What is the solution for h^'=6-5cos(pi/(15)t) ?

The solution for h^'=6-5cos(pi/(15)t) is h(t)=6t-(75)/pi sin((pit)/(15))+c_{1}

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h^'=6-5cos(pi/(15)t)

h^{^{'}} = 6 - 5 cos (\frac{π}{15} t)

h (t) = 6 t - \frac{75}{π} sin (\frac{π t}{15}) + c_{1}

Solution steps

h^{'} (t) = 6 - 5 cos (\frac{π}{15} t)

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

then

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

h (t) = \int 6 - 5 cos (\frac{π}{15} t) d t

\int 6 - 5 cos (\frac{π}{15} t) d t = 6 t - \frac{75}{π} sin (\frac{π t}{15}) + c_{1}

h (t) = 6 t - \frac{75}{π} sin (\frac{π t}{15}) + c_{1}

\int \frac{1 - sin ( x )}{x + cos ( x )} d x

x \to + (- 1) lim (x)

\frac{d}{d x} (- \frac{16}{x ^{3}})

y = \frac{3}{x}, y = 12 x, y = \frac{1}{3} x

h \to 0 lim (\frac{( x - h ) - x}{h})

What is the solution for h^'=6-5cos(pi/(15)t) ?
The solution for h^'=6-5cos(pi/(15)t) is h(t)=6t-(75)/pi sin((pit)/(15))+c_{1}