Question

Expand the integrands of Kand E[see ( 12.3 )] in power series in${\mathit{k}}^{\mathbf{2}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }$(assuming small k ), and integrate term by term to find power series approximations for the complete elliptic integrals Kand E.

Short answer

The value of integral in elliptic integrals are given below.

$$\begin{gathered} K=\frac{\mathrm{\pi }}{2}+\frac{{\mathrm{\pi k}}^2}{2^3}+\frac{9{\mathrm{\pi k}}^4}{2^7}+.. \\ E=\frac{\mathrm{\pi }}{2}+\frac{{\mathrm{\pi k}}^2}{2^3}-\frac{3{\mathrm{\pi k}}^4}{2^7} \end{gathered}$$

Step-by-step solution

Given Information

The function is $k^2{\sin }^2\theta$.

Definition of elliptic form.

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

Find the value of K.

The function is $k^2{\sin }^2\theta$.

The formula states that $\frac{1}{\sqrt{1-x}}=1+\frac{x}{2}+\frac{3x^2}{8}+\frac{3x^3}{16}+...$.

$$\begin{gathered} K=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}d\theta +\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\sin }^2\theta d\theta +\frac{3k^4}{8}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\sin }^4\theta d\theta +.. \\ K=\frac{\mathrm{\pi }}{2}+\frac{{\mathrm{\pi k}}^2}{2^3}+\frac{9{\mathrm{\pi k}}^4}{2^7}+.. \end{gathered}$$

Find the value of E.

The function is $k^2{\sin }^2\theta$.

The formula states that $\sqrt{1+x}=1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}+...$.

$$\begin{gathered} E=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}d\theta -\frac{k^2}{2}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\sin }^2\theta d\theta -\frac{k^4}{8}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\sin }^4\theta d\theta +.. \\ E=\frac{\mathrm{\pi }}{2}-\frac{{\mathrm{\pi k}}^2}{2^3}-\frac{3{\mathrm{\pi k}}^4}{2^7}+.. \end{gathered}$$

Hence, the value of integral in elliptic integrals are given below.

$$\begin{gathered} K=\frac{\mathrm{\pi }}{2}-\frac{{\mathrm{\pi k}}^2}{2^3}-\frac{9{\mathrm{\pi k}}^4}{2^7}+.. \\ E=\frac{\mathrm{\pi }}{2}-\frac{{\mathrm{\pi k}}^2}{2^3}-\frac{3{\mathrm{\pi k}}^4}{2^7} \end{gathered}$$