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1/(sin2x+sinx)的不定积分怎么求
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1/(sin2x+sinx)的不定积分怎么求
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答案和解析
令cosx=u,则:d(cosx)=du.
∴∫[1/(sin2x+sinx)]dx
=∫{1/[sinx(2cosx+1)]}dx
=∫{sinx/[(sinx)^2(2cosx+1)]}dx
=-∫{1/[(1+cosx)(1-cosx)(2cosx+1)]}d(cosx)
=-∫{1/[(1+u)(1-u)(2u+1)]du
=-(1/2)∫[1/(1-u)+1/(1+u)]/(2u+1)}du
=-∫{1/[(2-2u)(2u+1)]}du-∫{1/[(2+2u)(2u+1)]}du
=-(1/3)∫[1/(2-2u)+1/(2u+1)]du-∫[1/(2u+1)-1/(2+2u)]du
=(1/6)ln|2-2u|+(1/6)ln|2u+1|-(1/2)ln|2u+1|-(1/2)ln|2+2u|+C
=(1/6)ln|1-u|-(1/3)ln|2u+1|-(1/2)ln|1+u|+C
=(1/6)ln(1-cosx)-(1/3)ln|1+2cosx|-(1/2)ln(1+cosx)+C.
∴∫[1/(sin2x+sinx)]dx
=∫{1/[sinx(2cosx+1)]}dx
=∫{sinx/[(sinx)^2(2cosx+1)]}dx
=-∫{1/[(1+cosx)(1-cosx)(2cosx+1)]}d(cosx)
=-∫{1/[(1+u)(1-u)(2u+1)]du
=-(1/2)∫[1/(1-u)+1/(1+u)]/(2u+1)}du
=-∫{1/[(2-2u)(2u+1)]}du-∫{1/[(2+2u)(2u+1)]}du
=-(1/3)∫[1/(2-2u)+1/(2u+1)]du-∫[1/(2u+1)-1/(2+2u)]du
=(1/6)ln|2-2u|+(1/6)ln|2u+1|-(1/2)ln|2u+1|-(1/2)ln|2+2u|+C
=(1/6)ln|1-u|-(1/3)ln|2u+1|-(1/2)ln|1+u|+C
=(1/6)ln(1-cosx)-(1/3)ln|1+2cosx|-(1/2)ln(1+cosx)+C.
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