xzb0909
幼苗
共回答了13个问题采纳率:76.9% 举报
(1)∫xsinxcosxdx
=1/8 ∫2xsin2x d2x
=-1/8 ∫2xd(cos2x)
=-1/8 (2xcos2x-∫cos2xd2x)
=-1/4 xcos2x- 1/8 sin2x+C
(2)∫xtanxtanxdx=∫x[(secx)^2-1]dx
=∫x(secx)^2dx-∫xdx=∫xd(tanx)-∫xdx
=xtanx-∫tanxdx-∫xdx
=xtanx-(-lncosx+C1)-(x^2/2+C2)
=xtanx+lncosx-x^2/2+C
1年前
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