You may also use the alternative method to find the antiderivative of the function `y=cos^3 x` , such that:

`int cos^3 x dx = int cos^2 x*cos x dx`

You may use the half angle identity, such that:

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

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

Using the linearity of integral yields:

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

You may use the double angle identity for cos 2x, such that:

`cos 2x = 1 - 2sin^2 x`

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

Using the linearity of integral yields:

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

You need to come up with the following substitution, such that:

`sin x = u => cos x dx = du`

`int cos^2 x*cos x dx = 2* (1/2)int cos x dx - int u^2 du`

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

Substituting back `sin x` for u yields:

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

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