∫1/sin^4x dx
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/25 21:49:24
∫1/sin^4xdx∫1/sin^4xdx∫1/sin^4xdx∫1/sin⁴xdx=∫csc⁴xdx=∫csc²xd(-cotx)=-∫(1+cot²x
∫1/sin^4x dx
∫1/sin^4x dx
∫1/sin^4x dx
∫ 1/sin⁴x dx
= ∫ csc⁴x dx
= ∫ csc²x d(- cotx)
= - ∫ (1 + cot²x) d(cotx)
= - (cotx + 1/3 * cot³x) + C
∫sinxcosx/(1+sin^4x)dx
∫1/sin^4x dx
∫sin(1/x)dx
求∫1/(sin^4x+cos^4x)dx,
∫[1/(sin^2(x)cos^4(x)]dx
求解∫1/(cos^4(x)sin^2(x))dx
∫ x sin(x+1) dx
∫sin（6x）sin（4x）dx
∫sin²x(1-sin²x)dx是多少?
∫sin^2x/(1+sin^2x )dx求解,
∫sixcosx/1+sin^4x dx请你详解
∫(tanx)^1/4 dx/sin^2x
求∫In[1/sin^4(x)]dx~不定积分啊~
∫cosx/(1+4sin^2x)dx
∫sin^4 (x) dx 求不定积分
∫ 4/sin^2x dx
∫(1-sin/x+cos)dx不定积分
∫（1-sin^3x）dx