What is the solution for (dy)/(dx)= 1/(2cos^2(x)) ?

The solution for (dy)/(dx)= 1/(2cos^2(x)) is y= 1/2 tan(x)+c_{1}

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(dy)/(dx)= 1/(2cos^2(x))

\frac{d y}{d x} = \frac{1}{2 cos ^{2} ( x )}

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

Solution steps

\frac{d y}{d x} = \frac{1}{2 cos ^{2} ( x )}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

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

then

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

y = \int \frac{1}{2 cos ^{2} ( x )} d x

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

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

\frac{d y}{d x} = y (1 - 2 y)

\int_{0}^{1} x (x^{2} - 1) e^{- x^{2}} d x

\int_{0}^{3} (x) d x

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\int 8 x (4 x^{2} + 5)^{6} d x

What is the solution for (dy)/(dx)= 1/(2cos^2(x)) ?
The solution for (dy)/(dx)= 1/(2cos^2(x)) is y= 1/2 tan(x)+c_{1}