tan^n(x) =
sin^2(x)/cos^2(x)tan^(n-2)(x)
sin^2(x)/cos^2(x) =
[1-cos^2(x)]/cos^2(x) =
1/cos^2(x) - 1
So:
tan^n(x) = tan^(n-2)(x)/cos^2(x) -
tan^(n-2)(x)
Integral of tan^(n-2)(x)/cos^2(x)dx =
tan^(n-1)(x)/(n-1)
How do I derive the integration reduction formula for tangent?
Integral of (tan x)^n dx = ...
I can do the derivations for sin/cosine, but I'm getting stuck on tan.
Thanks!
2 answers
makes perfect sense. Thanks Count!