丿1/(sin^2x+2cos^2x)dx不定积分求过 程

来源：学生作业帮助网 编辑：作业帮 时间：2024/11/18 08:26:36
丿1/(sin^2x+2cos^2x)dx不定积分求过 程
x){cFqf^QQr~1LxYϗlcӋ Wt$铨CΆ[QjFCm 8#M k'}:%Ov/(NM kXچ ph2 %y(`ӥӟPQ,#Ģd.~qAb($

丿1/(sin^2x+2cos^2x)dx不定积分求过 程
丿1/(sin^2x+2cos^2x)dx不定积分求过 程

丿1/(sin^2x+2cos^2x)dx不定积分求过 程
原式=∫1/(1+(cosx)^2) dx 分子分母同除以(cosx)^2
=∫(secx)^2/((secx)^2+1) dx
=∫1/((secx)^2+1) d (tanx)
=∫1/((tanx)^2+2) d (tanx)
套公式
=1/√2*arctan((tanx)/√2)+C

化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x 2cos x (sin x -cos x)+1 怎么证明2(1-sin x)(1+cos x)=(1-sin x+cos x)^2 化简 1/2cos x-根号3/2sin x 根3sin x+cos x 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x 已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x 已知(1+sin x)/cos x=-1/2,则cos/(sin x-1)=? 求证：tan x/2=sin x/(1+cos x) 求证：1-sin(x)=cos(x)*tan(x/2) lim(sin(x^2*cos(1/x)))/x怎么做? (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 y=2cos x (sin x+cos x) 化简sin(x)cos(x)cos(2x) 化简sin(x)cos(x)cos(2x) sin x cos x cos 2x 化简 (sin^2x+cos^2x )(sin^2x-cos^2x)怎么得sin^2x-cos^2x?