Chapter 8.7, Problem 70E

To determine

To Find: The expanded form of the binomial series and use ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx}=\frac{1\cdot 3\cdot 5\cdot \cdot \cdot \left(2n-1\right)}{2\cdot 4\cdot 6\cdot \cdot \cdot 2n}\cdot \frac{\pi }{2}$ to express $L$ as the series in the power of the eccentricity up the term in $e^6$ .

Answer to Problem 70E

The required expression is $L=4a\left[\frac{\pi }{2}-\frac{\pi }{8}{e}^2-\frac{3\pi }{128}{e}^4+\frac{5\pi }{512}{e}^6\right]$ .

Explanation of Solution

Given:

The given formula is ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx}=\frac{1\cdot 3\cdot 5\cdot \cdot \cdot \left(2n-1\right)}{2\cdot 4\cdot 6\cdot \cdot \cdot 2n}\cdot \frac{\pi }{2}$ .

Calculation:

Consider the given expression is,

  ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx}=\frac{1\cdot 3\cdot 5\cdot \cdot \cdot \left(2n-1\right)}{2\cdot 4\cdot 6\cdot \cdot \cdot 2n}\cdot \frac{\pi }{2}$

Consider the given series is,

  ${\left(1+x\right)}^r=1+rx+\frac{\left(r\right)\left(r-1\right)}{2!}{x}^2+\frac{\left(r\right)\left(r-2\right)}{3!}{x}^3+\mathrm{....}$

Consider the given expression is,

  $L=4a{\displaystyle {\int }_0^{\frac{\pi }{2}}\left(\sqrt{1-e^2{\sin }^2\theta }\right)d\theta }$

Then, by the binomial series,

  $\begin{array}{c}L=4a{\displaystyle {\int }_0^{\frac{\pi }{2}}{\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}\infty \\ n\end{array}\right)}{\left(-e^2{\sin }^2\theta \right)}^n}d\theta \\ =4a{\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}\frac{1}{2}\\ n\end{array}\right)}{\left(-1\right)}^n{e}^{2n}\left(\frac{1\cdot 3\cdot 5\cdot \cdot \cdot \left(2n-1\right)}{2\cdot 4\cdot 5\cdot 6\cdot \cdot \cdot \cdot 2n}\cdot \frac{\pi }{2}\right)\end{array}$

Consider that only six term are needed then,

  $\begin{array}{c}L=4a\left[\left(\begin{array}{c}\frac{1}{2}\\ 0\end{array}\right){\left(-1\right)}^0{e}^0+\left(\begin{array}{c}\frac{1}{2}\\ 1\end{array}\right){\left(-1\right)}^1{e}^2\left(\frac{1}{2}\cdot \frac{\pi }{2}\right)+\left(\begin{array}{c}\frac{1}{2}\\ 2\end{array}\right){\left(-1\right)}^1{e}^4\left(\frac{1}{2}\cdot \frac{2}{4}\cdot \frac{\pi }{2}\right)++\left(\begin{array}{c}\frac{1}{2}\\ 3\end{array}\right){\left(-1\right)}^3{e}^6\left(\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{\pi }{2}\right)\right]\\ =4a\left[\frac{\pi }{2}-\frac{\pi }{8}{e}^2-\frac{3\pi }{128}{e}^4+\frac{5\pi }{512}{e}^6\right]\end{array}$