Indefinite Integral: $\int sec^{n}x dx$

Question

How do I show that

$$\int \sec^{n}(x)\,\mathrm{d}x = \frac{1}{n-1}\sec^{n-2}(x)\tan(x) + \frac{n-2}{n-1}\int \sec^{n-2}(x)\,\mathrm{d}x$$

Answer 1

Peel off two factors of secant: $$ \sec^n x = \sec^{n-2} x \cdot\sec^2 x $$ Now integrate by parts with $u=\sec^{n-2} x$ and $dv = \sec^2 x\,dx$. Rearrange making use of the formula $$ \tan^2 x = \sec^2 x - 1 $$