Area inside loop of polar equation, unsolvable problem? Is this problem solvable?
"Please find the area inside the first loop of the following equation (using polar coordinates): r = cos$(\theta)$ - sec$(\theta)$." 
From what I can tell, this function does not loop at all: 
https://www.wolframalpha.com/input/?i=r+%3D+cos%28%CE%B8%29+-+sec%28%CE%B8%29 
 A: I agree there is no loop. Such a curve is a Cissoid of Diocles.
A: It doesn't loop, and centered at the origin, the area swept out between $\theta=-\frac\pi2$ and $\theta=\frac\pi2$ would be infinite. However, the area swept out between the curve and the asymptote at $x=-1$, would be
$$
\begin{align}
&\int_{-\pi/2}^{\pi/2}\left[\frac12\sec^2(\theta)-\frac12\left(\cos(\theta)-\sec(\theta)\right)^2\right]\,\mathrm{d}\theta\\
&=\frac12\int_{-\pi/2}^{\pi/2}\left[2-\cos^2(\theta)\right]\,\mathrm{d}\theta\\
&=\frac12\int_{-\pi/2}^{\pi/2}\left[2-\frac{1+\cos(2\theta)}2\right]\,\mathrm{d}\theta\\
&=\frac14\int_{-\pi/2}^{\pi/2}\left[3-\cos(2\theta)\right]\,\mathrm{d}\theta\\
&=\frac{3\pi}4
\end{align}
$$


The implicit equation for this curve is
$$
y^2(1+x)+x^3=0
$$
We can verify the area
$$
\begin{align}
2\int_{-1}^0\sqrt{\frac{-x^3}{1+x}}\,\mathrm{d}x
&=2\int_0^1\sqrt{\frac{x^3}{1-x}}\,\mathrm{d}x\\
&=2\int_0^1x^{3/2}(1-x)^{-1/2}\,\mathrm{d}x\\[9pt]
&=2\,\mathrm{B}(\tfrac52,\tfrac12)\\[6pt]
&=2\,\frac{\Gamma(\frac52)\Gamma(\frac12)}{\Gamma(3)}\\
&=2\,\frac{\frac34\sqrt\pi\cdot\sqrt\pi}2\\[9pt]
&=\frac{3\pi}4
\end{align}
$$
