∫ dx/[sin^(2) x·cos^(2) x]
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/19 16:18:42
∫dx/[sin^(2)x·cos^(2)x]∫dx/[sin^(2)x·cos^(2)x]∫dx/[sin^(2)x·cos^(2)x]∫dx/(sin²xcos²x)=∫dx/
∫ dx/[sin^(2) x·cos^(2) x]
∫ dx/[sin^(2) x·cos^(2) x] 
∫ dx/[sin^(2) x·cos^(2) x]
∫ dx/(sin²xcos²x)
= ∫ dx/(sinxcosx)²
= ∫ dx/(1/2 * sin2x)²
= ∫ dx/[(1/4)sin²2x]
= 4∫ csc²2x dx
= 4(1/2)∫ csc²2x d(2x)
= 2(-cot2x) + C
= -2cot2x + C
∫sin(x) cos^2(x)dx
∫ dx/[sin^(2) x·cos^(2) x]
∫dx/[sin^2(x/2)cos^2(x/2)]
∫(1+sin^2x)/(cos^2x)dx
∫dx/cos^2X+4sin^2X
∫(cos^3x/sin^2x)dx
求微积分 ∫sin^2(x)cos^4(x) dx
求不定积分 ∫ (ln sin x) / (cos^2 x) dx
∫ [cos^3(x)]/[sin^2 (x)]dx
∫sin^3(x)cos^2(x)dx=
求积分：∫sin^2 (x) /cos^3 (x) dx
∫[1/(sin^2(x)cos^4(x)]dx
求解∫1/(cos^4(x)sin^2(x))dx
求不定积分∫(cos2x)/(sin^2x)(cos^2x)dx∫(cos2x)/(sin^2x)(cos^2x)dx
求不定积分∫[1/(sin^2 cos^2(x)]dx
求不定积分∫[1/sin^2cos^2 (x)]dx
∫(1-sin/x+cos)dx不定积分
∫dx/（sin²2x）=¼∫dx/（sin²x·cos²x）是如何推导的?为什么不是等于½∫dx/（sin²x·cos²x）?