What is the solution for (dy)/(dx)=tan(x)+sec(x) ?

The solution for (dy)/(dx)=tan(x)+sec(x) is y=-ln(cos(x))+ln(tan(x)+sec(x))+c_{1}

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(dy)/(dx)=tan(x)+sec(x)

\frac{d y}{d x} = tan (x) + sec (x)

y = - ln (cos (x)) + ln (tan (x) + sec (x)) + c_{1}

Solution steps

\frac{d y}{d x} = tan (x) + sec (x)

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = tan (x) + sec (x)

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

then

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

y = \int tan (x) + sec (x) d x

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

y = - ln (cos (x)) + ln (tan (x) + sec (x)) + c_{1}

\frac{cos ( x )}{2 sin ( x )}

\frac{\partial}{\partial v} (4 u sin (v))

\int x x - 15 d x

\int x e^{x y} d y

\int e^{x + 1} d x

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