∫[tan(3x+1)/cos^2(3x+1)]dx如何计算,

化简tan(3x/2)-tan(x/2)-sinx/(cos(3x/2)*cos(x/2)) 1/cos(x) - tan(x) = 3 求sin(x)1/cos(x) - tan(x) = 3求sin(x) 求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x ∫[tan(3x+1)/cos^2(3x+1)]dx如何计算, 两道不定积分的题(1)∫(1-tan x)/(1+tan x)dx(2)∫1/[5(cos x)^2+3(sin x)^2]dx (tan x+1/tan)cos^2 x等于? tan(3x/2)-tan(x/2)-2sinx/(cosx+cos(2x))=? 1+cos(2x)/sin(2x)=3/4,求tan(x/2) sin(2x)/1+cos(2x)=3/4,求tan(x/2)的数学题 化简1+sin(x-2π)*cos[x-(7π/2)]-tan(π-x)*tan(3π/2-x)-cos^2(-x)1+sin(x-2π)*cos[x-(7π/2)]-tan(π-x)*tan(3π/2-x)-cos^2(-x) 怎么化啊 化简:sin(2派-x)tan(派+x)cot(-x-派)/tan(3派-x)cos(派-x) ∫cos^3x dx ∫tan^3xsecx dx tan(x-π/4)=2,求sin2x+2cos2x/2cos^x-3sin2x-1 求证：tan x/2=sin x/(1+cos x) 求证：1-sin(x)=cos(x)*tan(x/2) (2sin^2- 3sinxcox)/sin^2 x+cos^2 x=(2tan^2 x-3tanx)/tan^2x+1这一步骤怎么得到? 已知（1-tan x）/(2+tan x )=1,求证3 sin 2x =-4cos 2x（反证法、分析法、综合法都可以哦!） 用凑微分法解答tan(x+3)/cos^2(x+3) dxtan(x+3)/[cos(x+3)]^2 dx