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Chapter 11: Q12.4P (page 559)

${鈭�}_{0}^{1} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - \frac{t^{2}}{4}}} d t$

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

01

Given Information

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

02

Definition of elliptic form.

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

03

Find the value of Integral.

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

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

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

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

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

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Most popular questions from this chapter

The figure is part of a cycloid with parametric equations $x = a (胃 + s i n 胃), y = a (1 - c o s 胃)$ (The graph shown is like Figure 4.4 of Chapter 9 with the origin shifted to P2.) Show that the time for a particle to slide without friction along the curve from (x₁, y₁) to the origin is given by the differential equation for 胃(t) is $t = \sqrt{\frac{a^{y}}{g}} {鈭�}_{0}^{1} \frac{d y}{\sqrt{y (y_{1} - y)}}$ .

Hint: Show that the arc length element is $d s = \sqrt{\frac{2 a}{y}} d y$ . Evaluate the integral to show that the time is independent of the starting height y₁ .

In Problem 4 to 13, identify each of the integral as an elliptic (see Example 1 and 2). Learn the notation of your computer program (see Problem 3) and then evaluate the integral by computer.

9. ${鈭�}_{\frac{- 1}{2}}^{\frac{1}{2}} \frac{d t}{\sqrt{1 - t^{2}} \sqrt{4 - 3 t^{2}}}$ .

Show that for $k = 0$ , $u = F (蠒, 0) = 蠒, s n u = s i n u, c n u = c o s u, d n u = 1 .$ and for $k = 1$ $u = F (蠒, 1) = l n (s e c 蠒 + t a n 蠒), 蠒 = g d u, s n u = t a n h u, c n u = d n u = sech u$ .

Use the recursion relation (3.4), and if needed, equation (3.2) to simplify:

$螕 (\frac{2}{3}) / 螕 (\frac{8}{3})$

Use the recursion relation (3.4), and if needed, equation (3.2) to simplify:

$\frac{螕 (\frac{2}{3})}{螕 (\frac{5}{3})}$

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