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求不定积分∫TANX/(3SINX^2+COSX^2)DX
人气:233 ℃ 时间:2020-04-07 19:45:40
解答
∫tanxdx/(3sinx^2+cosx^2)
=∫tanxdx/(3-2cosx^2)
=∫tanxdx/cosx^2(3/cosx^2-2)
=∫tanxdtanx/(3/cosx^2-2)
=(1/2)∫dtanx^2/(3tanx^2+1)
=(1/6)ln(3tanx^2+1)+C
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