Question

∫1/(1-sin^4)dx

Answer 1

∫1/[1-(sinx)^4]*dx
=1/2*∫{[1+(sinx)^2]+[1-(sinx)^2]}/{[1+(sinx)^2][1-(sinx)^2]}*dx
=1/2*{∫1/[1+(sinx)^2]+∫1/[1-(sinx)^2]}*dx
=1/2*{∫1/[1+(sinx)^2]+∫1/(cosx)^2}*dx
=1/2*∫1/[1+(sinx)^2]*dx+1/2*∫1/(cosx)^2*dx
=1/2*∫1/[1+(sinx)^2]*dx+1/2*tanx
设u=tan(x/2),sinx=2u/(u^2+1),dx=2du/(u^2+1)
∫1/[1+(sinx)^2]*dx=arctan(√2*tanx)/√2+C
原式=1/2*∫1/[1+(sinx)^2]*dx+1/2*tanx
=1/2*arctan(√2*tanx)/√2+1/2*tanx+C
=1/(2√2)*arctan(√2*tanx)+1/2*tanx+C