计算积分∫1/1+(sinx)^2 dx上下限0到2π

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

∫[0,π]√(1-sinx) dx ＝∫[0,π]√(sin^2(x/2)+cos^2(x/2)-2sin(x/2)cos(x/2)) dx =∫[0,π]√[（sin(x/2)-cos(x/2)]^2dx =∫[0,π/4][cos(x/2)-sin(x/2)]dx+∫[π/4,π][sin(x/2)-cos(x/2)]dx =2[sin(x/2)+cos(x/2)][0,π/4]-2[sin(x/2)+cos(x/2)][π/4,π]

如以上回答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！

点此我要举报以上问答信息