∫1/(1+cos^2(x)) dx

来源：学生作业帮编辑：灵鹊做题网作业帮分类：数学作业时间：2024/06/13 10:55:59

∫1/(1+cos^2(x)) dx

∫dx/{1+[cos(x)]^2}
= ∫[sec(x)]^2dx/{1+[sec(x)]^2}
= ∫[sec(x)]^2dx/{2+ [tan(x)]^2}
= ∫2^(-1/2)d[tan(x)/2^(1/2)]/{1+ [tan(x)/2^(1/2)]^2}
= 2^(-1/2)arctan[tan(x)/2^(1/2)] + C
C 为任意常数.

∫ (1+cos^2 x)/cos^2 x dx = ∫1/(1+cos^2(x)) dx ∫(1-cos^(2)2x)dx ∫(1+sinx) / cos^2 x dx / (cos x + 1 ) dx ∫[1/(sin^2(x)cos^4(x)]dx 求不定积分 1.∫ x/(1+（x^2))dx 2.∫cos^2 x sinx dx ∫1/1-cos x dx ∫cos((x/3)-1)dx=? ∫(1-sin/x+cos)dx不定积分微积分1/cos^2(x)dx=? ∫x(cos^2x-cos^4x)^1/2dx