# What is antiderivative of function cos^3x-1/1+cos2x?

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You need to remember that you may find the antiderivative if you evaluate the indefinite integral of the given function.

You should transform the denominator using the half angle formula such that:

`1+ cos 2x = 2cos^2((2x)/2) => 1 + cos 2x = 2cos^2 x`

Hence, substituting `2cos^2 x` for `1 +cos 2x` yields:

`int (cos^3x-1)/(1+ cos 2x) dx = int (cos^3x-1)/(2cos^2 x) dx`

Using the property of linearity of integrals yields:

`int (cos^3x-1)/(2cos^2 x) dx = int (cos^3x)/(2cos^2 x) dx - int (dx)/(2cos^2 x)`

Reducing like terms yields:

`int (cos^3x-1)/(2cos^2 x) dx = (1/2)intcos x dx - (1/2)tan x`

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

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

**Hence, evaluating the antiderivative of the given function yields `int (cos^3x-1)/(2cos^2 x) dx = (1/2)(sin x - tan x) + c.` **

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