Posted by john on Thursday, June 3, 2010 at 11:22pm.
Yo can find a reduction formula for the integral of tan^n(x) as follows. We have that
tan^n(x) = sin^n(x)/cos^n(x) =
sin^(n-2)(x)/cos^n(x) sin^2(x) =
sin^(n-2)(x)/cos^n(x) [1-cos^2(x)] =
sin^(n-2)(x)/cos^n(x) -
sin^(n-2)(x)/cos^(n-2)(x) =
sin^(n-2)(x)/cos^(n-2)(x) 1/cos^2(x) -
sin^(n-2)(x)/cos^(n-2)(x) =
tan^(n-2)(x) 1/cos^2(x) - tan^(n-2)(x)
Now, 1/cos^2(x) is the derivate of
tan(x), so you can immediately integrate the first term:
Integral of tan^(n-2)(x) 1/cos^2(x)dx =
1/(n-1) tan^(n-1)(x)
The integral of the second term is, of course, a similar problem as the original problem, but with a lower value for n, so you can interate the formula until you end up at n = 1 or n = 0.
So, denoting the integral of tan^n(x)dx by I_n, we have:
I_n = 1/(n-1) tan^(n-1)(x) - I_{n-2}
For n = 3, we get:
I_3 = 1/2 tan^2(x) - I_1
and I_1 is the integral of tan(x)dx, which is -Log|cos(x)|.