Question:

<div>In this case, we note that $(\tan x)' = \sec^2 x$ and we write the given integral as $$\int \sec^n x \, dx = \int \sec^{n-2} x \cdot \sec^2 x \, dx$$ If we take $dv = \sec^2 x \, dx$, then we have $v = \tan x$ and we may integrate by parts with $u = \sec^{n-2} x, du = (n-2) \sec^{n-3} x \cdot \sec x \tan x = (n-2) \sec^{n-2} x \tan x$. Using the fact that $1 + \tan^2 x = \sec^2 x$, one may thus establish the identity $$\int \sec^n x \, dx = \sec^{n-2} x \tan x - (n-2) \int \sec^{n-2} x \tan^2 x \, dx$$ $$= \sec^{n-2} x \tan x - (n-2) \int \sec^{n-2} x (\sec^2 x - 1) \, dx$$ $$= \sec^{n-2} x \tan x - (n-2) \int \sec^n x \, dx + (n-2) \int \sec^{n-2} x \, dx$$ Since the integral on the left hand side also appears on the right hand side, this gives $$(n-1) \int \sec^n x \, dx = \sec^{n-2} x \tan x + (n-2) \int \sec^{n-2} x \, dx$$</div>