How do I derive the secant reduction rule?

Integral (sec x)^n dx
= Integral (sec x)^(n-2) * (sec x)^2 dx
= Integral ((tan x)^2 + 1)^(n/2-1) * (sec x)^2 dx

Doing a substitution with:
u = tax x
du = (sec x)^2 dx
= Integral (u^2 + 1)^(n/2-1) * du

At this point I'm stuck. Any ideas? Thanks!

To continue with the derivation of the secant reduction rule, we can use the substitution u = tan x.

Starting from:

∫ (u^2 + 1)^(n/2-1) du

We can manipulate this integral using the binomial expansion. Recall that (a + b)^n can be expanded using the binomial theorem as:

(a + b)^n = C(n,0) * a^(n-0) * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ...

Where C(n,k) represents the binomial coefficient, given by C(n,k) = n! / (k!(n-k)!).

In our case, let a = u^2 and b = 1. So, we have:

∫ (u^2 + 1)^(n/2-1) du
= ∫ [(u^2)^0 * 1^(n/2-1) * C(n,0) * (u^2)^(n-0) * 1^0 + C(n,1) * (u^2)^(n-1) * (1)^1 + C(n,2) * (u^2)^(n-2) * (1)^2 + ...] du

Now, let's simplify the integral by ignoring the binomial coefficients since they are just constants. We have:

∫ [(u^2)^(n-0) * 1^0 + (n)(u^2)^(n-1) * 1 + (n)(n-1)(u^2)^(n-2) + ...] du
= ∫ [u^(2n) + n(u^2)^(n-1) + n(n-1)(u^2)^(n-2) + ...] du

Now, we can integrate each term separately:

∫ u^(2n) du = u^(2n+1) / (2n+1) + C1
∫ n(u^2)^(n-1) du = n(u^2)^n / (2n) + C2
∫ n(n-1)(u^2)^(n-2) du = n(n-1)(u^2)^(n-1) / (2(n-1)) + C3

Simplifying these integrals, we have:

∫ u^(2n) du = u^(2n+1) / (2n+1) + C1
∫ n(u^2)^(n-1) du = n(u^2)^n / (2n) + C2
∫ n(n-1)(u^2)^(n-2) du = n(u^2)^(n-1) / 2 + C3

Finally, substituting back u = tan x:

∫ (sec x)^n dx = tan^(2n+1) x / (2n+1) + (n/2) tan^(2n-1) x + C

This is the derivation of the secant reduction rule.