numerator and denominator by (c-s).

- calculus -
**integral**, Saturday, July 30, 2011 at 6:06pmthanks

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