While solving a boundary value problem, I reached the following expression for a Fourier constant (CC01) after simplification using Simplify. The expression is still huge. Is there any way to simplify this further in Mathematica using some definitions I mention below (or without using them)
CC01 = (32 E^(
w γ) (-E^(-w γ)
pc (Ta - tci) δ (NTUC^2 + δ^2) Sin[α/
2] Sin[α/2 + β] (-2 E^NTUH NTUH^3 ph α +
2 E^NTUH NTUH ph α^3 -
2 E^NTUH NTUH^2 ph α^3 - 2 E^NTUH ph α^5 +
2 E^NTUH NTUH^4 α γ +
4 E^NTUH NTUH^2 α^3 γ +
2 E^NTUH α^5 γ +
2 NTUH ph α (NTUH^2 - α^2) Cos[α] +
E^NTUH NTUH ph α (NTUH^2 + α^2) Cos[
2 β] +
E^NTUH NTUH^3 ph α Cos[2 (α + β)] +
E^NTUH NTUH ph α^3 Cos[2 (α + β)] -
2 NTUH^3 ph α Cos[α + 2 β] -
2 NTUH ph α^3 Cos[α + 2 β] -
4 NTUH^2 ph α^2 Sin[α] -
E^NTUH NTUH^2 ph α^2 Sin[2 β] -
E^NTUH ph α^4 Sin[2 β] +
E^NTUH NTUH^4 γ Sin[2 β] +
2 E^NTUH NTUH^2 α^2 γ Sin[2 β] +
E^NTUH α^4 γ Sin[2 β] +
E^NTUH NTUH^2 ph α^2 Sin[2 (α + β)] +
E^NTUH ph α^4 Sin[2 (α + β)] -
E^NTUH NTUH^4 γ Sin[2 (α + β)] -
2 E^NTUH NTUH^2 α^2 γ Sin[
2 (α + β)] -
E^NTUH α^4 γ Sin[2 (α + β)]) (E^
NTUC δ Cos[θ] - δ Cos[δ + \
θ] + E^NTUC NTUC Sin[θ] -
NTUC Sin[δ + θ]) (2 δ +
Sin[2 θ] - Sin[2 (δ + θ)]) -
ph (Ta -
thi) α (NTUH^2 + α^2) (E^
NTUH α Cos[β] - α Cos[α + \
β] + E^NTUH NTUH Sin[β] -
NTUH Sin[α + β]) (2 α + Sin[2 β] -
Sin[2 (α + β)]) Sin[δ/
2] Sin[δ/2 + θ] (2 E^NTUC NTUC^3 pc δ +
2 E^NTUC NTUC^4 γ δ -
2 E^NTUC NTUC pc δ^3 +
2 E^NTUC NTUC^2 pc δ^3 +
4 E^NTUC NTUC^2 γ δ^3 +
2 E^NTUC pc δ^5 + 2 E^NTUC γ δ^5 -
2 NTUC pc δ (NTUC^2 - δ^2) Cos[δ] -
E^NTUC NTUC pc δ (NTUC^2 + δ^2) Cos[
2 θ] -
E^NTUC NTUC^3 pc δ Cos[2 (δ + θ)] -
E^NTUC NTUC pc δ^3 Cos[2 (δ + θ)] +
2 NTUC^3 pc δ Cos[δ + 2 θ] +
2 NTUC pc δ^3 Cos[δ + 2 θ] +
4 NTUC^2 pc δ^2 Sin[δ] +
E^NTUC NTUC^4 γ Sin[2 θ] +
E^NTUC NTUC^2 pc δ^2 Sin[2 θ] +
2 E^NTUC NTUC^2 γ δ^2 Sin[2 θ] +
E^NTUC pc δ^4 Sin[2 θ] +
E^NTUC γ δ^4 Sin[2 θ] -
E^NTUC NTUC^4 γ Sin[2 (δ + θ)] -
E^NTUC NTUC^2 pc δ^2 Sin[2 (δ + θ)] -
2 E^NTUC NTUC^2 γ δ^2 Sin[
2 (δ + θ)] -
E^NTUC pc δ^4 Sin[2 (δ + θ)] -
E^NTUC γ δ^4 Sin[
2 (δ + θ)])))/((2 α +
Sin[2 β] - Sin[2 (α + β)]) (2 δ +
Sin[2 θ] -
Sin[2 (δ + θ)]) (E^(
2 w γ) (2 E^NTUH NTUH^3 ph α -
2 E^NTUH NTUH ph α^3 +
2 E^NTUH NTUH^2 ph α^3 + 2 E^NTUH ph α^5 +
2 E^NTUH NTUH^4 α γ +
4 E^NTUH NTUH^2 α^3 γ +
2 E^NTUH α^5 γ -
2 NTUH ph α (NTUH^2 - α^2) Cos[α] -
E^NTUH NTUH ph α (NTUH^2 + α^2) Cos[
2 β] -
E^NTUH NTUH^3 ph α Cos[2 (α + β)] -
E^NTUH NTUH ph α^3 Cos[2 (α + β)] +
2 NTUH^3 ph α Cos[α + 2 β] +
2 NTUH ph α^3 Cos[α + 2 β] +
4 NTUH^2 ph α^2 Sin[α] +
E^NTUH NTUH^2 ph α^2 Sin[2 β] +
E^NTUH ph α^4 Sin[2 β] +
E^NTUH NTUH^4 γ Sin[2 β] +
2 E^NTUH NTUH^2 α^2 γ Sin[2 β] +
E^NTUH α^4 γ Sin[2 β] -
