# calculate antiderivative of (sin^3 x-4)/(1-cos 2x)

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### 1 Answer

You should rewrite the integrand using the double angle formula of `cos 2x` such that:

`(sin^3 x-4)/(1-cos 2x) = (sin^3 x-4)/(1-(1 - 2sin^2 x))`

`(sin^3 x-4)/(1-cos 2x) = (sin^3 x-4)/(1 - 1+ 2sin^2 x)`

`(sin^3 x-4)/(1-cos 2x) = (sin^3 x-4)/(2sin^2 x)`

You may integrate now, using the transformed integrand, such that:

`int (sin^3 x-4)/(1-cos 2x) dx= int (sin^3 x-4)/(2sin^2 x) dx`

`int (sin^3 x-4)/(1-cos 2x) dx = int (sin^3 x)/(2sin^2 x) dx - int 4/(2sin^2 x) dx`

`int (sin^3 x-4)/(1-cos 2x) dx = (1/2)int (sin x) dx - 4/2 int (dx)/(sin^2 x)`

`int (sin^3 x-4)/(1-cos 2x) dx = -(1/2)cos x - 2*(-cot x) + c`

`int (sin^3 x-4)/(1-cos 2x) dx = 2 cot x - (cos x)/2 + c`

**Hence, evaluating the antiderivative of the function yields `int (sin^3 x-4)/(1-cos 2x) dx = 2 cot x - (cos x)/2 + c.` **

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