(dr)/(dθ)=(sin(piθ))^4

解

\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}

解答ステップ

\frac{d r}{d θ} = (sin (π θ))^{4}

代用

\frac{d r}{d θ}

と

r^{'} (θ)

r^{^{'}} (θ) = (sin (π θ))^{4}

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

であれば

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}