# Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 50

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Prove $\displaystyle \int \sec^n x dx = \frac{\tan x \sec^{n-2} x}{n - 1} + \frac{n-2}{n-1} \int \sec^{n-2} x dx ( n \neq 1)$

Notice that $\displaystyle \int \sec^n x dx = \int \sec^{n - 2} x \cdot \sec^2 x dx$

So if we let $n = \sec^{n-2} x (\sec x \tan x) dx$ and $dv = \sec^2 x dx$, then

$du = (n-2) \sec^{n-2} x ( \sec x \tan x ) dx$ and $v = \int \sec^2 x dx = \tan x$

Thus,

\begin{aligned} \int \sec^{n-2} x \cdot \sec^2 x dx = uv - \int v du &= \tan x \left( \sec^{n-2} x \right) - \int (\tan x) \left[ (n-2) \sec^{n-3} x (\sec x \tan x) \right]\\ \\ &= \tan \left( \sec^{n-2} x \right) - (n-2) \int \sec^{n-2} x \tan^2 x \end{aligned}

Recall for the identity $\tan^x = \sec^ x - 1$

So,

$\displaystyle [(n-2)+1] \int \sec^{n-2} x \cdot \sec^2 x dx = \tan x \left( \sec^{n-2} x \right) + (n-2) \int \sec^{n-2} x dx$

$\displaystyle (n-1) \int \sec^{n-2} x \cdot \sec^2 x dx = \tan x \left( \sec^{n-2} x\right) + (n-2) \int \sec^{n-2} x dx$

Dividing with sides by $(n-1)$, we get

$\displaystyle \int \sec^{n-2} x \cdot \sec^2 x dx = \frac{\tan x \left(\sec^{n-2}x\right)}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2} x dx$