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∫((x+sinx)/(1+cosx))dx
=∫[(x+sinx)/2cos²(x/2)]dx
=∫(x+sinx)d(tan(x/2))
=(x+sinx)*tan(x/2)-∫tan(x/2)d(x+sinx)
=xtan(x/2)+sinx*tan(x/2)-∫tan(x/2)(1+cosx)dx
=xtan(x/2)+2sin(x/2)cos(x/2)*tan(x/2)-∫tan(x/2)*2cos²(x/2)dx
=xtan(x/2)+2sin²(x/2)-∫sin(x)dx
=xtan(x/2)+2sin²(x/2)-cos(x)
因此定积分的值为pi/2
=∫[(x+sinx)/2cos²(x/2)]dx
=∫(x+sinx)d(tan(x/2))
=(x+sinx)*tan(x/2)-∫tan(x/2)d(x+sinx)
=xtan(x/2)+sinx*tan(x/2)-∫tan(x/2)(1+cosx)dx
=xtan(x/2)+2sin(x/2)cos(x/2)*tan(x/2)-∫tan(x/2)*2cos²(x/2)dx
=xtan(x/2)+2sin²(x/2)-∫sin(x)dx
=xtan(x/2)+2sin²(x/2)-cos(x)
因此定积分的值为pi/2
更多追问追答
追答
这样做也是可以的,左边为:
∫xdx/(1+cosx)=∫xdx/[2cos²(x/2)]=∫xd[tan(x/2)]=x*tan(x/2)-∫[tan(x/2)]dx=x*tan(x/2)+2*ln|cos(x/2)|
关键在于化成tan(x/2)后用分部积分法求解
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