求不定积分：∫(sinx)^2dx

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/08/24 02:45:45

求不定积分：∫(sinx)^2dx

∫sin²xdx
=1/2 ∫(1-cos2x) dx
=1/2 (x-∫cos2xdx)
=1/2(x-1/2∫cos2xd2x)
=1/2(x-1/2sin2x)
=x/2-(sin2x)/4
验算：
x/2-(sin2x)/4 '
=1/2-2/4 cos2x
=1/2-1/2(cos²x-sin²x)
=1/2(1-cos²x+sin²x)
=1/2(2sin²x)
=sin²x

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