∫x^2/(1+^2)dx,∫sin^2xdx计算不定积分

来源：学生作业帮助网编辑：作业帮时间：2024/05/07 04:48:43

∫x^2/(1+^2)dx,∫sin^2xdx计算不定积分
∫x^2/(1+^2)dx,∫sin^2xdx计算不定积分

∫x^2/(1+^2)dx,∫sin^2xdx计算不定积分
∫ x^2/(1+x^2)dx
= ∫ (1 - 1/(1+x^2))dx
= x - arctan(x) + C
∫ sin^2xdx
= 1/2 ∫ (1 - cos(2x))dx
= 1/2 (x - 1/2 sin(2x)) + C

∫sin^2x/(1+sin^2x )dx求解, ∫(1-sin^2( x/2))dx ∫(cosx／1+sin^2x)dx ∫sin^2x(1+tanx)dx ∫1/sin(x/2)dx ∫1+sin^2x/1-cos2x dx∫1+sin^2x/1-cos2x dx ∫x/sin^2(x) dx ∫sin(x) cos^2(x)dx ∫(2/sin^2x)dx= 求∫ sin^2 x/2 dx .. ∫sin(6x)sin(2x)dx=? 求不定积分：∫sin(x^2)dx 求不定积分∫sin(x/2)dx ∫[(sin^2)x]dx求不定积分 ∫sinx/(cosx-sin^2x)dx ∫sin(√2x)dx ∫ 4/sin^2x dx 求∫sin(x^2)dx