（4）怎么做呢？高等数学，考研数学
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(4) I = ∫dx/[2sinx(1+cosx)]
= ∫dx/{4cos(x/2)sin(x/2)]^3}
= (1/2)∫cos(x/2)d(x/2)/{[cos(x/2)]^2sin(x/2)]^3}
= (1/2)∫dsin(x/2)/【sin(x/2)]^3{1-[sin(x/2)]^2}】
= (1/4)∫{2/sin(x/2)+2/[sin(x/2)]^3+1/[1-sin(x/2)]-1/[1+sin(x/2)]}dsin(x/2)
= (1/4)∫{2ln|sin(x/2)|-1/[sin(x/2)]^2-ln|1-[sin(x/2)]^2|+C
= (1/4)∫{2ln|sin(x/2)|-1/[sin(x/2)]^2-2ln|cos(x/2)|+C
= ∫dx/{4cos(x/2)sin(x/2)]^3}
= (1/2)∫cos(x/2)d(x/2)/{[cos(x/2)]^2sin(x/2)]^3}
= (1/2)∫dsin(x/2)/【sin(x/2)]^3{1-[sin(x/2)]^2}】
= (1/4)∫{2/sin(x/2)+2/[sin(x/2)]^3+1/[1-sin(x/2)]-1/[1+sin(x/2)]}dsin(x/2)
= (1/4)∫{2ln|sin(x/2)|-1/[sin(x/2)]^2-ln|1-[sin(x/2)]^2|+C
= (1/4)∫{2ln|sin(x/2)|-1/[sin(x/2)]^2-2ln|cos(x/2)|+C
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