Answer to Question #260495 in Calculus for Roots

Question

An example of the complete elliptic integral of the second kind, which is defined by:

E(k)=∫ π/2(upper),0(lower) (1−k^2sin^2⁡x dx), 0 ≤k ≤1.

Note that the definite integral is over the variable x, and that the result is a function of k, which is the parameter inside the integral.

(a) Consider the elliptic integral for k=0, that is E(0). Evaluate the definite integral to show that E(0)=π/2.

(b) Consider the elliptic integral for k=1, that is E(1). Evaluate the definition integral to show that E(1)=1.

(c) Now consider the elliptic integral for k=1/2, that is E(1/2). In this case we cannot find an explicit antiderivative to evalute the definite integral using the Fundamental Theorem of Calculus. We can however resort to comparison properties for definite integrals to bound the value of the integral. Starting with the fact that 0 ≤sin^(2)x ≤ 1, show that (√3)/4*π ≤ E(1/2) ≤1/2π.

Answer 1

a)

$E(0)=\int^{\pi/2}_0dx=\pi/2$

b)

$E(1)=\int^{\pi/2}_0\sqrt{1-sin^2x}dx=\int^{\pi/2}_0cosxdx=sinx|^{\pi/2}_0=1$

c)

$E(1/2)=\int^{\pi/2}_0\sqrt{1-sin^2x/4}dx$

since $0 ≤sin^2x ≤ 1$ :

$\sqrt3/2\le\sqrt{1-sin^2x/4}\le1$

then:

$\sqrt3/2\cdot\pi/2\le\int^{\pi/2}_0\sqrt{1-sin^2x/4}dx\le1\cdot\pi/2$

$\pi\sqrt3/4\le\int^{\pi/2}_0\sqrt{1-sin^2x/4}dx\le\pi/2$

$\pi\sqrt3/4\le E(1/2)\le\pi/2$