What is the solution for 2y^'=((1-sin(x))^{-1})/2 ?

The solution for 2y^'=((1-sin(x))^{-1})/2 is y=-1/(2(tan(x/2)-1))+c_{1}

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2y^'=((1-sin(x))^{-1})/2

2 y^{^{'}} = \frac{( 1 - sin ( x ) ) ^{- 1}}{2}

y = - \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} + c_{1}

Solution steps

2 y^{'} (x) = \frac{( 1 - sin ( x ) ) ^{- 1}}{2}

Isolate

y^{'} (x) : y^{'} (x) = \frac{1}{4 ( 1 - sin ( x ) )}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{1}{4 ( 1 - sin ( x ) )} d x

\int \frac{1}{4 ( 1 - sin ( x ) )} d x = - \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} + c_{1}

y = - \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} + c_{1}

What is the solution for 2y^'=((1-sin(x))^{-1})/2 ?
The solution for 2y^'=((1-sin(x))^{-1})/2 is y=-1/(2(tan(x/2)-1))+c_{1}