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求不定积分∫xtanx(sec^2)xdx!
人气:279 ℃ 时间:2020-06-09 16:40:47
解答
原式=∫xsinx/cos^3(x)*dx
=-∫x/cos^3(x)*d(cosx)
=1/2∫xd(1/cos^2(x))
=x/(2cos^2(x))-1/2∫dx/cos^2(x)
=x/(2cos^2(x))-tanx/2+C
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