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An integral problem with Sin and Cos

Mathematica Asked by Qin-Tao Song on April 23, 2021

I have a difficulty in calculating the following integral with Mathematica:

$$
int^{pi}_{0}rm dtheta_1frac{ sin^2theta_1[2(2-costheta_1costheta)^2+(sinthetasintheta_1)^2]}{[(2-costheta_1costheta)^2-(sinthetasintheta_1)^2]^{5/2}},
$$

where $ pi>theta>0 $.

Here is the code:

Integrate[Sin[th1]^2*(2*(2 - Cos[th]*Cos[th1])^2 + (Sin[th]*Sin[th1])^2)/((2 - Cos[th]*Cos[th1])^2 - (Sin[th]*Sin[th1])^2)^(5/2), {th1, 0, Pi}]

3 Answers

If numerical solution is sufficient:

int[th_?NumericQ] :=NIntegrate[Sin[th1]^2*(2*(2 - Cos[th]*Cos[th1])^2 +(Sin[th]*Sin[th1])^2)/((2 - Cos[th]*Cos[th1])^2 - (Sin[th]*Sin[th1])^2)^(5/2)
, {th1, 0, Pi}]

Plot[int[th], {th, 0, Pi}, PlotRange -> {0, 1}]

enter image description here

Answered by Ulrich Neumann on April 23, 2021

Making use of

f[x_?NumericQ] =  NIntegrate[ Sin[th1]^2*(2*(2 - Cos[x]*Cos[th1])^2 + (Sin[x]*
     Sin[th1])^2)/((2 - Cos[x]*Cos[th1])^2 - (Sin[x]*Sin[th1])^2)^(5/2), {th1, 0, Pi}]

,one obtains

f[Pi/4]

0.676581

and

Plot[f[x], {x, 0, Pi}]

enter image description here

Answered by user64494 on April 23, 2021

The simplest answer I could come up with is

((Sqrt[2]*
  Sqrt[5 + Cos[2*[Theta]] + 4*I*Sqrt[3]*Sin[[Theta]]])/(9*(7 - 
    Cos[2*[Theta]])))*(2*(3 - 2*I*Sqrt[3]*Sin[[Theta]])*
 EllipticK[-((I*8*Sqrt[3]*Sin[[Theta]])/(5 + Cos[2*[Theta]] - 
       I*4*Sqrt[3]*Sin[[Theta]]))] - 
    I*(7 - Cos[2*[Theta]])*
 EllipticE[
  1 - (I*8*Sqrt[3]*Sin[[Theta]])/(5 + Cos[2*[Theta]] + 
      I*4*Sqrt[3]*Sin[[Theta]])]) - ((Sqrt[2]*
  Sqrt[5 + Cos[2*[Theta]] - 4*Sqrt[3]*I*Sin[[Theta]]])/(9*(7 - 
    Cos[2*[Theta]])))*
 (8*Sqrt[3]*Sin[[Theta]]*
 EllipticK[
  1 - (I*8*Sqrt[3]*Sin[[Theta]])/(5 + Cos[2*[Theta]] + 
      I*4*Sqrt[3]*Sin[[Theta]])] - 
I*(7 - Cos[2*[Theta]])*
 EllipticE[
  1 + (I*8*Sqrt[3]*Sin[[Theta]])/(5 + Cos[2*[Theta]] - 
      I*4*Sqrt[3]*Sin[[Theta]])])

I have no idea how to treat the complex moduli further...

Answered by Andreas on April 23, 2021

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