∫sin^2x/(1+sin^2x )dx求解,

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/07 00:54:32

/>原式=∫[1-1/(2sin²x+cos²x)]dx
=∫dx - ∫[sec²x/(2tan²x+1)]dx
=∫dx - ∫1/[(√2tanx)²+1)]d(tanx)
=∫dx - √2/2* ∫1/[(√2tanx)²+1)]d(√2tanx)
=x - ∫dx - √2/2*arctan(√2tanx)+c

求不定积分 ∫sin2x/(1+sin^2x)dx 求解! 微积分求解：∫sin^3 (x) cos^2 (x) dx cos^3(2x)sin(2x)dx求解 ∫sin(x) cos^2(x)dx ∫e^(-2x)sin(1/2x)dx ∫1/x^2 * sin 1/x dx ∫[1/(sin^2(x)cos^4(x)]dx ∫x sin(x^2+1)dx怎么做求解 ∫sin(6x)^3*cos(6x)^8 dx=? 范围[pi/2,0] 不定积分∫sin(x^2)dx 求不定积分∫sin(x/2)dx 求不定积分：∫sin(x^2)dx