∫1/(1-sin^4)dx
1个回答

∫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

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