Area between $\cfrac{x}{\sin(x)}$ and $x\sin(x)$ close to (0,0)

Question

I am trying to calculate the area between $\cfrac{x}{\sin(x)}$ and $x\sin(x)$ that is close to point $(0,0)$ as seen in this picture:

The first step is to find the two points that limit that area. These two curves are interesting because they have many points in common, this is a zoom out of that section:

So the area I am looking for is: $$ \int_a^b \cfrac{x}{\sin(x)} - x\sin(x) dx $$

In order to find $a$ and $b$, I did: $$ \cfrac{x}{\sin(x)} = x\sin(x) \therefore \sin^2(x) = 1 \therefore \sin(x) = 1 $$ So I assume $a = -\pi/2$ and $b = \pi/2$

Is this correct?

Answer 1

Just added for your curiosity.

As Rushabh Mehta answered, there is no simple closed form for the result.

In fact, using rather comples function $$I=\int \frac x {\sin(x)}\,dx=x \left(\log \left(1-e^{i x}\right)-\log \left(1+e^{i x}\right)\right)+i \left(\text{Li}_2\left(-e^{i x}\right)-\text{Li}_2\left(e^{i x}\right)\right)$$ where appears the polylogarithm function.

This makes $$\int_{-\frac \pi 2}^{+\frac \pi 2} \frac x {\sin(x)}\,dx=\left( 2 C-\frac{i \pi ^2}{4}\right)-\left(-2 C-\frac{i \pi ^2}{4} \right)=4C$$ and then the result $4C-2 \approx 1.66386$ already mentioned by @Toby Mak in comments.

For the fun of it, a good approximation could be obtained using a series expansion built at $x=0$. This would give $$\frac x {\sin(x)}=1+\frac{x^2}{6}+\frac{7 x^4}{360}+\frac{31 x^6}{15120}+\frac{127 x^8}{604800}+\frac{73 x^{10}}{3421440}+O\left(x^{12}\right)$$ Integrating termwise and using the bounds $$\int_{-\frac \pi 2}^{+\frac \pi 2} \frac x {\sin(x)}\,dx=\pi +\frac{\pi ^3}{72}+\frac{7 \pi ^5}{28800}+\frac{31 \pi ^7}{6773760}+\frac{127 \pi ^9}{1393459200}+\frac{73 \pi ^{11}}{38539100160}$$ which is $\approx 3.66371$ and the a final value of $\approx 1.66371$.

Edit

Looking for an easy to get approximate value, I wondered what could give the magnificent approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ which was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (hat is to say more than $1400$ years ago).

It would give $$I=\int \frac x {\sin(x)}\,dx \approx \int \frac{5 \pi ^2-4 (\pi -x) x}{16 (\pi -x)}\,dx =\frac{1}{16} \left(-2 x^2-5 \pi ^2 \log (16 (\pi -x))+2 \pi ^2\right)$$ $$\int_{0}^{\frac \pi 2} \frac x {\sin(x)}\,dx\approx \frac{1}{32} \pi ^2 (10\log (2)-1)\approx 1.82942$$ and then, for the whole problem $\frac{1}{16} \pi ^2 (10\log (2)-1)-2\approx 1.65883$

Answer 2

To find the points of intersection, begin by setting $\frac x{\sin{x}}=x\sin{x}$, which gives us when $x\neq n\pi$ for $n\in\mathbb{N}$:$$\sin^2{x}=1\to\sin{x}=\pm1\to x=(2n+1)\cdot\frac\pi2\quad\forall n\in\mathbb{N}$$

In this case, you solely care about the area containing $(0,0)$, whose limit points are $\pm\frac\pi2$. So your integral is $$\int_{-\frac\pi2}^{\frac\pi2} \frac x{\sin{x}}-x\sin{x}dx\sim\color{red}{1.66386}$$

Unfortunately, there is no simple closed form for the indefinite integral, so this approximation will have to do.