What is the solution for f^'(x)=cos(x),f(0)=-5 ?

The solution for f^'(x)=cos(x),f(0)=-5 is f(x)=sin(x)-5

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f^'(x)=cos(x),f(0)=-5

f^{^{'}} (x) = cos (x), f (0) = - 5

f (x) = sin (x) - 5

Solution steps

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

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

then

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

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

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

f (x) = sin (x) + c_{1}

Apply initial conditions:

f (x) = sin (x) - 5

f (x) = sin (x) - 5

\frac{d}{d x} (sin^{2} (- 3 x + 6))

(- 10, 3), (- 8, - 8)

n = 0 \sum \infty (\frac{1}{5})^{n + 1}

\int (e^{x} + 2 x)^{2} d x

\int \frac{x ^{3}}{1 - x ^{4}} d x

What is the solution for f^'(x)=cos(x),f(0)=-5 ?
The solution for f^'(x)=cos(x),f(0)=-5 is f(x)=sin(x)-5