∫1/(1-sin^4)dx

来源：学生作业帮助网编辑：六六作业网时间：2024/12/18 14:41:44

∫1/(1-sin^4)dx∫1/(1-sin^4)dx∫1/(1-sin^4)dx∫1/[1-(sinx)^4]*dx=1/2*∫{[1+(sinx)^2]+[1-(sinx)^2]}/{[1+(s

∫1/(1-sin^4)dx
∫1/(1-sin^4)dx

∫1/(1-sin^4)dx
∫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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