What is integral of a function of a differential?

Question

What is $$\int \dfrac{d\theta}{\sin\frac{d\theta}{2}}$$

I thought of approximating the $\sin$ term $$\sin \dfrac{d\theta}{2} \approx\dfrac{d\theta}{2}$$ and so the integral evaluates to be $$\int 2$$

which is wrong as comments have highlighted.

---

Where it was encountered

This is a problem I encountered while finding the tension force on a ring due to rotation about its central axis.

From figure:

• The dark grey vector is infinitesimal Tension $dT$
• The angle between the dark blue radii is $d\theta$, and half it is $\frac{d\theta}{2}$
• light gray vector $ = dT \cos\frac{d\theta}{2}$
• red vector $ = dT \sin\frac{d\theta}{2}$

Observe that $\cos$ component of $dT$ will cancel, and $\sin$ components will add up.

These $\sin$ components will provide the necessary centripetal force for rotation, therefore:

$$2\rm{dT}\sin\frac{\mathbb{d\theta}}{2} = \dfrac{v^2 \mathbb{dm}}{r}$$

But $\rm dm = \lambda dl \ $ and $ \rm dl = rd\theta$ ($\lambda$ = linear mass density)

So we finally get:

$$\rm 2dT\sin\frac{d\theta}{2} = \frac{\lambda v^2 r d\theta}{r}$$

$$\rm \frac{2}{\lambda v^2}\int dT = \int \frac{ d\theta}{\sin\frac{d\theta}{2}}$$

$$???$$

Answer 1

I think the problem is that it shouldn't be $dT$ but rather just T.

Answer 2

I think thats not right the reason is dimensional analysis. Dimensions of force are $[MLT^{-2}]$ While if you somehow integrate it (which i dont know how) you will get only an a constant as integral of $\sin(x)$ is just another trigo ratio here $-\cos(x)$ which is dimensionless even if $x$ has dimensions like those of phase in wave optics ie $sin(kx-\omega t)$ thus i think you have made a mistake in your derivation.