求不定积分∫cos2x/(sinx)^2 dx
求不定积分∫cos2x/(sinx)^2 dx
数学人气：910 ℃时间：2019-12-01 13:55:56
优质解答
∫cos2x/sin²xdx=∫(cos²x-sin²x)/sin²xdx=∫(cos²x+sin²x-2sin²x)/sin²xdx=∫(1-2sin²x)/sin²xdx=∫(1/sin²x-2)dx=∫1/sin²xdx-∫2dx=-cotx-2x+C...=∫(1/sin²x-2)dx这没懂cos²x+sin²x=1知道吧 ∫(cos²x+sin²x-2sin²x)/sin²xdx=∫(1-2sin²x)/sin²xdx=∫(1/sin²x-2)dx【分子分开】=∫1/sin²xdx-∫2dx=-cotx-2x+C
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