What is the solution for (dr)/(dθ)=8(cos(θ/3))^3 ?

The solution for (dr)/(dθ)=8(cos(θ/3))^3 is r(θ)=24(sin(θ/3)-1/3 sin^3(θ/3))+c_{1}

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(dr)/(dθ)=8(cos(θ/3))^3

\frac{d r}{d θ} = 8 (cos (\frac{θ}{3}))^{3}

r (θ) = 24 (sin (\frac{θ}{3}) - \frac{1}{3} sin^{3} (\frac{θ}{3})) + c_{1}

Solution steps

\frac{d r}{d θ} = 8 (cos (\frac{θ}{3}))^{3}

Substitute

\frac{d r}{d θ}

with

r^{'} (θ)

r^{^{'}} (θ) = 8 (cos (\frac{θ}{3}))^{3}

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

then

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

r (θ) = \int 8 (cos (\frac{θ}{3}))^{3} d θ

\int 8 (cos (\frac{θ}{3}))^{3} d θ = 24 (sin (\frac{θ}{3}) - \frac{1}{3} sin^{3} (\frac{θ}{3})) + c_{1}

r (θ) = 24 (sin (\frac{θ}{3}) - \frac{1}{3} sin^{3} (\frac{θ}{3})) + c_{1}

x \to 0 + lim ((e^{x} + x)^{\frac{2}{x}})

y^{^{''}} + 10 y^{^{'}} + 2500 = 500 sin (100 x)

\frac{d}{d x} ((\frac{x + 1}{x - 1})^{\frac{1}{2}})

\int \frac{1}{3 x ^{2} + 4 x + 5} d x

\frac{\partial}{\partial y} (3 y e^{3 x})

What is the solution for (dr)/(dθ)=8(cos(θ/3))^3 ?
The solution for (dr)/(dθ)=8(cos(θ/3))^3 is r(θ)=24(sin(θ/3)-1/3 sin^3(θ/3))+c_{1}