Let α∈0,π2 be fixed. If ∫sin⁡(x+α)sin⁡(x−α)dx=f1(x)cos⁡2α+f2(x)sin⁡2α+c where c constant of integration then

∫sin⁡(x+α)sin⁡(x−α)dx=∫sin⁡(x−α+2α)sin⁡(x−α)dx Put x−α=θ⇒dx=dθ=∫sin⁡(θ+2α)sin⁡θdθ =cos⁡2α∫dθ+sin⁡2α∫cos⁡θsin⁡θdθ=cos⁡2α⋅(x−α)+sin⁡2αlog⁡|sin⁡(x−α)|+c