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x*tan^2x+cos^2x 求不定积分
人气:327 ℃ 时间:2020-05-28 11:28:34
解答
∫(x*tan^2x+cos^2x)dx
=∫(x(sec^2x-1)+(1/2)(1+cos2x)dx
=∫xsec^2xdx-x^2/+(1/2)x+(1/2)sin2x
=∫xdtanx-x^2/2+(1/2)x+(1/2)sin2x
=xtanx-∫tanxdx-x^2/2+(1/2)x+(1/2)sin2x
=xtanx+ln|cosx|-x^2/2+(1/2)x+(1/2)sin2x+C
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