What is the solution for f^'(θ)=cos(θ)-sin(θ) ?

The solution for f^'(θ)=cos(θ)-sin(θ) is f(θ)=sin(θ)+cos(θ)+c_{1}

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f^'(θ)=cos(θ)-sin(θ)

f^{^{'}} (θ) = cos (θ) - sin (θ)

f (θ) = sin (θ) + cos (θ) + c_{1}

Solution steps

f^{'} (θ) = cos (θ) - sin (θ)

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

then

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

f (θ) = \int cos (θ) - sin (θ) d θ

\int cos (θ) - sin (θ) d θ = sin (θ) + cos (θ) + c_{1}

f (θ) = sin (θ) + cos (θ) + c_{1}

\frac{\partial}{\partial y} (4 x^{3} y^{3} + 4)

\int_{- 1}^{3} 4 - x^{2} d x

y = \frac{3}{x}, y = 7 x, y = \frac{1}{7} x

f (x) = \frac{( x ^{2} - 6 x )}{x + 1}

\frac{d}{d x} (e^{\frac{x}{a}})

What is the solution for f^'(θ)=cos(θ)-sin(θ) ?
The solution for f^'(θ)=cos(θ)-sin(θ) is f(θ)=sin(θ)+cos(θ)+c_{1}