What is the solution for (dy)/(dx)=(4cos(2x))/(3-sin(2x)) ?

The solution for (dy)/(dx)=(4cos(2x))/(3-sin(2x)) is y=-2ln(3-sin(2x))+c_{1}

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(dy)/(dx)=(4cos(2x))/(3-sin(2x))

\frac{d y}{d x} = \frac{4 cos ( 2 x )}{3 - sin ( 2 x )}

y = - 2 ln (3 - sin (2 x)) + c_{1}

Solution steps

\frac{d y}{d x} = \frac{4 cos ( 2 x )}{3 - sin ( 2 x )}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{4 cos ( 2 x )}{3 - sin ( 2 x )}

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

then

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

y = \int \frac{4 cos ( 2 x )}{3 - sin ( 2 x )} d x

\int \frac{4 cos ( 2 x )}{3 - sin ( 2 x )} d x = - 2 ln (3 - sin (2 x)) + c_{1}

y = - 2 ln (3 - sin (2 x)) + c_{1}

\frac{10}{s ( s ^{2} + 8 s + 25 )}

\int x e^{- \frac{x}{5}} d x

(\frac{1}{x ^{2} + 1})^{^{'}}

\frac{d y}{d x}, y^{2} - x = 0

\int \frac{x ^{3} + x + 1}{x ^{2} + 2 x + 1} d x

What is the solution for (dy)/(dx)=(4cos(2x))/(3-sin(2x)) ?
The solution for (dy)/(dx)=(4cos(2x))/(3-sin(2x)) is y=-2ln(3-sin(2x))+c_{1}