Find the arc length of the curve given by x=(1+ sin phi)

Find the arc length of the curve given by $x=(1+\sin\varphi)\cos\varphi$, $y=(1+\sin\varphi)\sin\varphi+\dfrac14$.

The constant $\frac14$ added to $y$ is a vertical shift and does not affect any derivative, so it plays no role in the arc length.

Step 1: Differentiate

\[\frac{dx}{d\varphi}=\cos^2\varphi-(1+\sin\varphi)\sin\varphi = \cos2\varphi-\sin\varphi\]

\[\frac{dy}{d\varphi}=\sin\varphi\cos\varphi+(1+\sin\varphi)\cos\varphi = \sin2\varphi+\cos\varphi\]

Step 2: Compute the speed squared

\[\left(\frac{dx}{d\varphi}\right)^2+\left(\frac{dy}{d\varphi}\right)^2 = \cos^22\varphi+\sin^22\varphi+\sin^2\varphi+\cos^2\varphi + 2(\sin2\varphi\cos\varphi-\cos2\varphi\sin\varphi)\]

Using $\sin(A-B)=\sin A\cos B-\cos A\sin B$ with $A=2\varphi,\,B=\varphi$:

\[=1+1+2\sin(2\varphi-\varphi)=2+2\sin\varphi = 2(1+\sin\varphi)\]

Step 3: Arc-length element

\[\sqrt{2(1+\sin\varphi)}\]

Step 4: Integrate over a full period

With $u=\varphi-\frac\pi2$ (so $\sin\varphi=\cos u$) and $1+\cos u=2\cos^2\frac u2$:

\[L=\int_0^{2\pi}\sqrt{2(1+\sin\varphi)}\,d\varphi = \int_{-\pi/2}^{3\pi/2}2\left|\cos\frac u2\right|du\]

Step 5: Evaluate accounting for the sign of $\cos(u/2)$

On $u\in[-\frac\pi2,\pi]$, $\cos\frac u2\ge0$; on $u\in(\pi,\frac{3\pi}2]$, $\cos\frac u2<0$. Splitting the integral there:

\[L=2\int_{-\pi/2}^{\pi}\cos\frac u2\,du - 2\int_{\pi}^{3\pi/2}\cos\frac u2\,du = 4\left[1+\frac{\sqrt2}2\right]-4\left[\frac{\sqrt2}2-1\right]=8\]

\[\boxed{L=8}\]