不定积分∫1/√2dx解释说明及步骤
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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
【数学辅导团】为您解答,不理解请追问,理解请及时选为满意回答!(*^__^*)谢谢!
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
【数学辅导团】为您解答,不理解请追问,理解请及时选为满意回答!(*^__^*)谢谢!
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