E^NTUH NTUH^2 ph α^2 Sin[2 (α + β)] -
E^NTUH ph α^4 Sin[2 (α + β)] -
E^NTUH NTUH^4 γ Sin[2 (α + β)] -
2 E^NTUH NTUH^2 α^2 γ Sin[
2 (α + β)] -
E^NTUH α^4 γ Sin[
2 (α + β)]) (2 E^NTUC NTUC^3 pc δ +
2 E^NTUC NTUC^4 γ δ -
2 E^NTUC NTUC pc δ^3 +
2 E^NTUC NTUC^2 pc δ^3 +
4 E^NTUC NTUC^2 γ δ^3 +
2 E^NTUC pc δ^5 + 2 E^NTUC γ δ^5 -
2 NTUC pc δ (NTUC^2 - δ^2) Cos[δ] -
E^NTUC NTUC pc δ (NTUC^2 + δ^2) Cos[
2 θ] -
E^NTUC NTUC^3 pc δ Cos[2 (δ + θ)] -
E^NTUC NTUC pc δ^3 Cos[2 (δ + θ)] +
2 NTUC^3 pc δ Cos[δ + 2 θ] +
2 NTUC pc δ^3 Cos[δ + 2 θ] +
4 NTUC^2 pc δ^2 Sin[δ] +
E^NTUC NTUC^4 γ Sin[2 θ] +
E^NTUC NTUC^2 pc δ^2 Sin[2 θ] +
2 E^NTUC NTUC^2 γ δ^2 Sin[2 θ] +
E^NTUC pc δ^4 Sin[2 θ] +
E^NTUC γ δ^4 Sin[2 θ] -
E^NTUC NTUC^4 γ Sin[2 (δ + θ)] -
E^NTUC NTUC^2 pc δ^2 Sin[2 (δ + θ)] -
2 E^NTUC NTUC^2 γ δ^2 Sin[
2 (δ + θ)] -
E^NTUC pc δ^4 Sin[2 (δ + θ)] -
E^NTUC γ δ^4 Sin[
2 (δ + θ)]) - (-2 E^
NTUH NTUH^3 ph α + 2 E^NTUH NTUH ph α^3 -
2 E^NTUH NTUH^2 ph α^3 - 2 E^NTUH ph α^5 +
2 E^NTUH NTUH^4 α γ +
4 E^NTUH NTUH^2 α^3 γ +
2 E^NTUH α^5 γ +
2 NTUH ph α (NTUH^2 - α^2) Cos[α] +
E^NTUH NTUH ph α (NTUH^2 + α^2) Cos[
2 β] +
E^NTUH NTUH^3 ph α Cos[2 (α + β)] +
E^NTUH NTUH ph α^3 Cos[2 (α + β)] -
2 NTUH^3 ph α Cos[α + 2 β] -
2 NTUH ph α^3 Cos[α + 2 β] -
4 NTUH^2 ph α^2 Sin[α] -
E^NTUH NTUH^2 ph α^2 Sin[2 β] -
E^NTUH ph α^4 Sin[2 β] +
E^NTUH NTUH^4 γ Sin[2 β] +
2 E^NTUH NTUH^2 α^2 γ Sin[2 β] +
E^NTUH α^4 γ Sin[2 β] +
E^NTUH NTUH^2 ph α^2 Sin[2 (α + β)] +
E^NTUH ph α^4 Sin[2 (α + β)] -
E^NTUH NTUH^4 γ Sin[2 (α + β)] -
2 E^NTUH NTUH^2 α^2 γ Sin[
2 (α + β)] -
E^NTUH α^4 γ Sin[
2 (α + β)]) (-2 E^NTUC NTUC^3 pc δ +
2 E^NTUC NTUC^4 γ δ +
2 E^NTUC NTUC pc δ^3 -
2 E^NTUC NTUC^2 pc δ^3 +
4 E^NTUC NTUC^2 γ δ^3 -
2 E^NTUC pc δ^5 + 2 E^NTUC γ δ^5 +
2 NTUC pc δ (NTUC^2 - δ^2) Cos[δ] +
E^NTUC NTUC pc δ (NTUC^2 + δ^2) Cos[
2 θ] +
E^NTUC NTUC^3 pc δ Cos[2 (δ + θ)] +
E^NTUC NTUC pc δ^3 Cos[2 (δ + θ)] -
2 NTUC^3 pc δ Cos[δ + 2 θ] -
2 NTUC pc δ^3 Cos[δ + 2 θ] -
4 NTUC^2 pc δ^2 Sin[δ] +
E^NTUC NTUC^4 γ Sin[2 θ] -
E^NTUC NTUC^2 pc δ^2 Sin[2 θ] +
2 E^NTUC NTUC^2 γ δ^2 Sin[2 θ] -
E^NTUC pc δ^4 Sin[2 θ] +
E^NTUC γ δ^4 Sin[2 θ] -
E^NTUC NTUC^4 γ Sin[2 (δ + θ)] +
E^NTUC NTUC^2 pc δ^2 Sin[2 (δ + θ)] -
2 E^NTUC NTUC^2 γ δ^2 Sin[
2 (δ + θ)] +
E^NTUC pc δ^4 Sin[2 (δ + θ)] -
E^NTUC γ δ^4 Sin[2 (δ + θ)])));
I mention below definitions that may be used but I failed in employing them using Simplify. The command kept running and finally spitted out the same expression. I have already tried some previous answers such as one here but failed. Any help is appreciated.
u == (L/2) - (L/4) (Sin[2 α + 2 β] - Sin[2 β])
v == (l/2) - (l/4) (Sin[2 δ + 2 θ] - Sin[2 θ])
I1 == (L/α) (Cos[β] - Cos[α + β])
I2 == (l/δ) (Cos[θ] - Cos[δ + θ])
