How do I derive the integration reduction formula for tangent?

Question

Integral of (tan x)^n dx = ...

I can do the derivations for sin/cosine, but I'm getting stuck on tan.

Thanks!

Answer 1

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)

Answer 2

makes perfect sense. Thanks Count!