(dr)/(dθ)=(sin(piθ))^4

Lösung

\frac{d r}{d θ} = (sin (π θ))^{4}

r = \frac{1}{π} (- \frac{1}{4} sin^{3} (π θ) cos (π θ) + \frac{3}{8} (π θ - \frac{1}{2} sin (2 π θ))) + c_{1}

Schritte zur Lösung

\frac{d r}{d θ} = (sin (π θ))^{4}

Ersetze

\frac{d r}{d θ}

mit

r^{'} (θ)

r^{^{'}} (θ) = (sin (π θ))^{4}

Wenn

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

dann

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

r = \int (sin (π θ))^{4} d θ

\int (sin (π θ))^{4} d θ = \frac{1}{π} (- \frac{1}{4} sin^{3} (π θ) cos (π θ) + \frac{3}{8} (π θ - \frac{1}{2} sin (2 π θ))) + c_{1}

r = \frac{1}{π} (- \frac{1}{4} sin^{3} (π θ) cos (π θ) + \frac{3}{8} (π θ - \frac{1}{2} sin (2 π θ))) + c_{1}

\int 3 e^{- \frac{x}{2}} d x

f (x) = 3 e^{x}

\frac{\partial}{\partial x} (3 x y^{2} + 16 x^{2} y^{3})

\int θ tan (8) θ^{2} d θ

\frac{\partial}{\partial x} (e^{x y} ln (9 y))