Q21P Show that the four answers given... [FREE SOLUTION]

Chapter 11: Q21P (page 559)

Show that the four answers given in Section 1 for ${鈭�}_{0}^{\frac{蟺}{2}} \frac{诲胃}{\sqrt{肠辞蝉胃}}$ are all correct. Hints: For the beta function result, use(6.4). Then get the gamma function results by using (7.1) and the various 螕 function formulas. For the elliptic integral, use the hint of Problem 17 with $伪 = \frac{蟺}{2}$ .

Short Answer

Expert verified

The solution is mentioned below.

$I = \sqrt{2} K (\frac{1}{\sqrt{2}}) I = 尾 (\frac{1}{2}, \frac{1}{4}) I = \frac{螕 \frac{1}{2} 螕 \frac{1}{4}}{螕 \frac{3}{4}}$

Step by step solution

Given Information

The value of integration is $伪 = {鈭�}_{0}^{\frac{蟺}{2}} \frac{1}{\sqrt{\cos 胃}} d 胃$ .

Definition of elliptic form.

The elliptic form of the integral is defined as $K (k) = {鈭�}_{0}^{\frac{蟺}{2}} \frac{1}{\sqrt{1 - k^{2} \sin^{2} 胃}} d 胃$ .

Find the value of Integral

The value of integration is $伪 = {鈭�}_{0}^{\frac{蟺}{2}} \frac{1}{\sqrt{\cos 胃}} d 胃$ .

Substitute $\cos 胃 = 1 - 2 \sin^{2} \frac{胃}{2}$

The integral becomes as follows

$伪 = {鈭�}_{0}^{\frac{蟺}{2}} \frac{1}{\sqrt{1 - 2 \sin^{2} \frac{胃}{2}}} d 胃$

Substitute the values given below in the above equation.

$x^{2} = 2 \sin^{2} \frac{胃}{2} d x = \frac{\sqrt{2}}{2} \cos \frac{胃}{2} d 胃$

The equation becomes as follows.

$I = \sqrt{2} K (\frac{1}{\sqrt{2}}) I = 尾 (\frac{1}{2}, \frac{1}{4}) I = \frac{螕 \frac{1}{2} 螕 \frac{1}{4}}{螕 \frac{3}{4}}$

Hence, The solution is mentioned below.

$I = \sqrt{2} K (\frac{1}{\sqrt{2}}) I = 尾 (\frac{1}{2}, \frac{1}{4}) I = \frac{螕 \frac{1}{2} 螕 \frac{1}{4}}{螕 \frac{3}{4}} I = \sqrt{2} {鈭�}_{0}^{1} \frac{1}{\sqrt{1 - x^{2}} \sqrt{1 - \frac{1}{2 x^{2}}}} d x$

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Short Answer

Step by step solution

Given Information

Definition of elliptic form.

Find the value of Integral

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