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

d er i v a t i v e f (x) = 12

d er i v a t i v e f (x) = 3 - \frac{18}{5 x ^{\frac{1}{10}}}

d er i v a t i v e g (u) = 6 u + 5 u

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}