solve part (d)Consider the ellipse += 1.(a) Show that the circumference of the ellipse is given by C= 4 f72 V4 cos? t + 9 sin? t dt.(b) Show that this integral can be simplified to C = 8 Jo" /1+sin? t dt.1+5u2/4(c) By taking u=sin(t), show that it further simplifies to C = 8du. (Be careful, as this is an improper1-u?integral.)(d) Now make a substitution x = ? to find C = 8 74a dx(Also improper.)Væ(x-1)(9-4x)We can think of the denominator here as related to the elliptic curve y2 = x(x – 1)(9 - 4), and in the sense ofdifferential geometry we are really integrating a differential form on this elliptic curve over a loop. Integrals such asthese are called elliptic integrals, and they cannot be evaluated in terms of elementary functions. But this is howelliptic curves received their name.

Suppose u=gx and g' be a continuous function on a,b and also f is continuous over range of the…

solve part (d) Consider the ellipse +5 = 1. (a) Show that the circumference of the ellipse is given by C = 4 2 V4 cos? t + 9 sin? t dt. 7/2 (b) Show that this integral can be simplified to C = 8 S2 /1+ sin? t dt. (c) By taking u=sin(t), show that it further simplifies to C = 8 1 1+5u²/4 1-u? du. (Be careful, as this is an improper integral.) (d) Now make a substitution x = ? to find C = 8 9/4 a dz (Also improper.) Vz(z-1)(9-42)

solve part (d) Consider the ellipse +5 = 1. (a) Show that the circumference of the ellipse is given by C = 4 2 V4 cos? t + 9 sin? t dt. 7/2 (b) Show that this integral can be simplified to C = 8 S2 /1+ sin? t dt. (c) By taking u=sin(t), show that it further simplifies to C = 8 1 1+5u²/4 1-u? du. (Be careful, as this is an improper integral.) (d) Now make a substitution x = ? to find C = 8 9/4 a dz (Also improper.) Vz(z-1)(9-42)

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

solve part (d)
Consider the ellipse +
= 1.
(a) Show that the circumference of the ellipse is given by C
= 4 f72 V4 cos? t + 9 sin? t dt.
(b) Show that this integral can be simplified to C = 8 Jo" /1+sin? t dt.
1+5u2/4
(c) By taking u=sin(t), show that it further simplifies to C = 8
du. (Be careful, as this is an improper
1-u?
integral.)
(d) Now make a substitution x = ? to find C = 8 74
a dx
(Also improper.)
Væ(x-1)(9-4x)
We can think of the denominator here as related to the elliptic curve y2 = x(x – 1)(9 - 4), and in the sense of
differential geometry we are really integrating a differential form on this elliptic curve over a loop. Integrals such as
these are called elliptic integrals, and they cannot be evaluated in terms of elementary functions. But this is how
elliptic curves received their name.

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