TITLE: Find the area of the region bounded by $\sin(x)\sin(y)=k$ where $0 \leq x \leq \pi$, $0 \leq y \leq \pi$, and $0 \leq k \leq 1$
QUESTION [5 upvotes]: On the coordinate plane, the equation $\sin(x)\sin(y)=k$ where $0 \leq x \leq \pi$, $0 \leq y \leq \pi$, and $0 \leq k \leq 1$ forms a closed region. Find the area of this region in terms of $k$.
I've had some progress, but I've hit a dead end with an integral that appears to not be solvable.
Solving in terms of $x$, you get that $y=\arcsin(\dfrac{k}{\sin(x)})$. When you integrate from the bounds given by the minimum and maximum $x$ of this region, you get $\int_{\arcsin(k)}^{\pi-\arcsin(k)} \arcsin(\dfrac{k}{\sin(x)}) dx$. Thus, considering the fact that this integral gives the region below the closed region, the expression for the closed region is $$\pi(\pi-2\arcsin(k))-2\int_{\arcsin(k)}^{\pi-\arcsin(k)} \arcsin\left(\dfrac{k}{\sin(x)}\right).$$ However, I need help with the integral as WolframAlpha cannot evaluate it.
Help would be appreciated (or perhaps even a better solution than this!)
Thanks

REPLY [2 votes]: Starting from @K.defaoite's answer,
$$A=2\int_{-\arccos(C)}^{\arccos(C)}\arccos\left(\frac{C}{\cos(x)}\right)\,dx$$
it seems that we could have a "reasonable" approximation expanding the integrand as a Taylor series around $x=0$
$$\arccos\left(\frac{C}{\cos(x)}\right)=\arccos(C)+\sum_{n=1}^\infty a_n\,x^{2n}$$ where
$$a_n= \frac {C\, P_{n}(C)}{(2n)! \,\,\left(1-C^2\right)^{n-\frac{1}{2}}}$$ The first polynomials are listed below
$$\left(
\begin{array}{cc}
n & P_n(C) \\
 1 & -1 \\
 2 & 2 C^2-5 \\
 3 & -16 C^4+32 C^2-61 \\
 4 & 272 C^6-984 C^4+522 C^2-1385 \\
 5 & -7936 C^8+35584 C^6-74256 C^4-2096 C^2-50521 \\
 6 & 353792 C^{10}-1946368 C^8+4184192 C^6-7738688 C^4-1973438 C^2-2702765
\end{array}
\right)$$
Using these terms, the results are
$$\left(
\begin{array}{ccc}
C & \text{approximation} & \text{exact} \\
 0.00 & 9.86960 & 9.86960 \\
 0.05 & 8.94015 & 8.79304 \\
 0.10 & 8.13310 & 7.99304 \\
 0.15 & 7.41526 & 7.29659 \\
 0.20 & 6.76475 & 6.66673 \\
 0.25 & 6.16666 & 6.08584 \\
 0.30 & 5.61055 & 5.54351 \\
 0.35 & 5.08889 & 5.03283 \\
 0.40 & 4.59611 & 4.54891 \\
 0.45 & 4.12799 & 4.08806 \\
 0.50 & 3.68129 & 3.64744 \\
 0.55 & 3.25342 & 3.22477 \\
 0.60 & 2.84231 & 2.81820 \\
 0.65 & 2.44627 & 2.42618 \\
 0.70 & 2.06389 & 2.04741 \\
 0.75 & 1.69397 & 1.68078 \\
 0.80 & 1.33550 & 1.32533 \\
 0.85 & 0.98759 & 0.98022 \\
 0.90 & 0.64947 & 0.64472 \\
 0.95 & 0.32048 & 0.31817 \\
 1.00 & 0.00000 & 0.00000
\end{array}
\right)$$ which seems to be quite decent.
Edit
When $c$ is small,expanding the integrand as a Taylor series
$$\arccos\left(\frac{C}{\cos(x)}\right)=\frac{\pi }{2}-C \sec (x)-\frac{1}{6} C^3 \sec ^3(x)-\frac{3}{40} C^5 \sec^5(x)+O\left(C^7\right)$$ we have for $A$ the approximation
$$A=2\pi  \cos ^{-1}(C)+$$ $$\frac{1}{240} \left(C \left(27 C^4+80 C^2+960\right) \log \left(\frac{\cos (t)-\sin
   (t)}{\sin (t)+\cos (t)}\right)-C \sqrt{1-C^2} \left(27 C^2+98\right)\right)$$ where $t=\frac{1}{2} \cos ^{-1}(C)$.
For $C=0.1$, this gives $8.00120$.