∫arcsinx*arccosxdx永不分积分法怎么求
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∫arcsinx*arccosxdx永不分积分法怎么求
先化简t=arcsin(x) x=sin(t)
arccos(x)=π/2 -t
∫t(π/2 -t)dsin(t)=t(π/2 -t)sin(t) -∫ sint d(t(π/2 -t))
=t(π/2 -t)sin(t) -∫ (π/2-2t)sint dt
=t(π/2 -t)sin(t) +∫ (π/2-2t) dcos(t)
=t(π/2 -t)sin(t) + (π/2-2t) cos(t)-∫cos(t)d (π/2-2t)
=t(π/2 -t)sin(t) + (π/2-2t) cos(t)+∫2cos(t)dt
=t(π/2 -t)sin(t) + (π/2-2t) cos(t)+2sin(t)+C
=arcsin(x)arccos(x)x+(arccos(x)-arcsin(x))√(1-x²)+2arcsin(x)+C
arccos(x)=π/2 -t
∫t(π/2 -t)dsin(t)=t(π/2 -t)sin(t) -∫ sint d(t(π/2 -t))
=t(π/2 -t)sin(t) -∫ (π/2-2t)sint dt
=t(π/2 -t)sin(t) +∫ (π/2-2t) dcos(t)
=t(π/2 -t)sin(t) + (π/2-2t) cos(t)-∫cos(t)d (π/2-2t)
=t(π/2 -t)sin(t) + (π/2-2t) cos(t)+∫2cos(t)dt
=t(π/2 -t)sin(t) + (π/2-2t) cos(t)+2sin(t)+C
=arcsin(x)arccos(x)x+(arccos(x)-arcsin(x))√(1-x²)+2arcsin(x)+C
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