limx→π2(∫x3(π/2)3(sin(2t1/3)+cos(t1/3)) dt(x-π2)2) is equal to [2024]
(4)
Let L=limx→π/2[∫x3(π/2)3(sin(2t1/3)+cos(t1/3)) dt(x-π2)2] 00 form, so by L'Hospital rule
=limx→π2ddx∫x3(π/2)3(sin(2t1/3)+cos(t1/3))dt2(x-π2) sin(2×π2)·ddx(π2)3-sin(2x)·ddxx3
=limx→π/2+(cosπ2ddx(π2)3-cosx·ddxx3)2(x-π2) (By Leibnitz rule of integration)
=limx→π/2-3x2sin2x-3x2cosx2(x-π2) (00 form)
=limx→π/2-6xsin2x-6x2cos2x-6xcosx+3x2sinx2
=6π24+3π242=9π28