What is the solution for f^'(x)=0.7sin(x)-1.1cos(x) ?

The solution for f^'(x)=0.7sin(x)-1.1cos(x) is f(x)=-0.7cos(x)-1.1sin(x)+c_{1}

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f^'(x)=0.7sin(x)-1.1cos(x)

f^{^{'}} (x) = 0.7 sin (x) - 1.1 cos (x)

f (x) = - 0.7 cos (x) - 1.1 sin (x) + c_{1}

Solution steps

f^{'} (x) = 0.7 sin (x) - 1.1 cos (x)

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

then

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

f (x) = \int 0.7 sin (x) - 1.1 cos (x) d x

\int 0.7 sin (x) - 1.1 cos (x) d x = - 0.7 cos (x) - 1.1 sin (x) + c_{1}

f (x) = - 0.7 cos (x) - 1.1 sin (x) + c_{1}

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What is the solution for f^'(x)=0.7sin(x)-1.1cos(x) ?
The solution for f^'(x)=0.7sin(x)-1.1cos(x) is f(x)=-0.7cos(x)-1.1sin(x)+c_{1}