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求不定积分∫x*sinx^2
∫x*sinx^2dx
人气:149 ℃ 时间:2020-04-07 19:11:50
解答
答:∫xsinx^2dx
=∫x(1-cos2x)/2dx
=1/2∫xdx-1/2∫xcos2xdx
=1/2∫xdx-1/4∫x d(sin2x)
=1.4x^2-1/4xsin2x-1/8cos2x+c
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