1/(2+sin^2x) dx cos根号下x dx 求高数大神指导
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/21 15:00:57
![1/(2+sin^2x) dx cos根号下x dx 求高数大神指导](/uploads/image/z/7238219-59-9.jpg?t=1%2F%282%2Bsin%5E2x%29+dx+cos%E6%A0%B9%E5%8F%B7%E4%B8%8Bx+dx+%E6%B1%82%E9%AB%98%E6%95%B0%E5%A4%A7%E7%A5%9E%E6%8C%87%E5%AF%BC)
x)30.̋3THPH/~`OvtWmlzzƳ.Y|g=OI*'W~
E>_Ɏ=lA+MtO9. 5z@jxֻikSu^ $¶Dy˶R*lJRJtƓn)?m/W<_Ȱ/.H̳y u
1/(2+sin^2x) dx cos根号下x dx 求高数大神指导
1/(2+sin^2x) dx cos根号下x dx 求高数大神指导
1/(2+sin^2x) dx cos根号下x dx 求高数大神指导
第一问的标注方式有歧义;第二问用换元法,记根号X=t,然后dx=2tdt,原式变为cost*2tdt进行分部积分
∫sin(x) cos^2(x)dx
∫(1+sin^2x)/(cos^2x)dx
∫[1/(sin^2(x)cos^4(x)]dx
求解∫1/(cos^4(x)sin^2(x))dx
求不定积分∫[1/(sin^2 cos^2(x)]dx
求不定积分∫[1/sin^2cos^2 (x)]dx
1/(2+sin^2x) dx cos根号下x dx 求高数大神指导
求不定积分:$(sin x +cos x)/(1+cos^2x) dx
∫3/(x*x*x+1)dx ∫1/(2+sin x)dx 和 ∫1/(2sin x-cos x+5)dx的解法
∫dx/[sin^2(x/2)cos^2(x/2)]
∫dx/cos^2X+4sin^2X
∫ dx/[sin^(2) x·cos^(2) x]
∫(cos^3x/sin^2x)dx
求不定积分 :| ( sin^2 x / cos^3x ) dx
求微分sin^2x*cos^5x*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