       Re: Can't integrate sqrt(a+b*cos(t)+c*cos(2t))

• To: mathgroup at smc.vnet.net
• Subject: [mg90798] Re: Can't integrate sqrt(a+b*cos(t)+c*cos(2t))
• From: Grischika at mail.ru
• Date: Thu, 24 Jul 2008 04:55:03 -0400 (EDT)
• References: <g6710s\$sb6\$1@smc.vnet.net>

```On 23 =C9=C0=CC, 13:26, Valeri Astanoff <astan... at gmail.com> wrote:
> Good day,
>
> Neither Mathematica 6 nor anyone here can integrate this:
>
> In:= Integrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
> Out= Integrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
>
> In:= NIntegrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], {t, 0, Pi}]
> Out= 6.72288
>
> I know the exact result:
>
> In:= =9A(1/5^(3/4))*(Sqrt*(10*EllipticE[(1/10)*(5 - Sqrt)] -
> =9A =9A =9A =9A 10*EllipticK[(1/10)*(5 - Sqrt)] + (5 + 3*Sqrt)*
> =9A =9A =9A =9A EllipticPi[(1/10)*(5 - 3*Sqrt), (1/10)*(5 - Sqrt)])=
)//N
> Out= 6.72288
>
> but I would like to prove it.
>
> Thanks in advance to the samaritan experts...
>
> V.Astanoff

Hello.
You can try to take indefinite integral:

eq=Integrate[Sqrt[5 - 4*Cos[t] + Cos[2*t]], t]

Here Mathematica gives you an answer:

((2/5 + (4*I)/5)*Cos[t/2]^4*((2 + I)*Sqrt[1 - 2*I]*
EllipticE[I*ArcSinh[Sqrt[1 - 2*I]*Tan[t/2]], -3/5 + (4*I)/5]*
Sec[t/2]^2*Sqrt[1 + (1 - 2*I)*Tan[t/2]^2]*
Sqrt[1 + (1 + 2*I)*Tan[t/2]^2] -
I*((6 - 2*I)*Sqrt[1 - 2*I]*EllipticF[
I*ArcSinh[Sqrt[1 - 2*I]*Tan[t/2]], -3/5 + (4*I)/5]*Sec[t/2]^2*
Sqrt[1 + (1 - 2*I)*Tan[t/2]^2]*Sqrt[1 + (1 + 2*I)*Tan[t/2]^2] -
4*Sqrt[1 - 2*I]*EllipticPi[1/5 + (2*I)/5,
I*ArcSinh[Sqrt[1 - 2*I]*Tan[t/2]], -3/5 + (4*I)/5]*Sec[t/2]^2*
Sqrt[1 + (1 - 2*I)*Tan[t/2]^2]*Sqrt[1 + (1 + 2*I)*Tan[t/2]^2] +
(2 + I)*(Tan[t/2] + 2*Tan[t/2]^3 + 5*Tan[t/2]^5))))/
Sqrt[5 - 4*Cos[t] + Cos[2*t]]

Then find Limits:
Limit[eq, t ->0]

gives 0,

Limit[eq, t -> Pi]

gives

(2/5 + I/5)*Sqrt[2/5 + (4*I)/5]*
((1 + 2*I)*Sqrt*EllipticE[-3/5 - (4*I)/5] -
5*EllipticE[-3/5 + (4*I)/5] + (4 - 4*I)*Sqrt*
EllipticK[-3/5 - (4*I)/5] - (10 - 10*I)*EllipticK[8/5 - (4*I)/5] -
(4 + 8*I)*EllipticPi[1/5 + (2*I)/5, -3/5 + (4*I)/5] -
4*Sqrt*EllipticPi[1 - 2*I, -3/5 - (4*I)/5])

so, the result is
N@%

-6.72287972344033 - 9.947771772989*^-15*I

The problem only with the sign of result.

```

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