I have an expression in multiple variables that is something like
4.85746*10^-7 Cos[ϕ] (1 +
1/2 (Abs[(-1.5782 Sqrt[1 - z^2 Sin[ϕ]^2] +
1.329 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])/(1.5782 Sqrt[
1 - z^2 Sin[ϕ]^2] +
1.329 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])]^2 +
Abs[(1.329 Sqrt[1 - z^2 Sin[ϕ]^2] -
1.5782 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])/(1.329 Sqrt[
1 - z^2 Sin[ϕ]^2] +
1.5782 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])]^2) Cos[
2 ArcSin[
z Sin[ϕ]]] - ((1 +
1/2 (-Abs[(-1.5782 Sqrt[1 - z^2 Sin[ϕ]^2] +
1.329 Sqrt[
1 - 0.70913 z^2 Sin[ϕ]^2])/(1.5782 Sqrt[
1 - z^2 Sin[ϕ]^2] +
1.329 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])]^2 -
Abs[(1.329 Sqrt[1 - z^2 Sin[ϕ]^2] -
1.5782 Sqrt[
1 - 0.70913 z^2 Sin[ϕ]^2])/(1.329 Sqrt[
1 - z^2 Sin[ϕ]^2] +
1.5782 Sqrt[
1 - 0.70913 z^2 Sin[ϕ]^2])]^2))^2 (Cos[
2 ArcSin[0.842099 z Sin[ϕ]] -
2 ArcSin[z Sin[ϕ]]] +
1/2 (Abs[(-1.5782 Sqrt[1 - z^2 Sin[ϕ]^2] +
1.329 Sqrt[
1 - 0.70913 z^2 Sin[ϕ]^2])/(1.5782 Sqrt[
1 - z^2 Sin[ϕ]^2] +
1.329 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])]^2 +
Abs[(1.329 Sqrt[1 - z^2 Sin[ϕ]^2] -
1.5782 Sqrt[
1 - 0.70913 z^2 Sin[ϕ]^2])/(1.329 Sqrt[
1 - z^2 Sin[ϕ]^2] +
1.5782 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])]^2) Cos[
2 ArcSin[z Sin[ϕ]]]))/(1 +
1/4 (Abs[(-1.5782 Sqrt[1 - z^2 Sin[ϕ]^2] +
1.329 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])/(1.5782 Sqrt[
1 - z^2 Sin[ϕ]^2] +
1.329 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])]^2 +
Abs[(1.329 Sqrt[1 - z^2 Sin[ϕ]^2] -
1.5782 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])/(1.329 Sqrt[
1 - z^2 Sin[ϕ]^2] +
1.5782 Sqrt[
1 - 0.70913 z^2 Sin[ϕ]^2])]^2)^2 + (Abs[(-1.5782 \
Sqrt[1 - z^2 Sin[ϕ]^2] +
1.329 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])/(1.5782 Sqrt[
1 - z^2 Sin[ϕ]^2] +
1.329 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])]^2 +
Abs[(1.329 Sqrt[1 - z^2 Sin[ϕ]^2] -
1.5782 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])/(1.329 Sqrt[
1 - z^2 Sin[ϕ]^2] +
1.5782 Sqrt[1 - 0.70913 z^2 Sin[ϕ]^2])]^2) Cos[
2 ArcSin[0.842099 z Sin[ϕ]]]))
I want to integrate this expression first w.r.t. ϕ with limits 0 to 7 π/18 and then indefinite integral Integrate[F, z]
w.r.t z
. I tried it in many ways, but it is very difficult for me to solve this integral. Can anyone help to find out the solution of this integral. I would be highly obliged.