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∫arccosx^1/2dx不定积分
人气:112 ℃ 时间:2020-10-02 03:41:55
解答
令x=cos²t,则dx=2cost*(-sint)=-2costsint=-sin2t,t=arccos(x^1/2)
∫arccosx^1/2dx
=∫t*(-sin2t)dt
=1/2∫td(cos2t)
=1/2t*cos2t-1/2∫-2sin2tdt
=1/2t*cos2t-1/2cos2t+C
=1/2cos2t(t-1)+C
=1/2(2cos²t-1)(t-1)+C
=1/2(2x-1)[arccos(x^1/2)-1]+C
=(x-1/2)[arccos(x^1/2)-1]+C
=xarccos(x^1/2)-1/2arccos(x^1/2)-x+1/2+C
=xarccos(x^1/2)-1/2arccos(x^1/2)-x+D
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