计算积分∫1/1+(sinx)^2 dx上下限0到2π

1/1+(sinx)^2
=1/[((sinx)^2+(cosx)^2)+(sinx)^2]
=1/[2(sinx)^2+(cosx)^2]
=(1/(cosx)^2) /[1+2(tanx)^2]
=(secx)^2 /[1+2(tanx)^2]

所以
原积分=4∫(0->π/2) (secx)^2 /[1+2(tanx)^2] dx
=4∫d(tanx)//[1+2(tanx)^2]
=2√2arctan[√2tanx] |(0,π/2)
=2√2(π/2-0)=√2π本回答被提问者采纳