求解∫1/(cos^4(x)sin^2(x))dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/05 15:49:26
xS=O0+LiXv"KI~FH%a`i#R$ X`f@JLD_pgN({7*zQ4$))_'1QC^;'&HK@KR4oP*0C&naVfF tf2:Ed
(d1/t9x{ضݚUѲ{"
;@'X
eCA:LF#6&lV[
@UE)UWwg.D(tLtBMdpRSRuE&dݖf4 m1`/n^ϯ-
求解∫1/(cos^4(x)sin^2(x))dx
求解三角函数 cos 2 X (1 - 2 sin ^2 2 X) + cos 4 X ( 1 + cos 2 X)cos 2 X (1 - 2 sin ^2 2 X) + cos 4 X ( 1 + cos 2 X) = cos 2 X cos4X (1 + cos 2 X) = cos 4 X (1 +2cos 2X) = 0
求解∫cos^2(1-2x)dx,∫(sin ax cos ax) 用第二积分换元法做得不对啊
高分求解有关两角和差的三角函数的数学题已知sin x-cos x-1=sin x*cos x,则sin 2x=___.
∫sin^2x/(1+sin^2x )dx求解,
∫[1/(sin^2(x)cos^4(x)]dx
微积分求解:∫sin^3 (x) cos^2 (x) dx 如题.
求解 ∫sin(6x)^3*cos(6x)^8 dx=? 范围[pi/2,0]
求解微分方程y'*cos(y)=x+1-sin(y)
5sin^2(X)+sin(2X)-cos^2(X)=1, 求解XX在0—360度之间
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
求下列不定积分.求详解(1)∫[(cos 2x)/(cos x﹢sin x)]dx; (2) ∫cot²xdx; (3) ∫{(1+2x²)/[x²(1+x²)]}dx; (4) ∫sin²(x/2)dx; (5) ∫[(cos2x)/(sin²xcos²x)]dx; (6) ∫[e^(x-4)]dx;求解
求证(cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x
化简sin(90度+x)+cos(270度-x)-2sin(派+x)-cos(90度-x) 求解过程
求解微分方程xy'ln(x)sin(y)+cos(y)(1-x*cos(y))=0
证明一个sin&cos的等式证明 1+sin^2(x)+sin^4(x)+sin^6(x)=[1-sin^8(x)]/[cos^2(x)]
(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x
化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x