in powers of (-epsilon^2sin^2(theta)) by using the binomial theorem.
Length of ellipse = L = 4a*integral of (sqrt(1--epsilon^2sin^2(theta)) from 0 to pi/2 = 4* integral of sqrt(a^2cos^2(theta)+b^2*sin^2(theta)) from 0 to pi/2.
where epsilon equals eccentricity of an ellipse: epsilon = sqrt(a^2-b^2)/a
The hint I was given: (Use integral of sin^2n(theta) from 0 to pi/2= (2n-1)/(2n) * pi/2) and use term by term integration to express L as a power series in powers of epsilon.
using the form (1+x)^k, x=-epsilon^2 * sin^2(theta), and k = 1/2
That would equal to the sum from n=1 to infinity (1/2 choose k) *x^n *(-1)^n.
I'm clueless on what to do now,
Length of ellipse = L = 4a*integral of (sqrt(1--epsilon^2sin^2(theta)) from 0 to pi/2 = 4* integral of sqrt(a^2cos^2(theta)+b^2*sin^2(theta)) from 0 to pi/2.
where epsilon equals eccentricity of an ellipse: epsilon = sqrt(a^2-b^2)/a
The hint I was given: (Use integral of sin^2n(theta) from 0 to pi/2= (2n-1)/(2n) * pi/2) and use term by term integration to express L as a power series in powers of epsilon.
using the form (1+x)^k, x=-epsilon^2 * sin^2(theta), and k = 1/2
That would equal to the sum from n=1 to infinity (1/2 choose k) *x^n *(-1)^n.
I'm clueless on what to do now,