∫ (1+cos^2 x)/cos^2 x dx = 我的算法：∫ (1+cos^2 x)/cos^2 x dx = 1/cos^2 x + 1 = 2/(cos2x +1 ) + 1 = 2（cos2x+1)^-1 +1=这么变成了 2(cos2x+1)^0/(0) 不对啊 ==

来源：学生作业帮助网 编辑：作业帮 时间：2024/08/01 15:23:22
∫ (1+cos^2 x)/cos^2 x dx = 我的算法：∫ (1+cos^2 x)/cos^2 x dx = 1/cos^2 x + 1 = 2/(cos2x +1 ) + 1 = 2（cos2x+1)^-1 +1=这么变成了 2(cos2x+1)^0/(0) 不对啊 ==
x){ԱZAP;98HBSPHPUx1?ڥ`kkT`v6-r(HkAOFM:$1B&hWh'$Ay7~> qjťF B-Rd R4 hPG8+(h8?O;ڞlM$#.T

∫ (1+cos^2 x)/cos^2 x dx = 我的算法：∫ (1+cos^2 x)/cos^2 x dx = 1/cos^2 x + 1 = 2/(cos2x +1 ) + 1 = 2（cos2x+1)^-1 +1=这么变成了 2(cos2x+1)^0/(0) 不对啊 ==
∫ (1+cos^2 x)/cos^2 x dx =
我的算法：∫ (1+cos^2 x)/cos^2 x dx = 1/cos^2 x + 1 = 2/(cos2x +1 ) + 1 = 2（cos2x+1)^-1 +1=这么变成了 2(cos2x+1)^0/(0) 不对啊 ==

∫ (1+cos^2 x)/cos^2 x dx = 我的算法：∫ (1+cos^2 x)/cos^2 x dx = 1/cos^2 x + 1 = 2/(cos2x +1 ) + 1 = 2（cos2x+1)^-1 +1=这么变成了 2(cos2x+1)^0/(0) 不对啊 ==
∫ (1+cos^2 x)/cos^2 x dx =∫1/cos^2 x + 1dx
=∫1/cos^2 xdx+x
=∫1d（tanx)+x
=tanx+x+c

原式=∫(1/cos²x+1)dx
=∫(sec²x+1)dx
=∫d(tanx)+∫dx
=tanx+x+C (C是积分常数)

cos x>-1/2 证明cos 2x cos x=1/2(cos x+cos 3x) 2cos x (sin x -cos x)+1 2（cos x）^2=1+cos 2x, y=(cos x-2)/(cos x-1)的值域 ∫(1-cos^(2)2x)dx ∫1/(1+cos^2(x)) dx ∫(1+sinx) / cos^2 x dx Cos x=-1/2 y=cos^2 x-cos x +1的值域y=cos^2 x+cos x +1的值域, 化简cos 2x (cos(x))^-2导数 cos(x^2)dx 求该不定积分 ∫ (1+cos^2x) / (1+cos 2x) 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 ∫ e^(3x)cos(2x) ∫x*(cos^2x)dx ∫x／[(cos^2)x]dx