Find the integral.

`int sec^3x dx`

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You should use the following formula such that:

`int sec^n x dx = (sin x*sec^(n-1) x)/(n-1) + (n-2)/(n-1) int sec^(n-2) x dx`

Reasoning by analogy yields:

`int sec^3 x dx = (sin x*sec^2 x)/2 + 1/2 int sec x dx`

`int sec^3 x dx = (1/2)(tan x*sec x + int sec x dx)`

`int sec x dx = int 1/(cos 2(x/2))dx`

`int sec x dx = int 1/(cos^2(x/2) - sin^2(x/2)) dx`

`int sec x dx = int 1/((cos(x/2) - sin(x/2))(cos(x/2) + sin(x/2)))`

`int sec x dx = ln(cos(x/2) + sin(x/2)) - ln(cos(x/2)- sin(x/2))`

`int sec^3 x dx = (1/2)(tan x*sec x + ln(cos(x/2) + sin(x/2)) - ln(cos(x/2) - sin(x/2))) + c`

**Hence, evaluating the indefinite integral yields `int sec^3 x dx = (1/2)(tan x*sec x + ln(cos(x/2) + sin(x/2)) - ln(cos(x/2) - sin(x/2))) + c.` **

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