令y=0.75sint,则y²/(0.75)²=sin²t, √[1-y²/(0.75)²]=cost
dy=0.75costdt, t=arcsin(4y/3)
∫(0.75-y)√[1-y²/(0.75)²]dy
=∫0.75(1-sint)cost(0.75cost)dt
=(9/16)∫cos²t-sintcos²t dt
=(9/16)(∫cos²tdt -∫sintcos²t dt)
=(9/16)[∫(cos2t+1)/2 dt +∫cos²t dcost]
=(9/16)(sin2t/4 + t/2 + cos³t/3)+C
=(9/16)(sintcost/2 + t/2 + cos³t/3)+C
=(9/16)[(4y/3)√(1-16y²/9)/2 + arcsin(4y/3)/2 + (1-16y²/9)√(1-16y²/9]/3)+C
=y√(9-16y²)/8 + 9arcsin(4y/3)/32 + (9-16y²)√(9-16y²)/144 + C