Question

Write the integral in equation (12.7) as an elliptic integral and show that (12.8)gives its value. Hints: Write $\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{\right)}$ and a similar equation for$\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }$. Then make the change of variable$\mathit{x}\mathbf{=}\frac{\mathbf{sin}\mathbf{\left(}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{\right)}}{\mathbf{sin}\mathbf{\left(}\frac{\mathbf{\alpha }}{\mathbf{2}}\mathbf{\right)}}$.

Short answer

The elliptic form of the given function is $I=\sqrt{2}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-u^2}\sqrt{1-si{n}^2\frac{\alpha }{2}{u}^2}}d\theta$, and it is proved that $\sqrt{2}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-u^2}\sqrt{1-{\sin }^2\frac{\alpha }{2}{u}^2}}d\theta =\sqrt{2}K\left(\sin \frac{\alpha }{2}\right)$.

Step-by-step solution

Given Information

The value of integral from equation (12.7) is $\underset{0}{\overset{\alpha }{\int }}\frac{1}{\sqrt{\cos \theta -\cos \alpha }}d\theta$.

Definition of elliptic form

The elliptic form of the integral is defined as $\mathbf{K}\mathbf{\left(}\mathbf{k}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\frac{\mathbf{\pi }}{\mathbf{2}}}{\mathbf{\int }}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}}\mathbf{d}\mathbf{\theta }$.

Find the value of Integral

Substitute $\cos \theta =1-2{\sin }^2\frac{\theta }{2}$into $\underset{0}{\overset{\alpha }{\int }}\frac{1}{\sqrt{\cos \theta -\cos \alpha }}d\theta$.

$\frac{1}{\sqrt{2}}\underset{0}{\overset{\alpha }{\int }}\frac{1}{\sqrt{{\sin }^2\frac{\alpha }{2}-{\sin }^2\frac{\theta }{2}}}d\theta$

Factor out${\sin }^2\theta$in the equation mentioned above.

$I=\frac{1}{\sqrt{2}\sin \frac{\alpha }{2}}\underset{0}{\overset{\alpha }{\int }}\frac{1}{\sqrt{1-\frac{{\sin }^2\frac{\theta }{2}}{{\sin }^2\frac{\alpha }{2}}}}d\theta$

Substitute the values given below in the above equation.

$$\begin{gathered} u=\frac{\sin \frac{\theta }{2}}{\sin \frac{\alpha }{2}} \\ du=\frac{\cos \frac{\theta }{2}}{2\sin \frac{\alpha }{2}}d\alpha \end{gathered}$$

The equation becomes as follows:

role="math" localid="1664306147751" $I=\sqrt{2}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-u^2}\sqrt{1-{\sin }^2\frac{\alpha }{2}{u}^2}}d\theta$

The formula for the elliptic function is $K\left(k\right)=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{\sqrt{1-k^2{\sin }^2\theta }}d\theta$.

Equate the above equation with the value of I, the value of I becomes follows:

$I=\sqrt{2}K\left(\sin \frac{\alpha }{2}\right)$

Hence, the statement has been proven.