What is the solution for f^'(x)=((1-sec(x))/(tan(x))) ?

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

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f^'(x)=((1-sec(x))/(tan(x)))

f^{^{'}} (x) = (\frac{1 - sec ( x )}{tan ( x )})

f (x) = ln (sin (x)) - ln (tan (\frac{x}{2})) + c_{1}

Solution steps

f^{'} (x) = (\frac{1 - sec ( x )}{tan ( x )})

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

then

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

f (x) = \int \frac{1 - sec ( x )}{tan ( x )} d x

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

f (x) = ln (sin (x)) - ln (tan (\frac{x}{2})) + c_{1}

\frac{\partial}{\partial x} (1 - x^{2})

\frac{\partial}{\partial x} (r cos (t) sin (x))

\int \frac{x}{( x + 3 )} d x

\frac{\partial}{\partial y} (4 x^{2} y)

f (x) = \frac{2}{( x - 1 ) ^{3}}

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