What is the solution for f^'(x)=sec(2x)tan(2x) ?

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

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f^'(x)=sec(2x)tan(2x)

f^{^{'}} (x) = sec (2 x) tan (2 x)

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

Solution steps

f^{'} (x) = sec (2 x) tan (2 x)

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

then

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

f (x) = \int sec (2 x) tan (2 x) d x

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

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

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

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

x \to 4 lim (\frac{1}{x ^{2} - 16})

2 t^{3}

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

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