Chapter 11: Q12.4P (page 559)

$\int_{0}^{1} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - \frac{t^{2}}{4}}} d t$

Short Answer

Expert verified

The value of integral in elliptic form is $F (\sin^{- 1} (\frac{1}{2}), \frac{1}{2})$ .

Step by step solution

Given Information

The value of integration is $\int_{0}^{1} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - \frac{t^{2}}{4}}} d t$ .

Definition of elliptic form.

The elliptic form of the integral is defined as $F = \int_{0}^{1} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - k^{2} t^{2}}} d t$ .

Find the value of Integral.

The value of integration is $\int_{0}^{1} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - \frac{t^{2}}{4}}} d t$ .

The formula for the beta function is $F = \int_{0}^{1} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - k^{2} t^{2}}} d t$ .

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

$l = \int_{0}^{1} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - \frac{t^{2}}{4}}} d t l = F (\sin^{- 1} (\frac{1}{2}), \frac{1}{2})$

Hence, the value of integral in elliptic form is $l = F (\sin^{- 1} (\frac{1}{2}), \frac{1}{2})$ .

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$\int_{0}^{1} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - \frac{t^{2}}{4}}} d t$

Short Answer

Step by step solution

Given Information

Definition of elliptic form.

Find the value of Integral.

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∫0111-t21-t24dt

Short Answer

Step by step solution

Given Information

Definition of elliptic form.

Find the value of Integral.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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Study anywhere. Anytime. Across all devices.

$\int_{0}^{1} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - \frac{t^{2}}{4}}} d t$