Area inside polar curve Find the area of the region inside 
$r=7\sin\theta$ but outside $r=1$.
I have tried finding the area of both using $A=\frac{1}{2}\int_\alpha^\beta r^2 d\theta$, tried arc length...but I can't find the answer.

\begin{align*}
  A &= \frac{1}{2}
       \int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
       (r^{2}-1) \, d\theta \\
    &= \frac{1}{2} \int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
       (49\sin^{2}-1) \, d\theta \\
    &= \frac{1}{2} \int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
       \left[ 49\left( \frac{1-\cos 2\theta}{2} \right)-1 \right] \, d\theta \\
    &= \left[
         \frac{47}{4}\theta-\frac{49}{8} \sin 2\theta
       \right]_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}} \\
    &= \frac{47\pi}{4}-\frac{47\pi}{2}\sin^{-1} \frac{1}{7}+
       \frac{49}{4} \sin \left( 2\sin^{-1} \frac{1}{7} \right) \\
    &= \frac{47\pi}{4}-\frac{47\pi}{2}\sin^{-1} \frac{1}{7}+
       \frac{49}{2} \times \frac{1}{7} \times \sqrt{\frac{48}{49}} \\
    &= 2\sqrt{3}+\frac{47\pi}{2}-\frac{47}{2}\sin^{-1} \frac{1}{7} \\
    &= 2\sqrt{3}+\frac{47}{2}\cos^{-1} \frac{1}{7}
\end{align*}
