Q12.6P 鈭�0蟺319-sin2胃d胃.... [FREE SOLUTION]

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Chapter 11: Q12.6P (page 559)

${鈭�}_{0}^{\frac{蟺}{3}} \frac{1}{\sqrt{9 - s i n^{2} 胃}} d 胃 .$

Short Answer

Expert verified

The value of integral in elliptic form is $\frac{1}{3} F (\frac{蟺}{3}, \frac{1}{3}) 鈮� 0.35499$ .

Step by step solution

Given Information

The value of integration is ${鈭�}_{0}^{\frac{蟺}{3}} \frac{1}{\sqrt{9 - \sin^{2} 胃}} d 胃$ .

Definition of elliptic form.

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

Find the value of Integral.

The value of integration is ${鈭�}_{0}^{\frac{蟺}{3}} \frac{1}{\sqrt{9 - \sin^{2} 胃}} d 胃$ .Factor out 9 the equation becomes as follows.

$l = {鈭�}_{0}^{\frac{蟺}{3}} \frac{1}{\sqrt{9 (1 - \frac{\sin^{2} 胃}{9})}} d 胃 l = {鈭�}_{0}^{\frac{蟺}{3}} \frac{1}{3 \sqrt{(1 - \frac{\sin^{2} 胃}{9})}} d 胃$

The formula for the beta function is $F (\frac{蟺}{2}, k) = {鈭�}_{0}^{\frac{蟺}{2}} \frac{1}{\sqrt{1 - k^{2} \sin^{2} 胃}} d 胃$ .

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

$l = \frac{1}{3} {鈭�}_{0}^{\frac{蟺}{3}} \frac{1}{\sqrt{(1 - \frac{\sin^{2} 胃}{9})}} d 胃 l = \frac{1}{3} F (\frac{蟺}{3}, \frac{1}{3}) l 鈮� 0.35499$

Hence, The value of integral in elliptic form is $\frac{1}{3} F (\frac{蟺}{3}, \frac{1}{3}) 鈮� 0.35499$ .

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${鈭�}_{0}^{\frac{蟺}{3}} \frac{1}{\sqrt{9 - s i n^{2} 胃}} d 胃 .$

Short Answer

Step by step solution

Given Information

Definition of elliptic form.

Find the value of Integral.

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