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Expand the integrands of Kand E[see ( 12.3 )] in power series ink2sin2θ(assuming small k ), and integrate term by term to find power series approximations for the complete elliptic integrals Kand E.

Short Answer

Expert verified

The value of integral in elliptic integrals are given below.

K=Ï€2+Ï€°ì223+9Ï€°ì427+..E=Ï€2+Ï€°ì223-3Ï€°ì427

Step by step solution

01

Given Information

The function is k2sin2θ.

02

Definition of elliptic form.

The elliptic form of the integral is defined asE(π2,k)=∫0π21-k2sin2θdθ.

03

Find the value of K.

The function is k2sin2θ.

The formula states that 11-x=1+x2+3x28+3x316+....

K=∫0Ï€2dθ+∫0Ï€2sin2θdθ+3k48∫0Ï€2sin4θdθ+..K=Ï€2+Ï€°ì223+9Ï€°ì427+..

04

Find the value of E.

The function is k2sin2θ.

The formula states that 1+x=1-x2-x28-x316+....

E=∫0Ï€2dθ-k22∫0Ï€2sin2θdθ-k48∫0Ï€2sin4θdθ+..E=Ï€2-Ï€°ì223-3Ï€°ì427+..

Hence, the value of integral in elliptic integrals are given below.

K=Ï€2-Ï€°ì223-9Ï€°ì427+..E=Ï€2-Ï€°ì223-3Ï€°ì427

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