# Integral of sec^n(x)dx

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You need to use reduction formula to integrate the function, such that:

int sec^n x dx = int sec^(n-2) x* sec^2 x dx

You need to use integration by parts, such that:

int udv = uv - int vdu

u = sec^(n-2) x => du = (n-2)*sec^(n-3) x tan(x)

dv = sec^2 x => v = tan x

int sec^(n-2) x* sec^2 x dx = tan x*sec^(n-2) x - (n-2)*int sec^(n-3) x*tan x dx

int sec^(n-2) x* sec^2 x dx = tan x*sec^(n-2) x - (n-2)*int sec^(n-2) x*tan^2 x dx

You need to replace sec^2 x - 1 for tan^2 x , such that:

int sec^(n-2) x* sec^2 x dx = tan x*sec^(n-2) x - (n-2)*int sec^(n-2) x*(sec^2 x - 1) dx

int sec^(n-2) x* sec^2 x dx = tan x*sec^(n-2) x - (n-2)*int (sec^n x- sec^(n-2) x) dx

You need to put I_n = int sec^n x dx, such that:

I_n = tan x*sec^(n-2) x - (n-2)*I_n + (n-2)*int sec^(n-2) x dx

I_n + (n-2)*I_n = tan x*sec^(n-2) x + (n-2)*int sec^(n-2) x dx

I_n*(n - 2 + 1) = tan x*sec^(n-2) x + (n-2)*int sec^(n-2) x dx

Put int sec^(n-2) x dx = I_(n-2)

(n-1)*I_n = tan x*sec^(n-2) x + (n-2)*I_(n-2)

I_n = 1/(n-1)* tan x*sec^(n-2) x + (n-2)/(n-1)*I_(n-2)

Hence, evaluating the given integral, using reduction formula yields I_n = 1/(n-1)* tan x*sec^(n-2) x + (n-2)/(n-1)*I_(n-2).

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