不定积分 :∫ xcos^2xdx 求详细过程和答案 拜托大神.

2024-11-08 05:29:59
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网友(1):

∫ xcos^2xdx
=∫ x(1+cos2x/2)dx
=1/2∫ xdx+1/2∫xcos2xdx
=x²/4+1/4∫xdsin2x
=x²/4+1/4*xsin2x-1/4∫sin2xdx
=x²/4+1/4*xsin2x-1/8∫sin2xd2x
=x²/4+1/4*xsin2x+1/8*cos2x+C