∫ 4/sin^2x dx
来源:学生作业帮助网 编辑:六六作业网 时间:2025/02/01 17:44:06
∫4/sin^2xdx∫4/sin^2xdx∫4/sin^2xdx∫4/sin²xdx=4∫1/sin²xdx=4∫csc²xdx=-4cotx+C
∫ 4/sin^2x dx
∫ 4/sin^2x dx
∫ 4/sin^2x dx
∫ 4/sin²x dx =4∫ 1/sin²x dx =4∫ csc²x dx=- 4cotx +C
∫ 4/sin^2x dx
∫dx/cos^2X+4sin^2X
求微积分 ∫sin^2(x)cos^4(x) dx
∫x^2sin(4x)dx 怎么算啊
用换元法计算不定积分∫x sin[(x^2)+4] dx
∫[1/(sin^2(x)cos^4(x)]dx
求解∫1/(cos^4(x)sin^2(x))dx
∫x/sin^2(x) dx
∫sin(x) cos^2(x)dx
∫sin(6x)sin(4x)dx
∫sin^2x/(1+sin^2x )dx求解,
∫sin(6x)sin(2x)dx=?
∫[sin(2x+л/4]^-2 dx
∫(0,PI/2) sin^4 (x) dx
∫(tanx)^1/4 dx/sin^2x
∫cosx/(1+4sin^2x)dx
∫sinxcosx/(1+sin^4x)dx
∫sin^4 (x) dx 求不定积分