91Ó°ÊÓ

The given equations are$cy={\sin }^2\frac{\mathrm{θ}}{2}$ and $dx=\sqrt{\frac{cy}{1-cy}}dy$.

In the same way that geometry is the study of shape and algebra is the study of generalisations of arithmetic operations, calculus, sometimes known as infinitesimal calculus or "the calculus of infinitesimals," is the mathematical study of continuous change.. Differentiation and integration are the two main branches.

Differentiate the equation $cy={\sin }^2\frac{\mathrm{θ}}{2}$.

$$\begin{gathered} cy={\sin }^2\frac{\mathrm{θ}}{2} \\ cy=\frac{1}{2}\left(1-\cos \mathrm{θ}\right) \\ dy=\frac{1}{2c}\sin \mathrm{θ} \\ dy=\frac{1}{c}\sin \frac{\mathrm{θ}}{2}\cos \frac{\mathrm{θ}}{2} \end{gathered}$$

Rewrite the differential equation in term of dx.

$$\begin{gathered} \mathrm{dx}=\sqrt{\frac{\mathrm{cy}}{1-\mathrm{cy}}}\mathrm{dy} \\ \mathrm{dx}=\sqrt{\frac{{\sin }^2{\displaystyle \frac{\mathrm{θ}}{2}}}{1-{\sin }^2{\displaystyle \frac{\mathrm{θ}}{2}}}}\frac{1}{\mathrm{c}}\sin \frac{\mathrm{θ}}{2}\cos \frac{\mathrm{θ}}{2}\mathrm{»åθ} \\ \mathrm{dx}=\frac{1}{\mathrm{c}}\frac{\sin {\displaystyle \frac{\mathrm{θ}}{2}}}{\cos {\displaystyle \frac{\mathrm{θ}}{2}}}\sin \frac{\mathrm{θ}}{2}\cos \frac{\mathrm{θ}}{2}\mathrm{»åθ} \\ \mathrm{dx}=\frac{1}{\mathrm{c}}{\sin }^2\frac{\mathrm{θ}}{2}\mathrm{»åθ} \end{gathered}$$

Solve further,

$\mathrm{dx}=\frac{1}{2\mathrm{c}}\left(1-\mathrm{³¦´Ç²õθ}\right)\mathrm{»åθ}.....\left(1\right)$

Integrate the equation (1) with respect to $\mathrm{θ}$.

$$\begin{gathered} \mathrm{x}=∫\frac{1}{2\mathrm{c}}\left(1-\mathrm{³¦´Ç²õθ}\right)\mathrm{»åθ} \\ \mathrm{x}=\frac{1}{2\mathrm{c}}\left(\mathrm{θ}-\mathrm{²õ¾\pm ²Ôθ}\right)+\mathrm{c}\text{'} \end{gathered}$$

Therefore, the equations 4.2, $\mathrm{dx}=\frac{1}{2\mathrm{c}}\left(1-\mathrm{³¦´Ç²õθ}\right)\mathrm{»åθ}$ and $\mathrm{x}=\frac{1}{2\mathrm{c}}\left(\mathrm{θ}-\mathrm{²õ¾\pm ²Ôθ}\right)+\mathrm{c}\text{'}$, are verified.