Vtyjkyuk

I am trying to calculate the area between $cfracxsin(x)$ and $xsin(x)$ 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:
$$
int_a^b cfracxsin(x) - xsin(x) dx
$$

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

Is this correct?

Just added for your curiosity.

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

In fact, using rather comples function
$$I=int frac x sin(x),dx=x
left(log left(1-e^i xright)-log left(1+e^i xright)right)+i left(textLi_2left(-e^i xright)-textLi_2left(e^i xright)right)$$ where appears the polylogarithm function.

This makes
$$int_-frac pi 2^+frac pi 2 frac x sin(x),dx=left( 2 C-fraci pi ^24right)-left(-2 C-fraci pi ^24 right)=4C$$ 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
$$frac x sin(x)=1+fracx^26+frac7 x^4360+frac31 x^615120+frac127
x^8604800+frac73 x^103421440+Oleft(x^12right)$$ Integrating termwise and using the bounds
$$int_-frac pi 2^+frac pi 2 frac x sin(x),dx=pi +fracpi ^372+frac7 pi ^528800+frac31 pi ^76773760+frac127
pi ^91393459200+frac73 pi ^1138539100160$$ 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
$$sin(x) simeq frac16 (pi -x) x5 pi ^2-4 (pi -x) xqquad (0leq xleqpi)$$ 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
$$I=int frac x sin(x),dx approx int frac5 pi ^2-4 (pi -x) x16 (pi -x),dx =frac116 left(-2 x^2-5 pi ^2 log (16 (pi -x))+2 pi ^2right)$$
$$int_0^frac pi 2 frac x sin(x),dxapprox frac132 pi ^2 (10log (2)-1)approx 1.82942$$ and then, for the whole problem $frac116 pi ^2 (10log (2)-1)-2approx 1.65883$

To find the points of intersection, begin by setting $frac xsinx=xsinx$, which gives us when $xneq npi$ for $ninmathbbN$:$$sin^2x=1tosinx=pm1to x=(2n+1)cdotfracpi2quadforall ninmathbbN$$

In this case, you solely care about the area containing $(0,0)$, whose limit points are $pmfracpi2$. So your integral is $$int_-fracpi2^fracpi2 frac xsinx-xsinxdxsimcolorred1.66386$$

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